Review of Linear Systems Theory

I. Introduction

Physical optics is the study of the wave nature of light. Diffraction, polarization, coherence and interferometry are all physical optics effects, and play a key role in many modern optical systems. The first half of this course will be devoted to studying the key concepts of physical optics.

In the early 1960’s, the mathematics of linear systems was applied to physical optics, providing a new and powerful way of analyzing optical systems. This analysis technique is now known as Fourier optics. With the tools of Fourier optics we will be able to explain the intricacies of holography, understand how astronomers can get incredibly detailed images of distant galaxies using radio telescopes separated by thousands of kilometers (a technique called Very Long Baseline Interferometry), calculate the diffraction effects of microscopes which limit their resolution, understand Synthetic Aperture Radar, describe methods of “seeing through” a turbulent atmosphere, understand how optical pattern recognition machines work, and much more. Before we start, however, it is useful to review the most important aspects of linear systems theory.

II. Linear Systems

Many optical devices and effects can be described as linear

systems. A system is linear if we can apply superposition.

$$ \mathcal{L}\{a_1 s(\bar{x}) + a_2 t(\bar{x})\} = a_1 \mathcal{L}\{s(\bar{x})\} + a_2 \mathcal{L}\{t(\bar{x})\} $$

The operator $\mathcal{L}$ describes the linear system. In electronics for example, this system could be a filter. In optics it may be an imaging system of some kind.

The operator $\mathcal{L}$ can be described in various ways. One particularly useful way is by its impulse response (often called a point spread function in optics). Physically, it is the response of the linear system to a single point at a particular location in a plane. Consider the point (expressed as delta function) located at $x_0$. The response of the linear system to this point is given by:

$$ \mathcal{L}\{\delta(\bar{x}_1 - \bar{x}_0)\} = h(\bar{x}_2; \bar{x}_0) $$

$h(\bar{x}_2; \bar{x}_0)$ is the point spread function associated with the input point $\bar{x}_0$.

Note that the input to the linear system can always be described by a weighted sum of delta functions:

$$ g_1(\bar{x}_1) = \int g_1(\bar{x}_0)\delta(\bar{x}_1 - \bar{x}_0)d\bar{x}_0 $$

This is simply equivalent to treating the input $g_1(\bar{x}_1)$ as an infinite number of samples (delta functions) with varying strengths. By linear superposition, each one of these samples can be passed through the operator separately and added together at the output.

Superposition Integral

Input → Output:

$$ \mathcal{L}\{ g_1(\bar{x}_1) \} = g_2(\bar{x}_2) $$

Substitute the delta function representation of the input:

$$ g_2(\bar{x}_2) = \mathcal{L}\left\{ \int g_1(\bar{x}_0)\,\delta(\bar{x}_1 - \bar{x}_0)\, d\bar{x}_0 \right\} $$

By linearity, the operator moves inside the integral, and each delta function produces the PSF:

$$ = \int g_1(\bar{x}_0)\, h(\bar{x}_2; \bar{x}_0)\, d\bar{x}_0 $$

This is the Superposition Integral - the response to a single input point of strength $g_1(\bar{x}_0)$ located at $\bar{x}_0$.

[!note] Apply to point spread function

$$ \Rightarrow \mathcal{L}\{ \delta(\bar{x}_1) \} = \underbrace{h(\bar{x}_2;\,0)}_{\text{PSF}} $$

Then, by shift invariance,

$$ \mathcal{L}\{ \delta(\bar{x}_1 - \bar{x}_0) \} = h(\bar{x}_2 - \bar{x}_0;\,0) = \underbrace{h(\bar{x}_2 - \bar{x}_0)}_{\text{PSF}} $$

Plugging this back into the superposition integral gives the convolution integral:

$$ \underbrace{g_2(\bar{x}2)}{\text{output}}

\int \underbrace{g_1(\bar{x}0)}{\text{input}} ; \underbrace{h(\bar{x}_2 - \bar{x}0)}{\text{PSF}} ; d\bar{x}_0 $$

In optics we are often interested in a special kind of linear operator called a shift-invariant operator (completely analogous to the time-invariant operators in electronics).

If an operator is shift-invariant, then:

$$ \mathcal{L}\{g_1(\bar{x}_1)\} = g_2(\bar{x}_2) $$

Shift invariance implies:

$$ \mathcal{L}\{g_1(\bar{x}_1 - \bar{x}_0)\} = g_2(\bar{x}_2 - \bar{x}_0) $$

In other words, the behavior of the system is not a function of the independent variable (spatial position in this case).

### Shift-Invariant Systems

Assume $\mathcal{L}\{g_1(x_1, y_1)\} = g_2(x_2, y_2)$, where $g_1(x_1, y_1)$ is the object, $g_2(x_2, y_2)$ is the image, and $\mathcal{L}$ represents the imaging process.

Shift invariance implies:

$$ \mathcal{L}\{g_1(x_1 - x_0, y_1 - y_0)\} = g_2(x_2 - x_0, y_2 - y_0) $$

Physically, if we move the object around, the blurry image (from imperfect imaging) moves around also, but does not change appearance.

Several imaging effects are shift invariant:

• Diffraction
• Misfocussing
• Spherical aberration

Some are not:

• Coma
• Astigmatism
• Distortion

Point Spread Function of Linear Shift-Invariant Systems

Consider the point spread function of a linear shift-invariant (LSI) system. If we place a delta function at the origin, we have:

$$ \mathcal{L}\{\delta(\bar{x}_1)\} = h(\bar{x}_2; 0) $$

But from the definition of shift-invariance, we also have:

$$ \mathcal{L}\{\delta(\bar{x}_1 - \bar{x}_0)\} = h(\bar{x}_2 - \bar{x}_0; 0) $$

Thus, the entire system can be described by a single point spread function $h(\bar{x}_2)$. The superposition integral now becomes

Convolution Integral

$$ g_2(\bar{x}_2) = \int g_1(\bar{x}_0) \, h(\bar{x}_2 - \bar{x}_0) \, d\bar{x}_0 $$

This is the familiar convolution integral. It will become apparent that many optical phenomena (imaging, diffraction, etc.) can be described in terms of point spread functions and convolutions in two dimensions.

$$ s(x,y) = \iint g(x',y') \, p(x-x', y-y') \, dx' \, dy' $$

III. Fourier Transforms Applied to Linear Shift-Invariant Systems

In electronics, we know that a temporal Fourier transform is a useful device for studying the behavior of linear circuits. For example, filters are easily described by a transfer function $\tilde{H}(f)$, where $f$ is a temporal frequency variable. We shall see that optical propagation is conveniently described by a two-dimensional Fourier transform $\tilde{H}(f_x, f_y)$, where $f_x$ and $f_y$ are spatial frequencies.

To understand why the Fourier transform is so useful for describing linear shift invariant systems, we recall that the superposition property allows us to:

1. Express the input function as a weighted sum of simple waveforms (basis functions)
2. Pass each waveform through the operator separately
3. Sum the result at the end

But what do we choose for these simple waveforms? Ideally, we would like to choose a waveform that is unaffected by the linear operator (except to be multiplied by a complex constant). Then, as each waveform is passed through the operator the effect of the operator is particularly easy to calculate.

Eigenfunctions of Linear Operators

The set of functions which are unaffected by a linear operator (except to be multiplied by a complex constant) are called the eigenfunctions of the operator.

$$ \mathcal{L}\{\psi(\bar{x}; \bar{f}_0)\} = H(\bar{f}_0)\psi(\bar{x}; \bar{f}_0) $$

This is the Eigenfunction Equation.

There are in general an infinite number of eigenfunctions associated with an operator. In the above, $\psi(\bar{x}; \bar{f}_0)$ is the $\bar{f}_0$th eigenfunction which is a function of $\bar{x}$. Thus, $\psi(\bar{x}; \bar{f})$ describes a family of eigenfunctions. The eigenfunction $\psi(\bar{x}; \bar{f}_0)$ can pass through the operator unchanged except that it is multiplied by the complex constant $H(\bar{f}_0)$. $H(\bar{f}_0)$ is the eigenvalue associated with the $\bar{f}_0$th eigenfunction.

Finding Eigenfunctions of Linear Shift-Invariant Operators

We are interested in finding the eigenfunctions associated with linear shift invariant operators. We shall show that the complex exponentials are the eigenfunctions we desire. Let

$$ \mathcal{L}\{\exp(j2\pi\bar{f}_0 \cdot \bar{x})\} = g(\bar{x}; \bar{f}_0) $$

Now consider a shifted version of this

Complex Exponentials as Eigenfunctions

$$ \mathcal{L}\{\exp(j2\pi\bar{f}_0 \cdot (\bar{x} - \bar{x}_0))\} = \mathcal{L}\{\exp(-j2\pi\bar{f}_0 \cdot \bar{x}_0)\exp(j2\pi\bar{f}_0 \cdot \bar{x})\} $$$$ = \exp(-j2\pi\bar{f}_0 \cdot \bar{x}_0) \mathcal{L}\{\exp(j2\pi\bar{f}_0 \cdot \bar{x})\} $$$$ = \exp(-j2\pi\bar{f}_0 \cdot \bar{x}_0) g(\bar{x}; \bar{f}_0) $$

But by shift-invariance:

$$ \mathcal{L}\{\exp(j2\pi \bar{f}_0 \cdot (\bar{x} - \bar{x}_0))\} = g(\bar{x} - \bar{x}_0; \bar{f}_0) $$

Therefore:

$$ g(\bar{x} - \bar{x}_0; \bar{f}_0) = \exp(-j2\pi\bar{f}_0 \cdot \bar{x}_0) g(\bar{x}; \bar{f}_0) $$

Now this is only true if $g(\bar{x}; \bar{f}_0)$ is of the form:

$$ g(\bar{x}; \bar{f}_0) = H(\bar{f}_0)\exp(j2\pi\bar{f}_0 \cdot \bar{x}) $$

where $H(\bar{f}_0)$ = complex constant. This is the solution to the equation above.

Thus we conclude:

$$ \mathcal{L}\{\exp(j2\pi\bar{f}_0 \cdot \bar{x})\} = H(\bar{f}_0)\exp(j2\pi\bar{f}_0 \cdot \bar{x}) $$

which satisfies the eigenfunction equation.

Why Fourier Transform is Useful

It is now clear why the Fourier transform is so useful in analyzing a linear shift-invariant system. The input function can be decomposed into a linear combination of eigenfunctions $\exp(j2\pi\bar{f} \cdot \bar{x})$.

The amount of each eigenfunction contained in the input is given by the Forward Fourier Transform:

$$ \tilde{G}(\bar{f}) = \int g_1(\bar{x})\exp(-j2\pi\bar{f} \cdot \bar{x})d\bar{x} $$

Each weighted eigenfunction is run through the linear shift invariant system. The effect of the system is to multiply each of the eigenfunctions by a complex number $H(\bar{f})$, where H takes on different values for each different eigenfunction. The function $H(\bar{f})$ is called the transfer function of the linear system.

After passing through the operator a single eigenfunction waveform becomes

Individual Eigenfunction Response

$$ \tilde{G}(\bar{f}_0) \cdot H(\bar{f}_0)\exp(j2\pi\bar{f}_0 \cdot \bar{x}) $$

To calculate the output of the linear system we simply pass all weighted eigenfunctions through the operator and add up the result:

$$ g_2(\bar{x}) = \int \tilde{G}(\bar{f}) \cdot H(\bar{f})\exp(j2\pi\bar{f} \cdot \bar{x})d\bar{f} $$

This is the Inverse Fourier Transform (sum of results).

IV. Properties of the Spatial Fourier Transform

A. Spatial Frequency

The spatial Fourier transform is similar to the more familiar temporal transform in many respects. Mathematically, the only difference is that it is two-dimensional as opposed to its one-dimensional temporal counterpart. However, there are several conceptual differences which are worth exploring.

The concept of frequency is well understood for temporal signals. If a voltage has a temporal behavior expressed by:

$$ V(t) = A\sin(2\pi f_0 t) $$

we know that this corresponds to a sinusoid which repeats itself every $\frac{1}{f_0}$ seconds. The frequency has units of inverse seconds.

Equivalently, a spatial description of a particular image might be expressed as:

$$ A(x,y) = \sin(2\pi f_x x) \cdot \sin(2\pi f_y y) $$

This corresponds to a two-dimensional distribution which repeats itself every $\frac{1}{f_x}$ cm in the x direction and $\frac{1}{f_y}$ cm in the y direction. The spatial frequency has components in two directions, and units of inverse centimeters.

Relationship Between Angular Plane Waves and Fourier Transform

A solution to the Helmholtz equation for wave propagation is given by a plane wave:

$$ A \exp(j\phi)\exp(j\vec{k} \cdot \vec{r}) $$

where $A$ and $\phi$ are the amplitude and phase of the plane wave. In rectangular coordinates this becomes:

$$ u(x,y,z) = A \exp(j\phi)\exp(jk_z \cdot z)\exp(j(k_x \cdot x + k_y \cdot y)) $$

Now consider the field in a particular z plane $z = z_0$:

$$ u(x,y)_{z=z_0} = A \exp(j(\phi + k_z \cdot z_0))\exp(j(k_x \cdot x + k_y \cdot y)) $$

Notice that the field is described by a complex exponential. When one moves along the x-axis (or y-axis), the phase of this complex exponential changes. The rate at which it changes (frequency) is given by:

$$ k_x = k\cos\theta $$

Thus a complex spatial frequency component is physically equivalent to a plane wave, where the frequency of the component is proportional to the cosine of the propagation angle of the plane wave.

$$ \exp(j2\pi f_x x) $$$$ f_x = \frac{\cos\theta}{\lambda} = \text{spatial freq.} $$

Negative and positive frequencies correspond to plane waves travelling in opposite directions (e.g., upward vs downward tilted). The spatial Fourier transform of an arbitrary two-dimensional field can be thought of as a linear superposition of plane waves, where:

• The angle of the plane wave corresponds to the spatial frequency
• The amplitude and phase of the plane wave corresponds to the complex Fourier component associated with that spatial frequency

Causality (Spatial vs Temporal)

In time series analysis, causality states that the value of the output at time $t_0$ does not depend on future values of the input $t > t_0$. This relationship requires all impulse responses to be nonsymmetric, and thus the transfer function must be complex.

When time is the independent variable, causality is required since “future” cannot affect “present”, but “past” can.

$g(x)$	$\tilde{G}(f_x)$
Real	Hermitian
Hermitian	Real
Real/Even	Real/Even
Complex/Even	Complex/Even
Complex/Odd	Complex/Odd

When a linear system is analyzed as a function of spatial variables, causality no longer is required. Values to the “right” of a point can effect the value of the point just as easily as values to the “left”. The point spread function can then be symmetric, and real transfer functions are possible.

Two-Dimensionality and Separability

The most obvious difference between a temporal and spatial Fourier transform is the two-dimensional aspect of the latter. If we write the forward Fourier transform as:

$$ \tilde{U}(f_x, f_y) = \iint_{-\infty}^{\infty} u(x,y)\exp(-j2\pi(f_x \cdot x + f_y \cdot y))dxdy $$

and if $u(x,y) = u_1(x) \cdot u_2(y)$, then we can express the transform as the product of two one-dimensional transforms:

$$ \tilde{U}(f_x, f_y) = \iint u_1(x) \cdot u_2(y)\exp(-j2\pi(f_x \cdot x + f_y \cdot y))dxdy $$$$ = \int u_1(x)\exp(-j2\pi(f_x \cdot x))dx \cdot \int u_2(y)\exp(-j2\pi(f_y \cdot y))dy $$$$ = \tilde{U}(f_x) \cdot \tilde{U}(f_y) $$

Circular Symmetry and Polar Coordinates

Many problems in optics involve fields with circular symmetry. For these problems, it is useful to express the two-dimensional Fourier transform in polar coordinates. Consider a function with circular symmetry $u(\sqrt{x^2 + y^2})$.

The two-dimensional Fourier transform of u is given by:

$$ \tilde{V}(f_x, f_y) = \iint_{-\infty}^{\infty} u(\sqrt{x^2 + y^2})\exp(-j2\pi(f_x \cdot x + f_y \cdot y))dxdy $$

Let $r, \theta$ be radial and angular variables in real space such that:

$$ x = r\cos\theta, \quad y = r\sin\theta $$

Let $\rho, \phi$ be radial and angular variables in Fourier space such that:

$$ f_x = \rho\cos\phi, \quad f_y = \rho\sin\phi $$

The Fourier transform then becomes:

$$ \tilde{V}(\rho\cos\phi, \rho\sin\phi) = \int_0^{\infty} \int_0^{2\pi} u(r')\exp(-j2\pi(r'\cos\theta' \cdot \rho\cos\phi + r'\sin\theta' \cdot \rho\sin\phi))r'dr'd\theta' $$

Derivation of Hankel Transform

Since:

$$ \cos\theta'\cos\phi + \sin\theta'\sin\phi = \cos(\theta' - \phi) $$$$ \tilde{V}(\rho\cos\phi, \rho\sin\phi) = \int_0^{\infty} \int_0^{2\pi} u(r')\exp(-j2\pi\rho r'\cos(\theta' - \phi))r'dr'd\theta' $$

Using the Bessel function identity:

$$ \int_0^{2\pi} \exp(-ja\cos(\theta' - \phi))d\theta' = 2\pi J_0(a) $$

we have:

$$ \tilde{V}(\rho\cos\phi, \rho\sin\phi) = 2\pi \int_0^{\infty} u(r')J_0(2\pi\rho r')r'dr' $$

where $J_0()$ is the zero-order Bessel function. Notice that the right-hand side of the equation is only a function of $\rho$. Hence we can define a new one-dimensional function:

$$ \tilde{U}(\rho) = \tilde{V}(\rho\cos\phi, \rho\sin\phi) $$

such that:

$$ \tilde{U}(\rho) = 2\pi \int_0^{\infty} u(r')J_0(2\pi\rho r')r'dr' $$

This is called the zero order Hankel transform, or sometimes the Fourier-Bessel transform.

V. Sampling Theorem

Sampling Theorem Proof

A function $g(x,y)$ is band-limited if its Fourier transform $\tilde{G}(f_x, f_y)$ has compact support, i.e.:

$$ \tilde{G}(f_x, f_y) = 0 \text{ for } |f_x| > B_x \text{ and } |f_y| > B_y $$

where $B_x$ and $B_y$ are the bandlimits.

Proof of Sampling Theorem

Define a sampling function (the Dirac comb):

$$ \text{comb}(x) = \sum_{n=-\infty}^{\infty} \delta(x - n) $$

Assume a function $g(x,y)$ is bandlimited so that:

$$ \tilde{G}(f_x, f_y) = 0 \text{ for } |f_x| > B_x, |f_y| > B_y $$

Form a sampled version of $f(x,y)$ by multiplying with a sampling function:

$$ g_s(x,y) = g(x,y) \cdot \frac{\text{comb}(x/a)}{a} \cdot \frac{\text{comb}(y/b)}{b} $$

We can show that:

$$ \mathcal{F}\{\text{comb}(x)\} = \text{comb}(f_x) $$

where $\mathcal{F}$ is the Forward Fourier Transform.

Hence:

$$ \mathcal{F}\{g_s(x,y)\} = \tilde{G}_s(f_x, f_y) = \tilde{G}(f_x, f_y) ** \left(\frac{a \cdot \text{comb}(af_x)}{a} \cdot \frac{b \cdot \text{comb}(bf_y)}{b}\right) $$

where $**$ is a two-dimensional convolution.

The result shows:

• Continuous spectrum $\tilde{G}(f_x)$ gets replicated at intervals of $\frac{1}{a}$
• Sampled spectrum shows folding at the folding frequency $B_x$
• The Nyquist frequency is $\frac{1}{a}$ (the sampling rate)

Nyquist Criterion

It is clear from the figure that the aliases will not overlap if:

$$ \frac{1}{a} \geq 2B_x, \text{ or } a \leq \frac{1}{2B_x} $$

and

$$ \frac{1}{b} \geq 2B_y, \text{ or } b \leq \frac{1}{2B_y} $$

This is the Nyquist Criterion.

In real space, the Nyquist criterion says that the highest frequency component in the x direction (frequency $B_x$), must be sampled at least twice.

Recovery via Low-Pass Filtering

If we satisfy the Nyquist criterion, it is a simple matter to recover the original function $g(x,y)$ by low pass filtering the sampled result $g_s(x,y)$.

$$ \tilde{G}_s(f_x, f_y) \cdot \text{rect}\left(\frac{f_x}{2B_x}\right)\text{rect}\left(\frac{f_y}{2B_y}\right) = \tilde{G}(f_x, f_y) $$

In real space, this equation becomes:

$$ g(x,y) = g_s(x,y) ** [4B_x B_y \, \text{sinc}(2B_x x) \, \text{sinc}(2B_y y)] $$

where $\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$ and $\text{rect}(x) = \begin{cases} 1, & |x| \leq \frac{1}{2} \\ 0, & \text{otherwise} \end{cases}$

Note: $\text{rect}(x/a)$ is a rectangle of width $a$ centered at the origin.

This is a statement of the Whittaker-Shannon Sampling Theorem.

Aliasing

What happens when the Nyquist criterion is violated? Aliasing results, where high frequencies fold over and look like lower ones.

Notice the important additional results which can be seen from the above proof of the sampling theorem:

• Sampling in real space produces a repeated continuous function in Fourier space
• Sampling in Fourier space produces a repeated continuous function in real space

VI. Common Transform Pairs

Function	Transform
$\text{rect}(x)\text{rect}(y)$	$\text{sinc}(x)\text{sinc}(y)$
$\delta(x,y)$	$1$
$\delta(x \pm x_0, y \pm y_0)$	$\exp[\pm j2\pi(x_0 f_x + y_0 f_y)]$
$\text{comb}(x)\text{comb}(y)$	$\text{comb}(f_x)\text{comb}(f_y)$
$\text{circ}(r) = \begin{cases} 1, & r < 1 \\ 0, & \text{else} \end{cases}$	$\frac{J_1(2\pi\rho)}{\rho}$

### Aliasing in Practice

Anti-Aliasing Filter Chain

The standard signal processing chain to avoid aliasing:

Continuous   →   Low-Pass    →   Sampler   →   Digital
  Image          Filter           (CCD)        Signal
              (anti-alias)

In optical systems, several mechanisms naturally provide anti-aliasing:

Mechanism	How it works
Optical blur (PSF)	Finite aperture limits resolution, attenuating high frequencies
Detector integration	Each pixel averages over its area, acting as a spatial low-pass filter
Intentional defocus	Spreading the PSF further reduces bandwidth

Aliasing of Noise

Even if the signal satisfies Nyquist, broadband noise may not:

Component	Spectrum	After Sampling
Signal	Band-limited (within Nyquist)	Clean replicated copies
Noise	Broadband (extends beyond Nyquist)	Aliases fold into signal band

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VII. Fourier Transform Reference

Fourier Theorems

1. Linearity

$$ \mathcal{F}\{a \cdot g(x,y) + b \cdot h(x,y)\} = a \cdot \mathcal{F}\{g(x,y)\} + b \cdot \mathcal{F}\{h(x,y)\} $$

2. Similarity Theorem (Scaling Theorem)

If $\mathcal{F}\{g(x,y)\} = G(f_x, f_y)$

then $\mathcal{F}\{g(ax, by)\} = \frac{1}{|ab|} G\left(\frac{f_x}{a}, \frac{f_y}{b}\right)$

3. Shift Theorem

If $\mathcal{F}\{g(x,y)\} = G(f_x, f_y)$

then $\mathcal{F}\{g(x-a, y-b)\} = G(f_x, f_y) \exp[-j2\pi(f_x a + f_y b)]$

4. Parseval’s Theorem

If $\mathcal{F}\{g(x,y)\} = G(f_x, f_y)$

Then:

$$ \iint_{-\infty}^{\infty} |g(x,y)|^2 \, dxdy = \iint_{-\infty}^{\infty} |G(f_x, f_y)|^2 \, df_x df_y $$

5. Convolution Theorem

If $\mathcal{F}\{g(x,y)\} = G(f_x, f_y)$ and $\mathcal{F}\{h(x,y)\} = H(f_x, f_y)$

then:

$$ \mathcal{F}\{g * h\} = G \cdot H $$

or equivalently:

$$ \mathcal{F}\left\{\iint_{-\infty}^{\infty} g(\xi,\eta)h(x-\xi, y-\eta) \, d\xi d\eta\right\} = G(f_x, f_y) H(f_x, f_y) $$

6. Autocorrelation Theorem

If $\mathcal{F}\{g(x,y)\} = G(f_x, f_y)$, then:

$$ \mathcal{F}\{g \star g^*\} = |G(f_x, f_y)|^2 $$

where $\star$ denotes correlation (convolution with the conjugate-reversed function):

$$ \mathcal{F}\left\{\iint_{-\infty}^{\infty} g(\xi,\eta) g^*(\xi-x, \eta-y) \, d\xi d\eta\right\} = |G(f_x, f_y)|^2 $$

The dual relation (Wiener-Khinchin):

$$ \mathcal{F}\{|g(x,y)|^2\} = G \star G^* = \iint G(\xi,\eta) G^*(\xi-f_x, \eta-f_y) \, d\xi d\eta $$

7. Fourier Integral Theorem

$$ \mathcal{F}\mathcal{F}^{-1}\{g(x,y)\} = \mathcal{F}^{-1}\mathcal{F}\{g(x,y)\} = g(x,y) $$

for all points of $g(x,y)$ that are continuous.

Special Functions

Rectangle Function (rect)

$$ \text{rect}(x) = \begin{cases} 1, & |x| \leq \frac{1}{2} \\ 0, & \text{else} \end{cases} $$

Shape: A flat-topped pulse of width 1 centered at the origin.

Fourier transform: $\mathcal{F}\{\text{rect}(x)\} = \text{sinc}(f_x)$

In optics: Models slit apertures, uniform illumination, and ideal bandpass filters.

Sinc Function

$$ \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} $$

Shape: An oscillating function that decays as $1/x$, with zeros at all nonzero integers. The central lobe has width 2; sidelobes alternate in sign.

Fourier transform: $\mathcal{F}\{\text{sinc}(x)\} = \text{rect}(f_x)$

In optics: The diffraction pattern of a slit; the ideal interpolation kernel for bandlimited signals (Shannon reconstruction).

Triangle Function (Λ)

$$ \Lambda(x) = \begin{cases} 1 - |x|, & |x| \leq 1 \\ 0, & \text{else} \end{cases} $$

Shape: A symmetric triangle (tent) of width 2 and height 1.

Fourier transform: $\mathcal{F}\{\Lambda(x)\} = \text{sinc}^2(f_x)$

In optics: The autocorrelation of a rect (two rects convolved); appears in coherence theory and the MTF of defocused systems.

Comb Function (Shah)

$$ \text{comb}(x) = \sum_{n=-\infty}^{\infty} \delta(x - n) $$

Shape: An infinite train of equally-spaced delta functions at integer positions.

Fourier transform: $\mathcal{F}\{\text{comb}(x)\} = \text{comb}(f_x)$ — the comb is its own Fourier transform.

In optics: Represents sampling (multiplying by comb samples a function); diffraction gratings; the reason sampled spectra are periodic.

Circle Function (circ)

$$ \text{circ}(r) = \text{circ}\left(\sqrt{x^2 + y^2}\right) = \begin{cases} 1, & r \leq 1 \\ 0, & \text{else} \end{cases} $$

Shape: A uniform disk of radius 1 in 2D.

Fourier transform: $\mathcal{F}\{\text{circ}(r)\} = \dfrac{J_1(2\pi\rho)}{\rho}$ (the Airy pattern / jinc function)

In optics: Models circular apertures and lenses. Its transform is the Airy disk — the fundamental diffraction limit for circular pupils.

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Physical Significance

In Fourier optics, this result describes:

• The Fraunhofer diffraction pattern of a circular aperture
• The point spread function of a circular lens
• The fundamental limit of optical resolution (Airy disk)

Related Topics

• Bessel function
• Fraunhofer Diffraction
• Optical Transfer Function

Useful Fourier Pairs for Apertures

When apertures are scaled, the similarity theorem applies. These are the most common Fourier pairs in diffraction calculations:

Aperture	Function	Fourier Transform
Rectangular (width $a$, height $b$)	$\text{rect}\left(\dfrac{x}{a}, \dfrac{y}{b}\right)$	$ab \cdot \text{sinc}(a\xi) \cdot \text{sinc}(b\eta)$
Square (side $a$)	$\text{rect}\left(\dfrac{x}{a}, \dfrac{y}{a}\right)$	$a^2 \cdot \text{sinc}(a\xi, a\eta)$
Circular (radius $r_0$)	$\text{circ}\left(\dfrac{r}{r_0}\right)$	$\pi r_0^2 \cdot \dfrac{2J_1(2\pi r_0 \rho)}{2\pi r_0 \rho}$

---

Properties of Delta Functions

1. Defining Properties

$$ \delta(x - x_0) = 0, \quad x \neq x_0 $$$$ \int_{x_1}^{x_2} f(\alpha) \, \delta(\alpha - x_0) \, d\alpha = f(x_0), \quad x_1 < x_0 < x_2 $$

2. Scaling Properties

$$ \delta\left(\frac{x - x_0}{b}\right) = |b| \, \delta(x - x_0) $$$$ \delta(ax - x_0) = \frac{1}{|a|} \delta\left(x - \frac{x_0}{a}\right) $$$$ \delta(-x + x_0) = \delta(x - x_0) $$$$ \delta(-x) = \delta(x) \quad \text{(even function)} $$

3. Properties in Products

$$ f(x) \, \delta(x - x_0) = f(x_0) \, \delta(x - x_0) $$$$ x \, \delta(x - x_0) = x_0 \, \delta(x - x_0) $$$$ \delta(x) \, \delta(x - x_0) = 0, \quad x_0 \neq 0 $$$$ \delta(x - x_0) \, \delta(x - y_0) $$

is not defined

4. Integral Properties

$$ \int_{-\infty}^{\infty} A \, \delta(\alpha - x_0) \, d\alpha = A $$$$ \int_{-\infty}^{\infty} \delta(\alpha - x_0) \, \delta(x - \alpha) \, d\alpha = \delta(x - x_0) $$

Properties of Convolution

Commutative

$$ f(x) * h(x) = h(x) * f(x) $$

Distributive

$$ [a \, v(x) + b \, w(x)] * h(x) = a[v(x) * h(x)] + b[w(x) * h(x)] $$

Shift Invariance

If $f(x) * h(x) = g(x)$

then $f(x - x_0) * h(x) = g(x - x_0)$

Associative

$$ [v(x) * w(x)] * h(x) = v(x) * [w(x) * h(x)] $$

Identity Operator

$$ f(x) * \delta(x) = f(x) $$

Derivatives

$$ f(x) * \delta^{(k)}(x) = f^{(k)}(x) $$

where $\delta^{(k)} = \frac{d^k \delta(x)}{dx^k}$ ($k$ = derivative)

and $f^{(k)} = \frac{d^k f(x)}{dx^k}$

Repeated Convolution

Convolving multiple functions together produces a result that tends toward a Gaussian, regardless of the original shapes (provided they have finite variance):

$$ f_1(x) * f_2(x) * f_3(x) * \cdots * f_n(x) \xrightarrow{n \to \infty} \text{Gaussian} $$

Each convolution “smooths” the result further. Sharp features get rounded, asymmetries cancel out, and the characteristic bell curve emerges. After just a few convolutions, even rectangular or triangular functions look nearly Gaussian.

Scaling

If $f(x) * h(x) = g(x)$

then $f\left(\frac{x}{b}\right) * h\left(\frac{x}{b}\right) = |b| \, g\left(\frac{x}{b}\right)$

Example of Scaling Theorem

Let $f(x) = h(x) = \text{rect}(x)$

$$ g(x) = f(x) * h(x) = \text{tri}(x) $$

where $\text{tri} = \begin{cases} 1 - |x|, & |x| \leq 1 \\ 0, & \text{else} \end{cases}$

---

Let $b = 2$

$$ f\left(\frac{x}{b}\right) = \text{rect}\left(\frac{x}{2}\right) $$$$ g(x) = \text{rect}\left(\frac{x}{2}\right) * \text{rect}\left(\frac{x}{2}\right) = 2 \, \text{tri}\left(\frac{x}{2}\right) $$

Area with no shift $= 1 \times 2 = 2$

The triangle function now extends from $-2$ to $2$ with peak value of $2$ at $x = 0$.

Examples of Convolutions in Two Dimensions

i) $f(x,y) * \delta(x,y)$ - Perfect Sampling

Laser beam scanning negligible width

$$ g(x,y) = f(x,y) * \delta(x,y) = f(x,y) $$

Detector receives exact photograph $f(x,y)$.

---

ii) $f(x,y) * b(x,y)$ - Total Light Through Spot

$$ g(x',y') = \iint f(x,y) \, b(x-x', y-y') \, dx \, dy $$

Detector receives: $g(x,y) = f(x,y) * b(x,y) = f(x,y) \otimes b(x,y)$

---

iii) $f(x,y) * \delta(x)$ - Line Response

$$ g(x,y) = f(x,y) * \delta(x) \cdot 1 $$$$ = \int_{-\infty}^{\infty} f(x,\beta) \, d\beta = \ell_x(x) $$

Detector receives line response: y-dependence is averaged out, result is only a function of x.

Line Response and Transfer Function

$$ \xrightarrow{d\omega} \boxed{h(x,y)} \xrightarrow{\ell_\omega} \quad \mathcal{F}\{\ell_x(x)\} = H(f_x,0) $$

To find other profiles of the general transfer function, we can rotate the line response to any angle (not necessary) just along x or y axis.

$$ H(f_x,0) = \iint f(x,y) e^{-j2\pi f_x x} \, dx \, dy $$$$ = \int \left[\int f(x,\beta) \, d\beta\right] e^{-j2\pi f_x x} \, dx $$

where $\ell_x(x)$ = line response

---

Edge Response (Step Response)

Consider applying a step to the input:

$$ f(x,y) = \text{step}(x) \cdot 1 $$$$ e_x(x) = \mathcal{L}\{\text{step}(x) \cdot 1\} = h(x,y) * * \text{step}(x) \cdot 1 $$$$ = \iint h(\alpha,\beta) \, \text{step}(x-\alpha) \, d\alpha \, d\beta $$$$ = \int \left[\int h(\alpha,\beta) \, d\beta\right] \text{step}(x-\alpha) \, d\alpha $$$$ = \int_{-\infty}^{\infty} \ell_x(\alpha) \, \text{step}(x-\alpha) \, d\alpha = \int_{-\infty}^{x} \ell_x(\alpha) \, d\alpha $$$$ \implies \quad \ell_x(x) = \frac{d}{dx} e_x(x) $$### Transfer Function from Line and Edge Response

Finally:

$$ H(f_x,0) = \mathcal{F}\{\ell_x(x)\} = \mathcal{F}\left\{\frac{d}{dx} e_x(x)\right\} $$$$ = j2\pi f_x \, \mathcal{F}\{e_x(x)\} $$

Physical Interpretation

Impulse Response

$$ \delta(x) \rightarrow \boxed{\frac{h(\alpha)}{H(f_x)}} \rightarrow h(x) $$

• Spectrum of input: $1$
• Spectrum of output: $H(f_x) = \mathcal{F}\{h(x)\}$

Step Response

$$ \text{step}(x) \rightarrow \boxed{\frac{h(x)}{H(f_x)}} \rightarrow \int h(\alpha) \, d\alpha = e_x(x) $$

• Spectrum of input: $\frac{1}{j2\pi f_x}$ (plus DC)
• Transfer function: $H(f_x) = j2\pi f_x \, \mathcal{F}\{e_x(x)\}$

## VIII. Diffraction Theory Fundamentals

Lecture start: 2025-02-05

• Sound waves in gas require scalar values (scalar diff. theory)
• EM waves are vectors (vector theory)

Maxwell’s Equations (Current & Source-free)

Derivation of Wave Equation

We have:

$$ \nabla \times (\nabla \times \vec{E}) = -\mu \cdot \nabla \times \left(\frac{\partial \vec{H}}{\partial t}\right) = -\mu_0 \epsilon(r) \frac{\partial^2 \vec{E}}{\partial t^2} $$

Therefore:

$$ \nabla^2 \vec{E} - \nabla\left(\nabla \cdot \frac{\vec{D}}{\epsilon(r)}\right) - \mu_0 \epsilon(r) \frac{\partial^2 \vec{E}}{\partial t^2} = 0 $$

If medium is homogeneous, $\epsilon(r) = \text{constant}$, then:

$$ \nabla \cdot \frac{\vec{D}}{\epsilon(r)} = \frac{1}{\epsilon}[\nabla \cdot \vec{D}] = 0 $$### Scalar Wave Equation

Then, for each component of the vector $E_x, E_y, E_z$, we have:

$$ \nabla^2 E_i - \frac{n^2}{c^2} \frac{\partial^2 E_i}{\partial t^2} = 0, \quad i = x, y, \text{ or } z $$

where the wave propagation speed is $c/n$.

When these assumptions hold, the cross term can be neglected and we can use a scalar wave equation:

$$ \nabla^2 E - \frac{n^2}{c^2} \frac{\partial^2 E}{\partial t^2} = 0 $$

where $E$ is a scalar representing the amplitude of a single field component.

Helmholtz Equation (Frequency-Domain Form)

Starting from the wave equation:

$$ \nabla^2 E - \frac{n^2}{c_0^2} \frac{\partial^2 E}{\partial t^2} = 0 $$

Take the time Fourier transform of the equation.

Thus:

$$ \nabla^2 \hat{u}(\bar{r},\omega) + \frac{n^2 \omega^2}{c_0^2} \hat{u}(\bar{r},\omega) = 0 $$

Define the wavenumber:

$$ k = \frac{n\omega}{c_0} = \frac{2\pi n}{\lambda} $$

Therefore, the Helmholtz equation is:

Monochromatic Wave

$$ u(\bar{r},t) = a(x,y,z) \cos[2\pi\nu_0 t - \phi(x,y,z)] $$$$ = a(x,y,z) \cos\left[2\pi\nu_0 \left(t - \frac{\phi(x,y,z)}{2\pi\nu_0}\right)\right] $$$$ \Rightarrow \hat{u}(\bar{r},\omega_0) = a(x,y,z) \, \mathcal{F}\left\{\cos\left[2\pi\nu_0 \left(t - \frac{\phi(x,y,z)}{2\pi\nu_0}\right)\right]\right\} $$$$ = a(x,y,z) \, e^{-j2\pi\nu \frac{\phi(x,y,z)}{2\pi\nu_0}} \left[\frac{\delta(\nu-\nu_0)}{2} + \frac{\delta(\nu+\nu_0)}{2}\right] $$$$ = a(x,y,z) \, e^{-j\phi(x,y,z)} \frac{\delta(\nu-\nu_0)}{2} + a(x,y,z) \, e^{+j\phi(x,y,z)} \frac{\delta(\nu+\nu_0)}{2} $$

where $a(x,y,z)$ is real.

Complex Amplitude

We denote the complex amplitude as the amplitude of one of these delta functions (use $\delta(\nu - \nu_0)$):

$$ \boxed{\hat{u}(x,y,z) = \frac{a(x,y,z)}{2} \exp[-j\phi(x,y,z)]} $$

Complex Amplitude

Returning to Original Time Signal

To return to original time signal, we perform the inverse F.T. in time:

$$ u(x,y,z,t) = \mathcal{F}_t^{-1}\left\{\hat{u}(x,y,z)\delta(\nu+\nu_0) + \hat{u}^*(x,y,z)\delta(\nu-\nu_0)\right\} $$$$ = \int \frac{a(x,y,z)}{2} e^{+j\phi(x,y,z)} \delta(\nu+\nu_0) e^{+j2\pi\nu t} \, d\nu $$

$$

• \int \frac{a(x,y,z)}{2} e^{-j\phi(x,y,z)} \delta(\nu-\nu_0) e^{+j2\pi\nu t} , d\nu $$

$$ = \frac{a(x,y,z)}{2} e^{j\phi(x,y,z)} e^{-j2\pi\nu_0 t} + \frac{a(x,y,z)}{2} e^{-j\phi(x,y,z)} e^{j2\pi\nu_0 t} $$$$ = \frac{a(x,y,z)}{2}\left[e^{j(2\pi\nu_0 t - \phi(x,y,z))} + e^{-j(2\pi\nu_0 t - \phi(x,y,z))}\right] $$$$ = a(x,y,z) \cos(2\pi\nu_0 t - \phi(x,y,z)) $$

Returning to Time Signal (Alternative Method)

$$ u(x,y,z,t) = \text{Re}\left\{\hat{u}^*(x,y,z) e^{j2\pi\nu_0 t}\right\} $$

But this is simply:

$$ u(x,y,z,t) = \frac{\hat{u}^*(x,y,z) e^{j2\pi\nu_0 t}}{2} + \frac{\hat{u}(x,y,z) e^{-j2\pi\nu_0 t}}{2} $$

But note this is just a Fourier synthesis where we have added both positive & negative frequency components at frequency $\nu_0$ with weights given by the Fourier coefficients $\hat{u}(x,y,z)$.

---

Plane-Wave Solutions of the Helmholtz Equation

The Helmholtz equation is:

$$ \nabla^2 \hat{u}(\bar{r}) + k^2 \hat{u}(\bar{r}) = 0 $$

A plane-wave solution is:

$$ \hat{u}(\bar{r}) = \hat{C} e^{j\vec{k} \cdot \bar{r}} $$

where the wavevector is $\vec{k} = (k_x, k_y, k_z)$.

Alternative Expression

Let the complex amplitude be written as:

$$ \hat{C} = |A| e^{j\phi} $$

Write the wavevector as:

$$ \vec{k} = (k_x, k_y, k_z) = k(\gamma_x, \gamma_y, \gamma_z) $$

Plane Wave in Cartesian Coordinates

Thus:

$$ \hat{u}(x,y,z) = A e^{j\phi} e^{jk(\gamma_x x + \gamma_y y + \gamma_z z)} $$

Plane Wave Time Domain and Direction Cosines

The real field distribution is obtained by inverse Fourier transforming (in temporal frequency), this spatial distribution multiplied by $\frac{\delta(\nu+\nu_0) + \delta(\nu-\nu_0)}{2}$.

This gives:

$$ u(x,y,z,t) = \frac{A e^{+j(k(\gamma_x x + \gamma_y y + \gamma_z z) + \phi)}}{2} e^{-j2\pi\nu_0 t} + \frac{A e^{-j(k(\gamma_x x + \gamma_y y + \gamma_z z) + \phi)}}{2} e^{+j2\pi\nu_0 t} $$$$ = A \cos(2\pi\nu_0 t - k(\gamma_x x + \gamma_y y + \gamma_z z) - \phi) $$

Direction Cosines Normalization

Recall: $\vec{k} = |\vec{k}|(\gamma_x, \gamma_y, \gamma_z)$

Then:

$$ \vec{k} \cdot \vec{k} = |\vec{k}|^2 (\gamma_x, \gamma_y, \gamma_z) \cdot (\gamma_x, \gamma_y, \gamma_z) = |\vec{k}|^2 (\gamma_x^2 + \gamma_y^2 + \gamma_z^2) $$

But also $\vec{k} \cdot \vec{k} = |\vec{k}|^2$. Therefore:

Propagating vs. Evanescent Components

Case 1: Propagating wave

If $\gamma_x^2 + \gamma_y^2 < 1$, then $\gamma_z$ is real and $e^{jk\gamma_z z}$ represents a propagating wave in the $+z$ direction.

Case 2: Evanescent wave

If $\gamma_x^2 + \gamma_y^2 > 1$, then $\gamma_z$ is imaginary. Write $\gamma_z = j\alpha$ where $\alpha > 0$. Then:

$$ e^{jk\gamma_z z} = e^{jk(j\alpha)z} = e^{-k\alpha z} $$

Plane Wave Propagation

Eigenfunctions of Free-Space Propagation

The transverse plane wave factor $e^{jk(\gamma_x x + \gamma_y y)}$ is an eigenfunction of the free-space propagation operator.

Why This Works

Write the field at the input plane as:

$$ \hat{u}(x,y,z_0) = A e^{j\phi} e^{jk\gamma_z z_0} e^{jk(\gamma_x x + \gamma_y y)} $$

Propagating from $z_0$ to $z_1$ multiplies the field by the transfer function:

$$ H = e^{jk\gamma_z(z_1 - z_0)} $$

Thus the output is:

$$ \hat{u}(x,y,z_1) = e^{jk\gamma_z(z_1 - z_0)} \cdot \hat{u}(x,y,z_0) $$

Substituting the input field:

$$ \hat{u}(x,y,z_1) = A e^{j\phi} e^{jk\gamma_z z_1} e^{jk(\gamma_x x + \gamma_y y)} $$

Eigenvalue Interpretation

Since propagation acts as:

$$ e^{jk(\gamma_x x + \gamma_y y)} \to e^{jk\gamma_z(z_1 - z_0)} \cdot e^{jk(\gamma_x x + \gamma_y y)} $$

Transfer Function of Propagation

Since the eigenvalue describes the transfer function:

$$ H(\gamma_x, \gamma_y) = \exp\left[jk\gamma_z(z_1 - z_0)\right] = \exp\left[jk\sqrt{1-(\gamma_x^2 + \gamma_y^2)}(z_1-z_0)\right] $$

where $\gamma_x^2 + \gamma_y^2 \leq 1$ (non-evanescent)

Complex Amplitude and Amplitude Transmittance

Consider complex amplitude (monochromatic, with time dependence removed) at a plane $z = z_i$:

$$ u(x,y,z_i) = u_i(x,y) $$

We define the complex amplitude transmittance as:

$$ t_i(x,y) = \frac{u_i^+(x,y)}{u_i^-(x,y)} $$

where:

• $u_i^-(x,y)$ = field just before the aperture (incident)
• $u_i^+(x,y)$ = field just after the aperture (transmitted)

Why t = 1 Inside, t = 0 Outside

Therefore, for an ideal aperture:

$$ t_i(x,y) = \begin{cases} 1, & (x,y) \text{ inside aperture} \\ 0, & (x,y) \text{ outside aperture} \end{cases} $$

This is often called the aperture function or pupil function.

Other Types of Transmittance

• Film (amplitude mask): $t_i(x,y)$ describes the dark/light pattern; $0 \leq |t_i| \leq 1$

• Phase object (clear but not flat): $|t_i(x,y)| = 1$ but $t_i$ is complex, representing different phase delays from varying thickness

Angular Plane Wave Spectrum

The angular plane wave spectrum is the fundamental concept connecting spatial patterns to propagating waves. Any complex field distribution $u_i(x,y)$ at a plane $z = z_i$ can be decomposed into a superposition of plane waves traveling in different directions.

Fourier Transform Relationship

The angular spectrum $U_i(\xi, \eta)$ and the spatial field $u_i(x,y)$ form a Fourier transform pair:

$$ U_i(\xi, \eta) = \mathcal{F}\{u_i(x,y)\} = \iint u_i(x,y) \, e^{-j2\pi(\xi x + \eta y)} \, dx \, dy $$$$ u_i(x,y) = \mathcal{F}^{-1}\{U_i(\xi, \eta)\} = \iint U_i(\xi, \eta) \, e^{j2\pi(\xi x + \eta y)} \, d\xi \, d\eta $$

Connecting Spatial Frequency to Propagation Angle

A plane wave propagating in direction $(\gamma_x, \gamma_y, \gamma_z)$ has the form:

$$ C \exp\left[jk(\gamma_x x + \gamma_y y + \gamma_z z)\right] $$

At a fixed plane $z = z_i$, the transverse part is:

$$ C \exp\left[jk(\gamma_x x + \gamma_y y)\right] = C \exp\left[j2\pi\left(\frac{\gamma_x}{\lambda} x + \frac{\gamma_y}{\lambda} y\right)\right] $$

Comparing with the Fourier basis $e^{j2\pi(\xi x + \eta y)}$, the spatial frequencies are:

$$ \boxed{\xi = \frac{\gamma_x}{\lambda} = \frac{\cos\theta_x}{\lambda} = f_x} $$$$ \boxed{\eta = \frac{\gamma_y}{\lambda} = \frac{\cos\theta_y}{\lambda} = f_y} $$

where $\gamma_x = \cos\theta_x$ and $\gamma_y = \cos\theta_y$ are the direction cosines (angles measured from the $x$ and $y$ axes).

Why This Matters

The angular spectrum representation is powerful because:

1. Propagation becomes multiplication: Each plane wave component propagates independently, just multiplied by the transfer function $H(\xi, \eta) = e^{jk\gamma_z \Delta z}$

2. Diffraction is natural: The finite extent of an aperture spreads the angular spectrum, causing the beam to diverge

3. Computation is efficient: Propagation via FFT → multiply by $H$ → IFFT is $O(N \log N)$

Angular Spectrum Diffraction Formulation

The complete recipe for propagating a field from plane $z = z_i$ to plane $z = z_o$ using the angular spectrum method:

Step 1: Input field and its angular spectrum

The input field $u_i(x,y)$ at plane $z = z_i$ has angular spectrum:

$$ U_i(\xi, \eta) = \iint u_i(x,y) \, e^{-j2\pi(\xi x + \eta y)} \, dx \, dy $$

Step 2: Diffraction transfer function

The transfer function for propagation through distance $\Delta z = z_o - z_i$ is:

$$ H(\xi, \eta) = \exp\left[jk(z_o - z_i)\sqrt{1 - \lambda^2(\xi^2 + \eta^2)}\right] $$

Step 3: Propagation in frequency domain

Multiply the angular spectrum by the transfer function:

$$ U_o(\xi, \eta) = U_i(\xi, \eta) \cdot H(\xi, \eta) $$

Step 4: Output field via inverse transform

The output field at plane $z = z_o$ is:

$$ u_o(x,y) = \mathcal{F}^{-1}\{U_i(\xi, \eta) \cdot H(\xi, \eta)\} $$

Explicitly:

$$ u_o(x,y) = \iint U_i(\xi, \eta) \exp\left[jk(z_o - z_i)\sqrt{1 - \lambda^2(\xi^2 + \eta^2)}\right] e^{j2\pi(\xi x + \eta y)} \, d\xi \, d\eta $$

Convolution Form

By the convolution theorem, multiplication in the frequency domain corresponds to convolution in the spatial domain:

$$ u_o(x,y) = u_i(x,y) * \mathcal{F}^{-1}\{H(\xi,\eta)\} $$

The inverse Fourier transform of the transfer function $\mathcal{F}^{-1}\{H\}$ is the impulse response (point spread function) of free-space propagation.

Paraxial Approximation

When the angular spectrum is concentrated near the optical axis (low spatial frequencies dominate), the paraxial approximation simplifies the transfer function.

Condition: $\lambda^2(\xi^2 + \eta^2) \ll 1$ (spatial frequencies much smaller than $1/\lambda$)

Binomial expansion: For small $a$:

$$ (1 - a)^{1/2} \approx 1 - \frac{a}{2} $$

Applying this to the transfer function:

$$ \sqrt{1 - \lambda^2(\xi^2 + \eta^2)} \approx 1 - \frac{\lambda^2}{2}(\xi^2 + \eta^2) $$

The paraxial transfer function becomes:

$$ H_{\text{paraxial}}(\xi,\eta) = \exp\left[jk\Delta z\right] \cdot \exp\left[-j\pi\lambda\Delta z(\xi^2 + \eta^2)\right] $$

Transmitted Field and Angular Plane Wave Spectrum

If we know the amplitude transmittance $t_i(x,y)$ and the incident wave field $u_i(x,y)$, we obviously can calculate the transmitted complex field:

$$ \boxed{u_i^+(x,y) = u_i^-(x,y) \cdot t_i(x,y)} $$

Complete Example: Tilted Plane Wave Through Circular Aperture

This section provides a complete angular-spectrum analysis of an oblique plane wave passing through a circular aperture.

Setup and Notation

Step 1: Incident Plane Wave at Object Plane $z = z_0$

A plane wave incident obliquely arrives at the object plane with the transverse factor:

$$ u^-(x,y) = A e^{j\phi} e^{jk(\gamma_x x + \gamma_y y)} = A_0 e^{j2\pi \mathbf{f}_0 \cdot \mathbf{r}} $$

where $A_0 = A e^{j\phi}$ and $\mathbf{r} = (x, y)$.

Step 2: Aperture Transmittance

Take a circular aperture of radius $a = d/2$. The transmittance is:

$$ t(\mathbf{r}) = \text{circ}\left(\frac{\|\mathbf{r}\|}{a}\right) = \begin{cases} 1, & \|\mathbf{r}\| \leq a \\ 0, & \|\mathbf{r}\| > a \end{cases} $$

The transmitted field immediately after the aperture is:

$$ u^+(\mathbf{r}) = u^-(\mathbf{r}) \cdot t(\mathbf{r}) = A_0 e^{j2\pi \mathbf{f}_0 \cdot \mathbf{r}} \cdot t(\mathbf{r}) $$

Step 3: Angular Spectrum Representation

Using the modulation/shift property:

$$ \mathcal{F}\{e^{j2\pi \mathbf{f}_0 \cdot \mathbf{r}} \cdot t(\mathbf{r})\}(\mathbf{f}) = T(\mathbf{f} - \mathbf{f}_0) $$

where $T(\mathbf{f}) = \mathcal{F}\{t\}(\mathbf{f})$ is the aperture’s Fourier transform.

Therefore the angular spectrum of the transmitted field is:

$$ U^+(\mathbf{f}) = A_0 \, T(\mathbf{f} - \mathbf{f}_0) $$

Step 4: Fourier Transform of Circular Aperture

The 2D FT of a circular aperture of radius $a$ is radially symmetric:

$$ T(\mathbf{f}) = \mathcal{F}\{t\}(\mathbf{f}) = a^2 \frac{J_1(2\pi a \rho)}{\rho}, \quad \rho = \|\mathbf{f}\| $$

Thus:

$$ U^+(\mathbf{f}) = A_0 \, a^2 \frac{J_1(2\pi a \|\mathbf{f} - \mathbf{f}_0\|)}{\|\mathbf{f} - \mathbf{f}_0\|} $$

The angular spectrum is the Airy-like radial profile centered at $\mathbf{f}_0$.

Step 5: Free-Space Propagation from $z_0$ to $z_1$

Propagation multiplies each spectral component by the transfer factor:

$$ H(\mathbf{f}; \Delta z) = \exp(jk_z(\mathbf{f}) \Delta z), \quad \Delta z = z_1 - z_0 $$

with:

$$ k_z(\mathbf{f}) = \sqrt{k^2 - (2\pi f_x)^2 - (2\pi f_y)^2} $$

The propagated angular spectrum is:

$$ U(\mathbf{f}; z_1) = U^+(\mathbf{f}) \cdot H(\mathbf{f}; \Delta z) = A_0 \, T(\mathbf{f} - \mathbf{f}_0) \, e^{jk_z(\mathbf{f}) \Delta z} $$

Step 6: Field at Output Plane $z = z_1$

The field is the inverse 2D FT:

$$ u(\mathbf{r}, z_1) = \iint_{\mathbb{R}^2} U(\mathbf{f}; z_1) e^{j2\pi \mathbf{f} \cdot \mathbf{r}} \, d^2\mathbf{f} $$

Change variable $\mathbf{s} = \mathbf{f} - \mathbf{f}_0$:

$$ u(\mathbf{r}, z_1) = A_0 e^{j2\pi \mathbf{f}_0 \cdot \mathbf{r}} \iint_{\mathbb{R}^2} T(\mathbf{s}) e^{jk_z(\mathbf{s} + \mathbf{f}_0) \Delta z} e^{j2\pi \mathbf{s} \cdot \mathbf{r}} \, d^2\mathbf{s} $$

Step 7: Special/Limiting Cases

Case (a): Small aperture — If $T(\mathbf{s})$ is sharply peaked around $\mathbf{s} = 0$, approximate $k_z(\mathbf{s} + \mathbf{f}_0) \approx k_z(\mathbf{f}_0)$. The plane-wave factor acts almost like a true eigenfunction.

Case (b): Fraunhofer (far-field) — For large $\Delta z$, the field reduces to the FT of the exit pupil at scaled coordinates. The far-field intensity is $|T(\mathbf{f} - \mathbf{f}_0)|^2$ — the Airy disk pattern shifted by incidence angle.

Case (c): Evanescent components — If $\|\mathbf{f} - \mathbf{f}_0\| > k/(2\pi)$, then $k_z$ is imaginary and that component decays as $\exp(-|\text{Im}\, k_z| \Delta z)$. High spatial frequencies (fine details) are attenuated.

Step 8: Final Expression

Physics Summary

Propagation Transfer Function

$$ = \exp\left\{jk(z_e - z_i)\left[1 - \lambda^2(\xi^2 + \eta^2)\right]^{1/2}\right\} $$

Thus, the resultant magnitude and phase of the plane waves at $z_e$ are given by:

Propagation Condition

$$ \gamma_x^2 + \gamma_y^2 < 1 $$

True only when:

$$ \tilde{U}_i(\xi,\eta) \cdot H(\xi,\eta) = \tilde{U}_e(\xi,\eta) \cdot \exp\left\{jk(z_e - z_i)\left[1 - \lambda^2(\xi^2 + \eta^2)\right]^{1/2}\right\} $$$$ \Rightarrow (\lambda\xi)^2 + (\lambda\eta)^2 < 1 $$$$ \xi^2 + \eta^2 < \frac{1}{\lambda^2} $$

Finally, the amplitude distribution at $z_e$ can be calculated by an inverse Fourier transform:

$$ u_e(x,y) = \mathcal{F}^{-1}\left\{\tilde{U}_i(\xi,\eta) \cdot H(\xi,\eta)\right\} $$$$ = \iint_{-\infty}^{\infty} \tilde{U}_i(\xi,\eta) \exp\left\{jk(z_e - z_i)\left[1 - \lambda^2(\xi^2 + \eta^2)\right]^{1/2}\right\} e^{+j2\pi(\xi x + \eta y)} d\xi d\eta $$# Diffraction as a Linear Shift-Invariant System $$ \frac{u_i(x,y)}{\tilde{U}_i(\xi,\eta)} \xrightarrow{\boxed{\text{Free Space Propagation } H(\xi,\eta)}} \frac{u_e(x,y)}{\tilde{U}_e(\xi,\eta) = \tilde{U}_i(x,y) \cdot H(\xi,\eta)} $$

Diffraction from plane $z_i$ to plane $z_e$

Similarity Between 1-dim and 2-dim Fourier Synthesis

One dim:

$$ f(t) \rightarrow \boxed{\quad} \rightarrow \text{sinusoid} $$

Two dim:

• No $y$ dependence: $e^{j\pi\gamma_x x}$ gives different phases of the plane wave
• $\gamma_x$ = effective phase change in x direction (equivalent to frequency of complex sinusoid)

$$ \cos(\cos^{-1}(\gamma_x)) = \frac{\lambda}{\lambda_x} = \gamma_x, \quad \text{or} \quad \lambda_x = \frac{\lambda}{\gamma_x} = \frac{\lambda}{\cos\theta} $$

Square Wave Aperture (One-dim case)

Now, synthesis of a square aperture is identical to one-dim case:

$$ f(x,y) \xrightarrow{H(x,y)} g(x,y) $$# Simple Example of Angular Plane Wave Decomposition

Sine Wave Grating

$$ u_i(x,y) = \cos(2\pi f_0 x) $$

Angular plane wave spectrum:

$$ \tilde{U}_i(\xi,\eta) = \iint u_i(x,y) e^{-j2\pi(\xi x + \eta y)} \, dx \, dy $$$$ = \frac{1}{2}\left[\delta(\xi - f_0) + \delta(\xi + f_0)\right] \delta(\eta) $$

where $\gamma_x = \lambda\xi$, $\gamma_y = \lambda\eta$

Physically, this corresponds to 2 plane waves at angles $\cos^{-1}(f_0\lambda)$ and $\cos^{-1}(-f_0\lambda)$ in x direction, and angles $\cos^{-1}(0\lambda)$ in y direction (i.e., perpendicular to y).

$$ \alpha = \cos^{-1}(f_0\lambda) \quad \text{or} \quad \gamma_x = f_0\lambda $$$$ \beta = \cos^{-1}(-f_0\lambda) \quad \text{or} \quad \gamma_x = -f_0\lambda $$

Interference Pattern

But we know from physics I that the interference of two plane waves creates a sinusoidal interference pattern!

$$ \frac{1}{2}e^{jk_x x} + \frac{1}{2}e^{-jk_x x} = \cos(k_x x) = \cos(|k|\gamma_x x) $$$$ = \cos\left(\frac{2\pi}{\lambda}(f_0\lambda)x\right) = \cos(2\pi f_0 x) $$

Returning to Diffraction Formula

We have:

$$ \tilde{U}_e(\xi,\eta) = \tilde{U}_i(\xi,\eta) \cdot H(\xi,\eta) $$

where:

$$ \tilde{U}_e(\xi,\eta) = \iint u_e(x,y) e^{-j2\pi(\xi x + \eta y)} \, dx \, dy $$$$ \tilde{U}_i(\xi,\eta) = \iint u_i(x,y) e^{-j2\pi(\xi x + \eta y)} \, dx \, dy $$$$ H(\xi,\eta) = \exp\left\{jk(z_e - z_i)\left[1 - \lambda^2(\xi^2 + \eta^2)\right]^{1/2}\right\} $$

where $\gamma_x = \lambda\xi$, $\gamma_y = \lambda\eta$

We can write this in convolution form using the convolution theorem:

$$ \Rightarrow u_e(x,y) = u_i(x,y) ** \mathcal{F}^{-1}\{H(\xi,\eta)\} $$

Fresnel Approximation

This is the Fresnel Approach. However, to simplify things, we first restrict the angular plane waves to small angles ($\gamma_x \ll 1$, $\gamma_y \ll 1$):

$$ \Rightarrow \lambda^2\xi^2 \ll 1 \quad \text{and} \quad \lambda^2\eta^2 \ll 1 $$

Using the binomial expansion $(1-a)^{1/2} \approx 1 - \frac{a}{2}$, we have:

$$ H(\xi,\eta) \approx \exp\left\{jk(z_e - z_i)\left[1 - \frac{\lambda^2}{2}(\xi^2 + \eta^2)\right]\right\} $$$$ = \exp\{jk(z_e - z_i)\} \exp\{-j\pi\lambda(z_e - z_i)(\xi^2 + \eta^2)\} $$

We now need to calculate $h(x,y)$ by inverse Fourier transform:

$$ \mathcal{F}_\xi^{-1}\mathcal{F}_\eta^{-1}\left\{\exp\left[-j\pi\lambda(z_e - z_i)(\xi^2 + \eta^2)\right]\right\} $$$$ = \mathcal{F}_\xi^{-1}\left\{\exp\left[-j\pi\lambda(z_e - z_i)\xi^2\right]\right\} \cdot \mathcal{F}_\eta^{-1}\left\{\exp\left[-j\pi\lambda(z_e - z_i)\eta^2\right]\right\} $$

But $\exp\left[-j\pi\lambda(z_e - z_i)\xi^2\right] = \text{Gaus}(a\xi)$

where $a = \sqrt{j\lambda(z_e - z_i)}$

and $\text{Gaus}(\xi) = \exp(-\pi\xi^2)$

Fresnel Diffraction Formula

And $\mathcal{F}^{-1}\{\text{Gaus}(a\xi)\} = \frac{1}{|a|} \text{Gaus}\left(\frac{x}{a}\right)$

Recall Gaussian problem 2.4d: $\mathcal{B}\{e^{-\pi\xi^2}\} = e^{-\pi\xi^2}$

$$ \implies \mathcal{F}^{-1}\left\{\exp\left[-j\pi\lambda(z_e - z_i)\xi^2\right]\right\} $$$$ = \frac{1}{\sqrt{j\lambda(z_e - z_i)}} \exp\left[-\frac{\pi x^2}{j\lambda(z_e - z_i)}\right] $$

and $\mathcal{F}^{-1}\mathcal{F}^{-1}\left\{\exp\left[-j\pi\lambda(z_e - z_i)(\xi^2 + \eta^2)\right]\right\}$

$$ = \frac{1}{j\lambda(z_e - z_i)} \exp\left[\frac{j\pi(x^2 + y^2)}{\lambda(z_e - z_i)}\right] $$

Finally, $\mathcal{F}^{-1}\mathcal{F}^{-1}\{H(\xi,\eta)\} = \frac{\exp\{jk(z_e - z_i)\}}{j\lambda(z_e - z_i)} \exp\left[\frac{j\pi(x^2 + y^2)}{\lambda(z_e - z_i)}\right]$

We thus have the convolution form (with $\frac{\pi}{\lambda} = \frac{k}{2}$):

$$ \boxed{u_e(x,y) = u_i(x,y) ** \frac{\exp\{jk(z_e - z_i)\}}{j\lambda(z_e - z_i)} \exp\left[\frac{jk(x^2 + y^2)}{2(z_e - z_i)}\right]} $$$$ = \frac{\exp\{jk(z_e - z_i)\}}{j\lambda(z_e - z_i)} \iint u_i(\alpha,\beta) \exp\left\{\frac{jk[(x-\alpha)^2 + (y-\beta)^2]}{2(z_e - z_i)}\right\} d\alpha d\beta $$

Fresnel Formula (Angular Approximation / Fresnel Formula)

---

Huygens’ Principle

(Tie long awaited)

Recall from first quarter that Huygens’ principle stated that any field distribution could be thought of as a superposition of spherical waves:

$$ = \sum \text{superposition} $$

Mathematical Statement of Huygens’ Principle

$$ u_e(x_2, y_2) = \iint u_i(x_1, y_1) \frac{\exp(jkr)}{j\lambda r} \, dx_1 dy_1 $$

where $\exp\left(\frac{jkr}{j\lambda r}\right)$ is a spherical wave (multiplied by $\frac{1}{j\lambda}$)

Source of spherical waves in $x_1, y_1$

$$ r = \sqrt{(z_e - z_i)^2 + (x_1 - x_2)^2 + (y_1 - y_2)^2} $$$$ r = (z_e - z_i)\sqrt{1 + \frac{(x_1 - x_2)^2 + (y_1 - y_2)^2}{(z_e - z_i)^2}} $$

Approximations

(1) Fresnel Approximation

As before, we limit the observations to regions where:

$$ (z_e - z_i) >> |x_1 - x_2| \quad \text{and} \quad (z_e - z_i) >> |y_1 - y_2| $$

(2) $\frac{1}{r}$ Approximation

Then the binomial expansion for $r$ gives:

$$ r = (z_e - z_i) + \frac{1}{2(z_e - z_i)}\left[(x_1 - x_2)^2 + (y_1 - y_2)^2\right] $$

Use this approximation for exponential

(3) We also approximate $r \approx (z_e - z_i)$ in denominator of spherical wave. Reason: $\frac{1}{r} \approx \frac{1}{z_e - z_i}$, but we cannot use $\exp(jkr) \approx \exp(jkz)$ because $k$ may be large and this may cause the exponential to change from +1 to -1.

Proper Approximation for Huygens’ Principle

The proper approximation becomes:

$$ u_e(x_2, y_2) = \iint u_i(x_1, y_1) \frac{\exp\left\{jk\left[(z_e - z_i) + \frac{1}{2(z_e - z_i)}(x_1 - x_2)^2 + (y_1 - y_2)^2\right]\right\}}{j\lambda(z_e - z_i)} \, dx_1 dy_1 $$$$ \boxed{= \frac{\exp\{jk(z_e - z_i)\}}{j\lambda(z_e - z_i)} \iint u_i(x_1, y_1) \exp\left[\frac{jk}{2(z_e - z_i)}\left((x_1 - x_2)^2 + (y_1 - y_2)^2\right)\right] dx_1 dy_1} $$# Criterion for Validity of Fresnel Approximation

Homework: Goodman: 4-5, 4-7; Cathey: 1-3

(or quadratic form of Huygens’ Principle)

Recall that the Fresnel approximation was valid for “small” angular plane waves: $\gamma_x \ll 1$, $\gamma_y \ll 1$.

Specifically, the approximations were (for Huygens’ development):

i) $\frac{1}{r} \approx \frac{1}{(z_e - z_i)}$

ii) $\exp[jkr] \approx \exp\left\{jk\left[(z_e - z_i) + \frac{(x_1 - x_2)^2 + (y_1 - y_2)^2}{2(z_e - z_i)}\right]\right\}$

For i) to be true, we require:

$$ (z_e - z_i) >> |x_1 - x_2| $$

and

$$ (z_e - z_i) >> |y_1 - y_2| $$

For ii) to be true, we require the next term in the binomial expansion to be insignificant:

$$ r = \left[(z_e - z_i)^2 + (x_1 - x_2)^2 + (y_1 - y_2)^2\right]^{1/2} = (z_e - z_i)\left\{1 + \frac{(x_1 - x_2)^2 + (y_1 - y_2)^2}{(z_e - z_i)^2}\right\}^{1/2} $$$$ \exp(jkr) = \exp\left\{jk(z_e - z_i)\left[1 + \frac{(x_1 - x_2)^2 + (y_1 - y_2)^2}{2(z_e - z_i)^2} - \frac{[(x_1-x_2)^2 + (y_1-y_2)^2]^2}{8(z_e - z_i)^4} + \cdots\right]\right\} $$# Fresnel Condition

In order to neglect third term in expansion, we must have:

$$ \frac{k(z_e - z_i)\left[(x_1 - x_2)^2 + (y_1 - y_2)^2\right]^2}{8(z_e - z_i)^4} \ll 1 $$$$ \boxed{(z_e - z_i)^3 >> \frac{\pi}{4\lambda}\left[(x_1 - x_2)^2 + (y_1 - y_2)^2\right]^2_{\text{max}}} $$

This is a sufficient Fresnel Condition

(It may not actually be necessary)

Two Forms for the Fresnel Diffraction Formula

From here on, let $z$ simply be the distance between observation plane and aperture, i.e., $z = (z_e - z_i)$

Convolution Form

$$ U_2(x,y) = \frac{e^{jkz}}{j\lambda z} \iint U_1(\alpha,\beta)\, \exp\!\left[jk\frac{(x-\alpha)^2 + (y-\beta)^2}{2z}\right] d\alpha\, d\beta $$$$ U_2(x,y) = \frac{e^{jkz}}{j\lambda z} \bigl[U_1 * h\bigr](x,y), \quad h(x,y) = \exp\!\left[jk\frac{x^2+y^2}{2z}\right] $$

Transfer function:

$$ H(\xi,\eta) = \exp[jkz] \exp[-j\pi\lambda z(\xi^2 + \eta^2)] $$

By expanding out the quadratic phase in the integral and factoring, we arrive at a second form:

---

Fourier Transform Form (Quadratic-Phase / Fourier Form)

$$ U_2(x,y) = \frac{e^{jkz}}{j\lambda z} \exp\!\left[jk\frac{x^2+y^2}{2z}\right] \iint U_1(\alpha,\beta)\, \exp\!\left[jk\frac{\alpha^2+\beta^2}{2z}\right] \exp\!\left[-j\frac{2\pi}{\lambda z}(x\alpha+y\beta)\right] d\alpha\, d\beta $$

Recognizing the remaining exponential as the Fourier transform kernel, this can be written compactly as:

$$ U_2(x,y) = \frac{e^{jkz}}{j\lambda z} \exp\!\left[jk\frac{x^2+y^2}{2z}\right] \mathcal{F}\!\left\{U_1(\alpha,\beta)\, \exp\!\left[jk\frac{\alpha^2+\beta^2}{2z}\right]\right\}\Bigg|_{f_x = \frac{x}{\lambda z},\; f_y = \frac{y}{\lambda z}} $$

where the 2D Fourier transform is:

$$ \mathcal{F}\{g(\alpha,\beta)\}(f_x,f_y) = \iint g(\alpha,\beta)\, e^{-j2\pi(f_x\alpha + f_y\beta)}\, d\alpha\, d\beta $$# Fraunhofer Diffraction Formula (Far Field)

Note that this equation can be thought of as some phase factors multiplying the Fourier transform of the aperture transmittance and a quadratic phase factor $\exp\left\{\frac{jk[\alpha^2+\beta^2]}{2z}\right\}$

Fraunhofer Diffraction Formula (Far Field)

In the Fourier transform form of the Fresnel diffraction formula, the input field $U_1(\alpha,\beta)$ is multiplied by a quadratic-phase chirp factor $\exp\{jk(\alpha^2+\beta^2)/2z\}$ before being Fourier transformed. This chirp encodes the curvature of the Fresnel kernel — it’s what makes near-field diffraction different from a pure Fourier transform.

If the propagation distance $z$ is large enough, the phase variation of this chirp across the entire aperture becomes negligible, and the exponential can be replaced by unity:

$$ \exp\left\{\frac{jk[\alpha^2+\beta^2]}{2z}\right\} \approx 1 $$

This requires:

$$ \frac{k(\alpha^2+\beta^2)}{2z} \ll 1 \quad \text{for all } \alpha, \beta \text{ in the diffracting aperture} $$$$ \Rightarrow \boxed{z >> \frac{k(\alpha^2+\beta^2)_{\text{max}}}{2}} $$

Fraunhofer Condition

For an input aperture of radius $a$, the maximum phase variation of the chirp factor is:

$$ \Delta\phi = k\frac{a^2}{2z} = \pi\frac{a^2}{\lambda z} $$

Requiring $\Delta\phi \ll 1$ gives the equivalent condition:

$$ \boxed{\frac{a^2}{\lambda z} \ll 1} $$

Under this condition, the quadratic phase term in the input plane is negligible, so the Fresnel diffraction equation reduces to:

$$ \boxed{U_2(x,y) = \frac{e^{jkz}}{j\lambda z} \exp\!\left[jk\frac{x^2+y^2}{2z}\right] \iint U_1(\alpha,\beta)\, \exp\!\left[-j\frac{2\pi}{\lambda z}(x\alpha + y\beta)\right] d\alpha\, d\beta} $$

Recognizing the Fourier kernel, this may be written as:

$$ U_2(x,y) = \frac{e^{jkz}}{j\lambda z} \exp\!\left[jk\frac{x^2+y^2}{2z}\right] \mathcal{F}\{U_1(\alpha,\beta)\}\bigg|_{f_x = \frac{x}{\lambda z},\; f_y = \frac{y}{\lambda z}} $$

Fraunhofer Diffraction Formula

# Intensity in Fraunhofer Diffraction

Here $u_e(x,y)$ is the complex electric field at plane $z_e$. Most detectors of light (film, eye, photodiode) are only capable of measuring power.

We define the intensity (power/unit area) as:

$$ I_e(x,y) = u_e(x,y) \cdot u_e^*(x,y) $$

This gives an extremely simple diffraction formula in the Fraunhofer region:

$$ I_e(x,y) = \left| \frac{\exp(jkz)}{j\lambda z} \exp\left\{\frac{jk(x^2+y^2)}{2z}\right\} \mathcal{F}\{u_i(\alpha,\beta)\}\right|^2 $$$$ = \frac{1}{(\lambda z)^2} \left| \mathcal{F}\{u_i(\alpha,\beta)\}\bigg|_{\xi=\frac{x}{\lambda z}, \eta=\frac{y}{\lambda z}} \right|^2 $$

Or simply, the intensity pattern (as seen by eye, etc.) is the squared magnitude of the two-dim Fourier Transform (scaled by $\frac{1}{\lambda z}$).

$$ \text{Inverse Square Law} $$# Fraunhofer Condition

The Fraunhofer condition is fairly severe:

$$ z >> \frac{k(\alpha^2 + \beta^2)_{\text{max}}}{2} = \frac{\pi(\alpha^2 + \beta^2)_{\text{max}}}{\lambda} $$

Nevertheless, it is often a useful approximation.

Transfer Functions Associated with Approximations

$$ H(\xi,\eta) = \exp\left\{jkz\left[1 - \lambda^2(\xi^2+\eta^2)\right]^{1/2}\right\} $$

— No approximation

$$ H_F(\xi,\eta) \cong \exp\left\{jkz\left[1 - \frac{\lambda^2}{2}(\xi^2+\eta^2)\right]\right\} $$

— Fresnel Approximation

$H_F(\xi,\eta) \rightarrow$ does not exist, because the Fourier transform operator is not shift invariant and thus has no transfer function associated with it.

# Example of Fraunhofer Diffraction - Microwave Antenna

Microwave Antenna (Transmitter)

Setup:

• $\lambda = 1$ mm
• $d = 1$ meter
• $z = 10$ km

Question: How much power will be received by the receiving antenna?

Equivalent Set-up

$$ u^i(x,y) = A \cdot \text{circ}\left(\frac{r}{d/2}\right) $$

where $r$ = radius

Fraunhofer Condition

$$ z >> \frac{k(\alpha^2 + \beta^2)_{\text{max}}}{2} = \frac{\pi r^2}{\lambda} $$$$ z >> \frac{\pi(0.5)^2}{10^{-3}} = 750 \text{ m} $$

$\implies z = 10$ km is in the Fraunhofer region

Intensity Pattern

$$ I_e(x,y) = \frac{1}{(\lambda z)^2} \left| \mathcal{F}\mathcal{F}\left\{A \cdot \text{circ}\left(\frac{r}{d/2}\right)\right\}\right|^2_{\xi=\frac{x}{\lambda z}, \eta=\frac{y}{\lambda z}} $$$$ = \frac{1}{(\lambda z)^2} \left| \frac{A\pi d^2}{4} \text{somb}\left(\frac{d\rho}{\lambda z}\right) \right|^2 $$

or:

$$ = \frac{1}{(\lambda z)^2} \left| \frac{Ad}{2} \frac{J_1(\pi d\rho/\lambda z)}{\rho/\lambda z} \right|^2 $$

where $\rho = \sqrt{x^2 + y^2}$

Fraunhofer Diffraction - Airy Pattern

The intensity pattern follows a sombrero function:

$$ \text{somb}\left(\frac{d\rho}{\lambda z}\right) = \frac{2J_1\left(\frac{\pi d\rho}{\lambda z}\right)}{\frac{\pi d\rho}{\lambda z}} $$

and $J_1(\pi)(1.22) = 0$

The first zero of the Bessel function occurs at:

$$ \frac{d\rho_1}{\lambda z} = 1.22 $$$$ \rho_1 = \frac{(1.22)\lambda z}{d} $$$$ = \frac{(1.22)(10^{-3})(10^4)}{1} = 12.2 \text{ m} $$

Form of Result

From $\rho = 1.22 \frac{\lambda z}{d}$

and $\frac{\rho}{z} \approx \sin\theta$

we have:

$$ \boxed{\sin\theta = 1.22 \frac{\lambda}{d}} $$

Very Important Equation

# Far Field Diffraction from Square Aperture

For a square aperture, the result is almost the same:

$$ \text{rect}\left(\frac{x'}{d}, \frac{y'}{d}\right) \xrightarrow{z} I \propto \left| d^2 \text{sinc}\left(\frac{dx'}{\lambda z}, \frac{dy'}{\lambda z}\right) \right|^2 $$$$ \text{sinc}(1) = 0 $$$$ \Rightarrow \frac{da}{\lambda z} = 1, \text{ or } a = \frac{\lambda z}{d} $$

But $\frac{a}{z} = \sin\theta$

$$ \implies \boxed{\sin\theta = \frac{\lambda}{d}} $$

Thus we see a general result:

$$ \boxed{\sin\theta = K\frac{\lambda}{d}} $$

Far Field Diffraction from an Aperture

where $K$ is a value close to 1, and is dependent on the shape of the aperture.

Aperture	K
Square	$K = 1$
Circle	$K = 1.22$

Notice Effect of Rectangular Antennas

Rectangular Antenna Pattern

An antenna with a rectangular shape such as shown has less broadening of the beam in the horizontal direction ($\theta$) than the vertical direction ($\phi$).

$$ \sin\theta = \frac{\lambda}{d_1}, \quad \sin\phi = \frac{\lambda}{d_2} $$

Thus, a rectangular antenna has less beam broadening in the horizontal direction ($\theta$) than in the vertical direction ($\phi$).

Resolution Considerations

Consider a radar antenna that tries to locate an object by rotating with beam “illuminates” object.

Two objects can be distinguished separately only if beam is narrow. A broad antenna has better resolution in horizontal direction than vertical direction.

Diffraction Grating (Absorption Type)

Setup

Frequency of grating in $x$ is $\frac{1}{T} = f_0$

The depth of modulation is represented by $m$ where $0 < m \leq 1$.

$$ u_i^-(x,y) = A $$$$ t(x,y) = \left[\frac{1}{2} + \frac{m}{2}\cos(2\pi f_0 x)\right] \text{rect}\left(\frac{x}{\ell}\right) \text{rect}\left(\frac{y}{\ell}\right) $$$$ \implies u_i^+(x,y) = A\left[\frac{1}{2} + \frac{m}{2}\cos(2\pi f_0 x)\right] \text{rect}\left(\frac{x}{\ell}\right) \text{rect}\left(\frac{y}{\ell}\right) $$

Far Field Calculation

$$ \mathcal{FF}\{u_i^+(x,y)\} = A\left[\frac{1}{2}\delta(\xi,\eta) + \frac{m}{4}\delta(\xi+f_0,\eta) + \frac{m}{4}\delta(\xi-f_0,\eta)\right] ** \ell^2 \text{sinc}(\ell\xi)\text{sinc}(\ell\eta) $$$$ = \frac{A\ell^2}{2}\text{sinc}(\ell\eta)\left[\text{sinc}(\ell\xi) + \frac{m}{2}\text{sinc}(\ell(\xi+f_0)) + \frac{m}{2}\text{sinc}(\ell(\xi-f_0))\right] $$

Output Field

$$ u_e(x',y') = \frac{A\ell^2}{j\lambda z}\exp(jkz)\exp\left(j\frac{k(x'^2+y'^2)}{2z}\right)\text{sinc}\left(\frac{\ell y'}{\lambda z}\right)\left\{\text{sinc}\left(\frac{\ell x'}{\lambda z}\right) + \frac{m}{2}\text{sinc}\left[\frac{\ell}{\lambda z}(x'+\lambda z f_0)\right] + \frac{m}{2}\text{sinc}\left[\frac{\ell}{\lambda z}(x'-\lambda z f_0)\right]\right\} $$

The intensity pattern shows peaks at $x' = 0$, $x' = \pm f_0\lambda z$, with width $\frac{\ell x'}{\lambda z} = 1 \Rightarrow x' = \frac{\lambda z}{\ell}$.

Diffraction Grating - Resolution Analysis

Key Observations

Note:

1. Distance from origin to first zero of sinc function $= \frac{\lambda z}{\ell}$
2. Distance to next sinc function $= f_0\lambda z = \frac{\lambda z}{T}$

Since $f_0 = \frac{1}{T}$:

$$ \implies \text{If } T << \ell, $$$$ \frac{\lambda z}{\ell} << \frac{\lambda z}{T} \text{ and the overlap of sinc functions is small.} $$

Intensity Pattern

$$ I_e(x,y) \propto \frac{A^2\ell^4}{4\lambda^2 z^2}\text{sinc}^2\left(\frac{\ell y}{\lambda z}\right)\left\{\text{sinc}^2\left(\frac{\ell x}{\lambda z}\right) + \frac{m^2}{4}\text{sinc}^2\left[\frac{\ell}{\lambda z}(x'+\lambda z f_0)\right] + \frac{m^2}{4}\text{sinc}^2\left[\frac{\ell}{\lambda z}(x'-\lambda z f_0)\right]\right\} $$

Diffraction Orders

The position of the $\pm 1$ order components $= f_0\lambda z$

This device can be used to disperse white light into colors.

• The width of each order is $\frac{\lambda z}{\ell}$
• The spatial separation is $f_0\lambda z$

$$ \text{Total number of resolvable wavelengths} = \frac{(f_0\lambda z)}{\lambda z/\ell} = f_0\ell = \frac{\ell}{T} $$$$ = \text{number of cycles of sinc function (lines in grating)} $$

Spectral Resolution

# Diffraction Grating (Phase Type)

Setup

$$ u^-(x,y) = 1 $$$$ t(x,y) = \exp\left[j\frac{m}{2}\sin(2\pi f_0 x)\right] \text{rect}\left(\frac{x}{\ell}\right) \text{rect}\left(\frac{y}{\ell}\right) $$$$ u^+(x,y) = t(x,y) $$

The phase varies sinusoidally between $+\frac{m}{2}$ and $-\frac{m}{2}$.

Bessel Function Identity

$$ \exp\left[j\frac{m}{2}\sin(2\pi f_0 x)\right] = \sum_{q=-\infty}^{\infty} J_q\left(\frac{m}{2}\right)\exp(j2\pi q f_0 x) $$

Fourier Transform

$$ \implies \mathcal{FF}\{t(x,y)\} = \ell^2 \text{sinc}(\ell\xi)\text{sinc}(\ell\eta) * \left[\sum_{q=-\infty}^{\infty} J_q\left(\frac{m}{2}\right)\delta(\xi - qf_0, \eta)\right] $$$$ = \sum_{q=-\infty}^{\infty} \ell^2 J_q\left(\frac{m}{2}\right)\text{sinc}[\ell(\xi - qf_0)]\text{sinc}(\ell\eta) $$

Output Field

$$ \Rightarrow u(x',y') = \frac{\ell^2}{j\lambda z}\exp(jkz)\exp\left[j\frac{k}{2z}(x'^2+y'^2)\right]\sum_{q=-\infty}^{\infty} J_q\left(\frac{m}{2}\right)\text{sinc}\left[\frac{\ell}{\lambda z}(x'-qf_0\lambda z)\right]\text{sinc}\left(\frac{\ell y'}{\lambda z}\right) $$

Again if $f_0 >> \ell$ (or $T << \ell$), there is little overlap of sinc functions. Then:

$$ I(x',y') \cong \left(\frac{\ell^2}{\lambda z}\right)^2 \sum_{q=-\infty}^{\infty} J_q^2\left(\frac{m}{2}\right)\text{sinc}^2\left[\frac{\ell}{\lambda z}(x'-qf_0\lambda z)\right]\text{sinc}^2\left(\frac{\ell y'}{\lambda z}\right) $$

Behavior of $J_q^2\left(\frac{m}{2}\right)$ Terms

Bessel Function Squared Behavior

The functions $J_q^2\left(\frac{m}{2}\right)$ show characteristic behavior:

• $J_0^2$ starts at 1 and decreases
• Higher order terms $J_1^2, J_2^2, ...$ start at 0 and have peaks at increasing values of $\frac{m}{2}$

Eliminating Zeroth Order

Example: If we choose $\frac{m}{2} = 2.5$ (or $m = 5$ radians peak-to-peak phase change), $J_0^2(2.5) = 0$, and there is no zeroth order peak.

Diffraction Pattern

With $m = 8$ radians:

• Note $\frac{m}{2} = 4$ and $J_1(4) \approx 0$
• No first order terms

The resulting diffraction pattern shows:

• Orders at positions $q^{th}$, $3^{rd}$, $2^{nd}$, $1^{st}$, $0$, $1^{st}$, $2^{nd}$, $3^{rd}$, $4^{th}$…
• Relative intensities determined by $J_q^2\left(\frac{m}{2}\right)$

Antennas and Huygens Arrays

Far Field Diffraction Pattern for Square Antenna

Recall the far field diffraction pattern for a single square antenna (width = $d$):

$$ t(x,y) = \text{rect}\left(\frac{x}{d}, \frac{y}{d}\right) $$$$ I(\xi_x, \xi_y) = \left(\frac{1}{\lambda z}\right)^2 \left| \iint u_i(x,y) \exp\{-j2\pi(x\xi_x + y\xi_y)\} dx\,dy \right|^2 $$

where $\xi_x$, $\xi_y$ are direction cosines.

If we have a square antenna, this becomes:

$$ I(\xi_x, \xi_y) = I\left(\frac{\cos\theta}{\lambda}, \frac{\cos\phi}{\lambda}\right) = \left(\frac{1}{\lambda z}\right)^2 \left| d^2 \text{sinc}\left(\frac{d\cos\theta}{\lambda}\right) \text{sinc}\left(\frac{d\cos\phi}{\lambda}\right) \right|^2 $$

The intensity pattern shows angular sensitivity in the $\frac{\cos\theta}{\lambda}$ direction.

Antenna Array

Consider the effect of spacing $M$ antennas a distance $D$ apart ($M$ is odd, $D \geq d$):

$$ t(x,y) = \text{rect}\left(\frac{x}{d}, \frac{y}{d}\right) * \left[\frac{1}{D}\text{comb}\left(\frac{x}{D}\right) \cdot \text{rect}\left(\frac{x}{MD}\right)\right] $$

This creates a uniform linear array of $M$ antenna elements.

Antenna Array - Intensity Pattern

Array Intensity Formula

$$ \Rightarrow I\left(\frac{\cos\theta}{\lambda}, \frac{\cos\phi}{\lambda}\right) = \left(\frac{1}{\lambda z}\right)^2 \left| d^2 \text{sinc}\left(\frac{d\cos\theta}{\lambda}, \frac{d\cos\phi}{\lambda}\right) \right|^2 \cdot \left| \text{comb}\left(\frac{D\cos\theta}{\lambda}\right) * \text{sinc}\left(\frac{MD\cos\theta}{\lambda}\right) \right|^2 $$

The resulting pattern shows:

• Main narrow central lobe (labeled $I$)
• Grating lobes at regular intervals in $\frac{\cos\theta}{\lambda}$

Grating Lobes

These can cause some confusion in a receiving antenna in determining the actual angle at which a signal was sent.

Physical Reason for Grating Lobes

Two point sources make two sets of plane waves.

Phases received by antennas of both plane waves are identical.

• Array of Antennas
• Phases measured by antennas of both plane waves are identical

Antenna Arrays - Main Lobe Condition

Eliminating Grating Lobes

If we want to ensure that only the main lobe occurs, we require:

$$ \frac{\cos\theta}{\lambda} = \frac{1}{D} \text{ does not exist} $$$$ \Rightarrow \cos\theta = \frac{\lambda}{D} > 1 \quad (\Rightarrow \cos\theta \text{ does not exist}) $$$$ \implies D < \lambda $$

General Case - Linear Array of Arbitrary Antennas

For the general case of a linear array of arbitrary shaped antennas with shape $a(x,y)$ we have:

$$ \Rightarrow t(x,y) = a(x,y) * \left[\text{samp}(x)\right] $$

where $\text{samp}(x)$ is some sampling function (not necessarily equally spaced delta functions) of finite length.

From the convolution theorem, we simply have:

$$ I\left(\frac{\cos\theta}{\lambda}, \frac{\cos\phi}{\lambda}\right) = \left| A\left(\frac{\cos\theta}{\lambda}, \frac{\cos\phi}{\lambda}\right) \right|^2 \cdot \left| \mathcal{FF}\{\text{samp}(x)\} \right|^2 $$$$ = \underbrace{\text{Antenna Factor}}_{\text{Antenna Factor}} \times \underbrace{\text{Array Factor}}_{\text{Array Factor}} $$

Array Theorem

# Array Beam Steering

Intensity with Direction Cosines

Since we have:

$$ I\left(\frac{\cos\theta'}{\lambda}, \frac{\cos\phi}{\lambda}\right) = \left(\frac{1}{\lambda z}\right)^2 \left| T\mathcal{FF}\{t(x,y)\}\bigg|_{\xi=\frac{\cos\theta}{\lambda}, \eta=\frac{\cos\phi}{\lambda}} \right|^2 $$

And the Fourier shift theorem states:

$$ \mathcal{FF}\{t(x,y)e^{j2\pi\alpha x}\} = T\left(\frac{\cos\theta}{\lambda} - \alpha, \frac{\cos\phi}{\lambda}\right) $$$$ = T\left(\frac{\cos\theta - \cos\theta'}{\lambda}, \frac{\cos\phi}{\lambda}\right) $$

where $\alpha = \frac{\cos\theta'}{\lambda}$

Phase Delay Beam Steering

Then by simply providing a phase delay from one receiver (or transmitter) to the other, we can steer the beam.

The width of a beam (given by diffraction) is reduced by the overall array size (maximum spatial frequency at the edges) and the angular range (given by maximum angle in which aliasing has not occurred) are independent quantities.

Phased Array Implementation

Phase shifters
    ↓
[α] → [2α] → [3α] → [4α] →  Main lobe direction
         ↘    ↘    ↘    ↘
              (XMT R)

# Phased Array Receiving

Equivalent Large Antenna

In receiving, if we record a sum of all antennas with a given linear phase shift, we have the equivalent of a large antenna pointed at a specific direction.

Digital Beam Forming

        [φ₁]
θ' →  D-[φ₂]→ Σ → record
        [φ₃]
        [φ₄]

Main lobe
pattern angle corresponds
to phase shifts φ₁-φ₄ from e^(j2π cos θ' x), and beam is steered by cos θ

D → [rec] → [φ₁] ↘
D → [rec] → [φ₂] → Σ → result
D → [rec] → [φ₃] ↗
D → [rec] → [φ₄] ↗

Where $\phi_1 \to \phi_4$ come from $e^{j2\pi\frac{\cos\theta}{\lambda}x}$ and beam is steered by $\cos\theta$.

# Adaptive Beam Forming and Nulling

Angular Image of Energy

In this way, an angular image of the energy can be produced:

$$ \text{Signal at angle } \cos\theta_1: \sum_{k=1}^{N} f_k(t) e^{j2\pi \frac{\cos\theta_1}{\lambda} k\Delta} $$

where:

• $D$ = spacing between antennas
• $N$ = number of antennas

Adaptive Beam Forming and Nulling

Goal: Choose phases that place null on top of noise source

$$ \text{Signal at any angle } \cos\theta_p: \sum_{k=1}^{N} f_k(t) e^{j\pi \frac{\cos\theta_p}{\lambda} k\Delta} $$

System for “Imaging” Sources

Thus, we have a system for “imaging” sources which looks like:

Source B at angle θ_B    →  [  ]  → Signal at angle cos θ₁  (Source A) f_A(t)
f_B(t)                   → [FFT] → Signal at angle cos θ₂  (Source B) f_B(t)
Source A at angle θ_A    →  [  ]
f_A(t)

# Two-Dimensional Arrays

Array Configurations

Rectangular Array:

o o o o o
o o o o o
o o o o o
o o o o o
o o o o o

Mills Cross (9 antenna dishes):

      o
      o
o  o  o  o  o
      o
      o

Calculation for Mills Cross

$$ \rho(x,y) = \left[\delta(x) \cdot \text{comb}\left(\frac{y}{a}\right) \cdot \text{rect}\left(\frac{y}{5a}\right)\right] ** \text{cyl}\left(\frac{\sqrt{x^2+y^2}}{d}\right) \quad \leftarrow I $$

$$

• \left[\text{comb}\left(\frac{x}{a}\right)\delta(y) \cdot \text{rect}\left(\frac{x}{5a}\right)\right] ** \text{cyl}\left(\frac{\sqrt{x^2+y^2}}{d}\right) \quad \leftarrow II $$

$$ \text{-} \left[\delta(x)\delta(y)\right] ** \text{cyl}\left(\frac{\sqrt{x^2+y^2}}{d}\right) \quad \leftarrow III $$

Fourier Transforming

$$ I \Rightarrow \left[5a \cdot \text{comb}\left(\frac{a\cos\phi}{\lambda}\right) * \text{sinc}\left(\frac{5a\cos\phi}{\lambda}\right)\right] \cdot \frac{\pi d^2}{4}\text{somb}\left(d\sqrt{\left(\frac{\cos\theta}{\lambda}\right)^2 + \left(\frac{\cos\phi}{\lambda}\right)^2}\right) $$$$ II \Rightarrow \left[5a^2\text{comb}\left(\frac{a\cos\theta}{\lambda}\right) * \text{sinc}\left(\frac{5a\cos\theta}{\lambda}\right)\right] \cdot \frac{\pi d^2}{4}\text{somb}\left(d\sqrt{\left(\frac{\cos\theta}{\lambda}\right)^2 + \left(\frac{\cos\phi}{\lambda}\right)^2}\right) $$$$ III \Rightarrow \left[\frac{\pi d^2}{4}\text{somb}\left(d\sqrt{\left(\frac{\cos\theta}{\lambda}\right)^2 + \left(\frac{\cos\phi}{\lambda}\right)^2}\right)\right] $$

Mills Cross - Complete Solution

Combined Result

$$ \implies I + II - III \Rightarrow \left[5a^2 \text{comb}\left(\frac{a\cos\phi}{\lambda}\right) * \text{sinc}\left(\frac{5a\cos\phi}{\lambda}\right) + 5a^2 \text{comb}\left(\frac{a\cos\theta}{\lambda}\right) * \text{sinc}\left(\frac{5a\cos\theta}{\lambda}\right) - 1\right] $$$$ = \left\{5a\sum_{n=-\infty}^{\infty} \text{sinc}\left[5a\left(\frac{\cos\phi}{\lambda} - \frac{n}{a}\right)\right] + 5a\sum_{n=-\infty}^{\infty} \text{sinc}\left[5a\left(\frac{\cos\theta}{\lambda} - \frac{n}{a}\right)\right] - 1\right\} $$$$ \cdot \frac{\pi d^2}{4}\text{somb}\left(d\sqrt{\left(\frac{\cos\theta}{\lambda}\right)^2 + \left(\frac{\cos\phi}{\lambda}\right)^2}\right) $$

Pattern Analysis

The pattern shows:

• Antenna factor: Overall envelope from somb function
• Array Factor: Grid of peaks from the cross pattern
• Width of line = $\frac{1}{5a}$

The diagram shows the pattern in $\frac{\cos\theta}{\lambda}$ vs $\frac{\cos\phi}{\lambda}$ space, with the antenna factor providing the overall envelope.

---

Non-Uniform Arrays

Arrays chosen with random position to reduce sidelobes:

Regular array of $N$ radiators:

• Sharp peaks with regular sidelobe pattern
• Predictable grating lobes

Randomly spaced array of $N$ radiators:

• Main beam preserved
• Sidelobes reduced and spread out
• No distinct grating lobes

Mills Cross (Goodman’s Notation)

Configuration

        o
        o ↑a
    d ↗ o  o  o  o  o
        o
        o

Calculation

$$ \rho(x,y) = \left[\delta(x) \cdot \text{comb}\left(\frac{y}{a}\right) \cdot \text{rect}\left(\frac{y}{5a}\right)\right] * \text{circ}\left(\frac{\sqrt{x^2+y^2}}{d/2}\right) \quad \leftarrow I $$

$$

• \left[\text{comb}\left(\frac{x}{a}\right)\delta(y) \cdot \text{rect}\left(\frac{x}{5a}\right)\right] * \text{circ}\left(\frac{\sqrt{x^2+y^2}}{d/2}\right) \quad \leftarrow II $$

$$ \text{-} \left[\delta(x)\delta(y)\right] * \text{circ}\left(\frac{\sqrt{x^2+y^2}}{d/2}\right) \quad \leftarrow III $$

Fourier Transforming

$$ I: \left\{5a^2 \text{comb}\left(\frac{a\cos\phi}{\lambda}\right) * \text{sinc}\left[\frac{5a\cos\phi}{\lambda}\right]\right\} \cdot \frac{dJ_1(\pi d\rho')}{2\rho'} $$

where $\rho' = \sqrt{\left(\frac{\cos\theta}{\lambda}\right)^2 + \left(\frac{\cos\phi}{\lambda}\right)^2}$

$$ II: \left\{5a^2 \text{comb}\left[\frac{a\cos\theta}{\lambda}\right] * \text{sinc}\left[\frac{5a\cos\theta}{\lambda}\right]\right\} \cdot \frac{dJ_1(\pi d\rho')}{2\rho'} $$$$ III: \left\{\frac{dJ_1(\pi d\rho')}{2\rho'}\right\} $$$$ I + II - III \Rightarrow \left\{5a\sum_{n=-\infty}^{\infty}\text{sinc}\left[5a\left(\frac{\cos\phi}{\lambda} - \frac{n}{a}\right)\right] + 5a\sum_{n=-\infty}^{\infty}\text{sinc}\left[5a\left(\frac{\cos\theta}{\lambda} - \frac{n}{a}\right)\right] - 1\right\} \cdot \frac{dJ_1(\pi d\rho')}{2\rho'} $$

Mills Cross - Pattern Diagram

Two-Dimensional Pattern

The Mills Cross produces a characteristic pattern in $\left(\frac{\cos\theta}{\lambda}, \frac{\cos\phi}{\lambda}\right)$ space:

              ↑ cos φ/λ
              |
         ─────┼─────  Array Factor
        /  |  |  |  \
       /   |  |  |   \
      (────┼──┼──┼────)──→ cos θ/λ
       \   |  |  |   /
        \  |  |  |  /
         ─────┼─────
              |

     Antenna ↗     ↖ Line width = 1/5a
     Factor
              ←1/a→

Key Features

• Array Factor: Grid pattern from the cross arrangement of antennas
• Antenna Factor: Circular envelope from individual dish antennas
• Line width: $\frac{1}{5a}$ (determined by the total extent of the array)
• Line spacing: $\frac{1}{a}$ (determined by the spacing between antennas)

The resulting beam pattern is the product of the array factor (grid) and antenna factor (circular envelope).

Fourier Transform Property of a Lens

Setup

Consider a thickness spherical lens:

Parameter	Value
Gaskill	10-3
Goodman	5-9
Cathey	5-9

• Maximum thickness = $\Delta_0$
• Index of refraction = $n$
• Thickness at $(x,y)$ = $\Delta(x,y)$

Phase Delay Through a Lens

Phase delay of wave through a region of refractive index $n$:

$$ \text{Recall } e^{jk'r} \Rightarrow k'r \text{ gives the phase of propagation} $$$$ \text{But } k' = \frac{\omega}{c'} = \frac{\omega n}{c_0} \quad \text{since } n = \frac{c_0}{c'} = \frac{\text{free space velocity}}{\text{velocity}} $$$$ \implies \underbrace{\phi}_{\text{phase}} = k'r = \frac{\omega n}{c_0} \cdot r = kn \cdot r, \quad \text{where } k = \frac{\omega}{c_0} $$$$ \phi(x,y) = kn\Delta(x,y) + k[\Delta_0 - \Delta(x,y)] $$$$ = \underbrace{k n \Delta(x,y)}_{\text{delay from lens}} + \underbrace{k[\Delta_0 - \Delta(x,y)]}_{\text{delay from free space}} $$

where $\Delta(x,y)$ is thickness function of lens

Lens Thickness Function

Phase-Only Transmittance Function

$$ \Rightarrow t_\ell(x,y) \triangleq e^{j\phi(x,y)} = \exp[jk\Delta_0]\exp[jk(n-1)\Delta(x,y)] $$

This is a phase-only function.

We assume a “thin lens”, so that we have the field behind the lens given by:

$$ u^+(x,y) = u^-(x,y)t(x,y) = u^-(x,y)\exp[jk\Delta_0]\exp[jk(n-1)\Delta(x,y)] $$

Calculating the Thickness Function $\Delta(x,y)$

Split lens into two parts:

Convention: Convex surfaces have positive radii of curvature; concave surfaces have negative radii - for light traveling from left to right.

Left Half of Lens

• Radius $R_1$
• Center at $R_1 - x_1 - y_1$

Right Half of Lens

• Radius $-R_2$
• Center at $\sqrt{R_2^2 - x^2 - y^2}$

(Recall $R_2$ is a negative number by convention)

$$ \Rightarrow \Delta(x,y) = \Delta_1(x,y) + \Delta_2(x,y) $$

and:

$$ \Delta_1(x,y) = \Delta_{01} - \left(R_1 - \sqrt{R_1^2 - x^2 - y^2}\right) $$$$ = \Delta_{01} - R_1\left(1 - \sqrt{1 - \frac{x^2+y^2}{R_1^2}}\right) $$$$ \Delta_2(x,y) = \Delta_{02} - \left(-R_2 - \sqrt{R_2^2 - x^2 - y^2}\right) $$$$ = \Delta_{02} - \left(-R_2 + R_2\sqrt{\frac{R_2^2 - x^2 - y^2}{(-R_2)^2}}\right) \leftarrow \text{factor out } -R_2 \text{ and this is a positive number} $$$$ = \Delta_{02} + R_2\left(1 - \sqrt{1 - \frac{x^2+y^2}{R_2^2}}\right) $$

Lens Thickness Function - Complete Derivation

Total Thickness Function

$$ \boxed{\Delta(x,y) = \Delta_1(x,y) + \Delta_2(x,y) \approx \Delta_0 - R_1\left(1 - \sqrt{1 - \frac{x^2+y^2}{R_1^2}}\right) + R_2\left(1 - \sqrt{1 - \frac{x^2+y^2}{R_2^2}}\right)} $$

where $\Delta_0 = \Delta_{01} + \Delta_{02}$

Paraxial Ray Approximation

Assume deviation from center $(x,y)$ is small, so $x^2 + y^2 << R^2$:

$$ \Rightarrow \sqrt{1 - \frac{x^2+y^2}{R_1^2}} \approx 1 - \frac{x^2+y^2}{2R_1^2} $$

and

$$ \sqrt{1 - \frac{x^2+y^2}{R_2^2}} \approx 1 - \frac{x^2+y^2}{2R_2^2} $$

(Paraxial rays)

Approximate Thickness Function

Then we have:

$$ \boxed{\Delta(x,y) = \Delta_0 - R_1\left(\frac{x^2+y^2}{2R_1^2}\right) + R_2\left(\frac{x^2+y^2}{2R_2^2}\right) = \Delta_0 - \frac{x^2+y^2}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)} $$

Lens Transmittance Function

$$ \implies t_\ell(x,y) = \exp[jk\Delta_0]\exp\left[jk(n-1)\left(\Delta_0 - \frac{x^2+y^2}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\right)\right] $$$$ = \exp[jk\Delta_0]\exp\left[jk(n-1)\Delta_0\right]\exp\left[-jk(n-1)\frac{(x^2+y^2)}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\right] $$$$ = \exp[jkn\Delta_0]\exp\left[-jk(n-1)\frac{(x^2+y^2)}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\right] $$

Lens Focal Length and Transmittance Function

Lens Maker’s Formula

Define the focal length of a lens as:

$$ \boxed{\frac{1}{f} \triangleq (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)} $$

Lens Maker’s Formula

Transmittance Function for a Thin Lens

Then we have:

$$ \boxed{t_\ell(x,y) = \exp[jkn\Delta_0]\exp\left[-j\frac{k}{2f}(x^2+y^2)\right]} $$

Transmittance Function For a Thin Lens Using The Paraxial Ray Approximation

Meaning of Transmittance Function $t_\ell(x,y)$

• $\exp[jkn\Delta_0]$ is a constant phase delay of little interest

• $\exp\left[-j\frac{k}{2f}(x^2+y^2)\right]$ is a quadratic approximation to a converging spherical wave, converging at the point $z = f$.

         ← z₂ parabolic
           surface
    ↗  (    )
   (  ( z=f )  → z
    ↘  (    )
   z₁

# Object Placed Against the Lens

Setup

Normal       ↗            u_f(x',y')
Incidence  [lens]           ↗
Plane    → t_e+t_ℓ(x,y)     z = f
wave A   →
         ↗ p(x,y)
      t_0(x,y)

$$ u^-(x,y) = A $$$$ p(x,y) = \begin{cases} 1 & \text{inside lens aperture} \\ 0 & \text{outside} \end{cases} $$$$ u^+(x,y) = A\,t_0(x,y)\,t_\ell(x,y)\,p(x,y) = A\,t_0(x,y)\,p(x,y)\exp\left[-j\frac{k}{2f}(x^2+y^2)\right] $$

where we have dropped the constant phase term $t_\ell(x,y)$.

Field at the Focal Plane

To calculate the field at $z = f$ we use the Fresnel diffraction formula:

$$ u_f(x',y')\bigg|_{z_e=f} = \underbrace{\frac{\exp\{jkf\}}{j\lambda f}}_{z_e=f} \exp\left\{j\frac{k(x'^2+y'^2)}{2f}\right\} \iint u^+(x,y)\exp\left\{j\frac{k(x^2+y^2)}{2f}\right\} \exp\left\{-j\frac{2\pi}{\lambda f}(xx'+yy')\right\} dx\,dy $$

(Fourier Transform Form)

Substituting for $u^+(x,y)$:

$$ u_f(x',y')\bigg|_{z_e=f} = \frac{\exp\left\{j\frac{k(x'^2+y'^2)}{2f}\right\}}{j\lambda f} \iint A\,t_0(x,y)\,p(x,y)\exp\left\{-j\frac{k}{2f}(x^2+y^2)\right\}\exp\left\{j\frac{k(x^2+y^2)}{2f}\right\} \exp\left\{-j\frac{2\pi}{\lambda f}(xx'+yy')\right\} dx\,dy $$$$ = A\frac{\exp\left\{jk\frac{(x'^2+y'^2)}{2f}\right\}}{j\lambda f} \iint_{-\infty}^{\infty} t_0(x,y)\,p(x,y)\exp\left\{-j\frac{2\pi}{\lambda f}(xx'+yy')\right\} dx\,dy $$

Lens as Fourier Transform System

Simplified Result

If the object $t_0(x,y)$ is smaller than the lens aperture, we can neglect $p(x,y)$. Then:

$$ u_f(x',y')\bigg|_{z_e=f} = A\frac{\exp\left\{jk\frac{(x'^2+y'^2)}{2f}\right\}}{j\lambda f} \mathcal{FF}\{t_0(x,y)\}\bigg|_{\xi=\frac{x'}{\lambda f}, \eta=\frac{y'}{\lambda f}} $$

If we measure the intensity at $z_0 = f$ we have:

$$ I_f(x',y')\bigg|_{z=f} = u_e u_e^* = \frac{A^2}{\lambda^2 f^2}\left|\mathcal{FF}\{t_0(x,y)\}\right|^2_{\xi=\frac{x'}{\lambda f}, \eta=\frac{y'}{\lambda f}} $$

---

Object Placed in Front of the Lens

(at distance $d$)

      d_0
    ↗    ↖← f →↗ y'
  (      )      x'
t_0(x,y)

• Object: $t_0(x,y)$
• Angular plane wave spectrum of object: $\mathcal{FF}\{At_0(x,y)\} = F(\xi,\eta)$

We can calculate the plane wave spectrum at the lens by multiplying by the Fresnel operator to transfer from $\xi = \frac{x}{\lambda}$, $\eta = \frac{y}{\lambda}$:

Object in Front of Lens - Field Calculation

Fresnel Transfer to Lens

$$ \implies F(\xi,\eta) = \left[F_0(\xi,\eta) \cdot \exp\left\{jkd_0\left[1 - \frac{\lambda^2}{2}(\xi^2+\eta^2)\right]\right\}\right] $$$$ = F_0(\xi,\eta) \cdot \exp[-j\pi\lambda d_0(\xi^2+\eta^2)] $$

where we leave out $\exp[jkd_0]$ term since it is a constant.

Field at Focal Plane

From the object against the lens calculation, we have:

$$ u_f(x',y')\bigg|_{\substack{\text{Field at focal} \\ \text{pt. at lens}}} = \exp\left[j\frac{k}{2f}(x'^2+y'^2)\right] \cdot \frac{\mathcal{FF}\{At_0(x,y)\}}{j\lambda f}\bigg|_{\xi=\frac{x'}{\lambda f}, \eta=\frac{y'}{\lambda f}} $$$$ = \frac{\exp\left[j\frac{k}{2f}(x'^2+y'^2)\right]}{j\lambda f} F_0\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right) $$

where $F_0$ is the angular plane wave spectrum.

$$ \implies u_f(x',y') = \frac{\exp\left[j\frac{k}{2f}(x'^2+y'^2)\right]}{j\lambda f} \exp\left[-j\pi\lambda d_0\left(\frac{x'^2+y'^2}{\lambda^2 f^2}\right)\right] F_0\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right) $$$$ = \frac{1}{j\lambda f}\exp\left[j\frac{k}{2f}\left(1 - \frac{d_0}{f}\right)(x'^2+y'^2)\right] F_0\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right) $$$$ \Rightarrow u_f(x',y') = \frac{A}{j\lambda f}\exp\left[j\frac{k}{2f}\left(1 - \frac{d_0}{f}\right)(x'^2+y'^2)\right] \iint t_0(x,y)\exp\left[j\frac{2\pi}{\lambda f}(xx'+yy')\right] dx\,dy $$

Special Case: $f = d_0$

# Lenses and Fourier Transforms

Fourier Transform Pairs

    1                              δ(x,y)
    ↓     [lens]                    ↓     [lens]
   ←f→    δ(x,y)                   ←f→      1

𝓕{1} = δ(x,y)     Fourier Transform     𝓕{δ(x,y)} = 1
                       Pairs

Shift Theorem

       y
    x ↙  h ↘  f
         δ(x-h)  φ

$$ \mathcal{F}\{\delta(x-h)\}\bigg|_{\xi=\frac{x'}{\lambda f}} = 1 \cdot \exp[-j2\pi h\xi] $$$$ = \exp\left[-j2\pi h\frac{x'}{\lambda f}\right] $$$$ \implies \sin\phi = \frac{h}{f} \approx \phi \leftarrow \text{Propagation angle of plane wave} $$

Connection to Geometrical Optics

From geometrical optics, we note that the ray parallel to the optical axis must be directed through the back focal point.

Hence, we see immediately that $\tan\phi = \frac{h}{f} \approx \phi$.

# Spherical Wave and Imaging Formula

Setup

         ↗x
      (  )f(  )     x'
    ←x→     f    →x'←
      d₀

Spherical Wave in Focal Plane

$$ \exp\left[-j\frac{\pi r'^2}{\lambda R}\right] = \exp\left[-j\frac{\pi r'^2}{\lambda x'}\right] $$

where $r' = \sqrt{x'^2 + y'^2}$

Imaging Formula

Use imaging formulae to locate focal point:

$$ \frac{1}{d_0} + \frac{1}{d_i} = \frac{1}{f} $$

or: $\quad xx' = f^2$ (Newtonian imaging formula)

$$ \implies x' = \frac{f^2}{x} = R \quad (x = d_0 - f, \quad x' = d_i - f) $$

Spherical Wave Expression

And spherical wave becomes:

$$ \exp\left[-j\frac{\pi r'^2 x}{\lambda f^2}\right] $$$$ = \exp\left[-j\frac{\pi r'^2(d_0 - f)}{\lambda f \cdot f}\right] $$$$ = \exp\left[+j\frac{\pi r'^2}{\lambda f}\left(1 - \frac{d_0}{f}\right)\right] = \exp\left[j\frac{k}{2f}\left(1 - \frac{d_0}{f}\right)(x'^2+y'^2)\right] $$# Heuristic Explanation of F.T. Property of Lens

Physical Interpretation

Limiting Case

Of course,

$$ \frac{1}{d_0} + \frac{1}{d_i} = \frac{1}{f} $$

So if $d_0 = \infty$:

• $d_i = f$
• The Fraunhofer pattern at $\infty$ is reimaged at the back focal plane of the lens.

# Power Spectrum Analysis

Example: Particle Size Detection

Consider two particles (circles) at points $(a,b)$ and $(c,d)$.

  y↑
   |  ⊚(c,d)
   | ⊚(a,b)
   +────────→ x
            λ

Fourier Power Spectrum

Fourier power spectrum gives by:

$$ \left| \frac{\alpha J_1(2\pi\alpha\sqrt{f_x^2+f_y^2})}{\sqrt{f_x^2+f_y^2}} e^{j\pi(af_x+bf_y)} + \frac{\alpha J_1(2\pi\alpha\sqrt{f_x^2+f_y^2})}{\sqrt{f_x^2+f_y^2}} e^{j\pi(cf_x+df_y)} \right|^2 $$$$ = 2\left(\frac{\alpha J_1(2\pi\alpha\sqrt{f_x^2+f_y^2})}{\sqrt{f_x^2+f_y^2}}\right)^2 + 2\left(\frac{\alpha J_1(2\pi\alpha\sqrt{f_x^2+f_y^2})}{\sqrt{f_x^2+f_y^2}}\right) \cos(2\pi[(a-c)f_x + (b-d)f_y]) $$$$ = \underbrace{\text{independent of particle position}}_{\text{of particle position}} + \underbrace{\text{sum of random cosines}}_{\Rightarrow \text{noise}} $$

Distribution Vector

If scene consists of:

• $N_1$ particles of size $d_1$
• $N_2$ of size $d_2$
• …
• $N_L$ of size $d_L$

We can express this answer by a distribution vector:

$$ \underline{N} = (N_1, \cdots, N_L)^T $$

Particle Size Distribution - Matrix Formulation

Total Power Spectrum

If $G_i(f_x)$, $i = 1, L$ is the diffraction pattern produced by a single particle of size $d_i$, total power spectrum is:

$$ H(f_x) = \sum_{i=1}^{L} N_i G_i(f_x) + \underbrace{\eta(f_x)}_{\text{noise}} $$

Matrix Notation

Sample $f_x$ at discrete frequencies $f_{x_1}, f_{x_2}, \ldots, f_{x_N}$

$$ \Rightarrow \underline{H} = \underline{\underline{G}}\underline{N} + \underline{\eta} \quad \leftarrow \text{(Shaw Stark, 177, 178)} $$

Solution Methods

1. Filter result to remove $\underline{\eta}$
2. Invert by pseudoinverse to solve for $\underline{N}$

$$ \underline{H} = \underline{\underline{G}}\underline{N} $$

$\underline{\underline{G}}$ is not square and has no inverse

$$ \underline{\underline{G}}^+\underline{H} = (\underline{\underline{G}}^+\underline{\underline{G}})\underline{N} $$$$ (\underline{\underline{G}}^+\underline{\underline{G}})^{-1}\underline{\underline{G}}^+\underline{H} = \underline{N}_{LS} $$# Statistical Pattern Recognition

Feature Selection

Goal: We desire to choose a small number of features that are important for pattern recognition.

      ┌─────────┐     ┌──────────────┐
  N   │  Image  │     │   Feature    │    Features:
 ───► │         │────►│   Selector   │───► ∧ ∧  Ears (Number)
      │         │     │              │     ≋ ≋  Whiskers (Length)
      └─────────┘     └──────────────┘     ⌒   Tail (Length)
       N² data pts.                        Features (3)

Linear Transforms for Feature Extraction

Can we use linear transforms to extract features?

         ┌────────────┐  ┌───────────────┐
  f(x,y) │   Linear   │  │Dimensionality │  F'(u,v)
 ───────►│  Transform │─►│  Reduction    │────────►
         │   F̂(u,v)  │  │               │  M data pts
 N² data │            │  │               │  M << N²
   pts   └────────────┘  └───────────────┘
              ↑
         Unitary
        Transforms
         Preserve
          Energy

Dimensionality Reduction Example

Example of dimensionality reduction: Choose M data pts. with highest values.

Optimization of Linear Transform

Objective

Choose transform that minimizes the mean-square error when M features are used to regenerate (approx.) original image pts.

  F'(u,v)  ┌─────────┐  f'(x,y)
 ─────────►│ Inverse │─────────►
M features │ Linear  │ M data pts.
           │Transform│
           └─────────┘

Mean Square Error (MSE)

$$ \text{MSE} = \left\langle \left| \underbrace{f'(x,y)}_{\substack{\text{Reconstructed} \\ \text{image from} \\ \text{M features}}} - \underbrace{f(x,y)}_{\substack{\text{Original} \\ \text{image}}} \right|^2 \right\rangle $$

Karhunen-Loève Transform

Linear Transform that minimizes MSE:

$$ \rightarrow \text{Karhunen-Loève Transform} $$

(sometimes called Hotelling Transform or Principal Component Analysis)

Properties

• K-L transform basis functions depend on statistics of image
• Basis functions are given by the eigenvectors of the autocorrelation matrix of the statistical process

# Fourier Transform as Feature Extractor

Examples (Shaw Stark, pg. 172)

Applications where Fourier Transform is useful:

• Handwriting
• Printed matter
• Homogeneous aerosols
• Forests, crops (maybe)
• Targets with equal probability of occurring anywhere in the image

Conclusion

Examples Where F.T. is Useful

• Handwriting
• Train tracks, orchards
• Aerosol detection

---

Translational Invariance of Fourier Power Spectrum

  f(x,y)  ──────►  F.T.  ──────►  | |²  ──────► |F(f_x,f_y)|² ←─┐
                                                              │ Power
  f(x-a,y-b)      F(f_x,f_y)exp[j2π(af_x+bf_y)]  |F(f_x,f_y)|²←─┤ Spectrum
                                                              │ Does
                                                              │ Not
                                                              └─ Change

# Variations on Fourier Power Spectrum Sampling

Wedge-Ring Detector

(Shaw Stark, pg 163)

    ╱╲
   ╱  ╲
  ╱ ┌──┐╲
 ╱  │  │ ╲
╱   └──┘  ╲

The wedge-ring detector samples the Fourier plane in polar coordinates, providing:

• Rings for radial (frequency magnitude) information
• Wedges for angular (orientation) information

Key Properties

Thus, we can sample only half the transform without losing information.

(Shaw Stark, pg. 168)

Variations on Fourier Power Spectrum Sampling (Cont.)

Log-Polar Coordinate Distortion

         ┌───────────┐    ┌──────────┐
  f(r,θ) │ Log-polar │    │ Fourier  │     | |²
 ────────►│Coordinate │───►│Transform │────►
         │Distortion │    │          │
         └───────────┘    └──────────┘
              ↓               θ
         ┌────┴────┐          ↓
         │ log(r)  │        log(r)
         └─────────┘

Coordinate Transformation

Note: $f(ar, \theta + \theta_0)$ becomes:

$$ f(\underbrace{\log r + \log a}_{x\text{-axis}}, \underbrace{\theta + \theta_0}_{y\text{-axis}}) $$

Scale and rotation have been converted into an x-axis & y-axis shift.

Mellin Transform

• Transform in $\log(r)$ direction is a Mellin transform
• Features are no longer invariant to translation

(Shaw Stark, pg. 175 & 174)

Complete Invariance: Position, Scale, and Orientation

How to Include All Invariances?

Combining position invariance (as well as scale and orientation):

Original   ┌─────────┐      ┌───────────┐   ┌─────────┐
 Object    │ Fourier │  | |²│ Log-polar │   │ Fourier │  | |²   Feature
──────────►│Transform│─────►│Coordinate │──►│Transform│──────►
   to      │         │      │Distortion │   │         │
 Detect    └─────────┘      └───────────┘   └─────────┘

Invariance Properties

1. Translational Invariance: First Fourier transform + magnitude squared removes position dependence

2. Scale & Rotation Invariance:

   • Power spectrum will inversely scale with object
   • Power spectrum will rotate directly as object rotates
   • Log-polar distortion converts these to shifts
   • Second Fourier transform + magnitude squared removes scale and rotation dependence

# Telecentric Imaging System

Setup

Object          y₂              Image
  y₁         ↗  x₂           y₃↗
   ↗x₁      ╱    ╲             ╱ x₃
   ╲       ╱      ╲           ╱
    ╲─────╱────────╲─────────╱
      f₁      f₁       f₂      f₂

   └──────────┘    └──────────┘
   This stage produces    This stage produces
   an exact Fourier       an exact Fourier
   Transform (forward)    Transform (forward)

Field at Each Plane

Consider:

1. $t(x,y)$ in the object plane $(x_1, y_1)$
2. Plane wave illumination of object $A$

$$ \Rightarrow u_1(x_1, y_1) = At(x,y) $$

From previous development, we know that at focal plane of first lens:

$$ u_2(x_2, y_2) = \frac{A}{j\lambda f_1} \mathcal{FF}\{t(x,y)\}\bigg|_{\xi=\frac{x_2}{\lambda f_1}, \eta=\frac{y_2}{\lambda f_1}} $$$$ = \frac{A}{j\lambda f_1} T\left(\frac{x_2}{\lambda f_1}, \frac{y_2}{\lambda f_1}\right) $$

where $T\left(\frac{x_2}{\lambda f_1}, \frac{y_2}{\lambda f_1}\right) = \mathcal{FF}\{t(x,y)\}$

Field at Output Plane

We can equivalently calculate distribution in $(x_3, y_3)$ plane with same formula:

$$ u_3(x_3, y_3) = \left(\frac{A}{j\lambda f_1}\right)\left(\frac{1}{j\lambda f_2}\right) \iint T\left(\frac{x_2}{\lambda f_1}, \frac{y_2}{\lambda f_1}\right) \exp\left[-j\frac{2\pi}{\lambda f_2}(x_2 y_3 + y_2 y_3)\right] dx_2 dy_2 $$# Telecentric Imaging System - Magnification

Variable Substitution

$$ \Rightarrow i) \text{ Let } x_3 = -(-x_3) \text{ and } y_3 = -(-y_3) \text{ in exponential} $$$$ ii) \text{ Let } x_2 = f_1\left(\frac{x_2}{f_1}\right), \quad dx_2 = \lambda f_1 d\left(\frac{x_2}{\lambda f_1}\right), \quad y_2 = f_1\left(\frac{y_2}{f_1}\right) \text{ and } dy_2 = \lambda f_1 d\left(\frac{y_2}{\lambda f_1}\right) $$

Then:

$$ u_3(x_3, y_3) = \frac{Af_1}{f_2} \iint T\left(\frac{x_2}{\lambda f_1}, \frac{y_2}{\lambda f_1}\right) \exp\left\{j2\pi f_1\left(\frac{x_2}{\lambda f_1}\right) + (-y_3)\frac{f_1}{f_2}\left(\frac{x_2}{\lambda f_1}\right)\right\} d\left\{\frac{x_2}{\lambda f_1}\right\} d\left\{\frac{y_2}{\lambda f_1}\right\} $$$$ = A\frac{f_1}{f_2} u_1\left(\frac{-f_1 x_3}{f_2}, \frac{-f_1 y_3}{f_2}\right) $$

Magnification

Define a magnification $M = f_2/f_1$

Then:

$$ u_3(x_3, y_3) = \frac{A}{M} u_1\left(\frac{-x_3}{M}, \frac{-y_3}{M}\right) $$

Motion of Object & Image Planes

      ↑           ↑
     x₀          x_i
  ⟺  ╱ ╲    ╱ ╲  ⟺
    ←f→ ←f→

$$ x_i = x_o \text{ when } M = 1 $$$$ x_i = M^2 x_o \text{ in general} $$

Single Lens Imaging

Impulse Response Formulation

Every superposition, we can state the response of a linear system as the sum of impulse responses at point sources:

$$ u_i(x_i, y_i) = \iint h(x_i, y_i; x_o, y_o) u_o(x_o, y_o) dx_o dy_o $$

where $h$ is the response of the system to a point source at $x_o, y_o$

      u_o  u_o'         u_i
       ╲   ╱             │
       (  │  )           │
        ╲ │ ╱            │
    ←d_o→│←───d_i────→

Image Formation

For image formation, on impulse:

$$ h(x_i, y_i; x_o, y_o) = K\delta(x_i \pm Mx_o, y_i \pm My_o), \quad M = \frac{1}{\text{magnification}} $$

Consider a point source at $(x_o, y_o)$:

The wave at the lens is a spherical wave, which under the paraxial approximation is written as:

$$ u_\ell(x,y) = \frac{1}{j\lambda d_o} \exp\left\{j\frac{k}{2d_o}[(x-x_o)^2 + (y-y_o)^2]\right\} $$

After Passing Through the Lens

$$ u_\ell'(x,y) = u_\ell(x,y) \cdot \underbrace{p(x,y) \exp\left[-j\frac{k}{2f}(x^2+y^2)\right]}_{\text{transmittance of thin lens}} $$

where $p$ is the pupil function.

Using Fresnel diffraction from $u_\ell'$ to $u_i$:

$$ \text{Result} \Rightarrow h(x_i, y_i; x_o, y_o) = \frac{1}{j\lambda d_i} \iint u_\ell'(x,y) \exp\left\{j\frac{k}{2d_i}[(x_i-x)^2 + (y_i-y)^2]\right\} dx\,dy $$

Single Lens Imaging - Impulse Response

Combining Terms

$$ h(x_i, y_i; x_o, y_o) = \frac{1}{\lambda^2 d_o d_i} \exp\left[j\frac{k}{2d_i}(x_i^2+y_i^2)\right] \underbrace{\exp\left[j\frac{k}{2d_o}(x_o^2+y_o^2)\right]}_{\text{small object}} $$$$ \cdot \iint p(x,y) \exp\left[j\frac{k}{2}\left(\frac{1}{d_o} + \frac{1}{d_i} - \frac{1}{f}\right)(x^2+y^2)\right] $$$$ \cdot \exp\left\{-jk\left[\left(\frac{x_o}{d_o} + \frac{x_i}{d_i}\right)x + \left(\frac{y_o}{d_o} + \frac{y_i}{d_i}\right)y\right]\right\} dx\,dy $$

Simplifications

1. Since we are normally interested in image intensity, we can ignore $\exp\left[j\frac{k}{2d_i}(x_i^2+y_i^2)\right]$

2. $\exp\left[j\frac{k}{2d_o}(x^2+y^2)\right]$ cannot be ignored, since the superposition integral is taken over $x_o$ & $y_o$. However, if we are interested in an imaging condition, the light at $x_i, y_i$ must come from a small region of the object.

    (x_o,y_o)╱────────────╲
            ╱              ╲
           (                )──→ (x_i, y_i)

Small Region Approximation

If, within this small region, the phase does not change much, we can let:

$$ \exp\left[j\frac{k}{2d_o}(x_o^2+y_o^2)\right] \cong \exp\left[j\frac{k}{2d_o}\left(\frac{x_i^2+y_i^2}{M^2}\right)\right] $$

and the image plane phase can again be dropped.

Single Lens Imaging - Impulse Response (Continued)

Simplified Expression

We then have:

$$ h(x_i, y_i; x_o, y_o) = \frac{1}{\lambda^2 d_o d_i} \iint p(x,y) \exp\left[j\frac{k}{2}\left(\frac{1}{d_o} + \frac{1}{d_i} - \frac{1}{f}\right)(x^2+y^2)\right] $$$$ \cdot \exp\left\{-jk\left[\left(\frac{x_o}{d_o} + \frac{x_i}{d_i}\right)x + \left(\frac{y_o}{d_o} + \frac{y_i}{d_i}\right)y\right]\right\} dx\,dy $$

Finally, we restrict ourselves to the image plane:

$$ \boxed{\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}} $$

Lens Law

Space-Invariant Form

Then:

$$ h(x_i, y_i; x_o, y_o) = K \iint p(x,y) \exp\left\{-jk\left[\left(\frac{x_o}{d_o} + \frac{x_i}{d_i}\right)x + \left(\frac{y_o}{d_o} + \frac{y_i}{d_i}\right)y\right]\right\} dx\,dy $$

where $K = \frac{1}{\lambda^2 d_o d_i}$ and $M = \frac{-d_i}{d_o}$

$$ \Rightarrow h(x_i, y_i; x_o, y_o) = K \iint p(x,y) \exp\left\{j\frac{2\pi}{\lambda d_i}[(x_i - Mx_o)x + (y_i - My_o)y]\right\} dx\,dy $$

Change of Variables

Let $\tilde{x} = \frac{x}{\lambda d_i}$, $\tilde{y} = \frac{y}{\lambda d_i}$, $\tilde{x}_o = Mx_o$, $\tilde{y}_o = My_o$

$$ \rightarrow \boxed{h(x_i - \tilde{x}_o, y_i - \tilde{y}_o) = K\lambda^2 d_i^2 \iint p(\lambda d_i \tilde{x}, \lambda d_i \tilde{y}) \exp\left\{-j2\pi[(x_i - \tilde{x}_o)\tilde{x} + (y_i - \tilde{y}_o)\tilde{y}]\right\} d\tilde{x}\,d\tilde{y}} $$

(Space Invariant Form)

Finally, let $\tilde{h} = \frac{1}{K\lambda^2 d_i^2} h$

and:

$$ \underbrace{u_g(\tilde{x}_o, \tilde{y}_o)}_{\text{modified object}} = K\frac{\lambda^2 d_i^2}{M^2} \, u_o\!\left(\underbrace{\frac{-\tilde{x}_o}{M}, \frac{-\tilde{y}_o}{M}}_{\text{geometric optics expression}}\right) $$

Coherent Transfer Function (CTF)

Linear Shift-Invariant Superposition

Then the original superposition integral becomes:

$$ u_i(x_i, y_i) = \iint \tilde{h}(x_i - \tilde{x}_o, y_i - \tilde{y}_o) u_g(\tilde{x}_o, \tilde{y}_o) d\tilde{x}_o d\tilde{y}_o $$$$ \text{where } \tilde{h}(x_i, y_i) = \iint p(\lambda d_i \tilde{x}, \lambda d_i \tilde{y}) \exp[-j2\pi(\tilde{x}x_i + \tilde{y}y_i)] d\tilde{x}\,d\tilde{y} $$

What is the Transfer Function of the Lens?

Define frequency spectra of input & output:

• Note: Goodman notation $f_x = \xi$, $f_y = \eta$

Input:

$$ G_g(f_x, f_y) = \iint u_g(\tilde{x}_o, \tilde{y}_o) \exp[-j2\pi(f_x \tilde{x}_o + f_y \tilde{y}_o)] d\tilde{x}_o d\tilde{y}_o $$

Output:

$$ G_i(f_x, f_y) = \iint u_i(x_i, y_i) \exp[-j2\pi(f_x x_i + f_y y_i)] dx_i dy_i $$

Space-Invariant Transfer Function

$$ H(f_x, f_y) = \iint \tilde{h}(x_i, y_i) \exp[-j2\pi(f_x x_i + f_y y_i)] dx_i dy_i $$

From the convolution theorem, we have:

$$ G_i(f_x, f_y) = H(f_x, f_y) \cdot G_g(f_x, f_y) $$$$ H(f_x, f_y) = \text{coherent transfer function (CTF)} $$

CTF and Pupil Function

$$ \text{Recall: } \boxed{H(f_x, f_y) = \mathcal{FF}\{\tilde{h}(x_i, y_i)\} = \mathcal{FF}\{\mathcal{FF}^{-1}\{p(\lambda d_i \tilde{x}, \lambda d_i \tilde{y})\}\}} $$$$ = p(-\lambda d_i f_x, -\lambda d_i f_y) $$$$ \implies \boxed{\text{CTF} = \text{Pupil function in inverted coordinates}} $$

Example: Circular Lens Aperture

Lens of Diameter $\ell$

$$ \Rightarrow p(x,y) = \text{circ}\left(\frac{\sqrt{x^2+y^2}}{\ell/2}\right) $$$$ \Rightarrow H(f_x, f_y) = \text{circ}\left(\frac{\sqrt{f_x^2+f_y^2}}{1/(2\lambda d_i)}\right) = \begin{cases} 1 & 0 < \sqrt{f_x^2+f_y^2} < \frac{1}{2\lambda d_i} \\ \frac{1}{2} & \sqrt{\quad} = \frac{1}{2\lambda d_i} \\ 0 & \sqrt{\quad} > \frac{1}{2\lambda d_i} \end{cases} $$

Transfer Function Shape

        ┌─────────────┐
        │             │     $\sin\theta = \frac{\ell}{d}$ role:
        │             │     (used backwards) ↙
    ────┴──────┬──────┴────    f = 1/λ
              │
         ──►  │  ◄── at θ = λf_x
        1/2λd_i
              = cutoff

$$ \Rightarrow f_x = \frac{1}{2\lambda d_i} $$

Physical Interpretation

Physical Interpretation:

      ┌─────────────┐
      │   ←─d_i─→  │
      │      ╲     │
      │       ╲    │
      └────────╲───┘
            Huygens
            Sources

• Each Huygens source will interfere with any other source to form a grating $d_i$ away from lens
• Each grating is a spatial frequency component of the image
• Largest spatial frequency is formed from Huygens sources which are farthest from each other

Maximum Spatial Frequency and Numerical Aperture

Two Point Sources

Consider two point sources at $x = \ell/2$ and $x = -\ell/2$:

  ℓ┌─────────────┐
   │      d_i    │θ
   │    ╲   ╱    │
  ─┴─────┴─────┴─

Interference Pattern

$$ A = e^{j\left(2\pi\alpha x + \frac{x\ell\alpha\xi}{2d_i}\right)} + e^{j\left(-2\pi\alpha x + \frac{x\ell\alpha\xi}{2d_i}\right)}, \quad \alpha = \frac{\sin\theta}{\lambda}, \quad \cos\theta \approx \frac{\ell/2}{d_i} $$

Ignoring the spherical wave term gives:

$$ A = e^{j(2\pi\alpha x)} + e^{-j2\pi\alpha x} = 2\cos(2\pi\alpha x) $$$$ \Rightarrow \boxed{\text{Highest spatial freq. in image} = \alpha = \frac{\ell\sin\theta}{\lambda} = \frac{\ell}{2\lambda d_i}} $$

Numerical Aperture

$$ \boxed{NA = n\sin\theta = \text{numerical aperture}} $$

Note: Physical Significance (Object Space Resolution)

Object Space Resolution

$$ \left(f_{\max}\right)_{obj} = \frac{1}{2\lambda d_o} = \frac{NA}{\lambda} $$$$ \text{Resolution} = \frac{1}{f_{\max}} = \frac{1}{(NA)} $$

Illuminate with a pencil beam at Grating of freq. $f_x$

See smaller features if:

• $\lambda$ small
• NA large ($NA \leq 1$ in air → $(res)_{min} = \lambda$)

---

Sinusoidal Diffraction Grating

A sinusoidal diffraction grating (freq $f_o$) will diffract a pencil beam into 2 beams.

• If too high, both beams miss the lens, and no light passes through
• At some critical angle $\tan \theta = \frac{1}{2d_o}$, the two pencil beams just enter, and interference occurs in the output to generate a sinusoid

Since the angle is related to the grating freq by:

$$ \sin \theta = \lambda f_x $$

(note, this is just $\sin \theta = \frac{\lambda}{d}$, where $d = \frac{1}{f_x}$)

We have:

$$ \sin \theta_{\max} = (\lambda f_x)_{\max} \approx \tan \theta_{\max} = \frac{1}{2d_o} $$$$ \implies \boxed{(f_x)_{\max} = \frac{1}{2\lambda d_o}} $$

---

Notes

• High spatial frequencies pass through the edge of the lens, and low spatial frequencies through the center
• $(NA)_{object side} = n \sin \theta$ and $f_{\max} = \frac{NA}{\lambda}$
• A specific spatial freq. goes through a specific part of the lens

Coherence

Midterm: Next Monday

Types of Coherence

1. Temporal coherence (longitudinal)
2. Spatial coherence (transversal)

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A. Temporal Coherence

!temporal_coherence_wave.svg

→ Degree of correlation between $t_1$ & $t_2$

→ $\langle u(t_1) u^*(t_2) \rangle$ ← Time average

$$ \langle u(t_1) u^*(t_2) \rangle = \frac{1}{T} \int_0^T u(t+t_1) u^*(t+t_2) dt $$

If statistics are stationary, we can write the time average as:

$$ \langle u(t_1) u^*(t_2) \rangle = \lim_{T \to \infty} \frac{1}{T} \int_0^T u(t) u(t+\tau) dt, \quad \tau = t_2 - t_1 $$

Examples of high temporal coherence:

• Pure monochromatic source such as a single mode highly filtered white light
• Possibly low spatial coherence

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B. Spatial Coherence

!spatial_coherence_two_points.svg

Degree of correlation between $x_1$ & $x_2$

$$ \langle u(x_1) u^*(x_1) \rangle $$

← time average

$$ \Rightarrow |\langle u(x_1) u^*(x_2) \rangle| = 1 $$

→ perfect spatial coherence

If diffuse states $e^{i\phi(t)}$ ← function of time, and coherence is destroyed.

Examples of high spatial coherence:

1. Star (low temporal coherence)
2. Arc lamp (low-medium temporal coh.)
3. Gas laser (high temporal coh.)

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C. Mutual Coherence Function

$$ \Gamma(x_1, x_2, \tau) = \Gamma_{12}(\tau) = \langle u(x_1, t+\tau) u^*(x_2, t) \rangle $$

Cross correlation between waves at two different points and two different times.

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Detailed Analysis

Note: This does not imply that high spatial coherence means a plane wave illumination.

Example: !diffuser_spreading_light.svg

Consider $\langle u(x_1) u^*(x_2) \rangle$

$$ u(x_1) = e^{i(\omega t + \phi_1)} $$$$ u(x_2) = e^{i(\omega t + \phi_2)} $$

Note: phases are not changing in time.

$$ \langle u(x_1) u^*(x_2) \rangle = \frac{1}{T} \int_0^T e^{i(\phi_1 - \phi_2)} dt = e^{i(\phi_1 - \phi_2)} \frac{1}{T} \int_0^T dt $$$$ \Rightarrow |\langle u(x_1) u^*(x_2) \rangle| = 1 $$

→ Perfect spatial coherence

Mutual Coherence Function and Fringe Visibility

★★ Interference

Time-Average Intensity of two light fields $E_1$ & $E_2$ (at 2 places and 2 times)

$$ I = \langle (E_1 + E_2) \cdot (E_1^* + E_2^*) \rangle $$$$ = \langle |E_1|^2 \rangle + \langle |E_2|^2 \rangle + 2\text{Re}\langle E_1 \cdot E_2^* \rangle $$$$ = I_1 + I_2 + 2\text{Re}\langle E_1 E_2^* \rangle $$

where $I_1 = \langle |E_1|^2 \rangle$, and $I_2 = \langle |E_2|^2 \rangle$

---

Defining Mutual Coherence

We can define the mutual coherence as:

$$ \Gamma_{12}(\tau) = \langle E_1(t) E_2^*(t+\tau) \rangle $$$$ \gamma_{12}(\tau) = \frac{\Gamma_{12}(\tau)}{\sqrt{\Gamma_{11}(0) \Gamma_{22}(0)}} = \frac{\Gamma_{12}(\tau)}{\sqrt{I_1 I_2}} $$

!interference_spatial_coherence.svg

What is spatial coh. in this plane of pts $x_1$ & $x_2$?

$\Gamma_{12}(\tau=0)$

$$ \implies I = I_1 + I_2 + 2\text{Re}\{\Gamma_{12}(\tau)\}, \quad \text{where } \tau = \text{time difference} $$

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Define Degree of Coherence

$$ \gamma_{12}(\tau) = \frac{\Gamma_{12}(\tau)}{\sqrt{\Gamma_{11}(0)\Gamma_{22}(0)}} = \frac{\Gamma_{12}(\tau)}{\sqrt{I_1 I_2}} $$$$ \implies I = I_1 + I_2 + 2\sqrt{I_1 I_2} \text{Re}\{\gamma_{12}(\tau)\} $$$$ = I_1 + I_2 + 2\sqrt{I_1 I_2}|\gamma_{12}(\tau)|\cos[\alpha_{12}(\tau) + \delta] $$

where $|\gamma|$ ranges from $0 \to 1$ ← totally coherent $\alpha_{12}(\tau) + \delta = \arg\{\gamma_{12}(\tau)\}$

• If $|\gamma| = 1$, waves add in amplitude
• If $|\gamma| = 0$, waves add in intensity

$$ \Rightarrow I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2}|\gamma_{12}| $$$$ I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2}|\gamma_{12}| $$$$ \boxed{\gamma^2 = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{2\sqrt{I_1 I_2}|\gamma_{12}|}{I_1 + I_2}} $$

→ Visibility

(Note similarity to standing wave ratio)

Visibility and Coherence

$$ \rightarrow \text{If } I_1 = I_2 \text{ we have} $$$$ \boxed{\gamma = |\gamma_{12}|} $$$$ \implies \text{Complete coherence} \Rightarrow \text{visibility} = 1 $$$$ \text{Complete incoherence} \Rightarrow \text{visibility} = 0 $$

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D. Relation Between Temporal Coherence & Spectrum

Note:

$$ \langle u(t) u^*(t+\tau) \rangle = \lim_{T \to \infty} \frac{1}{T} \int_0^T u(t) u^*(t+\tau) dt $$$$ = u(t) * u^*(t) \leftarrow \text{autocorrelation} $$

Wiener-Khinchin Theorem ⇒

$$ \mathcal{F}\{u(t) * u^*(t)\} = \tilde{U}(\omega) \cdot \tilde{U}^*(\omega) = |U(\omega)|^2 $$

→ Spectral Power Density

⇒ Thus, the temporal coherence part of the mutual coherence function is F.T. related to the bandwidth of the source ⇒

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Examples:

| Source | Spectrum | $|\gamma_{11}(\tau)|$ | |——–|———-|———————-| | “Perfect” Laser (single freq.) | [Delta function at 5000 Å] | [Constant = 1] | | Discharge Tube (Low pressure) | [Peak at $\Delta\lambda \approx 1$ Å] | [Decaying curve, $2.3 \times 10^{-12}$ sec] |

Coherence Time and Length

Multi-Mode Lasers

| Source | Spectrum | R(τ) | $|\gamma_{11}(\tau)|$ | |——–|———-|——|———————-| | Two modes | [Two peaks separated by δf] | [Oscillating] | [Envelope with period ← δf →] | | Many modes | [Multiple peaks at δf₁, δf₂] | [Autocorrelation pattern] | [Decaying envelope with periods 1/δf₁, 1/δf₂] |

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★ Coherence Time

$$ \boxed{\langle \tau_o \rangle = \frac{1}{\Delta f}} $$

Since: [Spectrum diagram] → Coherence function with $\tau_o = \frac{1}{\Delta f}$]

⇒ Time delay when fringes no longer form

$$ \Delta \tau = t_2 - t_1 $$

If $\Delta \tau > \tau_o$, no fringes → Holography

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★ Coherence Length

$$ l_c = c\langle \tau_o \rangle = \frac{c}{\Delta f} $$

In terms of wavelengths, we have:

$$ \boxed{l_c = \frac{\lambda^2}{\Delta \lambda}} $$

Since:

$$ \frac{\Delta f}{f} = \frac{\Delta \lambda}{\lambda} $$$$ \frac{c}{\Delta f} = \frac{\bar{c}\lambda}{\bar{f}\Delta \lambda} $$

And:

$$ f = \frac{c}{\lambda} $$$$ \frac{df}{d\lambda} = -\frac{c}{\lambda^2} $$$$ \Rightarrow \Delta f = -\frac{c}{\lambda^2}\Delta \lambda $$

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Example: Coherence length of white light detectable by the eye

$$ \lambda = 5500 \text{ Å} $$$$ \lambda_{\min} = 4000 \text{ Å} $$$$ \lambda_{\max} = 7000 \text{ Å} $$$$ \Rightarrow l_c = \frac{\lambda^2}{\Delta \lambda} = \left(\frac{\lambda}{\Delta \lambda}\right)\lambda \approx 3\lambda $$

Measurement of Temporal Coherence

→ Measurement of Temporal Coherence

Michelson Interferometer

!michelson_interferometer_temporal.svg

$$ \tau = \frac{l_2 - l_1}{c} $$

Note: By adjusting $l_1$ w.r.t. $l_2$, any time can be correlated with any other. The fields are compared at the same point in space.

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Output of Michelson at a Single Point

$$ I(\tau) = \lim_{T \to \infty} \frac{1}{T} \int_0^T |E(t) + E(t+\tau)|^2 dt $$$$ = 2I_o + 2\text{Re}\{\Gamma_{11}(\tau)\} $$

(See previous derivation on pg. 3-3)

where $\tau = \frac{l_2 - l_1}{c}$

Derivation:

$$ I(\tau) = \frac{1}{T}\int_0^T |E(t) + E(t+\tau)|^2 dt $$$$ = \frac{1}{T}\left(\int_0^T |E(t)|dt + \frac{1}{T}\int_0^T |E(t+\tau)|dt\right) $$

$$

• \frac{1}{T}\int_0^T E(t)\bar{E}(t+\tau)dt + \frac{1}{T}\int_0^T E^*(t)E(t+\tau)dt $$

$$ = I_o + 2\text{Re}\left\{\frac{1}{T}\int_0^T E(t)E^*(t+\tau)dt\right\} $$

where $\tau = \frac{l_2 - l_1}{c}$

$$ = 2I_o + 2\text{Re}\{\Gamma_{11}^{Autocorrelation}\} $$

Thus, by moving one arm w.r.t. the other, we can obtain $\Gamma_{11}(\tau)$.

⇒ Since $\Gamma_{11}(\tau)$ is F.T. related to the spectrum of the source, we can measure $\Gamma_{11}(\tau)$ with this device and perform FFT to compute spectrum.

★ Fourier Transform Spectrometer

Fourier Spectrometer Advantages and Limitations

★ Advantages to Fourier Spectrometer

• All wavelengths are multiplexed onto a single detector - Use to be important in astronomy. Now used in IR (FTIR)

• High resolution possible with large instrument

• Angular acceptance is higher

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★ Limitations

• Resolution determined by:
  • A. Stability of large interferometers
  • B. SNR of detector measuring small correlations on a large bias

from very narrow lines

Spatial Coherence

!source_coherence_projection.svg

• Coherence measured in this plane
• Note: Incoherent source can become coherent through propagation

$$ \Gamma_{12}(0) = \text{measure of spatial coherence} $$

⇒ $r_1 = r_2$, we can vary $d$ until no fringes form. This is the coherence width.

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★ Effect of Source Size on Spatial Coherence

Consider fringes made by each point of an extended completely incoherent source and superimpose the result.

For a single point on S, we have a fringe pattern at P:

$$ I(P) = I_1(P) + I_2(P) + 2\sqrt{I_1(P)}\sqrt{I_2(P)}\cos\left[\bar{k}\left(\frac{d}{R}\right)x + \bar{k}\left(\frac{d}{r}\right)x''\right] $$

where $\bar{k} = \frac{2\pi}{\lambda}$

Geometry:

• $r_2 = \sqrt{r^2 + (\frac{d}{2} + x'')^2}$
• $r_1 = \sqrt{r^2 + (\frac{d}{2} - x'')^2}$
• $|r_2 - r_1| = \sqrt{r^2 + (\frac{d}{2} + x'')^2} - \sqrt{r^2 + (\frac{d}{2} - x'')^2}$

$$ = r\sqrt{1 + \frac{(\frac{d}{2} + x'')^2}{r^2}} - r\sqrt{1 + \frac{(\frac{d}{2} - x'')^2}{r^2}} $$$$ \approx \frac{r}{2}\left(1 + \frac{(\frac{d}{2} + x'')^2}{2r^2} - 1 - \frac{(\frac{d}{2} - x'')^2}{2r^2}\right) $$$$ = \frac{1}{2r^2}\left((\frac{d}{2} + x'')^2 - (\frac{d}{2} - x'')^2\right) $$$$ = \frac{1}{2r}\left((\frac{d}{2})^2 + dx'' + x''^2 - (\frac{d}{2})^2 + dx'' - x''^2\right) $$$$ = \frac{dx''}{r} $$

• Phase difference from s to top and bottom holes
• Phase difference from p to top and bottom holes

$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$

When extended source is represented by many point sources:

$$ I(P) = I_1(P) + I_2(P) + 2\sqrt{I_1(P)}\sqrt{I_2(P)}\int_{-S/2}^{S/2}\cos\left[\bar{k}\left(\frac{d}{R}\right)x + \bar{k}\left(\frac{d}{r}\right)x''\right]dx'' $$$$ = I_1(P) + I_2(P) + 2S\sqrt{I_1(P)}\sqrt{I_2(P)}\text{sinc}\left[\left(\frac{d}{\lambda R}\right)S\right]\cos\left[\bar{k}\left(\frac{d}{R}\right)x\right] $$

Spatial Coherence and Grating Shift

Note:

$$ \cos\left(\bar{k}\left(\frac{d}{R}\right)x + \bar{k}\left(\frac{d}{r}\right)x''\right) $$

• Phase shift φ (in radians)
• Grating freq. (in radians)

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⇒ When two different point sources are used, gratings are shifted by an amount (distance along source):

$$ \bar{k}\left(\frac{d}{R}\right)x \cdot \left(\frac{r}{\bar{k}d}\right) = \frac{xr}{R} $$

Term	Description
Phase shift in radians φ
Period of grating (for 2π radians)	⇒ Physical grating shift for φ phase shift

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Diagram

!two_source_fringe_shift.svg

• ← cannot to period of entire fringe →

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Condition for Negligible Degradation

If $\frac{xr}{R} \ll \frac{b}{2\pi} \cdot \frac{r}{\bar{k}d}$

shift ↓ ← Period (through one radium)

or $d \ll \frac{2\pi R}{kx} = \frac{R\lambda}{\pi x}$

$$ \frac{d}{R} \ll \frac{d}{\lambda} $$

For negligible degradation of fringe pattern: ⇒ simply require phase shift $\phi \ll \pi$ ⇒ $\bar{k}\left(\frac{d}{r}\right)x \ll \pi$

$$ \Rightarrow d \ll \frac{\lambda R}{2x} $$

For maximum x separation, $x_{\max} = S/2$: $\left[\text{sinc}\left(\frac{dS}{\lambda R}\right)\right]$

Van Cittert-Zernike Theorem

Fringe Visibility Condition

Notice: Fringes vanish if $d = \frac{\bar{\lambda}R}{S}$

or $\boxed{d = \frac{\bar{\lambda}}{\theta_s}}$ where $\theta_s = \frac{S}{R}$ = angular size of source

Note: From Fourier aperture produces: $\propto \frac{b}{\bar{\lambda}}$, $\sin\theta \approx \theta = \frac{\lambda}{b} = \tan\frac{\lambda}{b}$

$\implies b \approx \frac{R\bar{\lambda}}{S}$, exactly like above spatial coherence eqn.

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Generalization: Van Cittert-Zernike Theorem

Distances from $P_1$ & $P_2$ meas. pts. to source

$$ \text{Spatial Coherence} \Rightarrow \gamma_{12}(0) = \frac{\int_S I(x_s, y_s) e^{i\bar{k}(R_1-R_2)} dS}{\int I(x_s, y_s) dS} $$

where coherence function $\gamma = \frac{\Gamma_{12}(0)}{\sqrt{\Gamma_{11}(0)\Gamma_{22}(0)}}$

Same mathematics as diff pattern of coherent wave converging on $P_1$ after passing through transparency having amplitude transmittance proportional to intensity distribution of incoherent source.

$I(x_s, y_s)$ at $P_1$

Note: $\bar{k}(R_1-R_2)$ is the phase shift of the cosine fringe pattern.

---

This is exactly like the diffraction equation

⇒ If $P_1$ & $P_2$ are close to the axis & far from source, we have an equation like Fraunhofer diffraction

and hence $I(x_s, y_s)$ and $\gamma_{12}(0)$ are Fourier Transform related.

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Example: Measure the size of Betelgeuse (red giant)

!michelson_stellar_interferometer.svg

Measuring Stellar Angular Size

Assuming star is round, we have $\gamma_{12}(0)$ given as the F.T. of a circle ⇒

[Graph showing $\gamma_{12}(0)$ vs P with width of coherence area marked, showing $J_1(anl)$ behavior starting at 1 and decreasing to first zero at 1.22]

$$ \Rightarrow d_w = \frac{1.22\bar{\lambda}}{\theta_s} \leftarrow \text{Fringe disappear} $$

First zero of fringe visibility ← Angular size of source

Measurement: $d_w = 2.6$ meters

$$ \Rightarrow \theta_s = \frac{1.22\bar{\lambda}}{d_w} = 0.047 \text{ sec of arc} $$

From known distance ⇒ diameter is 280 times that of the sun — Red Giant

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Notes:

• Michelson in principle could also measure phase info. of $\gamma_{12}(0)$ and hence an inverse FT would reconstruct the intensity.

• Atmospheric turbulence prevents this from happening in practice. Hence, Michelson measured short-term visibility magnitude only

• Various techniques can be used to recover phase info in presence of random fluctuations ⇒ Speckle interferometry, speckle masking (triple product correlation), and phase retrieval (Gerschberg-Saxton algorithm)

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Incoherent Case: Optical Transfer Function

Optical Transfer Function (OTF)

★ Modulation Transfer Ftn. (MTF) is the modulus of OTF

[Block diagram showing: $A_1 + A_2$ → [1] → $\mathcal{F}\{A_1\} + \mathcal{F}\{A_2\}$]

System is linear for general amplitude ftn.

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Consider a case where $A_1$ & $A_2$ contain phase terms which are random functions of time

!incoherent_two_point_system.svg

$$ A = h(x-x_1)e^{i\phi_1(t)} + h(x-x_2)e^{i\phi_2(t)} $$

Average Intensity is the directly observable quantity:

$$ \langle I \rangle = \frac{1}{T}\int_0^T \left|h(x-x_1)e^{i\phi_1(t)} + h(x-x_2)e^{i\phi_2(t)}\right|^2 dt $$$$ = \frac{1}{T}\int_0^T |h(x-x_1)|^2 dt + \frac{1}{T}\int_0^T |h(x-x_2)|^2 dt + \frac{1}{T}\int_0^T h(x-x_1)h^*(x-x_2)e^{i(\phi_1-\phi_2)} dt $$

$$

• \frac{1}{T}\int_0^T h^*(x-x_1)h(x-x_2)e^{-i(\phi_1-\phi_2)} dt $$

Incoherent Imaging

Incoherent Case Analysis

If $\phi_1$ and $\phi_2$ are uncorrelated and uniformly distributed from 0 to $2\pi$, then:

$$ \frac{1}{T}\int_0^T e^{i(\phi_1-\phi_2)} dt = 0 $$

and

$$ \langle I \rangle = \langle |h(x-x_1)|^2 \rangle + \langle |h(x-x_2)|^2 \rangle $$$$ \implies \text{System is linear in intensity with impulse response} $$$$ \langle |h(x)|^2 \rangle $$

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Coherent vs Incoherent Case

Coherent case:

• $\langle E_1 \rangle = 0$
• $\langle E_2 \rangle = 0$ (only)
• $\langle E_1 + E_2 \rangle = 0$ (known)
• $\langle E_1 E_2^* \rangle = $ (statistics passes)
• $\langle |E|^2 \rangle + \langle |E|^2 \rangle$
• $= \langle E_1 \rangle * \langle E_2 \rangle$ zero fractions
• $= I_1 + I_2$

Incoherent light is therefore linear in intensity.

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Transfer Functions

We must use an intensity convolution $I(x, y)$:

Impulse response is $|h(x, y)|^2$ and every thing is coherent from a pt. source. Every thing is coherent from a pt. source.

$$ G_q(f_x, f_y) = K \iint |h(x_i - \bar{x}_o, y_i - \bar{y}_o)|^2 I_g(\bar{x}, \bar{y}) d\bar{x}_o d\bar{y}_i $$

!incoherent_optical_system.svg

If we define normalized frequencies: (“Normalize to zero-frequency” values)

$$ G_q(f_x, f_y) = \iint I_g(\bar{x}_o, \bar{y}_o) \exp[-j2\pi(f_x\bar{x}_o + f_y\bar{y}_o)] d\bar{x}_o d\bar{y}_i $$$$ \iint I_g(\bar{x}_o, \bar{y}_o) d\bar{x}_o d\bar{y}_i $$$$ G_i(f_x, f_y) = \iint I_i(x_i, y_i) \exp[-j2\pi(f_x x_i + f_y y_i)] dx_i dy_i $$$$ \iint I_i(x_i, y_i) dx_i dy_i $$

Optical Transfer Function

$$ \mathcal{H}(f_x, f_y) = \frac{\iint |\tilde{h}(x_i, y_i)|^2 \exp[-j2\pi(f_x x_i + f_y y_i)] dx_i dy_i}{\iint |h(x_i, y_i)|^2 dx_i dy_i} $$

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Then

$$ G_i(f_x, f_y) = \mathcal{H}(f_x, f_y) \cdot G_g(f_x, f_y) $$

Where:

• $\mathcal{H}$ = optical transfer function (OTF)
• $|\mathcal{H}|$ = modulation transfer function (MTF)

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Note:

$$ \mathcal{H}(f_x, f_y) = \frac{\mathcal{F}\{|\tilde{h}(x_i, y_i)|^2\}}{\mathcal{F}\{|\tilde{h}(x_i, y_i)|^2\}|_{f_x=0, f_y=0}} $$

From the autocorrelation theorem (Wiener-Khinchin):

$$ \mathcal{F}\{|\tilde{h}(x_i, y_i)|^2\} = H(f_x, f_y) \circledast H^*(f_x, f_y) $$$$ \implies \mathcal{H}(f_x, f_y) = \frac{H(f_x, f_y) \circledast H^*(f_x, f_y)}{(H(f_x, f_y) \circledast H^*(f_x, f_y))|_{f_x=0, f_y=0}} $$

---

Properties:

1. $\mathcal{H}(0, 0) = 1$

2. Fourier of a real ftn. ⟹ $2 \cdot \mathcal{H}(-f_x, -f_y) = \mathcal{H}^*(f_x, f_y)$

3. Schwarz inequality ⟹ $|\mathcal{H}(f_x, f_y)| \leq |\mathcal{H}(0, 0)|$

CTF and OTF Examples

Examples:

Consider the CTF and OTF of a system with a square pupil:

!square_pupil.svg

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Coherent Transfer Function (CTF):

$$ H(f_x, f_y) = \text{rect}\left(\frac{\lambda d_i f_x}{l}\right) \text{rect}\left(\frac{\lambda d_i f_y}{l}\right) $$

— Coherent Limit

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Optical Transfer Function (OTF):

$$ \mathcal{H}(f_x, f_y) = H(f_x, f_y) \circledast H(f_x, f_y) $$

— Incoherent Limit

$$ = \left[\text{rect}\left(\frac{\lambda d_i f_x}{l}\right) \circledast \text{rect}\left(\frac{\lambda d_i f_x}{l}\right)\right]\left[\text{rect}\left(\frac{\lambda d_i f_y}{l}\right) \circledast \text{rect}\left(\frac{\lambda d_i f_y}{l}\right)\right] $$$$ = \text{tri}\left(\frac{\lambda d_i f_x}{l}\right) \text{tri}\left(\frac{\lambda d_i f_y}{l}\right) $$

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[Graph showing CTF (rectangular) and OTF (triangular) curves plotted against $f_x$, with cutoff frequencies marked at $-\frac{1}{\lambda d_i}$, $-\frac{1}{2\lambda d_i}$, $\frac{1}{2\lambda d_i}$, $\frac{1}{\lambda d_i}$]

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⇒ An incoherent system has a cut-off at twice the freq. of a coherent system. However, this does not really mean it has twice the resolving power.

Two F numbers: A given distance range in two constraints: similarity to a certain spatial freq. in an incoherent system, propagated at this distance within the lines on wave. The number of points in biological equal threshold of finding a

Physical Interpretation of OTF

Physical Interpretation:

!otf_physical_interpretation.svg

• For this freq, the 2 pencil beams are shifted in angle depending on the illumination angle

• Grating is formed by any two pairs of Huygens sources that are spaced by a distance ($f_x \cdot \lambda d_i$)

• For low spatial frequencies, there are more pairs than at high frequencies (& thus large separations)

• Two Finger rule:

!two_finger_rule.svg

MTF is proportional to the number of locations two points occur at a specific separation and orientation ⇒ correlation

• Note redundancy in incoherent system which gives rise to noise suppression (coherent noise).

Pupil Shapes of Mammals

Pupil Shapes of Mammals:

Animal	Pupil Shape	MTF
Horse & goat	[Horizontal rectangular pupil]	[MTF graph showing vert/horiz curves, higher horizontal resolution]
Domestic cat	[Vertical slit pupil]	[MTF graph showing vert/horiz curves, higher vertical resolution]
Humans	[Circular pupil]	[MTF graph showing symmetric response]

Generalized Pupil Function

I. Generalized Pupil Function

!diffraction_limited_pupil.svg

Diffraction limited system ⇒ Impulse response = F.T. of pupil function (aperture)

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Aberrations

→ Aberrations are phase errors in optical system → If phase error at exit pupil is $kW(x,y)$, $k = 2\pi/\lambda$

We define a generalized pupil function:

$$ \boxed{P(x,y) = P(x,y) \exp[jkW(x,y)]} $$

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Coherent Case

The impulse response is the F.T. of $\mathbb{P}(x,y)$.

We also have the coherent transfer function = generalized pupil ftn.

$$ H(f_x, f_y) = \mathbb{P}(\lambda d_i f_x, \lambda d_i f_y) $$$$ = P(\lambda d_i f_x, \lambda d_i f_y) \exp[jkW(\lambda d_i f_x, \lambda d_i f_y)] $$

The effect of aberrations is simply to apply phase distortions to the pass band. It does not affect the band limit.

Incoherent Case OTF with Aberrations

Incoherent Case

$$ \mathcal{H}(f_x, f_y) = \frac{\mathbb{P}(\lambda d_i f_x, \lambda d_i f_y) \circledast \mathbb{P}^*(\lambda d_i f_x, \lambda d_i f_y)}{(\mathbb{P}(\lambda d_i f_x, \lambda d_i f_y) \circledast \mathbb{P}^*(\lambda d_i f_x, \lambda d_i f_y))|_{f_x=0, f_y=0}} $$

→ In general, aberrations lower the contrast of a certain spatial frequency

Aberrations and OTF

Examples:

!ctf_otf_comparison.svg

---

Aberrations and Their Effect on the OTF

Focusing Error (Square lens of width l)

Recall imaging relation: $\frac{1}{d_i} + \frac{1}{d_o} - \frac{1}{f} = 0$

Focusing error: $\frac{1}{d_i} + \frac{1}{d_o} - \frac{1}{f} = \epsilon$

Then, we have an extra term in the pupil function:

Recall: Point spread ftn for single lens imaging:

$$ h(x_o, y_o) = \frac{1}{\lambda d_o d_i} $$$$ \mathcal{F}\{P_{\text{pupil}} \cdot \exp[i\frac{k}{2}(\frac{1}{d_o} + \frac{1}{d_i})(x^2+y^2)]\} $$$$ = \exp[ik(\frac{1}{2d_o} + \frac{1}{2d_i})x \times (\frac{x_o}{d_o} + \frac{x_i}{d_i})] $$$$ P(x,y) \exp\left[jk\epsilon\frac{(x^2+y^2)}{2}\right] $$

and $W = \frac{\epsilon l^2}{8}$ where W is the maximum path length error along x or y axis.

$$ \mathcal{H}(f_x, f_y) = \Lambda\left(\frac{f_x}{2f_o}\right)\Lambda\left(\frac{f_y}{2f_o}\right) $$$$ \times \text{sinc}\left[\frac{8W}{\lambda}\left(\frac{f_x}{2f_o}\right)\left(1-\frac{|f_x|}{2f_o}\right)\right] \text{sinc}\left[\frac{8W}{\lambda}\left(\frac{f_y}{2f_o}\right)\left(1-\frac{|f_y|}{2f_o}\right)\right] $$

!lens_dimension.svg

$x = \frac{l}{2}$, $\epsilon\frac{(x^2+y^2)}{2} = \epsilon\frac{l^2}{8}$

Note: $W = 0$ ⇒ diffraction limit

$f_o = \frac{l}{2\lambda d_i}$ ← lens diameter

[Graph showing OTF curves for W=λ, W=λ/4, and W=0 (diffraction limit), with contrast reversal indicated]

Aberration Function Effects

Aberration Function

!aberration_pupil_zones.svg

⇒ Frequency is not represented

Aberration Ftn

---

Notes:

1. If complete cancellation of a freq. requires a $\lambda/2$ error

2. Contrast reversal requires > $\lambda/2$ error.

Coherent Imaging Effects

II. Coherent Imaging Effects

1. Ringing
2. Speckle

---

A. Ringing of a Knife Edge (Coherent)

$$ I_{im}(x') = \left|\int_{-\infty}^{\infty} \sqrt{I_{ob}(x)} \cdot K\left(\frac{x'}{q} + \frac{x}{p}\right) dx\right|^2 $$

where:

• $I_{ob}(x)$ = object intensity (assume no phase component)
• $K\left(\frac{x'}{q}\right)$ = amplitude impulse response of the lens
• $p$ = object distance
• $q$ = image distance

---

Consider a square aperture of size “2a”

Amplitude Impulse Response:

$$ K\left(\frac{x'}{q}\right) = \text{sinc}\left(ka\frac{x'}{q}\right), \quad k = \frac{2\pi}{\lambda} $$

and

$$ I_{im}(x') = \left|\int_{-\infty}^{\infty} \sqrt{I_{ob}(x)} \text{sinc}\left[ka\left(\frac{x'}{q} + \frac{x}{p}\right)\right] dx\right|^2 $$

---

Consider a Knife Edge Object

$$ \Rightarrow I_{ob}(\bar{x}) = \begin{cases} 1 & x > 0 \\ 0 & x < 0 \end{cases} = \text{step}(x) $$$$ I_{im}(x') = \left|\int_{-\infty}^{\infty} \text{step}(x) \text{sinc}\left[ka\left(\frac{x'}{q} + \frac{x}{p}\right)\right] dx\right|^2 $$$$ = \left|\int_0^{\infty} \text{sinc}\left[ka\left(\frac{x'}{q} + \frac{x}{p}\right)\right] dx\right|^2 $$$$ = \left[\frac{1}{2} - \frac{1}{\pi} Si\left(\frac{kax'}{q}\right)\right]^2 $$

Sine Integral Function

where: $Si(\theta) = \int_0^{\theta} \frac{\sin\theta}{\theta} d\theta$ = sine integral function

---

Physical Meaning (Coherent Case):

[Graph showing oscillating response with overshoot, then settling to steady state]

⇒ [Graph showing coherent edge response with ringing/oscillations]

---

Incoherent Case

$$ I_{im}(x') = \int_{-\infty}^{\infty} I_{ob}(x) \cdot K\left(\frac{x'}{q} + \frac{x}{p}\right) K^*\left(\frac{x'}{q} + \frac{x}{p}\right) dx $$

Intensity impulse response of square aperture (size “2a”):

$$ \left|K\left(\frac{x'}{q}\right)\right|^2 = \text{sinc}^2(kax'/q) $$$$ \Rightarrow I_{im}(x') = \frac{1}{2} - \frac{1}{\pi}\left[Si\left(\frac{2kax'}{q}\right) - \frac{1 - \cos^2\left(\frac{kax'}{q}\right)}{kax'/q}\right] $$

---

Physical Meaning (Incoherent):

Comparison between coh. & incoh.

[Graph showing step input → smooth monotonic response (incoherent)]

[Graph showing comparison with coherent response showing oscillations vs incoherent smooth response, with marking at $-\frac{1}{2}$ in amplitude and $= \frac{1}{4}$ in intensity]

Ref: Considine, P.S., JOSA 56, 1001, (1966)

Speckle

B. Speckle

(Viewpoint #1: Angular plane wave spectrum)

(Not just for laser light, but radar & sonar as well)

Coherent light reflects off of a “rough” white surface

“Rough” means surface varies in height by several wavelengths. Since screen is white, no amplitude modulation of illumination occurs.

!rough_surface_scattering.svg

At the plane, we have the original illum. $A(x,y)$ times a random phase $e^{j\phi(x,y)}$ due to the surface

→ We can decompose the amplitude at the surface into an angular plane wave spectrum. Thus, at this surface, all plane waves add together in such a way as to create a uniform amplitude function

→ If an optical system can capture all the plane waves and recombine them into an image, a uniform amplitude function results

!speckle_optical_system.svg

Image contains exactly the same set of plane waves as the object ⇒ uniform amplitude phase function

Speckle Formation

→ We can think of each point of the image (or object) as being a summation of many plane waves of random phases. Since light is coherent, plane waves add in amplitude as phasors.

!phasor_addition_speckle.svg

→ Now consider an optical system which had a small aperture so that some of the plane waves were not passed through.

!speckle_limited_aperture.svg

Histogram of speckle:

$$ p(I) = \frac{\exp(-I/I_o)}{I_o} $$

where $I_o = \langle I \rangle$ = mean intensity

[Graph showing histogram of speckle intensity - exponential decay curve from peak at 0, with marking showing “constant amplitude” and “from only 3 plane waves mixed” at points, and “Unit Circle” marker]

Phasor Diagram

The amplitude is no longer a constant, but rather is a random value which changes with every point on surface.

This is speckle. Since the human eye captures only a very small angle of plane waves, speckle is usually seen with any diffuse object (diffuse means “rough” is scattered into a wide angular plane wave excitation).

Speckle Statistics

→ We can think of each point of the image (or object) as being a summation of many plane waves of random phases. Since light is coherent, plane waves add in amplitude as phasors.

!phasor_speckle_statistics.svg

Every point in image is composed of a sum of phasors which always add up to 1

---

→ Now consider an optical system which had a small aperture so that some of the plane waves were not passed through.

[Diagram showing angular plane wave spectrum with limited aperture]

Histogram of speckle:

$$ p(I) = \frac{\exp(-I/I_o)}{I_o} $$

where $I_o = \langle I \rangle$ = mean intensity

[Histogram showing exponential decay of intensity probability, with annotations:

• “constant amplitude from only 3 plane waves mixed”
• Unit circle diagram showing phasor addition]

Phasor Diagram

The amplitude is no longer a constant, but rather is a random value which changes with every point on surface.

This is speckle. Since the human eye captures only a very small angle of plane waves, speckle is usually seen with any diffuse object (diffuse means “rough” is scattered into a wide angular plane wave excitation).

Speckle (Viewpoint #2: Impulse response of optical system)

→ The impulse response of an imaging system (sometimes called its point-spread function) is given by the Fourier transform of the aperture: $\mathcal{F}\{P(\lambda d_i \bar{x}, \lambda d_i \bar{y})\}$

Thus, if the aperture is a circle, the impulse response is $\text{somb}(\rho)$

[Diagram showing phase front $e^{i\phi_o}$ passing through aperture, creating $\text{somb}(r) \propto \frac{J_1(ar\bar{s})}{s}$]

→ A rough surface now consists of many points, each with its own random phase. This gives rise to many $\text{somb}(r)$ ftns., each with random phase.

[Diagram showing overlapping somb functions]

When somb ftns overlap, amplitudes = sum. However, due to the random phase of each, this sum can either add or subtract. Thus, overlap points are not uniform, and result in speckle

Notes:

★ 1) As lens gets larger (better resolution) size of speckle gets smaller 2) Size of speckle is ≈ diffraction limit of lens

Apodization

Apodization - What is it?

[Diagram showing aperture with tapered edges] ↔ [Corresponding F.T. pattern]

A - pod ← eq: podiatrist, pseudopod ↑ ↑ no feet ⇒ To cut off or remove the feet

This is like an F.T. system, except that the aperture has been passed up to the lens plane.

Apodization is the shaping of the aperture of an optical system to control side lobes (or sometimes enhance them to provide improved resolution).

---

→ Consider a typical astronomy problem:

Resolve a double star system where one star is $10^2$ times brighter than the other:

[Diagram showing bright star and dark companion with diffraction patterns, showing side lobes at $-\frac{2}{\lambda}$ and $\frac{2}{\lambda}$]

• We desire to reduce the side lobes as much as possible

Amplitude: $\text{rect}\left(\frac{x}{a}\right) \xrightarrow{\mathcal{F}} a\,\text{sinc}(au)$

Intensity: $I(u) = |a\,\text{sinc}(au)|^2 \propto \frac{1}{u^2}$ ← Spatial freq = sinθ

$I(u) \sim \frac{1}{u^2}$

---

Try using an amplitude mask with a (near coherent) triangular shading (Bartlett window)

[Diagram showing triangular aperture function with width a]

Amplitude Impulse Response:

$$ \mathcal{F}\left\{\text{tri}\left(\frac{x}{a}\right)\right\} = \mathcal{F}\left\{\text{rect}\left(\frac{2x}{a}\right) \circledast \text{rect}\left(\frac{2x}{a}\right)\right\} = \left(\frac{a}{2}\right)^2 \text{sinc}^2\left(\frac{au}{2}\right) $$

where $\text{tri}(x) = \begin{cases} 1-|x| & |x| \leq 1 \\ 0 & \text{else} \end{cases}$

$I(u)$ — Intensity impulse: $I(u) = \left(\frac{a}{2}\right)^4 \text{sinc}^4\left(\frac{au}{2}\right)$

$\propto \frac{1}{u^4}$

Note: Main lobe is increased by 2× because some high spatial freq’s removed.

[Graph comparing intensity responses, showing $\frac{1}{u^2}$ vs $\frac{1}{u^4}$ decay]

Apodization - Parzen Window

Try Parzen Window

$$ \mathcal{F}\left\{\text{tri}\left(\frac{ux}{a}\right) \circledast \text{tri}\left(\frac{ux}{a}\right)\right\} $$

Choose half size so autocorrelation fits in same space

$$ \mathcal{F}\left\{\text{tri}\left(\frac{ux}{a}\right) \circledast \text{tri}\left(\frac{ux}{a}\right)\right\} = \left(\frac{a}{4}\right)^4 \text{sinc}^4\left(\frac{au}{4}\right) $$

[Diagram showing Parzen window shape]

• 1st zero at $\frac{4}{a}$

⇒ Intensity fall off: $\propto \frac{1}{u^8}$

Note: As window gets “smoother”, side lobes decrease faster.

[Note: convolution represents a low-pass filter operation (in the case), reducing the high spatial freq’s and hence the sidelobes]

---

Theorem:

The asymptotic behavior of a F.T. is given by the order of the discontinuity in the aperture.

Discontinuity	Fourier
⌐⌐ ← Discontinuous in 0th derivation	⇒ $\frac{1}{x}$ ← Spectr
/\ " " 1st derivation	⇒ $\left(\frac{1}{x}\right)^2$

In general, if the discontinuity is of order n, the side lobes fall off as $\left(\frac{1}{x^{(n+1)}}\right)^2$

---

Note:

Central limit theorem states that in general, if we convolve something enough times, we end up with a Gaussian:

$$ g(x) \circledast g_2(x) \circledast g_3(x) \circledast \cdots \to \text{Gauss}(x) $$

This function is continuous in all derivatives (but also is ∞ in extent), has no sidelobes, and falls off faster than a power series ($e^{-x^2}$)

• Gaussian apertures can be created by evaporating onto a substrate using a spinning aperture that looks like: [spiral diagram]

or

Absorber: $e^{-x^2}$ But $\frac{ar}{2} r^2$ ⇒ $t(r) = e^{-\alpha r^2}$

FFT Windows and Mask Apodizers

Note similarity to FFT Windows

• Bartlett - Triangular
• Blackman
• Hamming
• Hanning - Raised cosine
• Kaiser Bessel
• Parzen

---

Mask Apodizers

It is often difficult to shade a large telescope mirror with a variable density apodizer. Hence, rectangular masks are sometimes used.

[Diagram showing telescope mirror with square apodizer]

$$ \text{rect}(x)\text{rect}(y) \xrightarrow{\mathcal{F}} A = \frac{\sin(\pi x)\sin(\pi y)}{\pi^2 xy} $$$$ I \propto \frac{\sin^2(\pi x)\sin^2(\pi y)}{x^2 y^2} $$$$ I_{(x\text{-axis})} \sim \frac{1}{x^2} $$$$ I_{(y\text{-axis})} \sim \frac{1}{y^2} $$$$ I_{(45° \text{ axis})} \sim \frac{1}{r^4} $$

[Diagram showing circular aperture]

$$ \text{circ}(r) \xrightarrow{\mathcal{F}} A = \frac{J_1(2\pi r)}{r}, \quad I = \frac{J_1^2(2\pi r)}{r^2} $$

For large r, $J_1(r) \sim \frac{1}{\sqrt{r}}$

$$ \Rightarrow I \sim \frac{1}{r^3} $$

Blocking Aperture Apodizers

Note:

This apodizer redistributes the $\frac{1}{r^3}$ sidelobes of a circle (which are uniformly distributed in angle) into $\frac{1}{r^2}$ and $\frac{1}{r^4}$ regions (x-axis/y-axis and 45°). It is of course only useful if observation is to be only along a line (e.g., double stars), and the orientation is known.

[Diagram showing square aperture pattern with low sidelobes from square aperture ($\frac{1}{r^4}$) marked]

---

Blocking Aperture “Apodizers”

Circular Blocks:

[Two diagrams showing circular apertures with central obscuration]

$$ g(r, \theta) = \text{circ}\left(\frac{r}{r_1}\right) - \text{circ}\left(\frac{r}{r_2}\right) $$$$ f(r, \theta) = \text{circ}\left(\frac{r}{r_1}\right) $$$$ G(\rho) = 2\pi \int_0^{\infty} g(r') J_o(2\pi\rho r') r' dr' = \mathcal{H}\{g(r)\} $$

(Hankel Transform)

Hankel Transform Examples

Hankel Transform

$$ \mathcal{H}\left\{\text{circ}\left(\frac{r}{r_1}\right)\right\} = r_1 \frac{J_1(2\pi r_1 \rho)}{\rho} $$$$ \mathcal{H}\left\{\text{circ}\left(\frac{r}{r_1}\right) - \text{circ}\left(\frac{r}{r_2}\right)\right\} = r_1 \frac{J_1(2\pi r_1 \rho)}{\rho} - r_2 \frac{J_1(2\pi r_2 \rho)}{\rho} $$

---

Notice what happens in the limit $r_2 \to r_1$:

$$ f(r, \theta) = \delta(r - r_1) \leftarrow \text{Thin ring} $$

[Diagram showing thin ring aperture]

$$ \mathcal{H}\{\delta(r - r_1)\} = 2\pi \int \delta(r' - r_1) J_o(2\pi\rho r') r' dr' $$$$ = 2\pi r_1 J_o(2\pi r_1 \rho) $$

[Graph showing $J_1(\rho)$ with oscillations, and $J_o(\rho)$ below it]

Note:

• Main lobe decreases!
• Side lobes increase

$$ |J_1(\rho)/\rho|^2 \sim \frac{1}{\rho^3} $$$$ |J_o(\rho)|^2 \sim \frac{1}{\rho} $$

(Recall: $J_n(x) \sim \sqrt{\frac{2}{\pi x}} \cos\left(x - \frac{n\pi}{2} - \frac{\pi}{4}\right)$ for large x)

Wavefront Modulation - Film

Wavefront Modulation

Film:

[Diagram showing film structure with protective layer, emulsion (gelatin), base (plastic, glass), and film grains (Silver Halide)]

• Photon is incident on film grain
• Electron is promoted to conduction band and gets trapped in crystal dislocation
• Electron electrostatically attracts silver ions to form metallic silver site (Unstable)
• If this process is repeated within a few seconds, a two-atom unit is formed (stable).

Reciprocity { Silver

Chemical development converts the entire grain (~10⁹ atoms) to silver if it has been exposed

Photographic gain (10⁹)

Stability { Development will not take place if there are less than a threshold number (~4) of silver atoms.

• Fixing removes excess silver halide (which will eventually turn to silver)

(Show film grain pictures in Cathey)

Film Characteristics (B&W Film)

Wavelength Sensitivity

Silver halide will only form an electron-hole pair for $\lambda \leq 0.5\mu m$ (green, blue, violet)

⇒ To achieve a panchromatic response (B&W film), sensitizing dyes must be added.

---

Exposure

$$ E(x, y) = I(x, y) \cdot T \quad (mJ/cm^2) $$

(Total # of photons per cm²)

Intensity × Time

---

Reciprocity

Simply implies film is affected by exposure only. Decrease in intensity can be compensated for by increase in time, etc.

---

Intensity Transmittance

$$ T(x, y) = \text{local average}\left\{\frac{I_{\text{trans at }(x,y)}}{I_{\text{inc at }(x,y)}}\right\} $$

---

Density

$$ D = \log_{10}\left(\frac{1}{T}\right) $$

Note: Transmittances multiply, but densities add

[Diagram showing three filters with transmittances $t_1$, $t_2$, $t_3$ and densities $D_1$, $D_2$, $D_3$]

$$ t_t = t_1 \cdot t_2 \cdot t_3 $$$$ D_t = D_1 + D_2 + D_3 $$

H-D Curve for Film

[Graph showing Density vs log E (Exposure) with slope = γ marked on linear region]

For linear region of H-D curve:

$$ D = \gamma_n \log_{10} E - D_o = \gamma_n \log(IT) - D_o $$

where γₙ = slope for Negative film

Recall definition of density:

$$ D = \log_{10}\left(\frac{1}{T_n}\right) $$

← Intensity transmittance

$$ \Rightarrow \log(T_n) = -\gamma_n \log(IT) + D_o $$

or

$$ T_n = 10^{D_o}(IT)^{-\gamma_n} = K \cdot I^{-\gamma_n} $$

← positive constant

---

Note: Because $\gamma_n$ is positive, this relation is never linear. However, with two photographic steps, we can achieve:

$$ T_p = K_p I^{-\gamma_p}, \quad \text{where } \gamma_p = -\gamma_{n_1}\gamma_{n_2} $$

is a negative number

Thus, if $\gamma_p = 1$, relationship is linear in int.

MTF of Film

Simple Model of Film:

[Block diagram: E → Linear System → E’ → Nonlinear Response (H&D eqn) → D]

What is MTF of linear system

---

Exposure Pattern:

$$ E = E_o + E_1 \cos(2\pi f x) $$

Modulation def:

$$ M_i = \frac{E_1}{E_o} $$

(Peak variation to background exposure)

---

From film, measure density variation, and infer log (exposure) through H&D curve

[Graph showing D vs log E with measured density and effective (log of) exposure marked]

$$ \implies M(f) = \frac{M_{\text{eff}}(f)}{M_i(f)} $$

(Show MTF of real film in Cathey)

Spatial Light Modulators

Liquid Crystals

(Show Goodman pg 168)

• Mechanical properties of liquids
• Optical properties of solids
• Ordering takes place along some planes but not others

Nematic: Ordering in one dimension only (all molecules are pointed in same direction)

Smectic: Ordering is in two dimensions. Randomness only within single layer.

Cholesteric: Molecules form long helical chains that are aligned in one dimension only. (Liquid crystal thermometers)

[Diagram showing helical structure with pitch ≈ λ₀ → white light reflects at ★ pitch ftn of temp.]

---

★ Consider Nematic Liquid Crystals Only (Twisted Nematic)

• Alignment performed by grooving substrate with polishing cloth.

[Diagram showing grooves with L.C. molecules aligned, showing 90° twist]

Properties of Twisted Nematics

★ Properties of Twisted Nematics

1) Birefringence

(Light of one linear polarization state travels at a different velocity than the other: elliptic polarization results)

2) Optical Rotatory Power (Optical Activity)

(Light of one circular polarization state travels at a different velocity than the other: Linear polarization is simply rotated).

• This occurs in twisted nematics because the polarization state follows the twist, induced.

3) Large dipole moment along molecule axis

---

★ Operation of LCD (Displays)

• Ignore birefringence effects

[Diagram showing LCD operation in OFF state:

• Light → Polarizer → LC with 90° twist → polarization rotates → polarizer → off state]

[Diagram showing LCD operation in ON state:

• Light → Polarizer → Transp. Electrode → Electric Field → Transp. Electrodes → polarization unchanged → polarizer → Light ON state]

Optical SLM (Hybrid Field Effect)

Electrical-to-Optical SLM

Can be fabricated by matrix-addressing liquid crystal

---

★ Optical-to-Optical SLM (Hybrid-Field-Effect)

[Detailed diagram showing:

• Alignment layer with light blocking layer

• Photosensor (CdS, etc.)

• Pol., B.S.

• Input/Read

• L.C.

• Glass substrate (fiber-optic face plate)

• Mirror/ITO (trans. conductor)

• A-C field

• Write Light (Incoherent Image)]

• Crystal is twisted nematic with 45° twist

• Optical rotatory power is cancelled by reflection

• Light incident from write side causes resistance of photoconductor to reduce (bright parts of image)

• Voltage falls across L.C. and causes molecules to rotate. Speed increases since stress.

• Birefringence is maximized by a partial rotation of the molecules. If thickness is chosen to result in half-wave plate, polarization is rotated from trans, want to heriz (or vice versa).

SLM Uses

Uses:

1. Incoherent-to-coherent light conversion

2. Wavelength conversion ($\lambda_1 \to \lambda_2$)

3. Optical amplification of image (projection TV)

---

Speed:

$\sim \frac{1}{30}$ sec cycle time for nematics

Somewhat faster for ferroelectric L.C.’s

Acoustooptic Effect (12.7)

Acoustooptic Effect

[Diagram showing:

• Absorber (Matched impedance)
• Travelling Acoustic Wave
• Acoustooptic Device (H₂O, TeO₂, Glass, SiO₂ (Quartz), etc.)
• Transducer
• hν input]

---

AO Effect:

→ Transducer produces an acoustical wave in material

→ Acoustic compression & rarefaction produces small index of refraction change.

→ Light sees a travelling phase grating with period ($\sim 10^5$ cm/sec)

$$ \Lambda = \frac{V_s}{f} \leftarrow \text{Speed of acoustic propagation in material } (\sim 10^5 \text{ cm/sec}) $$

For Acoustic f up to 100 MHz ⇒ sinθ = 0.003 ($\lambda = 0.5\mu m$)

$$ \Lambda = \frac{V_s}{f} \approx 10\mu m $$

⇒ $\Lambda \approx 10\mu m$

$\sin\theta = \frac{\lambda}{\Lambda}$ ⇒ θ ≈ 3° ($\lambda = 5\mu m$)

---

→ Interaction of Light and Thick Phase Grating

Bragg Condition:

[Diagram showing Bragg diffraction geometry with angles θᵢ, θᵣ and grating spacing Λ]

• Acoustic waves look like mirrors with very small reflectances due to index discontinuity.
• Prime indicates inside material

★ $\theta_i = \theta_r$ because of law of reflection at “mirror”

★ For all reflections to add in-phase:

$$ AO + \overline{OB} = m(\lambda_n) \leftarrow \text{Integer} $$$$ \Rightarrow \boxed{2\Lambda\sin\theta = m(\lambda/n)} $$

— Bragg Condition (inside substrate)

By Snell’s laws: $\sin\theta = n\sin\theta'$ ⇒ $\boxed{2\Lambda\sin\theta = m\lambda}$ — Bragg condition (outside)

Bragg Condition Notes

Note:

1. Bragg condition true only for “thick” gratings

2. Only one diffraction order (plus zero order from unreflected light) is produced

3. Diffraction order occurs only at specific input angles (Bragg condition)

---

→ What happens when grating is finite thickness?

[Diagram showing finite thickness grating with length L]

Recall: Gaussian confined in space bc ω₀ exists of plane waves: given by $\theta = \frac{\lambda}{\pi\omega_o}$

[Diagram showing angular spectrum with K(θ) showing main lobe and side lobes at $\frac{\lambda}{L}$]

Acoustic wave of finite extent can be expressed as an angular spectrum of infinite extent plane waves (Angular Spectrum Decomposition)

---

★ Several specific acoustic plane waves may satisfy Bragg equation

[Diagram showing multiple diffraction orders at angles with:

• Off-axis “Acoustic plane” wave due to finite extent of transducer
• Acoustic → “plane” wave component
• Orders: 0, +1, +2 marked
• Angles: -2α, -α, α, 2α]

$$ 2\Lambda\sin\theta = m\lambda $$$$ \sin\theta = m\left(\frac{\lambda}{2\Lambda}\right) $$$$ \Rightarrow \theta_B \approx \frac{m\lambda}{2\Lambda} $$

Let $\alpha = \frac{\lambda}{2\Lambda}$ = Primary Bragg angle

— High order Bragg angles = mα

Angular Plane Wave Spectrum (68)

Angular Plane Wave Spectrum of Finite Acoustic Transducer

[Diagram showing possible Bragg angles (components of acoustic wave that satisfy Bragg condition) with:

• Peaks at -3α, -2α, -α, 0, α, 2α, +3 positions
• Light wave vectors at angles
• Transducer of width L shown]

$$ \alpha = \frac{\lambda}{2\Lambda} $$

---

Number of diffraction orders under main lobe of sinc ftn. (strong orders)

$$ N \approx \frac{2\left(\frac{\lambda}{L}\right)}{\frac{\lambda}{2\Lambda}} = \frac{4\Lambda^2}{\lambda L} $$

(Not counting undiffracted order)

---

Define Q parameter:

$$ \boxed{Q = \frac{2\pi\lambda L}{\Lambda^2}} $$

← Inversely related to number of diffraction orders present

$$ \Rightarrow Q \cdot N = (2\pi\lambda L / \Lambda^2)(4\Lambda^2 / \lambda L) = 8\pi $$

---

1) $Q \gg 8\pi$ ⇒ Single diffraction order present (+ zero order) { Bragg region

2) $Q \ll 8\pi$ ⇒ Many orders present { Raman-Nath or Debye-Sears region

Thin Grating (Raman-Nath)

[Diagram showing thin grating with thickness d ≪ λ]

$$ t(x, y) = \exp\left[i m \sin\left(\frac{2\pi}{\Lambda}x\right)\right], \quad m = \text{modulation index} $$$$ = \sum_{g=-\infty}^{\infty} J_g\left(\frac{m}{2}\right) \exp\left(i2\pi g\frac{x}{\Lambda}\right) $$

↑ gth order Bessel ftn

$$ \mathcal{F}\{t(x,y)\} = \sum_{g=-\infty}^{\infty} J_g\left(\frac{m}{2}\right) \delta\left(\frac{\sin\theta}{\lambda} - \frac{g}{\Lambda}\right) $$

where $\theta_o = \lambda/\Lambda$

[Graph showing $J_o(m/2)$, $J_1(m/2)$, $J_2(m/2)$ vs θ at positions θₒ, 2θₒ]

[Graph showing Bessel Function Behavior with $J_o$, $J_1$, $J_2$ curves vs m]

Note: At certain modulation index m, the zero order can disappear.

Decrease of intensity at higher orders as order of Bessel ftn. increases

---

Amplitude modulation:

$$ u(t) = m(t)\cos(\omega t) $$

Freq. mod.: ← Note: This changes the diffraction angle

$$ u(t) = \cos((\omega + m\Omega)t) $$

⇒ Scanner

Particle Interaction Picture

Photon:

• Energy = $\hbar\omega$
• Momentum = $\hbar \vec{k}$

Phonon:

• Energy = $\hbar\omega_s$
• Momentum = $\hbar \vec{k}_s$

---

Photon-Phonon Collision:

[Diagram showing k-vector triangle with:

• $\vec{k}_d$ (diffracted)
• $-\theta$
• $\vec{k}_s$ (scattered)
• $\vec{k}_i$ (incident)
• Conservation of Momentum notation]

From picture:

$$ k_s = 2k_d\sin\theta $$

($|\vec{k}_d| \approx |\vec{k}_i|$ since $\omega_d = \omega_i + \omega_s \approx \omega_i$ and $|k| = \frac{\omega}{c}$)

$$ \Rightarrow 2\Lambda\sin\theta = (\Lambda/n) $$

(m₁ corresponds to multiple phonon interactions) — Bragg Law

---

Multiple Phonon Collisions

[Diagram showing multiple phonon interactions with:

• Acoustic wave at new angle due to spread of plane waves from finite acoustic aperture (Second Phonon)
• $|k_1| \approx |k_i| \approx |k_2|$
• Light vectors in a circle
• Acoustic wave at original Bragg angle (First phonon)]

⇒ Angle of phonon k vector corresponds to direction of propagation of acoustic wave

Frequency Shift

Also notice: Conservation of energy implies

$$ \hbar\omega_i + \hbar\omega_s = \hbar\omega_d $$$$ \Rightarrow \omega_d = \omega_i + \omega_s $$

Freq. of light has been shifted by sound freq.

---

★ Physical Interpretation of Frequency Shift

1) Doppler Shift:

[Diagram showing acoustic wave ωs, travelling at velocity Vs with angle θ]

$$ \omega_d - \omega_i + \omega_s $$$$ \Delta\omega = 2\omega \frac{V_s}{c/n} \cdot \frac{\omega_i}{c/n} $$

← velocity parallel to light

↑ Reflected wave

{Recall doppler shift given by $\omega_2 = \omega_1(1 - v_z/c_{/n})$ for source moving ⇒ $\omega_2 - \omega_i = \Delta\omega = \frac{\omega_i v_z}{c_m}$}

But $V_{\parallel} = V_s \cdot \sin\theta$ ⇒ $\Delta\omega = 2\omega\frac{V_s\sin\theta}{c/n}$

Also from Bragg Condition inside material:

$$ \sin\theta = \frac{m\lambda/n}{2\Lambda} $$

← ($2\Lambda\sin\theta = (m\frac{\lambda}{n})$)

Bragg Condition.

$$ \Rightarrow \Delta\omega = \frac{2\omega V_s \cdot m\lambda/n}{2\Lambda \cdot c/n} = m\frac{2\pi V_s}{\Lambda} = m\omega_s $$

← Freq. shift = Acoustic wavelength = Acoustic Freq.

(Higher order harmonics for higher diffraction orders)

---

2) Moving Grating (thin)

[Diagram showing grating moving at velocity Vs with diffraction order produced at $\sin\theta = \frac{\lambda}{\Lambda}$]

★ Shift theorem: $f(x) \overset{\mathcal{F}}{\longleftrightarrow} F(u)$ : $f(x-\alpha) \overset{\mathcal{F}}{\longleftrightarrow} F(u)e^{-j2\pi\alpha u}$

Shift by one acoustic period and observe phase at $\sin\theta = \frac{1}{\Lambda}$

⇒ $\alpha = \Lambda$, $u = \frac{\sin\theta}{\lambda} = \frac{1}{\Lambda}$

$$ \implies f(x-\Lambda) \overset{\mathcal{F}}{\longleftrightarrow} F(u)e^{-j2\pi(\Lambda)(\frac{1}{\Lambda})} = F(u)e^{-j2\pi} $$

↑ 2π phase shift

Frequency Shift Summary (70)

★ When grating is shifted one acoustic wavelength, phase shifts by $2\pi$ = 1 cycle

$$ \text{Acoustic frequency} $$

⇒ Travelling wave shifts grating $V_s$ wavelengths/sec

$$ \implies \text{Phase of diffracted light shifts } \frac{V_s \text{ cycles}}{\text{sec}} / 2\pi $$

⇒ Frequency shift of $V_s$

On-axis Kinoforms

• Phase-only structures
• Phase functions are recorded modulo $2\pi$

Examples:

Conventional Element	Kinoform Equivalent
Prism [linear phase ramp with period λ]	[Sawtooth pattern with period $\frac{1}{T}\lambda$]
Conventional Element ↓	Kinoform Equivalent ↓
Lens [parabolic phase]	Kinoform Lens [wrapped phase pattern]

---

Consider a simple blazed grating

[Diagram showing blazed grating with incident light λ, orders m=-2, m=-1, m=0, m=1, m=2, angle θ, and period |d|←]

Blazed Grating Analysis

$$ t(x) = \sum_{m=-\infty}^{\infty} \delta(x - mT) * \text{rect}\left(\frac{x}{T}\right) \exp(j2\pi\beta x) $$

where $\beta = \frac{(n-1)d}{\lambda T}$

[Diagram showing phase profile with $\Delta\phi_{total} = (k' - k)\Delta h = \frac{2\pi(n-1)\Delta h}{\lambda}$]

$\Delta h = \frac{\alpha(\Delta x)}{T}$

The far-field amplitude is given by:

$$ F(u) = \sum_{m=-\infty}^{\infty} \delta\left(u - \frac{m}{T}\right) \cdot \frac{\sin(\pi T(\beta - u))}{\pi T(\beta - u)} $$

where: $u = \frac{\sin\theta}{\lambda}$

↑ Diffraction orders

$\beta = \frac{1}{2\pi} \frac{\partial\phi}{\partial x} = \frac{(n-1)d}{\lambda T}$

$= \frac{(n-1)d\Delta x}{T\lambda\Delta x}$

$= \frac{(n-1)d}{\lambda T}$

---

⇒ Intensity of mth diffraction order ($u = \frac{m}{T}$)

$$ \eta_m = |\alpha_m|^2 = \left|\frac{\sin(\pi T(\beta - \frac{m}{T}))}{\pi T(\beta - \frac{m}{T})}\right|^2 $$

We are usually interested in the first diffraction order $(m=1)$

$$ \eta_1 = \left|\frac{\sin(\pi(\beta T - 1))}{\pi(\beta T - 1)}\right|^2 $$$$ \implies \eta_1 = 1 \quad \text{when } \beta = \frac{1}{T} $$

⇒ Blaze goes from 0 to $2\pi$

$$ \exp\{j2\pi\beta x\} = \exp\left\{j2\pi\frac{x}{T}\right\}\bigg|_{x=0 \to T} $$

Also: If $\beta = \frac{1}{T}$, $\frac{1}{nT} = \frac{(n-1)d}{\lambda T}$ ⇒ $d = \frac{\lambda}{n-1}$

Blazed Grating Wavelength Dependence

Note:

• Blazed grating is only 100% efficient for a single wavelength $\lambda_o$. (Let $d = \frac{\lambda_o}{n-1}$)

• For another wavelength $\lambda$, we have

$$ \beta = \frac{(n-1)d}{\lambda T} = \frac{\lambda_o}{\lambda T} $$

(Recall $d = \frac{\lambda_o}{n-1}$ for max. efficiency)

and

$$ \eta_1 = \left|\frac{\sin\left(\pi\left(\frac{\lambda_o}{\lambda} - 1\right)\right)}{\pi\left(\frac{\lambda_o}{\lambda} - 1\right)}\right|^2 $$

[Graph showing efficiency $\eta_1$ vs wavelength $\lambda$, with peak of 1.0 at $\lambda_o$, dropping to ~0.5 at $0.5\lambda_o$ and $1.5\lambda_o$, approaching 0 at $2\lambda_o$]

• Efficiency reduced at other wavelengths
• Residual light occurs as higher orders, often adding a background haze (similar to scatter, but deterministic)

Arbitrary Phase Profiles

Consider an arbitrary phase function

[Graph showing smooth phase function $\phi(x)$ vs x]

→ [Graph showing wrapped phase $\phi'(x)$ with discontinuities at $\pm\frac{1}{2}$, ranging between $-\frac{1}{2}$ and $\frac{1}{2}$]

$$ t_c(x) = \exp[j2\pi\phi(x)] $$$$ t_d = \exp[j2\pi\phi'(x)] $$

where $\phi'(x) = |\phi(x)|_{\text{mod } \alpha}$

---

★ How do we analyze this structure?

• Use nonlinear (limiter) analysis

• Plot diffractive phase $\phi'(x)$ as a function of refractive phase $\phi(k)$

[Graph showing sawtooth relationship between $\phi'(x)$ and $\phi(x)$, with slopes at -1.5, -0.5, 0.5, 1.0, 1.5]

Note: Phase ranges from $-\frac{1}{2}$ to $\frac{1}{2}$ for generality

---

Note: $\phi'(x)$ is periodic in $\phi(x)$ with period = 1

⇒ $\exp[j2\pi\phi'(x)]$ is also periodic in $\phi(x)$

⇒ we can write this as a generalized Fourier Transform

Fourier Analysis of Kinoforms

$$ \exp[j2\pi\phi'(x)] = \sum_{m=-\infty}^{\infty} C_m \exp[j2\pi m\phi(x)] $$

↑ Transform variable is a function instead of a single variable.

---

Solve for $C_m$ coefficients:

$$ C_m = \int_{-1/2}^{1/2} \exp[j2\pi\phi'(x)] \exp[-j2\pi m\phi(x)] \cdot d\phi(x) $$

But $\phi'(x) = \alpha\phi(x)$ in the region $(-1/2, 1/2)$

$$ \implies C_m = \int_{-1/2}^{1/2} \exp[j2\pi(\alpha - m)\phi(x)] \, d\phi(x) $$$$ = \frac{\sin(\pi(\alpha - m))}{\pi(\alpha - m)} $$

---

If $\alpha = 1$, then $C_1 = 1$ and $C_m = 0$ for $m \neq 1$.

If $\alpha \neq 1$, then additional orders exist.

---

Note:

• For blazed grating, these orders are plane waves (of varying angle)

• For Fresnel zone plate, these orders are spherical waves (of varying ROC)

• For arbitrary profile, these orders are more complex wavefronts given by:

$$ \sum_{m=-\infty}^{\infty} C_m \exp[j2\pi m\phi(x)] $$

Binary Optics Fabrication

• Use microelectronics tools to fabricate step-wise approximation to desired profile

Process Steps:

(1) [Expose pattern on resist layer on substrate]

(2) [Etch ($\frac{\lambda}{4}$) to create steps] ← P.R.

(3) [Expose second pattern]

(4) [Etch ($\frac{\lambda}{2}$) to create finer steps]

(5) [Result: Multi-level staircase pattern]

---

Note: N masks generate $2^N$ phase levels.

Diffraction Efficiency of Step-Wise Phase Pattern

★ Diffraction Efficiency of Step-Wise Phase Pattern:

• Multi-level structure can be considered as the desired profile minus an error:

Example:

[Diagram showing: N-levels continuous phase profile + off-set phase → Error → Desired minus sawtooth error pattern with period $\frac{\alpha}{N}$ and amplitude $\frac{1}{T}$, $\frac{2\pi}{N}$]

$$ t(x) = \exp\left(j2\pi\frac{x}{\alpha}\right) \cdot \left\{\left[\exp\left(-j2\pi\frac{x}{\alpha}\right) \cdot \text{rect}\left(\frac{Nx}{\alpha}\right)\right] * \frac{N}{\alpha}\text{comb}\left(\frac{Nx}{\alpha}\right)\right\} $$

---

Fourier transforming: (Angular plane wave spectrum where $u = \frac{\cos\theta}{\lambda}$)

$$ T(u) = \delta\left(u - \frac{1}{\alpha}\right) * \frac{\alpha}{N}\left\{\left[\delta\left(u + \frac{1}{\alpha}\right) * \text{sinc}\left(\frac{\alpha u}{N}\right)\right] \cdot \text{comb}\left(\frac{\alpha u}{N}\right)\right\} $$$$ = \delta\left(u - \frac{1}{\alpha}\right) * \left[\text{sinc}\left(\frac{\alpha u + 1}{N}\right) \cdot \sum_{n=-\infty}^{\infty} \delta\left(u - \frac{nN}{\alpha}\right)\right] $$$$ = \text{sinc}\left(\frac{u\alpha}{N}\right) \cdot \sum_{n=-\infty}^{\infty} \delta\left(u - \frac{nN+1}{\alpha}\right) $$

---

Inverse Transforming:

$$ \int_{-\infty}^{\infty} \text{sinc}\left(\frac{u\alpha}{N}\right) \sum_{n=-\infty}^{\infty} \delta\left(u - \frac{nN+1}{\alpha}\right) \exp(j2\pi ux) \, du $$$$ = \sum_{n=-\infty}^{\infty} \text{sinc}\left(\frac{nN+1}{N}\right) \exp\left(j2\pi\left(\frac{nN+1}{\alpha}\right)x\right) $$

↑ Amplitude of plane-wave component

plane waves at angle $\frac{\sin\theta}{\lambda} = \frac{nN+1}{\alpha}$

↑ Integer # of phase levels

↑ 2 period

Step-Wise Phase Efficiency

Consider case where n = 0 (Diffraction component from error term)

[Diagram showing staircase approximation with n=0 marked, height λ]

Plane wave component: $\exp\left[j2\pi\frac{x}{\alpha}\right]$

Diffraction Efficiency: $\eta = \left|\text{sinc}\left(\frac{1}{N}\right)\right|^2$

(true binary phase grating)

---

Efficiency values:

N	η
N = 2 (true binary phase grating)	η = 40.5%
N = 4	η = 81%
N = 8	η = 95%
N = 16	η = 98.7%

Simple Spatial Filtering (Amplitude-only Filters) (30)

Simple Spatial Filtering: (Amplitude-only Filters)

Recall Optical set-up for spatial filtering:

[Diagram showing optical system with planes P1, P2, P3, focal lengths f between each]

$$ A \to f(x,y) \xrightarrow{H(x',y')} g(x,y) = f(x,y) * h(x,y) $$

where: $h(x,y) = \mathcal{F}\mathcal{F}\{H(x,y)\}$

---

Abbe-Porter Experiment:

$$ f(x,y) = \text{Ronchi ruling} $$

[Grid pattern with period a]

(gratings are 2 freq of filter)

$$ \Rightarrow f(x,y) = \left\{\left[\frac{1}{b}\text{comb}\left(\frac{x}{b}\right) \cdot \delta(y)\right] * \text{rect}\left(\frac{x}{a}\right)\right\} * \left\{\left[\frac{1}{b}\text{comb}\left(\frac{x}{b}\right) \cdot \delta(x)\right] * \text{rect}\left(\frac{y}{a}\right)\right\} $$$$ H(x',y') = \text{rect}\left(\frac{x'}{\epsilon}\right)\text{rect}\left(\frac{y'}{\epsilon}\right) $$

[Diagram showing square filter with dimensions $\frac{1}{\epsilon} \times 2$, labeled “Two types of filter”]

---

In plane P2, (ignoring constant phase factors and attenuation)

$$ F(x', y') = \mathcal{F}\mathcal{F}\{f(x,y)\}\bigg|_{z=\frac{x'}{\lambda f}, \eta=\frac{y'}{\lambda f}} $$$$ = \left[\text{comb}(bz) \cdot a\,\text{sinc}(az)\delta(\eta)\right] * \left[\text{comb}(b\eta) \cdot a\,\text{sinc}(a\eta) \cdot \delta(z)\right] $$$$ = \left[\text{comb}\left(\frac{bx'}{\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ax'}{\lambda f}\right)\delta\left(\frac{y'}{\lambda f}\right)\right] * \left[\text{comb}\left(\frac{by'}{\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ay'}{\lambda f}\right)\delta\left(\frac{x'}{\lambda f}\right)\right] $$

Simple Spatial Filtering Continued

$$ H(x',y') \cdot F(x',y') = \left[\text{comb}\left(\frac{bx'}{\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ax'}{\lambda f}\right) \cdot \delta\left(\frac{y'}{\lambda f}\right)\right] * \left[\text{comb}\left(\frac{by'}{\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ay'}{\lambda f}\right) \cdot \delta\left(\frac{x'}{\lambda f}\right)\right] $$

Use a spatial filter: $H(x',y') = \text{rect}\left(\frac{x'}{c}\right)\text{rect}\left(\frac{y'}{d}\right)$

---

Case 1

Let: $c = N\frac{\lambda f}{b}$, where $N >> 1$

$d = \frac{\lambda f}{b}$

[Diagram showing filter passing N harmonics in x, 1 in y]

$$ \implies H(x',y') \cdot F(x',y') \cong \text{comb}\left(\frac{bx'}{\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ax'}{\lambda f}\right) \cdot \delta(y') $$

Image in P₃:

$$ \Rightarrow g(x,y) = \mathcal{F}^{-1}\mathcal{F}^{-1}\{H(x',y') \cdot F(x',y')\} $$$$ = \frac{1}{a}\left[\text{comb}\left(\frac{x}{b}\right) \cdot \delta(y)\right] ** \text{rect}\left(\frac{x}{a}\right) $$

[Diagram showing output with horizontal lines - “Neglect all constants, Horizontal lines have been removed”]

---

Case 2

Let $c = \frac{\lambda f}{b}$, $d = N\frac{\lambda f}{b}$ where $N >> 1$

[Diagram showing filter passing 1 harmonic in x, N in y]

$$ H(x',y') \cdot F(x',y') \cong \text{comb}\left(\frac{by'}{\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ay'}{\lambda f}\right) \cdot \delta\left(\frac{x'}{\lambda f}\right) $$$$ \Rightarrow g(x,y) = \mathcal{F}^{-1}\mathcal{F}^{-1}\{H(x',y') \cdot F(x',y')\} = \left[\text{comb}\left(-\frac{y}{b}\right) \cdot \delta(x)\right] ** \text{rect}\left(\frac{y}{a}\right) $$

Vertical lines have been removed.

Additional Filters in Abbe-Porter Experiment

Additional filters of interest in Abbe-Porter Experiment

Case I

Let $c = \frac{3\lambda f}{b}$, $d = \lambda f$

[Diagram showing filter with width c passing 3 orders]

$$ \Rightarrow H(x',y') \cdot F(x',y') = $$$$ \left[\text{comb}\left(\frac{bx'}{\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ax'}{\lambda f}\right) \cdot \delta\left(\frac{y'}{\lambda f}\right) \cdot \text{rect}\left(\frac{bx'}{3\lambda f}\right)\right] $$$$ = \frac{\lambda f}{b} \sum_{n=-\infty}^{\infty} \delta\left(x' - \frac{\lambda f}{b}n\right) \cdot \text{rect}\left(\frac{bx'}{3\lambda f}\right) \cdot a\,\text{sinc}\left(\frac{ax'}{\lambda f}\right) \cdot \delta\left(\frac{y'}{\lambda f}\right) $$

Only lets through $-1, 0, +1$ orders

$$ = \left(\frac{a\lambda f}{b}\right)\left[\delta(x') + \delta\left(x' - \frac{\lambda f}{b}\right) + \delta\left(x' + \frac{\lambda f}{b}\right)\right] \cdot \text{sinc}\left(\frac{ax'}{\lambda f}\right)\delta\left(\frac{y'}{\lambda f}\right) $$$$ = \left(\frac{a\lambda f}{b}\right)\left[\delta(x') + \delta\left(x' - \frac{\lambda f}{b}\right)\text{sinc}\left(\frac{a}{b}\right) + \delta\left(x' + \frac{\lambda f}{b}\right)\text{sinc}\left(\frac{a}{b}\right)\right]\delta\left(\frac{y'}{\lambda f}\right) $$

---

Image (neglecting constants) is given by:

$$ \mathcal{F}^{-1}\mathcal{F}^{-1}\{H(x',y') \cdot F(x',y')\} $$$$ = 1 + 2\,\text{sinc}\left(\frac{a}{b}\right)\cos\left(\frac{2\pi}{b}x\right) $$

Raised cosine:

• i) frequency $\frac{1}{b}$ $\Rightarrow$ period = b
• ii) amplitude $2\,\text{sinc}\left(\frac{a}{b}\right)$

[Diagram showing raised cosine waveform g(x,0) with period b]

[Diagram showing 2D output pattern with vertical stripes]

High-Pass Filtering

Case II: Consider a spatial filter shaped like:

[Diagram showing square blocking filter at center, same as case 1 except that F(0,0) is also removed]

Also assume $a \approx b$ (i.e. thin line)

[Diagram showing function f(x,y) with varying values around average value]

$$ \rightarrow \text{Recall } F(0,0) = \iint f(x,y) e^{-j2\pi(x \cdot 2 + y \cdot \eta)} dx\,dy \bigg|_{z=0,\eta=0} $$$$ = \iint f(x,y)\,dx\,dy = \text{average value of } f(x,y) $$$$ \rightarrow \text{By removing } F(0,0), \text{ we subtract the average value of } f(x,y) $$

from the original $f(x,y)$. Hence, we have:

[Diagram showing f(x,y) - f̄(x,y) with values oscillating around zero]

$$ \rightarrow \text{The intensity is actually observed, and } I = |f(x,y) - \overline{f(x,y)}|^2 $$

[Diagram showing I(x,y) with squared values]

Thus, the resultant image is a contrast reversed version of the original!

Applications of Spatial Filters

★★ Applications of Spatial Filters

★ High Pass, Low Pass, and Differentiation Filters

High Pass Transfer Function:

[Diagram showing 1D transfer function t(ξ) with rectangular passband, labeled “one dim”]

[Diagram showing 2D circular high-pass filter with hole in center, labeled “two-dim”]

---

Low Pass Transfer Function:

[Diagram showing 1D transfer function t(ξ) with rectangular passband centered at origin, labeled “one-dim”]

[Diagram showing 2D circular low-pass filter (filled circle), labeled “two-dim”]

---

Differentiation:

$$ g(x,y) = \frac{\partial f(x,y)}{\partial x} $$

Recall: $\mathcal{F}\left\{\frac{\partial f(x,y)}{\partial x}\right\} = j2\pi\xi \cdot F(\xi,\eta)$

$$ \implies H(\xi,\eta) = j2\pi\xi \rightarrow \xi $$

[Diagram showing 1D transfer function t(ξ) = ξ (linear ramp), labeled “one dim”]

[Diagram showing 2D differentiation filter with phase variation φ = π on left half, φ = 0 on right half, labeled “two-dim filter (differentiation in one-dimension)”]

Another Application of Spatial Filters

★★ Another Application of Spatial Filters

★ Raster-Line Removal

[Diagram showing TV/monitor screen with scan lines]

• Two-dim pictures are encoded on a one-dim time signal for transmission.
• This encoding involves sampling the picture in the vertical direction.

[Diagram showing video signal f(t) with line 1, line 2, line 3 marked, labeled “525 of these lines = 1 frame and occurs every 1/30 sec.”]

The reconstructed image is thus a sampled image in the $y$ (vertical) direction:

$$ g_s(x,y) = g(x,y) \cdot \frac{1}{\gamma}\text{comb}\left(\frac{y}{\gamma}\right) $$

where:

• $\gamma$ = sampling interval
• $g(x,y)$ = original picture
• $g_s(x,y)$ = sampled picture

The spectrum of the sampled picture is $G_s(\xi,\eta) = G(\xi,\eta) * \text{comb}(\gamma\eta)$.

[Diagram showing G(ξη) spectrum with triangle shape] $\rightarrow$ [Diagram showing Gs(ξ,η) with replicated spectra at intervals 1/γ]

Whittaker-Shannon Reconstruction and Half-Tone Printing

Using an optical system to perform Whittaker-Shannon reconstruction:

$$ G(\xi,\eta) = G_s(\xi,\eta) \cdot \text{rect}(\gamma\xi) $$

[Diagram showing optical system: Sampled (Raster scanned) Object in → Low pass filter → Continuous Image Out]

This is a low-pass filter in the $\eta$ direction, as required by the Whittaker-Shannon theorem.

---

Half-Tone Method of Picture Printing

[Diagram showing original grayscale image with varying intensities] → [Diagram showing × (Thru) operator] → [Diagram showing half-tone pattern with dots of varying sizes]

Original gray scale image → Half tone

[Diagram showing exposure vs density curve with clip level for film]

(i.e., if exposure is > clip, film is opaque, unit is clear. Unit 21 M is a dot, film is then a screen)

Result f(x,y):

[Diagram showing F(ξ,η) spectrum] → [Diagram showing output with note about low pass filter]

The human eye acts as a low-pass filter — only the 1st lobe image is perceived, so a continuous gray-scale image has been passed.

Phase Contrast Methods

★ Phase Contrast Methods (observing phase-only functions)

Consider a general object transmittance:

$$ A(x,y) = A(x,y) \cdot \text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right) $$

Define an average picture transmittance:

$$ A_{\text{ave}}(x,y) = \frac{1}{\ell^2} \iint_{-\ell/2}^{\ell/2} A(x,y)\,dx\,dy \cdot \text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right) $$$$ = \bar{A} \cdot \text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right) \quad \text{where } \bar{A} = \frac{1}{\ell^2}\iint_{-\ell/2}^{\ell/2} A(x,y)\,dx\,dy $$

We can define the picture as:

$$ A(x,y) = A_{\text{ave}}(x,y) + A(x,y) - A_{\text{ave}}(x,y) $$

We now have a spectrum in two parts:

$$ \mathcal{F}\mathcal{F}\{A(x,y)\} = \mathcal{F}\mathcal{F}\{A_{\text{ave}}(x,y)\} + \mathcal{F}\mathcal{F}\{A(x,y) - A_{\text{ave}}(x,y)\} $$$$ = \iint \bar{A}\,\text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right)\exp(-i2\pi(x\xi + y\eta))\,dx\,dy $$

$$

• \iint (A(x,y) - \bar{A})\text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right)\exp(-i2\pi(x\xi + y\eta)),dx,dy $$

$$ = U_0(\xi,\eta) + U_1(\xi,\eta) $$

---

Now, $U_0(\xi,\eta) = \bar{A}\ell^2\,\text{sinc}(\ell\xi)\,\text{sinc}(\ell\eta)$

$$ = \left(\iint A(x,y)\,dx\,dy\right)\text{sinc}(\ell\xi)\,\text{sinc}(\ell\eta) $$$$ U_1(\xi,\eta) = \iint A(x,y)\,\text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right)\exp(-i2\pi(x\xi + y\eta))\,dx\,dy $$$$ \text{-} \bar{A}\iint\text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right)\exp(-i2\pi(x\xi + y\eta))\,dx\,dy $$$$ \Rightarrow U_1(0,0) = \iint A(x,y)\,\text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right)\,dx\,dy - \bar{A}\iint\text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right)\,dx\,dy = \bar{A}\ell^2 - \bar{A}\ell^2 = 0 $$

Phase Contrast Methods Continued

Note:

$$ \lim_{\ell \to \infty} U_0(\xi,\eta) = \bar{A} \lim_{\ell \to \infty} \ell\,\text{sinc}(\ell\xi) \cdot \lim_{\ell \to \infty} \ell\,\text{sinc}(\ell\eta) $$$$ = \bar{A}\,\delta(\xi,\eta) $$

Also we have shown that for any $\ell$,

$$ U_1(0,0) = 0 $$

[Diagram showing U₁(ξ,η) and U₀(ξ,η) as separate components in frequency domain]

We now make a filter which is located only in the $\xi = 0, \eta = 0$ region of the transform, with transmittance $F(\xi,\eta) = ae^{i\alpha}$

$$ \implies U'(\xi,\eta) = U_0(\xi,\eta)F(\xi,\eta) + U_1(\xi,\eta) $$

---

Image created is proportional to $\mathcal{F}^{-1}\mathcal{F}^{-1}\{U'(\xi,\eta)\}$

$$ \mathcal{F}^{-1}\mathcal{F}^{-1}\{U_0(\xi,\eta)F(\xi,\eta)\} = ae^{i\alpha}\mathcal{F}^{-1}\mathcal{F}^{-1}\{U_0(\xi,\eta)\} $$$$ = ae^{i\alpha}A_{\text{ave}}(x,y) $$

Image $\Rightarrow$ $P(x',y') = \mathcal{F}^{-1}\mathcal{F}^{-1}\{U'(\xi,\eta)\} = ae^{i\alpha}A_{\text{ave}}(x',y') + A(x',y') - A_{\text{ave}}(x',y')$ (coordinates)

---

Consider phase object with small modulation:

$$ A(x,y) = e^{i\phi(x,y)}, \quad \phi(x,y) << 1 $$$$ \implies A_{\text{ave}}(x,y) = \frac{1}{\ell^2}\iint_{-\ell/2}^{\ell/2} A(x,y)\,dx\,dy \cdot \text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right) $$$$ \approx 1 \cdot \text{rect}\left(\frac{x}{\ell}\right)\text{rect}\left(\frac{y}{\ell}\right) $$

Phase Contrast - Dark Ground Technique

Then image amplitude becomes:

$$ P(x',y') \cong \left[ae^{i\alpha} + A(x',y') - 1\right]\text{rect}\left(\frac{x'}{\ell},\frac{y'}{\ell}\right) $$

and image intensity becomes (neglecting finite size from here on):

$$ I(x',y') = |P(x',y')|^2 = |ae^{i\alpha} + A(x',y') - 1|^2 $$

Substituting for $A(x',y') = e^{i\phi(x',y')}$:

$$ \boxed{I(x',y') = |ae^{i\alpha} + e^{i\phi(x',y')} - 1|^2} $$$$ = a^2 + 2\left[1 + a\cos(\alpha - \phi(x',y')) - a\cos(\alpha) - \cos(\phi(x',y'))\right] $$

---

★ Method 1 - Dark Ground Technique

Homework: Due 22:Apr:81 Cathey: 7-2, 7-3, 7-4

Filter: $F(\xi,\eta) = ae^{i\alpha}$, where $a = 0$, $\alpha = 0$

(black spot) [Diagram showing small black spot at center blocking F(ξ,η)]

$$ \Rightarrow I(x',y') = 2\left[1 - \cos\phi(x',y')\right] $$

For small angles $\phi(x',y') << 1$: $\cos\phi(x',y') \approx 1 - \frac{\phi^2(x',y')}{2}$

$$ \implies I(x',y') \approx 2\left[1 - 1 + \frac{\phi^2(x',y')}{2}\right] = \phi^2(x',y') $$

$e^{i\phi(x,y)} \approx 1 + j\phi(x,y)$

Remove constant term: $\Rightarrow$ Intensity is proportional to square of phase variation

[Diagram showing phase input with linear variation] $\cong$ [Diagram showing I = A² with parabolic intensity] $\rightarrow$ [Diagram showing one-sided intensity result $A \propto \phi(x,y)$]

$$ I = A^2 \qquad I + A^2 = (\phi(x,y))^2 $$

Phase Contrast - Zernike Methods

Method 2: Positive Phase Contrast (Zernike)

Filter: $F(\xi,\eta) = ae^{i\alpha}$, where $a = 1$, $\alpha = \pi/2$

[Diagram showing filter with dielectric to retard phase by π/2]

$$ \Rightarrow I(x',y') = 1 + 2\left[1 + \cos\left(\frac{\pi}{2} - \phi(x',y')\right) - \cos\left(\frac{\pi}{2}\right) - \cos(\phi(x',y'))\right] $$

If $\phi(x',y') << 1$:

Then $\cos\phi(x',y') \approx 1$

$\cos\left(\frac{\pi}{2} - \phi(x',y')\right) = \sin(\phi(x',y')) \approx \phi(x',y')$

$\cos\left(\frac{\pi}{2}\right) = 0$

and $\boxed{I(x',y') \approx 1 + 2\phi(x',y')}$

$e^{j\phi(x,y)} \cong 1 + j\phi(x,y)$

Shift constant term by $\pi/2$

$I = |j + j\phi(x,y)|^2$

$= |j + 2\phi(x,y) + \phi^2(x,y)|$

$\approx 1 + 2\phi(x,y)$

Intensity is linearly proportional to phase variation.

[Diagram showing phase input] [Diagram showing intensity output with linear relationship: $I = (1+\phi(x,y))^2 \approx 1 + 2\phi + \phi^2 \approx 1 + 2\phi$]

---

Method 3: Negative Phase Contrast

$$ F(\xi,\eta) = ae^{i\alpha}, \quad a = 1, \quad \alpha = \frac{3\pi}{2} $$

Then $I(x',y') = 1 + 2\left[1 + \cos\left[\frac{3\pi}{2} - \phi(x',y')\right] - \cos\left(\frac{3\pi}{2}\right) - \cos(\phi(x',y'))\right]$

Again, $\phi(x',y') << 1$

$\Rightarrow \cos\phi(x',y') \approx 1$

$\cos\left(\frac{3\pi}{2} - \phi(x',y')\right) = -\sin(\phi(x',y')) = -\phi(x',y')$

$\cos\left(\frac{3\pi}{2}\right) = 0$

and $\boxed{I(x',y') \cong 1 - 2\phi(x',y')}$

[Diagram showing phase input] [Diagram showing inverted intensity output: $I = (1-\phi(x,y))^2 \approx 1 - 2\phi + \phi^2 \approx 1 - 2\phi$]

Phase Contrast - Schlieren Method

Method 4

★ Schlieren Method (knife edge method)

[Diagram showing four different intensity distributions as knife edge cuts off different portions of light]

Recall: $\text{sgn}(\xi) = \begin{cases} 1 & \xi > 0 \\ 0 & \xi = 0 \\ -1 & \xi < 0 \end{cases}$

Similar to Heaviside Step Function.

Filter: $H(\xi,\eta) = \frac{1}{2}(1 + \text{sgn}(\xi))$

Input: $f(x,y) = e^{j\phi(x,y)}$

Small angle approx: $f(x,y) \approx 1 + j\phi(x,y)$

$$ \Rightarrow F(\xi,\eta) = \delta(\xi,\eta) + j\,\Phi(\xi,\eta) $$

where $\Phi(\xi,\eta) = \mathcal{F}\mathcal{F}\{\phi(x,y)\}$

---

Note:

If the knife edge is adjusted to completely remove the zero freq. $\delta(\xi,\eta)$, then the result is

$$ F(\xi,\eta) \cdot H(\xi,\eta) = \frac{1}{2}(1 + \text{sgn}(\xi)) \cdot j\,\Phi(\xi,\eta) $$$$ \mathcal{F}^{-1}\{F(\xi,\eta) \cdot H(\xi,\eta)\} = \frac{1}{2}\left(\delta(x,y) + \frac{j}{\pi x}\delta(y)\right) ** j\,\phi(x,y) $$$$ = \frac{j}{2}\left[\phi(x,y) + \frac{j}{\pi}\int\frac{\phi(\alpha,y)}{x-\alpha}\,d\alpha\right] $$$$ = \frac{j}{2}\left[\phi(x,y) + j\,H_x\{\phi(x,y)\}\right] $$

where $H_x\{\phi(x,y)\} = \frac{1}{\pi}\int\frac{\phi(\alpha,y)}{x-\alpha}\,d\alpha$

$$ = \underline{\text{One-Dim Hilbert Transform}} $$

Complex Spatial Filters

Recall telecentric system:

[Diagram showing optical system with planes P1, P2, P3 and focal lengths f]

$$ f(x,y) \xrightarrow{H(\xi,\eta)} g(x,y) $$

The transfer function $H(\xi,\eta)$ is physically present in plane P2.

If we use film to record $H(\xi,\eta)$, then

$$ t(x,y) = H(x',y') \rightarrow \text{positive real} $$

To perform convolutions $\mathcal{F}$ correlations, we require

convolution $\rightarrow$ $g(x,y) = f(x,y) * h(x,y)$ — convolution

or $G(\xi,\eta) = F(\xi,\eta) \cdot H(\xi,\eta)$

where $H(\xi,\eta) = \mathcal{F}\mathcal{F}\{h(x,y)\}$

correlation $\rightarrow$ and $g(x,y) = f(x,y) \star h^*(x,y) = f(x,y) * h^*(-x,-y)$ — correlation

or $G(\xi,\eta) = F(\xi,\eta) \cdot H^*(\xi,\eta)$

where $H^*(\xi,\eta) = \mathcal{F}\mathcal{F}\{h^*(-x,-y)\}$

---

★ This means we have to make a filter with transmittance:

$$ t(x',y') = H(x',y') = \mathcal{F}\mathcal{F}\{h(x,y)\} $$

or $t(x',y') = H^*(x',y') = \mathcal{F}\mathcal{F}\{h^*(-x,-y)\}$

Vander Lugt Filters

But we know that $H(x',y')$ is real only if $h(x,y)$ is even.

Since this is not usually the case, we conclude $H(x',y')$ is in general complex.

Homework: Cathey 7-6, 7-11, 7-14, Goodman 7-7

---

Vander Lugt Filters

[Diagram showing recording geometry with P1 and P2 planes, object wave h(x,y), and reference plane wave at angle θ]

Recording geometry:

Form optical Fourier transform of $h(x,y)$ in P2 $\Rightarrow$ $H\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)$

Add off-axis reference wave at angle $\theta$, $\Rightarrow e^{-j2\pi\alpha x'}$, where $\alpha = \frac{\sin\theta}{\lambda}$

$$ \implies U(x',y') = H\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right) + e^{-j2\pi\alpha x'} $$

Film responds to the intensity. Therefore

$$ t(x',y') \propto I(x',y') = \left|H\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right) + e^{-j2\pi\alpha x'}\right|^2 $$$$ = \left|H\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)\right|^2 + 1 + H\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)e^{j2\pi\alpha x'} + H^*\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)e^{-j2\pi\alpha x'} $$

Note: Although we recorded the intensity, we still have the complex function $H(x',y')$ recorded. However, we also had to record $H^*(x',y')$ to cancel out all phase terms.

Vander Lugt Filter Impulse Response

Calculate impulse response of this filter: $v(x,y)$

(Neglect all constants)

[Diagram showing plane wave input going through filter with spectrum components, set up to observe impulse response]

$$ t(x',y') $$

(proportional to transfer function)

$$ v(x,y) = \mathcal{F}^{-1}\mathcal{F}^{-1}\left\{t\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)\right\}\bigg|_{x=-\frac{x'}{\lambda f}, y=-\frac{y'}{\lambda f}} = \mathcal{F}^{-1}\mathcal{F}^{-1}\left\{H\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)H^*\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right) + 1 + H\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)e^{j2\pi\alpha x'} + H^*\left(\frac{x'}{\lambda f}, \frac{y'}{\lambda f}\right)e^{-j2\pi\alpha x'}\right\} $$$$ = h(x,y) ** h^*(-x,-y) + \delta\left(\frac{x}{\lambda f}, \frac{y}{\lambda f}\right) + h(x,y) * \delta\left(\frac{x+\alpha}{\lambda f}\right) + h^*(-x,-y) * \delta\left(\frac{x-\alpha}{\lambda f}\right) $$

But recall: $\delta\left(\frac{x}{a}+\alpha\right) = |a|\delta(x+a\alpha)$ $= \lambda f\,\delta(x+\alpha\lambda f) * \delta(y+\alpha \cdot nothing)$ since $\alpha = +a \cdot 0/\lambda$

$$ = h(x,y) ** h^*(x,y) + \delta(x,y) + h(x,y) * \delta(x+f\sin\theta) + h^*(-x,-y) * \delta(x-f\sin\theta) $$

---

★ Note: $h^*(x-f\sin\theta, -y)$ and $h(x+f\sin\theta, y)$ are physically separated from each other two terms.

(See above diagram)

Thus, we have recorded a filter whose impulse response is $h(x,y)$ (or $h^*(-x,-y)$).

---

Remember: Cathey 7-2: superposition problem 1: image

We have a filter: Using the Vander Lugt Filter to Perform Correlation

[Diagram showing optical system with P1, P2, P3 planes: Unit Amplitude Plane Wave → Input f(x,y) → Vander Lugt Filter t(x’,y’) → Output g(x,y)]

Vander Lugt Filter Output

$$ g(x,y) = f(x,y) ** v(x,y) $$

where $v(x,y) = \mathcal{F}^{-1}\mathcal{F}^{-1}\{t(x',y')\}$ and was previously calculated.

$$ \Rightarrow g(x,y) = f(x,y) ** h(x,y) ** h^*(x,y) $$

$$

• f(x,y) ** \delta(x,y) $$

$$

• f(x,y) ** h(x+f\sin\theta, y) $$

$$

• f(x,y) ** h^*(-x-f\sin\theta, -y) $$

$$ = f(x,y) ** \left[h(x,y) ** h^*(x,y)\right] \quad \text{(garbage)} $$

$$

• f(x,y) \quad \text{(input image)} $$

$$

• f(x,y) ** h(x+f\sin\theta, y) \quad \text{(convolution, centered at } x = -f\sin\theta\text{)} $$

$$

• f(x,y) ** h^*(x-f\sin\theta, y) \quad \text{(correlation, centered at } x = +f\sin\theta\text{)} $$

[Diagram showing output plane with separated terms: f(x,y)**h(x,y) at center, f(x,y) term, f(x,y)*h(x,y) correlation term at x=+fsinθ, with garbage/autocorrelation terms between]

---

★ Angle requirement for θ:

Consider:

1. Maximum width of $f(x,y) \rightarrow W_f$ (in x direction)
2. Maximum width of $h(x,y) \rightarrow W_h$ (in x direction)

The maximum width of a convolution (or correlation) of two functions equals the sum of individual maximum widths:

1. Max width of $f(x,y) ** [h(x,y) ** h^*(x,y)] = W_f + 2W_h$
2. Max width of $f(x,y) = W_f$
3. Max width of $f(x,y) ** h(x,y) = W_f + W_h$
4. Max width of $f(x,y) ** h^*(x,y) = W_f + W_h$

Vander Lugt Filter - Angle Requirements

[Diagram showing three terms with widths: $W_f + 2W_h$, $W_f$, $W_f + W_h$ separated by spacing $f\sin\theta$]

Requirement for no overlap:

$$ f\sin\theta \geq \frac{W_f + 2W_h}{2} + \frac{W_f + W_h}{2} $$

or $\sin\theta \geq \frac{3}{2}\frac{W_h}{f} + \frac{W_f}{\underbrace{f}_{\text{focal length of lens}}}$

For small angles, we have the simple result:

$$ \boxed{\theta \geq \frac{3}{2}\frac{W_h}{f} + \frac{W_f}{f}} $$

---

★ The Vander Lugt Filter Described In Terms of Superposition of Gratings

General Input Response:

Consider a complex function $|h(x,y)|e^{j\phi_h(x,y)}$ composed of the sum of point sources: $\iint |h(\alpha,\beta)|e^{j\phi_h(\alpha,\beta)}\delta(x-\alpha, y-\beta)\,d\alpha\,d\beta$

We study the result of recording a single point source $\delta(x-\alpha, y-\beta)$ with strength $|h(\alpha,\beta)|e^{j\phi_h(\alpha,\beta)}$

$$ \Rightarrow h(x,y)$ is a simple linear superposition of this result.

[Diagram showing P1 with point source $|h(\alpha,\beta)|e^{j\phi_h(\alpha,\beta)}\delta(x-\alpha,0)$ and P2 with reference plane wave $e^{j2\pi\alpha x'}$, labeled “Recording Geometry”]

Vander Lugt Filter - Two Plane Wave Interference

Note: We have in P2 the interference of two plane waves:

$$

t(x’,y’) \propto I(x’,y’) = \left|e^{-j2\pi\alpha x’} + |h(\alpha,0)|e^{j\phi_h(\alpha,0)}e^{-j2\pi\alpha x’/\lambda f}\right|^2

$$

$$

= 1 + |h(\alpha,0)|^2 + |h(\alpha,0)|e^{-j2\pi\alpha x’}e^{-j\phi_h(\alpha,0)}e^{+j2\pi\alpha x’/\lambda f}

$$

$$

• |h(\alpha,0)|e^{j2\pi\alpha x’}e^{j\phi_h(\alpha,0)}e^{-j2\pi\alpha x’/\lambda f}

$$

$$

= \underbrace{1 + |h(\alpha,0)|^2}{\text{bias}} + 2\underbrace{|h(\alpha,0)|}{AM}\cos!\left(\underbrace{2\pi!\left(\alpha - \frac{\alpha}{\lambda f}\right)}{FM}x’ + \underbrace{\phi_h(\alpha,0)}{\phi M}\right)

$$ \text{-}–

Result for single pt:

[Graph showing oscillating function $t(x',y')$ with DC bias level marked]

1. Amplitude of $h(x,y)$ at pt $(\alpha,0)$, $|h(\alpha,0)|$, controls amplitude modulation of carrier (or depth of modulating grating)

2. Phase of $h(x,y)$ at pt $(\alpha,0)$, $e^{j\phi_h(\alpha,0)}$, controls shift of carrier or phase modulation

3. Location of pt $(\alpha,0)$: controls frequency modulation. Notice: If $\alpha = 0$, carrier is $\cos(2\pi\alpha x' + \phi) \Rightarrow freq = \alpha$. If $\alpha \neq 0$, carrier is $\cos\left(2\pi\left(\alpha - \frac{\alpha}{\lambda f}\right)x' + \phi\right) \Rightarrow freq = \alpha - \frac{\alpha}{\lambda f}$.

Now general $h(x,y)$ is simply a superposition of many modulated gratings

Optical Interpretation of Matched Filtering

Optical Interpretation of Matched Filtering Operation

Recall: Frequency domain description of $f(x,y) ** h(x,y)$

and $f(x,y) ** h^*(x,y)$ $$

F(\xi,\eta) \cdot H(\xi,\eta)

$$ ([Convolution](/notes/areas/mathematics/real_analysis/definitions/convolution/)) $$

F(\xi,\eta) \cdot H^*(\xi,\eta) \quad \text{(correlation)}

$$ [Diagram showing optical system: Plane wave → f(x,y)=h(x,y) Input → Filter with H(ξ,η) and spectrum showing lines of constant phase, diffraction from input, H(ξ,η)·H*(ξ,η) removes phase component → Correlation point output]

Matched Filter when $f(x,y) = h(x,y)$ $$

\Rightarrow g(x,y) = h(x,y) ** h^*(x,y)

$$ \text{-}–

Notice:

1. Effect of filter $H^*(\xi,\eta)$ is to straighten out the lines of constant phase. These constant phase fronts are focused by last lens to a bright spot.

2. Effect of filter $H(\xi,\eta)$ (convolution) is to generate $H(\xi,\eta) \cdot H(\xi,\eta)$ which contains phase variation (not cancelled by $H^*(\xi,\eta)$). Thus, final lens does not focus this field to a bright spot. Thus, convolution is not useful for detecting patterns.

Signal Detection

Signal Detection: (type of pattern recognition)

Say we have a signal $S(x)$ to be detected and N signals $n_1(x), n_2(x), \ldots n_N(x)$ which could be present instead of $S(x)$.

We would like to design a filter

[Diagram showing: S(x), n_1(x), n_2(x), … n_N(x) → h(x) → signal to allow detection of S(x)]

\text{-}–

★ Imagine the problem of character recognition

Characters: A, B, C, D → [Detect A]

One obvious way to detect character A is to make a mask of A and place it over each input character,

[Images of characters A, B, C overlaid with mask]

and add up the common areas. Since all areas are common in the A-A match, but not all are common for the A-B and A-C match, the Total common area will be largest for A-A. Thus the largest number in this procedure detects the letter A. (Assuming all letters have equal total area)

Template Matching

$$

\Rightarrow \int_{-\infty}^{\infty}|S(x)|^2,d\alpha = \int_{-\infty}^{\infty}|n_i(x)|^2,d\alpha = K

$$ (We shall eliminate this constraint later by proper normalization)

\text{-}–

★ This procedure is called Template Matching for obvious reasons

Mathematically, we can write $$

\sigma = \iint f(x,y) \cdot \underbrace{R^*}_{g}(x,y),dx,dy \quad \xleftarrow{\text{input pattern “A” or “B” or…}} \xleftarrow{\text{test pattern “A”}} \quad \text{Template match}

$$ (Inner product)

We simply measure the value of $\sigma$ to see if it is large enough.

[Diagram showing: Input f(x,y) → Template Match → Threshold σ ≥ σ₀? → 1 = yes, 0 = No]

\text{-}–

Template Match observed in Function Space:

Recall function space: $f(x) = \{f(x_1), f(x_2), \ldots, f(x_3), \ldots\}$

N dimensional vector

$f(x)$ is a single point (vector) in N-dim space $h(x)$ is another vector Condition that $\int_{-\infty}^{\infty}|f(x)|^2\,dx = \int_{-\infty}^{\infty}|h(x)|^2\,dx$ $$

\Rightarrow f^* \cdot f = h^* \cdot h \Rightarrow \text{vector lengths are the same}

$$

Function Space and Template Match

In function space we have:

[Diagram showing vectors f and k* in N-dimensional space with axes x₁, x₂, x₃] $$

f \cdot k^*

$$ \text{-}–

★ It is now clear that the Template match of two functions (inner product of two vectors) simply calculates the projection of f onto k.

Thus, we compare two functions by their proximity in function space.

\text{-}–

★ How do we design a filter to perform this inner product?

Consider a filter with impulse response $$

h(x,y) = k^*(-x,-y) \quad \text{(Matched Filter)}

$$ [Diagram showing: f(x,y) → [h(x,y)] → g(x,y)] $$

g(x,y) = f(x,y) ** k^*(-x,-y)

$$

$$

= \iint f(\alpha,\beta) \cdot k^*(\alpha-x, \beta-y),d\alpha,d\beta

$$

$$

= \mathcal{X}_{fk}(x,y) \quad \underline{\text{Cross Correlation}}

$$

Cross Correlation and Inner Product

Note:

$$

\mathcal{X}_{fk}(0,0) = \iint f(\alpha,\beta) \cdot k^*(\alpha,\beta),d\alpha,d\beta = \sigma

$$ (Desired Inner Product) $$

\implies

$$ We simply use a filter with impulse response $k^*(-x,-y)$ and observe the output at $g(0,0)$.

\text{-}–

Note: If $f = k$ (pattern match) then output of filter $$

\mathcal{X}_{ff}(x,y)

$$ is the autocorrelation of $f(x,y)$.

We have shown that $$

\mathcal{X}{ff}(0,0) \geq \mathcal{X}{ff}(x,y)

$$ (Autocorrelation has a maximum value at center)

We can also show that $$

\mathcal{X}{fk}(0,0) \geq \mathcal{X}{fk}(x,y)

$$ (Cross correlation)

since in signal space, this corresponds to computing inner product of f with all other vectors k compared to k(x,y) and shifted versions: k(x-x’, y-y')

[Diagram showing vectors k(x,y), f(x,y), k(x-x’,y’), h(x-x’,y’) in function space]

Matched Filter for Pattern Detection

★ It is natural to ask what center values $g(a,b) = \mathcal{X}_{fk}(a,b)$ correspond to in the above pattern recognition system.

$$

\rightarrow g(a,b) = \mathcal{X}_{fk}(a,b) = \iint f(\alpha,\beta) \cdot k^*(\alpha-a, \beta-b),d\alpha,d\beta

$$ Recall $k(\alpha,\beta)$ was the function describing the pattern “A”

Thus $\mathcal{X}_{fk}(a,b)$ is the inner product between input $f(\alpha,\beta)$ and a shifted pattern “A”, described by $k(\alpha-a, \beta-b)$

Thus $\mathcal{X}_{fk}(a,b)$ tests for the presence of “A” shifted by an amount $(a,b)$.

\text{-}–

★ This type of filter is called a Matched Filter

★ Transfer function of matched filter:

$$

h(x,y) = k^*(-x,-y)

$$ ([Impulse response](/notes/areas/electrical_engineering/signals_systems/definitions/impulse_response/)) $$

\Rightarrow H(\xi,\eta) = \mathcal{F}{k^(-x,-y)} = K^(\xi,\eta)

$$ Thus, matched filter can be viewed as

[Diagram showing: F(ξ,η) → [K*(ξ,η)] → G(ξ,η)] $$

G(\xi,\eta) = F(\xi,\eta) \cdot K^*(\xi,\eta) \leftarrow \text{Product of F.T.}

$$ and when input matches filter, $G(\xi,\eta) = |K(\xi,\eta)|^2$

↳ $F(\xi,\eta) = K(\xi,\eta)$

Character Recognition Machine

Application 1 ←

Character Recognition Machine

Problem: Detect all the letters of the alphabet

Solution: Use a bank of Vander Lugt filters

[Diagram showing parallel matched filter bank architecture: \text{-} Input g(x,y) splits to multiple paths \text{-} Each path has filter H₁*, H₂*, H₃*, H₄*, H₅* etc. \text{-} Each filter output goes through integrator (∫) \text{-} Outputs are normalized by √(∬|h₁|²dxdy), √(∬|h₂|²dxdy), etc. \text{-} Final outputs go to comparator]

Letter “A”: $h_1(x,y) \rightarrow$ Matched filter is $H_1^*(\xi,\eta)$

Letter “B”: $h_2(x,y) \rightarrow$ " " $H_2^*(\xi,\eta)$

⋮

\text{-}–

★ Trouble with using outputs of matched filters directly:

Filter for “F” cannot tell the difference between “E” and “F”

[Images showing letters F filter, E input, F input]

Inner product (template match) is the same for both inputs.

Matched Filter Normalization

Solution to Trouble:

Normalize all outputs by $\sqrt{\iint |h_i(x,y)|^2\,dx\,dy}$

Then, largest output corresponds to input

Proof of normalization solution:

★ Peak Intensity of correctly matched filter

$h_k$ = input $h_k^*$ = filter impulse response $$

|U_k|^2 = \left[\iint |h_k|^2,dx,dy\right]^2 \xleftarrow{\text{because intensity is measured}}

$$

$$

\iint |h_k|^2,dx,dy \leftarrow \text{Normalization}

$$

$$

= \iint |h_k|^2,dx,dy

$$ \text{-}–

★ Peak of incorrectly matched filter (intensity) $|U_n|^2$, $(n \neq k)$

$h_k$ = input $h_n^*$ = filter impulse response $$

|U_n|^2 = \frac{\left|\iint h_k h_n^*,dx,dy\right|^2}{\iint |h_n|^2,dx,dy} \leftarrow \text{Normalization}

$$ \text{-}–

From Schwarz inequality:

$$

\left|\iint h_k h_n^*,dx,dy\right|^2 \leq \iint |h_k|^2,dx,dy \cdot \iint |h_n|^2,dx,dy

$$ To see this inequality is true, remember vector space interpretation of inner product: $\iint h_k h_n^*\,dx\,dy = h_k \cdot h_n$ (vectors)

We know $h_k \cdot h_n = \|h_k\| \|h_n\| \cos\theta$

Since $-1 < \cos\theta < 1$, we have $\|h_k \cdot h_n\|^2 = \|h_k\|^2 \|h_n\|^2 \cos^2\theta$ $$

\leq |h_k|^2 |h_n|^2

$$

Matched Filter Proof

$$

\implies |U_n|^2 = \frac{\left|\iint h_k h_n^*,dx,dy\right|^2}{\iint |h_n|^2,dx,dy} \leq \frac{\iint |h_k|^2,dx,dy \cdot \iint |h_n|^2,dx,dy}{\iint |h_n|^2,dx,dy}

$$

$$

= \iint |h_k|^2,dx,dy = |U_k|^2

$$ or $|U_n|^2 \leq |U_k|^2$

We also know that the equality only holds if $$

\underbrace{h_n(x,y)}{\text{filter}} = K \cdot \underbrace{h_k(x,y)}{\text{input}}

$$

$$

\implies

$$ All filter outputs $|U_n|^2$ are less than the proper filter output $|U_k|^2$, and selection can be based on the highest value.

\text{-}–

Problems with matched filtering techniques:

1. Rotation sensitive
2. Scale sensitive (size change)
3. Statistical variation sensitivity

Inverse Spatial Filtering

Application 2 ←

Inverse Spatial Filtering

★ Consider the problem of image blur

[Diagram showing: object with dimension d₀ → blurred arrow → blurred image g(x,y) with dimension d_i, Exposure time = Δt] $$

f(x,y) \xrightarrow{h(x,y)/H(\xi,\eta)} g(x,y)

$$ Equivalent linear system

\text{-}–

★ Blurring operation can be expressed as a filtering by LSI filter with impulse response $h(x,y)$.

Impulse response (response to a point object) is $$

h(x,y) = \text{rect}\left(\frac{x}{v\Delta t}\right)\delta(y)

$$ since streak in x direction is $v\Delta t$ long. $$

\implies H(\xi,\eta) = v\Delta t,\text{sinc}(v\Delta t,\xi)

$$

Inverse Filtering Analysis

[Diagram showing sinc function with first zero at 1/vΔt on ξ axis]

Note:

1. Blurring is a low pass type filter
2. Sinusoids with freq $n\left(\frac{1}{v\Delta t}\right)$ are completely blocked.

[Diagram showing: vΔt width → blur distance]

This is because the blur averages over exactly one cycle (or n cycles for $\frac{n}{v\Delta t}$) of this sinusoid, so it looks like a constant.

We can restore $F(\xi,\eta)$ (and hence the unblurred picture $f(x,y)$) by a filter with transfer function: $$

\frac{H^*(\xi,\eta)}{|H(\xi,\eta)|^2} \quad \text{sinc}

$$

$$

G(\xi,\eta) = F(\xi,\eta) \cdot H(\xi,\eta) \quad \text{(distorted)}

$$ and so $F(\xi,\eta) = \underbrace{F(\xi,\eta) \cdot H(\xi,\eta)}_{\text{input}} \cdot \left[\frac{H^*(\xi,\eta)}{|H(\xi,\eta)|^2}\right] \quad \text{(restored)}$

Generating the Inverse Filter

We would like to generate this filter: $$

\frac{H^*(\xi,\eta)}{|H(\xi,\eta)|^2}

$$ Part 1: $H^*(\xi,\eta) \cdot \frac{1}{|H(\xi,\eta)|^2}$ Part 2

\text{-}–

Part 1:

[Diagram showing optical setup with two focal lengths f, input h(x,y), impulse response = rect(x/vΔt), and output showing $H(\xi,\eta) + e^{-j2\pi\alpha x}$ with reference wave] $$

I(\xi,\eta) = |H(\xi,\eta)|^2 + 1 + \underbrace{H(\xi,\eta)e^{j2\pi\alpha x} + H^*(\xi,\eta)e^{-j2\pi\alpha x}}_{\text{on carrier}}

$$ \text{-}–

Part 2:

[Diagram showing h(x,y) input with no reference wave] $$

I = |H(\xi,\eta)|^2

$$

$$

t(\xi,\eta) = \left[|H(\xi,\eta)|^2\right]^{-\gamma} \xleftarrow{\text{Negative film response}}

$$

$$

= |H(\xi,\eta)|^{-2}

$$

Inverse Filtering - Restored Image

By placing these two filters in contact with one another, we have $$

\frac{H^*(\xi,\eta)}{|H(\xi,\eta)|^2}

$$ Now, using an optical correlator

(which is just our basic telecentric imaging system)

[Diagram showing optical system: blurred object → lenses → filter $\frac{H^*(\xi,\eta)e^{-j2\pi\alpha x}}{|H(\xi,\eta)|^2}$ + other terms → restored image at $f\sin\theta$]

Fresnel Zones and the Fresnel Zone Plate

Consider light (plane wave, monochromatic) illuminating a screen:

[Diagram showing plane wave illuminating screen with point P at distance R, showing zones a, b, c, d with radii r₁, r₂, etc., and path lengths R+λ/2, R+2λ/2, etc.]

[Diagram showing concentric Fresnel Zones viewed from point P]

\text{-}–

Compute radius of zones:

Right Triangle ΔPOa gives: $$

r_1^2 + R^2 = \left(R + \frac{\lambda}{2}\right)^2 = R^2 + R\lambda + \left(\frac{\lambda}{2}\right)^2

$$

$$

\Rightarrow r_1^2 = R\lambda + \left(\frac{\lambda}{2}\right)^2

$$ Since $R >> \lambda$ $$

\Rightarrow r_1^2 \cong R\lambda, \quad r_1 = \sqrt{R\lambda}

$$

Fresnel Zone Calculations

Example:

Let $R = 50$ cm

$\lambda = 0.5\,\mu$m $$

r_1 = \sqrt{50 \times 5 \times 10^{-5}},\text{cm} = 0.5,\text{mm}

$$ By the same method, we have: $$

r_2 = \sqrt{2R\lambda} = \sqrt{2},r_1

$$

$$

r_3 = \sqrt{3R\lambda} = \sqrt{3},r_1, \quad \text{etc.}

$$ \text{-}–

Phase

We approximate a zone as being constant phase.

Thus, if zone 1 has a certain phase, zone 2 is $\pi$ out of phase, zone 3 is $2\pi$ out, or in phase, etc.

\text{-}–

Amplitude

Amplitude is proportional to the area of zone.

\text{-} Area of central zone = $\pi r_1^2$ \text{-} Area of second zone = $\pi r_2^2 - \pi r_1^2 = \pi(2r_1^2 - r_1^2) = \pi r_1^2$ \text{-} Area of 3rd zone = $\pi r_3^2 - \pi r_2^2 = \pi(3r_1^2 - 2r_1^2) = \pi r_1^2$

etc.

Thus, all zones are of equal strength

\text{-}–

Superposition:

Amplitude at P is given by sum of zones.

(Larger zones contribute slightly less due to larger distance and angle - ignore this)

Fresnel Zone Superposition

Result at P is

$$

E = E_1 - E_2 + E_3 - \ldots

$$ ($E_n$ is amplitude from $n^{th}$ zone)

We can replace the value of a zone by the mean of the zone in front and behind: $$

\Rightarrow E_n = \frac{E_{n+1} + E_{n-1}}{2}

$$ Thus: $$

E = \frac{E_1}{2} + \underbrace{\left(\frac{E_1}{2} - E_2 + \frac{E_3}{2}\right)}{\approx,0} + \underbrace{\left(\frac{E_3}{2} - E_4 + \frac{E_5}{2}\right)}{\approx,0} + \ldots

$$ From above expression

However, if a circular aperture is used so that only the first zone is exposed, $$

E = E_1

$$ (amplitude is twice as great).

\text{-}–

If 2 zones are exposed, $$

E = E_1 - E_2 \approx 0

$$ (dark spot in center)

(See 3a) ←

\text{-}–

Zone plate

Block all regions with $-\pi$ phase.

Thus, point P adds all light in phase and is very bright.

Fresnel Diffraction Integral for Zone Plate

Recall solution to on-axis Fresnel Transform of circle:

[Diagram showing circular aperture with radius r₀, finding intensity on-axis] $$

U_0(x,y) = \text{circ}\left(\frac{\sqrt{x^2+y^2}}{r_0}\right)

$$ **Fresnel diffraction integral gives:** $$

U(0,0,z) = \exp\left[j\frac{2\pi}{\lambda}z\right] \iint \text{circ}\left(\frac{r}{r_0}\right) \exp\left[j\frac{\pi}{\lambda z}r^2\right] r,dr,d\theta

$$

$$

= \frac{2\pi\exp\left[j\frac{2\pi}{\lambda}z\right]}{j\lambda z} \exp\left[j\frac{\pi}{\lambda z}r^2\right] \frac{z\lambda}{j2\pi}\bigg|_0^{r_0}

$$

$$

= -2j\exp\left[j\frac{2\pi}{\lambda}z\right] \exp\left[j\frac{\pi r_0^2}{2\lambda z}\right] \sin\left[\frac{\pi r_0^2}{2\lambda z}\right]

$$

$$

\Rightarrow I(0,0,z) = 4\sin^2\left[\frac{\pi r_0^2}{2\lambda z}\right]

$$ [Graph showing I(0,0,z) vs z with oscillations, peaks at $\frac{\pi r_0^2}{2\lambda z} = 2\pi$ and $\frac{\pi r_0^2}{2\lambda z} = \pi$]

[Diagram showing Zone 2 and Zone 1 regions] $$

z = \frac{r_0^2}{2\lambda}

$$ This is "R" in our current development $$

\boxed{r_0 = \sqrt{nR\lambda}}

$$ Formula for size of $n^{th}$ Fresnel zone $$

\frac{r_0^2}{n\lambda} = R

$$ $n = 2 \Rightarrow$ Two Fresnel Zones

Zone Plate

[Diagram showing zone plate with concentric rings: φ=0 outer ring, φ=π, φ=0, φ=π, φ=0 at center, with radii r₁, r₂, r₃ marked] $$

\begin{cases} r_1 = \sqrt{R\lambda}, & R = \text{distance from plate to screen} \ r_n = \sqrt{n},r_1 \end{cases}

$$ \text{-}–

[Diagram showing zone plate with observation point P at distance R, and markings at ξ/3 and ξ/2]

Consider Fresnel Zones from distances $\frac{R}{2}$, $\frac{R}{3}$, etc.

P at $\frac{R}{2}$ $\Rightarrow$ $r_1' = \sqrt{\frac{R}{2}\lambda} = \frac{1}{\sqrt{2}}r_1$

P’’ at $\frac{R}{3}$ $\Rightarrow$ $r_1'' = \sqrt{\frac{R}{3}\lambda} = \frac{1}{\sqrt{3}}r_1$

Fresnel Zone Plate Focal Points

$$

\implies P’ = \frac{R}{2} \Rightarrow

$$ [Diagram showing zone plate at P’ = R/2 with zones marked: φ=2, φ=s₁, φ=0, φ=π, φ=π, where r₁’ = 1/√2 r₁ and r₂’ = r₁]

\text{-} Original Zone plate

Contributions from $\phi = 0$ and $\phi = \pi$

This produces a dark spot.

\text{-}–

P’’ ⇒ R/3

[Diagram showing zone plate at P’’ = R/3 with zones 1 and 2 canceling, zones 4-6 blocked, zones 7&8 cancel, zone 1 is a not positive phase, etc.]

\text{-} Zone 1 and zone 2 cancel \text{-} but zone 3 leaves a net positive phase. Zones 4-6 are blocked. Zone 7 & 8 cancel, Zone 1 is a not positive phase, etc.

This produces a light spot. The foci occur at $\frac{1}{n}R$, but $n$ even produces a dark spot.

$n =$ odd $\Rightarrow$ light spot

On-Axis Hologram as a Fresnel Zone Plate

Consider an object as consisting of a large number of point sources

A single pt. source of strength A is a spherical wave: $$

U_0 = \frac{A_0}{r}\exp(jkr), \quad r = \sqrt{x^2 + y^2 + z^2}

$$ [Diagram showing point source creating spherical wave with $U_0 = \frac{A_0}{r}\exp(jkr)$]

If $z >> \sqrt{x^2+y^2}$, $r \approx z_0\left(1 + \frac{x^2+y^2}{2z_0^2}\right)$

and $U_0 = \frac{A_0}{z_0}\exp(jkz_0)\exp\left(j\frac{k}{2z_0}(x^2+y^2)\right)$

Add reference plane wave: $U_R = A_R\exp(jkz_0)$

Transmittance of film becomes (proportional to intensity) $$

t(x,y) = I(x,y) = |U_R + U_0|^2 = |U_R|^2 + |U_0|^2 + U_R^U_0 + U_RU_0^

$$

$$

= A_R^2 + \left(\frac{A_0}{z_0}\right)^2 + \frac{A_RA_0}{z_0}\left[\exp\left{j\frac{k}{2z_0}(x^2+y^2)\right} + \exp\left{-j\frac{k}{2z_0}(x^2+y^2)\right}\right]

$$

$$

= A_R^2 + \left(\frac{A_0}{z_0}\right)^2 + \frac{2A_RA_0}{z_0}\cos\left(\frac{k}{2z_0}(x^2+y^2)\right)

$$

Continuous Zone Plate and Hologram Lenses

Note: $t(x,y) \propto \cos\left(\frac{k}{2z_0}(x^2+y^2)\right)$ is a continuous zone plate $$

\left[t’(x,y) = \left[\text{step}\left(\cos\frac{k}{2z_0}(x^2+y^2)\right)\right] \text{ is an actual zone plate}\right]

$$

$$

\text{step}(x) = \begin{cases} 1, & x > 0 \ 0, & x \leq 0 \end{cases}

$$

$$

\implies

$$ If we illuminate $t(x,y)$ with a plane wave, we expect a bright point to occur at $z_0$

[Diagram showing continuous zone plate illuminated by plane wave, creating bright point at z₀]

\text{-}–

★ Consider the hologram as Superposition of 8 lenses

We have $t(x,y) = A_R^2 + \left(\frac{A_0}{z_0}\right)^2 + \frac{A_RA_0}{z_0}\left[\exp\left\{j\frac{k}{2z_0}(x^2+y^2)\right\} + \exp\left\{-\frac{jk}{2z_0}(x^2+y^2)\right\}\right]$

Recall: transmittance function of a lens: $$

t(x,y) = \exp\left[-j\frac{k}{2f}(x^2+y^2)\right]

$$ $f$ = focal length of lens $$

\implies t(x,y) = \left{A_R^2 + \left(\frac{A_0}{z_0}\right)^2\right} \cdot (\text{lens of } \infty \text{ focal length})

$$

$$

• \left{\frac{A_RA_0}{z_0}\right} \cdot (\text{lens of focal length } -z_0)

$$

$$

• \left{\frac{A_RA_0}{z_0}\right} \cdot (\text{lens of focal length } +z_0)

$$

General Holographic Imaging

[Diagram showing plane wave passing through object with multiple lenses (lens f₁-z₁, lens f₂-∞, lens f₃+∞)]

Transmittance same as 3 lenses → {0, I, 0}

General Object + plane wave

[Diagram showing plane wave illuminating general object creating superposition of zone plates]

Sum of point sources

— See 6(a —

\text{-}–

★ General Treatment of Holography

Consider a one-lens imaging system:

[Diagram showing holographic imaging system with object (only care about intensity), information about object is contained in this plane, also contained in this plane since we see an image, and observation plane noting “Care about intensity and phase”]

This implies that the information is contained in all intermediate planes, although it may not be recognizably

Three-Dimensional Nature of Holograms

Three-dimensional nature:

Record:

[Diagram showing object with points A, B, C being recorded as hologram plate with concentric rings for each point]

\text{-} Each point source forms its own zone plate. \text{-} Focal length of zone plate is given by distance between object and hologram plate

\text{-}–

Play back:

[Diagram showing plane wave illuminating hologram, with observe points A, B, C visible - Negative Focal Pts. “Virtual Image”]

[Diagram showing plane wave illuminating hologram, with observe points showing Positive Focal Pts. “Real Image”]

\text{-}–

Note: Real image is pseudoscopic, i.e., it goes out where original object went in, & vice versa. It looks like a death mask.

Recording Phase Information

Consider recording this wave front:

[Diagram showing reference wave and object wave interfering at film, with indicatrices representing complex numbers]

Wave at film from object: $\hat{a}(x,y) = |a(x,y)|\exp[-j\phi(x,y)]$

★ → How do we record phase?

Recall technique of interferometry in radio:

Technique 1 (base banding): Original signal $f_1(t) = \cos(\omega_1 t + \phi(t))$ Add reference signal (local oscillator) $f_{\text{ref}}(t) = \cos(\omega_1 t)$ Mix sum through non-linear mixer

Recall: $\cos(\alpha)\cos(\beta) = \frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)]$ $$

|f_1(t) + f_2(t)|^2 = |\cos(\omega_1 t + \phi(t)) + \cos(\omega_1 t)|^2

$$

$$

= \cos^2(\omega_1 t + \phi(t)) + \cos^2(\omega_1 t) + 2\cos(\omega_1 t + \phi(t))\cos(\omega_1 t)

$$

$$

= \cos^2(\omega_1 t + \phi(t)) + \cos^2(\omega_1 t) + \cos(2\omega_1 t + \phi(t)) + \cos(\phi(t))

$$ Low pass result with cutoff $< \omega_1$ $$

\text{L.P.}{|f_1(t) + f_2(t)|^2} = \underbrace{K}_{\text{constant}} + \cos(\phi(t))

$$ \text{-}–

Technique 2 (heterodyning): Original signal: $f_1(t) = \cos(\omega_1 t + \phi(t))$ Add ref signal of different freq and known but not constant phase: $f_2(t) = \cos(\omega_2 t + \psi(t))$

Heterodyning Analysis

Mix:

$$

\Rightarrow |f_1(t) + f_2(t)|^2 = |\cos(\omega_1 t + \phi(t)) + \cos(\omega_2 t + \psi(t))|^2

$$

$$

= \cos^2(\omega_1 t + \phi(t)) + \cos^2(\omega_2 t + \psi(t)) + 2\cos(\omega_1 t + \phi(t))\cos(\omega_2 t + \psi(t))

$$ **Recall:** $\cos^2(\omega t) = \frac{1}{2} + \frac{1}{2}\cos(2\omega t)$ $$

= \cos^2(\omega_1 t + \phi(t)) + \cos^2(\omega_2 t + \psi(t)) + \cos((\omega_1 + \omega_2)t + \phi(t) + \psi(t))

$$

$$

• \cos((\omega_1 - \omega_2)t + \phi(t) - \psi(t))

$$ \text{-}–

Band pass:

$$

\text{B.P.}{|f_1(t) + f_2(t)|^2} = \cos(\underbrace{(\omega_1 - \omega_2)}_{\text{i.f. frequency}}t + \phi(t) - \psi(t))

$$ (i.f. = intermediate frequency)

High-freq. cut-off: $\omega_{c_1} < \omega_1$ or $\omega_2$

Low-freq. cut-off: $\omega_{c_2} > $ d.c.

Note: No d.c. term

\text{-}–

★ We use the exact same technique for a hologram, except that we substitute spatial frequency for temporal frequency

[Diagram showing frequency components at 0, $\omega_{c_1}$, $\omega_1$, $2\omega_1$, and $\omega_1 - \omega_2$]

Recall: Plane waves at different angles represent signals at different spatial frequencies: $$

f(x,y) = e^{-j2\pi\alpha x} \quad \quad \alpha = \text{spatial freq} = \frac{\sin\theta}{\lambda}

$$ Now, let object wave be $|a(x,y)|\exp[-j\phi(x,y)] = \hat{a}(x,y)$

Let reference wave be $|A(x,y)|\exp[-j\psi(x,y)] = \hat{A}(x,y)$

Mix: $I(x,y) = |A(x,y)|^2 + |a(x,y)|^2 + 2A(x,y)a(x,y)\cos(\underbrace{\psi(x,y) - \phi(x,y)}_{\text{phase is preserved}})$

Hologram Transmittance and Reconstruction

Transmittance is linearly proportional to intensity

Assume: i) $t(x,y) = \beta'(I(x,y))$

ii) Reference wave is constant amplitude $\Rightarrow |A(x,y)|^2 \cong |A|^2$ $$

\implies \underbrace{t(x,y)}{\text{transmittance of hologram}} = \underbrace{t_0}{\beta’|A|^2\text{ (bias)}} + \beta’\left[|a|^2 + \hat{A}^\hat{a} + \hat{A}\hat{a}^\right]

$$ \text{-}–

Reconstruction:

Let reconstruction wave = $\hat{B}(x,y)$

[Diagram showing $\hat{B}(x,y)$ passing through $t(x,y)$] $$

\hat{B}(x,y) \cdot t(x,y) = \underbrace{t_0\hat{B}}{U_1} + \underbrace{\beta’\hat{a}\hat{a}^*\hat{B}}{U_2} + \underbrace{\beta’\hat{A}^\hat{B}\hat{a}}_{U_3} + \underbrace{\beta’\hat{A}\hat{B}\hat{a}^}_{U_4}

$$ (off-axis correlation must be performed)

\text{-}–

Case 1: Let $\hat{B}(x,y) = \hat{A}(x,y)$

(Reconstruction wave is equal to reference wave)

Then $U_3$ becomes: $\beta'\hat{A}^*\hat{B}\hat{a} = \beta'\hat{A}^*\hat{A}\hat{a}$ $$

= \beta’(|A|^2\hat{a}(x,y))

$$

$$

\Rightarrow \boxed{\text{original object wave } \hat{a}(x,y) \text{ is recovered.}}

$$

Hologram Reconstruction - Case 2

Case 2: Let $\hat{B}(x,y) = \hat{A}^*(x,y)$

(Record wave is complex conj. of ref wave)

Then $U_4$ becomes: $\beta'\hat{A}\hat{B}\hat{a}^* = \beta'\hat{A}\hat{A}^*\hat{a}^*$ $$

= \underbrace{\beta’|A|^2}_{\text{constant}}\hat{a}^*

$$

$$

\boxed{\text{comp conj. of object wave } \hat{a}^*(x,y) \text{ is recovered}}

$$ \text{-}–

★ Image Formation

Consider image of a point source:

[Diagram showing ref wave A(x,y) illuminating point object $\hat{a}(x,y)$ at distance $z_0$ from film, creating hologram recording] $$

\implies \hat{a}(x,y) = \hat{a}_0\exp\left[jk\sqrt{z_0^2 + (x-x_0)^2 + (y-y_0)^2}\right]

$$ ↑ object wave — ↑ diverging spherical wave located at $(x_0, y_0, z_0)$

\text{-}–

Case 1 illumination: $\hat{B} = \hat{A}$

$$

\Rightarrow U_3(x,y) = \beta’|A|^2\hat{a}_0\exp\left[jk\sqrt{z_0^2 + (x-x_0)^2 + (y-y_0)^2}\right]

$$ This is a diverging spherical wave with apparent center at $(x_0, y_0, 0)$.

Virtual and Real Images

[Diagram showing ref $\hat{A}(x,y)$ wave passing through hologram creating virtual image at $z_0$]

\text{-}–

Case 2 illumination: $\hat{B} = \hat{A}^*$

$$

\Rightarrow U_4(x,y) = \beta’|A|^2\hat{a}^*

$$

$$

= \beta’|A|^2\hat{a}_0^*\exp\left[-jk\sqrt{z_0^2 + (x-x_0)^2 + (y-y_0)^2}\right]

$$ This is a converging spherical wave that converges at point $(x_0, y_0, z_0)$.

[Diagram showing converging wave creating real image]

Note:

[Diagram comparing original object view vs pseudoscopic real image view - “death mask”]

\text{-}–

Separation of Undesirable Terms (off-axis hologram)

Use Technique 2 in interferometry (pg. 62) — heterodyning — so information is contained on a carrier (or i.f. frequency).

Recall that heterodyning requires a reference signal (local oscillator) at a different frequency than the object signal. A different spatial frequency corresponds to a different angle.

Off-Axis Hologram Recording

[Diagram showing reference wave $\hat{A}(x,y) = A\exp[-j2\pi\alpha y]$, $\alpha = \frac{\sin\theta}{\lambda}$ at angle to film, with object wave $\hat{a}(x,y)$]

Amplitude at film: $U(x,y) = A\exp[-j2\pi\alpha y] + \hat{a}(x,y)$, $\alpha = \frac{\sin\theta}{\lambda}$

Intensity at film: $I(x,y) = A^2 + |\hat{a}(x,y)|^2 + A\hat{a}(x,y)\exp[j2\pi\alpha y]$ $$

• A\hat{a}^*(x,y)\exp[-j2\pi\alpha y]

$$ \text{-}–

★ Notice: If we write object as $\hat{a}(x,y) = |a(x,y)|\exp[-j\phi(x,y)]$

We have intensity: $$

I(x,y) = A^2 + |a(x,y)|^2 + 2\underbrace{A|a(x,y)|}{AM}\cos[\underbrace{2\pi\alpha y}{\text{carrier}} - \underbrace{\phi(x,y)}_{\phi M}]

$$ This is identical to heterodyning, where: $$

\underbrace{(\omega_1 - \omega_2)}{\text{temporal}} = \text{i.f.} = \underbrace{2\pi\alpha_y}{\text{spatial}}

$$

$$

\psi = 0

$$ because plane wave is local osc.

The local oscillator is a pure $\cos\omega t$ with no $\psi(t)$ component.

Reconstruction of Off-Axis Holograms

$$

t(x,y) = \underbrace{t_0}{t_1} + \beta’\left[\underbrace{|a|^2}{t_2} + \underbrace{A\hat{a}(x,y)\exp[j2\pi\alpha y]}{t_3} + \underbrace{A\hat{a}^*(x,y)\exp[-j2\pi\alpha y]}{t_4}\right]

$$ $\beta'$ = constant related to Film Transmittance

where $t_0 = \beta'A^2$

Assume reconstruction illumination is a normally incident plane wave: $$

B(x,y) = B

$$ Then: \text{-} $U_1(x,y) = t_0 B$ \text{-} $U_2(x,y) = \beta'B|a(x,y)|^2$ \text{-} $U_3(x,y) = \beta'BA\hat{a}(x,y)\exp[j2\pi\alpha y]$ \text{-} $U_4(x,y) = \beta'BA\hat{a}^*(x,y)\exp[-j2\pi\alpha y]$

\text{-}–

★ Notice:

\text{-} $U_1 \Rightarrow$ proportional to on-axis plane wave \text{-} $U_2 \Rightarrow$ garbage, also centered on-axis \text{-} $U_3 \Rightarrow$ proportional to object wave $\hat{a}(x,y)$ times a linear phase factor $\exp[j2\pi\alpha y]$. This is equivalent to tilting the object wave $\hat{a}(x,y)$ by angle $\theta$. \text{-} $U_4 \Rightarrow$ proportional to complex conj. of object wave $\hat{a}^*(x,y)$ times linear phase factor $\exp[-j2\pi\alpha y] \Rightarrow$ tilt is in $-\theta$ direction.

Reflection Holograms and Multiplex Holograms

Reflection Holograms

[Diagram showing object wave and ref wave entering from opposite sides of thick film with λ/2 spacing between layers, creating Bragg stack]

Play back:

[Diagram showing Bragg stack reflecting proper wavelength λ]

Bragg stack selects proper wavelength λ. Thus, it can work in white light.

\text{-}–

Multiplex Holograms

[Diagram showing object being spun while conventional photograph taken at many angles]

[Diagram showing single photograph #1 being recorded as thin hologram with object wave, ref wave at angles #2, #3, #4]

[Diagram showing cylindrical hologram playback with View 1 (Photo 1-m) and View 2 (Window m+1→n)]

Holographic Laser Cavities

Holographic Laser Cavities:

[Diagram showing laser cavity with diffractive optic mirror 1 (T = Mode), mirror 2 (diffractive optic), and transmittance functions $t_1(x',y')$ and $t_2(x',y')$]

Let $a(x,y)$ be complex field of desired mode at mirror 1: $$

\hat{a}(x,y) = \iint A(u,v)\exp[-j2\pi(xu+yv)],du,dv

$$ (Plane wave expansion)

At Mirror 2: $$

\hat{b}(x’,y’) = \iint A(u,v)\exp[-j2\pi(xu+yv)]\underbrace{\exp\left[jkl\left[1-(\lambda u)^2-(\lambda v)^2\right]^{1/2}\right]}_{\text{free-space propagator}},du,dv

$$ \text{-}–

Let $t_2(x',y') = \frac{\hat{b}^*(x',y')}{\hat{b}(x',y')}$ (All-phase function)

Then $\hat{b}(x',y') \cdot t_2(x',y') = \hat{b}^*(x',y')$ $$

= \iint A^*(u,v)\exp[-j2\pi(xu+yv)] \times \exp\left{-jkl\left[1-(\lambda u)^2-(\lambda v)^2\right]^{1/2}\right},du,dv

$$ Prop back to mirror 1: $$

\Rightarrow \iint A^(u,v)\exp[-j2\pi(xu+yv)],du,dv = a^(x,y).

$$ If we let $t_1(x,y) = \frac{a(x,y)}{a^*(x,y)}$, we have

the final field regenerating the original field $a(x,y)$

This is the laser mode.

Essentials of Holography

Recording:

Obj wave: $a(x,y)$

Ref wave: $b(x,y)$ $$

I(x,y) = |a(x,y) + b(x,y)|^2 = |b(x,y)|^2 + |a(x,y)|^2

$$

$$

• a(x,y)b^(x,y) + a^(x,y)b(x,y) \propto \underbrace{t(x,y)}_{\text{trans. of hologram}}

$$ \text{-}–

Play back:

Play back wave = $C(x,y)$ $$

A(x,y) = C(x,y) \cdot t(x,y)

$$

$$

= C(x,y)|b(x,y)|^2 + C(x,y)|a(x,y)|^2 + C(x,y)a(x,y)b^*(x,y)

$$

$$

• C(x,y)a^*(x,y)b(x,y)

$$ \text{-}–

Special Case: $b = e^{-j2\pi\alpha x}$, $c = e^{-j2\pi dx}$

($b = c$) $$

A \propto \underbrace{e^{-j2\pi\alpha x}}{t_1} + \underbrace{|a(x,y)|^2 e^{-j2\pi\alpha x}}{t_2} + \underbrace{a(x,y)}{t_3} + \underbrace{a^*(x,y)e^{-j2\pi(2\alpha)x}}{t_4}

$$ [Diagram showing recording geometry with object wave and reference wave at angle]

[Diagram showing playback geometry with reconstructed waves $t_1$, $t_2$, $t_3$, $t_4$ separating at different angles]

Special Case Hologram

Special Case: $b = e^{-j2\pi\alpha x}$, $c = 1$

$$

A \propto \underbrace{1}{t_1} + \underbrace{|a(x,y)|^2}{t_2} + \underbrace{a(x,y)e^{j2\pi\alpha x}}{t_3} + \underbrace{a^*(x,y)e^{-j2\pi\alpha x}}{t_4}

$$ [Diagram showing recording geometry with reference wave b at angle to film and object wave c]

[Diagram showing playback geometry with hologram producing four terms: $t_1$, $t_2$ on-axis, $t_3$ (Virtual), $t_4$ (Real)]

\text{-}–

Record → Playback

!hologram_spectrum.svg

In the frequency domain, the four hologram terms separate: \text{-} $t_1$, $t_2$ centered at the origin \text{-} $t_3$ centered at $\alpha$ (contains $\mathscr{F}\{a(x,y)\}$) \text{-} $t_4$ centered at $-\alpha$ (contains $\mathscr{F}\{a^*(x,y)\}$) $$

\text{Hologram} = t_1 + t_2 + a \cdot (\text{carrier}) + a^* \cdot (\text{carrier})

$$

Minimum Reference Wave Angle

!hologram_reconstruction.svg

\text{-}–

★★ Minimum Reference Wave Angle

We need to find the minimum carrier frequency “$\alpha$” so that the spectra of $t_3$ and $t_4$ do not overlap each other or $t_1$, $t_2$.

(This is very similar to the sampling theorem)

!hologram_spectrum.svg

\text{-} The DC terms ($t_1 + t_2$, autocorrelation of $|a|^2$) sit at the origin \text{-} $t_3 = \mathscr{F}\{\hat{a}\}$ is shifted to carrier frequency $+\alpha$ \text{-} $t_4 = \mathscr{F}\{\hat{a}^*\}$ is shifted to $-\alpha$ \text{-} To recover $\hat{a}$, the spectra must not overlap — this sets the minimum reference angle $$

\text{hologram} = t_1 + t_2 + \hat{a} \cdot \text{carrier} + \hat{a}^* \cdot (-\text{carrier})

$$ We could recover $\hat{a}$’s carrier by Fourier transforming $t(x,y)$ and spatial filtering, assuming $\mathscr{F}\{\hat{a}\}$’s carriers do not overlap any other spectral components.

\text{-}–

Define: $$

G_1(\xi,\eta) = \mathscr{F}\mathscr{F}{t_1(x,y)} = t_0,\delta(\xi,\eta)

$$

$$

G_2(\xi,\eta) = \mathscr{F}\mathscr{F}{t_2(x,y)} = \beta’ G_a(\xi,\eta) * G_a(\xi,\eta)

$$ where $G_a(\xi,\eta) = \mathscr{F}\mathscr{F}\{\hat{a}(x,y)\}$ $$

G_3(\xi,\eta) = \mathscr{F}\mathscr{F}{t_3(x,y)} = \beta’ A,G_a(\xi,\eta-\alpha)

$$ and $$

G_4(\xi,\eta) = \mathscr{F}\mathscr{F}{t_4(x,y)} = \beta’ A,G_a^*(-\xi,-\eta-\alpha)

$$

Fourier Transform of Hologram

Note: $G_a(\xi,\eta)$ is the Fourier transform of $\hat{a}(x,y)$

[Diagram showing object wave evaluated at hologram plane]

$\hat{a}(x,y)$ is the object wave evaluated at hologram plane.

The actual object distribution and the distribution at the hologram $\hat{a}(x,y)$ are related by free space diffraction (Fresnel diffraction formula). This means the F.T. of $\hat{a}(x,y)$ equals the F.T. of the object times the free space transfer function: $$

H(\xi,\eta) = \exp\left{jkz_0\left[1-\lambda^2(\xi^2+\eta^2)\right]^{1/2}\right}

$$ The diffraction from object to hologram only changes the phase of the Fourier components, so the bandwidth of $\hat{a}(x,y)$ at the hologram is the same as the bandwidth of the object. Let this bandwidth be $B$.

[Diagram showing $|G_a|$ with bandwidth $B$ centered at origin - “Transform of object”]

[Diagram showing $|G_1|$ spike, convolved with $|G_2|$ of width $2B$, and $|G_3|$ with bandwidth $B$ - “Transform of hologram transmittance”]

Separation Requirement and Minimum Angle

For separation, we require: $$

\alpha \geq 3B

$$ or since $\alpha = \frac{\sin\theta}{\lambda}$, $\sin\theta \geq 3B\lambda$ $$

\Rightarrow \boxed{\theta_{\min} = \arcsin(3B\lambda)}

$$ Minimum Reference Angle

\text{-}–

Example:

[Diagram showing reference wave at angle to hologram, with object (Beethoven) to the left]

Bandwidth of object = highest freq = $\frac{1}{\text{smallest change}}$

Assume smallest variation = $10\mu$ $$

\Rightarrow B = \frac{1}{10\mu} = 100 \text{ /mm}

$$

$$

\boxed{\theta_{\min} = \arcsin\left(3(100)(0.5 \times 10^{-3})\right) = 9°}

$$

$$

\boxed{\text{Carrier frequency } \alpha = 3B = 300 \text{ lines/mm}}

$$ \text{-}–

★ Note: This means very high resolution film (300 lines/mm is a lot!) must be used. Such film does exist. Kodak 649F has a resolution of 1500-2000 lines/mm (and an ASA of 0.06!). Note also that motion more than a fraction of a carrier period ($3\mu$ in above example) during exposure will wash out carrier $\Rightarrow$ no hologram.

Holographic Image Formation and Magnification

[Diagram showing optical setup with: \text{-} Reference wave at angle \text{-} Object at distance $d_1$ from hologram \text{-} Point $e_v$ on object \text{-} Hologram plane \text{-} Image plane at $(x_2, y_2)$ \text{-} Final observation point at $\phi_1$ and coordinates $(x_3, y_3)$]

Consider a diffuse object such as a statue, so the light reflecting off the object is: $$

\boxed{\text{This is the object}} \Rightarrow U_1(x_1,y_1) = |A(x_1,y_1)|e^{j\psi(x_1,y_1)} \leftarrow \text{Random phase}

$$ The reference beam is produced by a point source off-axis, $\rho_v$ away. In the hologram plane the reference produces: $$

\boxed{\text{Reference Beam}} \quad U_2(x_2,y_2) = V_0 \exp\left[jk_1\left(x_2\sin\phi + \frac{x_2^2+y_2^2}{2\rho_v}\right)\right]

$$ Notice: i) $k_1 \Rightarrow$ recording wavelength

ii) If $\rho_v \to \infty$, we have the usual off-axis plane wave $\exp(j2\pi\alpha x_2)$, $\alpha = \frac{\sin\phi}{\lambda}$

To calculate the light field created by the object at the hologram, we use the Fresnel diffraction formula: $$

\boxed{\text{Object Beam}} \quad U_2(x_2,y_2) = \frac{\exp(jk_1d_1)}{j\lambda_1 d_1} h_1(x_2,y_2;d_1)

$$

$$

\iint U_1(x_1,y_1) h_1(x_1,y_1;d_1) \exp\left[-jk_1 \frac{(x_1x_2+y_1y_2)}{d_1}\right] dx_1 dy_1

$$ where: $h_1(x_1,y;d) = \exp\left[jk_1\frac{(x^2+y^2)}{2d}\right]$ ← secondary wavelet

Hologram Interference and Transmittance

Recall that interference gives an intensity (and thus transmittance of hologram) which is in 4 parts: $$

\begin{cases} a = \text{obj} \ A = \text{ref} \end{cases} \quad I = |a+A|^2 = |A|^2 + |a|^2 + \underbrace{aA^*}{\text{primary}} + \underbrace{a^*A}{\text{secondary}}

$$

1. We define 3rd term as the primary wave since it contains the object “$a(x,y)$”

2. We define 4th term as the secondary wave since it contains the conjugate of the object $a^*(x,y)$

Now consider only the primary wave term resulting from $U_2(x_2,y_2)$ and $V_2(x_2,y_2)$ (i.e. $UV^*$): $$

\boxed{t(x_2,y_2) = U(x_2,y_2)V^*(x_2,y_2) = V_0 U_2(x_2,y_2) \exp\left[-jk_1\left(x_2\sin\phi + \frac{x_2^2+y_2^2}{2\rho_v}\right)\right]}

$$ Now illuminate the hologram with another off-axis spherical wave of wavelength $\lambda_2$: $$

\boxed{W(x_2,y_2) = W_0 \exp\left[jk_2\left(x_2\sin\psi + \frac{x_2^2+y_2^2}{2\rho_w}\right)\right]} \quad \boxed{\text{Read out beam}}

$$ After passing through the hologram and propagating a distance $d$: $$

U_3(x_3,y_3) = V_0 W_0 \frac{\exp(jk_2d_2)}{j\lambda_2 d_2} h_2(x_3,y_3;d_2)

$$ ← Fresnel diffraction formula $$

\times \iint U_2(x_2,y_2) h_1(x_2,y_2;-\rho_v) \exp[-jk_1 x_2 \sin\phi]

$$

$$

\times h_2(x_2,y_2;\rho_w) \exp[jk_2 x_2 \sin\psi] h_2(x_2,y_2;d_2)

$$ ← more Fresnel formulas $$

\times \exp\left[-jk_2 \frac{(x_2x_3+y_2y_3)}{d_2}\right] dx_2 dy_2

$$ ← diff. formula

Focusing Condition

Substituting for $U_2(x_2,y_2)$ yields: $$

U_3(x_3,y_3) = -V_0 W_0 \frac{\exp[j(k_1d_1+k_2d_2)]}{\lambda_1\lambda_2 d_1 d_2} h_2(x_3,y_3;d_2)

$$

$$

\times \iint \exp\left{j\pi(x_2^2+y_2^2)\left[\frac{1}{\lambda_2 d_2} + \frac{1}{\lambda_2 d_1} - \frac{1}{\lambda_1\rho_v} + \frac{1}{\lambda_2\rho_w}\right]\right}

$$

$$

\times \exp\left[j2\pi x_2\left(\frac{\sin\psi}{\lambda_2} - \frac{\sin\phi}{\lambda_1}\right)\right] \exp\left[-\frac{k_2(x_2x_3+y_2y_3)}{d_2}\right]

$$

$$

\times \iint U_1(x_1,y_1) h_1(x_1,y_1;d_1) \exp\left[-\frac{j2\pi}{\lambda_1}(x_1x_2+y_1y_2)\right] dx_1 dy_1 dx_2 dy_2

$$ Reversing the order of integration and dropping the minor sign and $\exp[j(k_1d_1+k_2d_2)]$: $$

U_3(x_3,y_3) = \frac{V_0 W_0}{\lambda_1\lambda_2 d_1 d_2} h_2(x_3,y_3;d_2) \iint U_1(x_1,y_1) h_1(x_1,y_1;d_1)

$$

$$

\times \iint \exp\left{j\pi(x_2^2+y_2^2)\left[\frac{1}{\lambda_2 d_2} + \frac{1}{\lambda_2 d_1} - \frac{1}{\lambda_1\rho_v} + \frac{1}{\lambda_2\rho_w}\right]\right}

$$

$$

\times \exp\left[-j2\pi x_2\left(\frac{\sin\phi}{\lambda_1} - \frac{\sin\psi}{\lambda_2} + \frac{x_1}{\lambda_1 d_1} + \frac{x_3}{\lambda_2 d_2}\right)\right]

$$

$$

\times \exp\left[-j2\pi y_2\left(\frac{y_1}{\lambda_1 d_1} + \frac{y_3}{\lambda_2 d_2}\right)\right] dx_2 dy_2 dx_1 dy_1

$$ \text{-}-- $$

\boxed{\text{Focusing Condition}: \quad \frac{1}{\lambda_1 d_1} + \frac{1}{\lambda_2 d_2} - \frac{1}{\lambda_1\rho_v} + \frac{1}{\lambda_2\rho_w} = 0}

$$ We shall show that if $d_1$ (object distance) and $d_2$ (image distance) satisfy this condition, the above expression reduces to an image.

With this condition applied, the inner integral becomes: $$

\iint \exp\left[-j2\pi x_2\left(\frac{\sin\phi}{\lambda_1} - \frac{\sin\psi}{\lambda_2} + \frac{x_1}{\lambda_1 d_1} + \frac{x_3}{\lambda_2 d_2}\right)\right] \exp\left[-j2\pi y_2\left(\frac{y_1}{\lambda_1 d_1} + \frac{y_3}{\lambda_2 d_2}\right)\right] dx_2 dy_2

$$

$$

= \iint \exp\left[-j2\pi(x_2\xi + y_2\eta)\right] dx_2 dy_2, \quad \text{where} \quad \xi = \left(\frac{\sin\phi}{\lambda_1} - \frac{\sin\psi}{\lambda_2} + \frac{x_1}{\lambda_1 d_1} + \frac{x_3}{\lambda_2 d_2}\right)

$$

$$

\eta = \left(\frac{y_1}{\lambda_1 d_1} + \frac{y_3}{\lambda_2 d_2}\right)

$$

Total Integral and Magnification

But this is just the Fourier transform of unity.

Hence $\displaystyle\iint \exp[-j2\pi(x_2\xi + y_2\eta)] dx_2 dy_2 = \delta(\xi,\eta)$: $$

= \delta\left(\frac{\sin\phi}{\lambda_1} - \frac{\sin\psi}{\lambda_2} + \frac{x_1}{\lambda_1 d_1} + \frac{x_3}{\lambda_2 d_2}\right) \delta\left(\frac{y_1}{\lambda_1 d_1} + \frac{y_3}{\lambda_2 d_2}\right)

$$ and the **total integral** becomes: $$

U_3(x_3,y_3) = \frac{V_0 W_0}{\lambda_1\lambda_2 d_1 d_2} h_2(x_3,y_3;d_2) \iint U_1(x_1,y_1) h_1(x_1,y_1;d_1) ,\delta\left(\frac{\sin\phi}{\lambda_1} - \frac{\sin\psi}{\lambda_2} + \frac{x_1}{\lambda_1 d_1} + \frac{x_3}{\lambda_2 d_2}\right)

$$

$$

\times \delta\left(\frac{y_1}{\lambda_1 d_1} + \frac{y_3}{\lambda_2 d_2}\right) dx_1 dy_1

$$

$$

= \frac{\lambda_1 d_1}{\lambda_2 d_2} V_0 W_0 h_2(x_3,y_3;d_2) \iint U_1(x_1,y_1) h_1(x_1,y_1;d_1)

$$

$$

\times \delta\left(x_1 + \frac{\lambda_1 d_1}{\lambda_2 d_2}x_3 + d_1\sin\phi - \frac{\lambda_1 d_1}{\lambda_2}\sin\psi\right) \delta\left(y_1 + \frac{\lambda_1 d_1}{\lambda_2 d_2}y_3\right) dx_1 dy_1

$$ \text{-}–

Define magnification (x-y plane) as $$

M = \frac{\lambda_2 d_2}{\lambda_1 d_1}

$$ Then $$

\boxed{U_3(x_3,y_3) = \frac{V_0 W_0}{M} h_2(x_3,y_3;d_2) , U_1\left(-\frac{x_3}{M} - d_1\sin\phi + d_2\sin\psi, -\frac{y_3}{M}\right)}

$$

$$

\times h_1\left(-\frac{x_3}{M} - d_1\sin\phi + \frac{d_2\sin\psi}{M}, -\frac{y_3}{M}; d_1\right)

$$ A scaled version of $U_1$ results, shifted by an amount $-d_1\sin\phi + d_2\frac{\sin\psi}{M}$. There are also two quadratic phase factors $h_2$ and $h_1$.

Magnification: Transverse and Longitudinal

Magnification (Transverses and Longitudinal)

The transverse magnification is $M = \frac{\lambda_2 d_2}{\lambda_1 d_1}$, which appears linearly proportional to the ratio of wavelengths.

★ However, $d_2/d_1$ is a function of wavelength

From the focusing condition (primary wave): $$

d_2 = \left(-\frac{1}{\rho_w} + \frac{\lambda_2}{\lambda_1\rho_v} - \frac{\lambda_2}{\lambda_1 d_1}\right)^{-1}

$$ (Image reconstruction / Distance for primary wave)

Performing the same analysis for the secondary wave: $$

d_2 = \left(-\frac{1}{\rho_w} - \frac{\lambda_2}{\lambda_1\rho_v} + \frac{\lambda_2}{\lambda_1 d_1}\right)^{-1}

$$ (Image reconstruction / Distance for secondary wave)

If $d_2$ is negative (behind hologram), we have a virtual image.

If $d_2$ is positive (in front of hologram), image is real.

★ From the above, either the primary or secondary image can be virtual or real, depending on $\lambda_1$, $\lambda_2$, $\rho_w$, $\rho_v$, $d_1$.

Notice: If $\lambda_1 = \lambda_2$ and $\rho_w = \rho_v$ (and $\phi = \psi$) $d_2$ must be $d_1$ for $M = +1$ and indicates that: $$

d_2 = -d_1 \quad (\text{primary wave}) \Rightarrow \text{virtual image}

$$ and $M = -1 \Rightarrow |U_3(x_3,y_3)|^2 = |U_1(x_3-d_1\sin\phi, y_3)|^2$ $$

d_2 = \left(\frac{1}{d_1} - \frac{2}{\rho_w}\right)^{-1}

$$ (secondary wave)

In this case, the primary wave forms an image just like the original (only shifted) on the opposite side of the viewer. The secondary image, however, depends on $d_1$ and $\rho_w$.

When $\rho_w = \infty$ (plane wave), $d_2 = d_1$ (real image), and $M = 1$

Magnification as a Function of Object Distance

Magnification as a function of object distance:

Recall: $M = \frac{\lambda_2 d_2}{\lambda_1 d_1}$

and $$

d_2 = \left(-\frac{1}{\rho_w} + \frac{\lambda_2}{\lambda_1\rho_v} - \frac{\lambda_2}{\lambda_1 d_1}\right)^{-1}

$$ (primary wave) $$

d_2 = \left(-\frac{1}{\rho_w} - \frac{\lambda_2}{\lambda_1\rho_v} + \frac{\lambda_2}{\lambda_1 d_1}\right)^{-1}

$$ (secondary wave) $$

\Rightarrow \boxed{M_{\pm} = \left(-\frac{\lambda_1 d_1}{\lambda_2\rho_w} \pm \frac{d_1}{\rho_v} + 1\right)^{-1}} \begin{array}{l} \leftarrow \text{primary} \ \leftarrow \text{secondary} \end{array} \quad \boxed{\text{Transverse Magnification}}

$$ \text{-}–

★ We can calculate longitudinal magnification by change in image distance vs. change in object distance: $$

M_{\text{long}} = \frac{d(d_2)}{d(d_1)}, \quad \text{where } d_2 = \frac{1}{\left(-\frac{1}{\rho_w} \pm \frac{\lambda_2}{\lambda_1\rho_v} - \frac{\lambda_2}{\lambda_1 d_1}\right)}

$$

$$

\Rightarrow M_{\text{long}} = \frac{d(d_2)}{d(d_1)} = \pm \frac{\lambda_2}{\lambda_1} \frac{d\left(\frac{1}{d_1}\right)}{d(d_1)} \cdot \frac{1}{\left(-\frac{1}{\rho_w} \pm \frac{\lambda_2}{\lambda_1\rho_v} - \frac{\lambda_2}{\lambda_1 d_1}\right)^2} = \pm \frac{\lambda_2}{\lambda_1}\left[\frac{1}{d_1^2}\right] \cdot \frac{1}{\left(-\frac{1}{\rho_w} \pm \frac{\lambda_2}{\lambda_1\rho_v} - \frac{\lambda_2}{\lambda_1 d_1}\right)^2}

$$

$$

= \pm \frac{\lambda_2}{\lambda_1} \cdot \frac{\left(\frac{\lambda_1}{\lambda_2}\right)^2}{\left(-\frac{d_1\lambda_1}{\lambda_2\rho_w} \pm \frac{d_1}{\rho_v} + 1\right)^2}

$$

$$

= \frac{\pm\lambda_1}{\lambda_2} \cdot \frac{1}{\left(-\frac{\lambda_1 d}{\lambda_2\rho_w} \pm \frac{d_1}{\rho_v} + 1\right)^2} = \pm \frac{\lambda_1}{\lambda_2} M_{\pm}^2

$$ \text{-}–

Note: It is possible to have significantly different transverse and longitudinal magnifications.

Magnification Example and Interferometry Application

Example: Plane wave read-out: $\rho_w = \infty$

$$

\Rightarrow M_{\pm} = \left(\pm\frac{d_1}{\rho_v} + 1\right)^{-1} \leq 2 \text{ not a function of } \lambda_1/\lambda_2

$$ But $M_{\text{long}} = \pm \frac{\lambda_1}{\lambda_2} M_{\pm}^2 = \pm \frac{\lambda_1}{\lambda_2}\left(\pm\frac{d_1}{\rho_v} + 1\right)^{-2}$

A large difference in wavelength $\frac{\lambda_1}{\lambda_2}$ has no effect on transverse magnification but a great effect on longitudinal magnification, causing severe distortions.

\text{-}–

★ Applications

Interferometry:

!interferometer_setup.svg

\text{-} Interferometer coherent beams (add amplitudes). Phase imbalance in one beam are observed as optical variations of the sum (of the sum in the form of intensity)

How can we interpret two beams which occur at different times?

(For example the two beams may come from the same object, see before a stress is placed on the object, and after the stress)

Holographic Interferometry

Using holography, record a multiple exposure (2 exposures in the obvious example): $$

E = \sum_{k=1}^{N} T_k I_k

$$ $T_k$ are exposure times, $I_k$ are intensities of different holograms $$

I_k = |\underbrace{A(x,y)}{\text{fixed reference}} + \underbrace{a_k(x,y)}{\text{many objects}}|^2

$$ Expanding: $$

E = \sum_{k=1}^{N} T_k |A|^2 + \sum_{k=1}^{N} T_k |a_k|^2 + \underbrace{\sum_{k=1}^{N} T_k A^* a_k}{\text{important term}} + \underbrace{\sum{k=1}^{N} T_k A a_k^*}_{\text{important term}}

$$ 3rd term gives transmittance: $$

t_3 = \beta \sum_{k=1}^{N} T_k A^* a_k

$$ 4th term: $t_4 = \beta \sum_{k=1}^{N} T_k A a_k^*$

If we illuminate 3rd term with $A$, we reconstruct $$

U_3 = \beta \sum_{k=1}^{N} T_k |A|^* A | a_k = \beta |A|^2 \underbrace{\sum_{k=1}^{N} T_k a_k}_{N \text{ virtual images superimposed}}

$$ \text{-}–

Example: Observing heat distribution

[Diagram showing metal block with reference wave, object wave through heated zone, and resulting interference pattern going to eye]

Perform a double exposure — once with the hot plate off, once with it on.

Non-Destructive Testing and Aberration Correction

Reconstruction:

[Diagram showing hologram with read-out beam producing interference pattern going to eye]

\text{-}–

Example: Non-destructive testing - How can we predict whether a mechanical component is going to fail at high stress without actually applying the full stress?

Solution: Apply a small stress and observe the strain using holographic interferometry.

[Diagram showing automobile tire with marked lines]

\text{-} Apply small amount of air in between exposures, and observe interference. Many interference lines near large strain $\Rightarrow$ weak spot in tire.

\text{-}–

Aberration Correction

Method 1: (Image Coding)

[Diagram showing recording geometry with illumination through aberrating medium onto object, with reference wave hitting film]

Recording Geometry

Aberration Correction Methods

Let the object wave (undistorted) at the aberrating medium be $U_1(x,y)$. The aberration introduces phase shifts of $\exp[jW(x,y)]$, so the resultant object wave is $U_1(x,y)\exp[jW(x,y)]$.

The hologram records this wave and its complex conjugate.

Play back using the conjugate read-out beam $A^*$ to produce a conjugate real image. The 4th term gives $|A^*A| \, a^* = |A|^2 U_1^*(x,y) \exp[-jW(x,y)]$, which occurs at the aberration plate a distance $d$ in front of the hologram.

[Diagram showing play back geometry with read out beam from $A^*$ (conjugate of reference beam), same aberrating medium, real image of object]

After passing through the same aberrating medium, we have: $$

\underbrace{|A|^2 U_1^(x,y)}{\text{real image}} \underbrace{\exp[-jW(x,y)] \exp[jW(x,y)]}{\text{aberration plate}} = |A|^2 U_1^(x,y)

$$ This produces an undistorted real image.

This technique has obvious applications in cryptography. Hologram is coded message. Aberration plate is the code.

Compensation Plate Method

Method 2: (Compensation plate)

Record the impulse response of the aberration:

[Diagram showing point source with spherical waves passing through aberrating medium onto film] $$

\text{Object wave} = a = \underbrace{\exp[jW(x,y)]}_{\text{aberration function}}

$$

$$

\text{Reference wave} = A = \exp[j2\pi\alpha y] \quad \text{(plane wave)}

$$ If we use a read-out of $a$, we generate $A$, which produces playback.

[Diagram showing general object passing through aberrating medium hitting film, then plane wave output]

Every point on object produces an aberrated spherical wave at the hologram. The component $a^*$ on the hologram can remove this aberration and produces a plane wave with angle proportional to the original point location. These angles are conjugate at a lens.

Lens Aberration Compensation

Application: Lens aberration compensation

Method 3: (Atmospheric compensation)

!atmospheric_compensation_recording.svg

Assume the aberrating medium is just in front of the film. $$

\text{Object wave: } a(x,y)\exp[jW(x,y)]

$$

$$

\text{Reference wave: } A(x,y)\exp[jW(x,y)]

$$ The hologram intensity is: $$

I \propto |a(x,y)\exp[jW(x,y)] + A(x,y)\exp[jW(x,y)]|^2

$$

$$

= |a(x,y)|^2 + |A(x,y)|^2 + a(x,y)A^(x,y) + a^(x,y)A(x,y)

$$ The aberration function $\exp[jW(x,y)]$ has not been recorded!

Reconstruction with wave $A$ selects the 3rd term: $$

|A^*A| , a(x,y)

$$ which reconstructs the original object. (See Goodman, p9267)

\text{-}–

Holographic data storage:

[Diagram showing object being recorded as hologram, then F.T. hologram shown going to F.T. into about 1 bit - Note: in F.T. hologram, is stored everywhere in plane.]

Photorefractive crystals act as real-time storage (3-dimensional).

Photorefractive Effect

[Diagram showing reference wave and object wave entering BaTiO₃ Crystal]

The photorefractive process occurs in three stages:

1. Intensity $I(x)$: The interference pattern generates photo-excited carriers. [Graph showing interference pattern — sinusoidal intensity variation with $x$]

2. Charge distribution $S(x)$: Free carriers migrate to dark regions, creating a spatial charge distribution. [Graph showing charge distribution with $+$ and $-$ regions]

3. Index variation $\Delta n(x)$: The linear electrooptic effect produces a volume phase grating proportional to the electric field. [Graph showing refractive index variation]

Holographic Association and Memory

Association

[Diagram showing associative memory network: \text{-} Input $(1,2,3,4)$ maps to output $(a,b,c,d)$ \text{-} Input $(26,25,24,23)$ maps to output $(10,5,2,6)$ \text{-} Input $(x,\beta,\gamma,\delta)$ maps to output $(z,\gamma,x,\omega)$]

Given $(1,2,3,4)$, retrieve $(a,b,c,d)$.

Given $(z,\gamma,x,\omega)$, retrieve $(26,25,24,23)$.

\text{-}–

[Diagram showing two beams A and B interfering]

The hologram transmittance has four terms: $$

t \propto \underbrace{|A|^2}{\text{I}} + \underbrace{|B|^2}{\text{II}} + \underbrace{AB^}_{\text{III}} + \underbrace{BA^}_{\text{IV}}

$$ \text{-} Illuminate with $B$: Term III yields $|B|^2 A$, reproducing the $A$ wavefront (with flat phase). \text{-} Illuminate with $A$: Term IV yields $|A|^2 B$, reproducing the $B$ wavefront (with flat phase).

\text{-}–

Partial data retrieval: A small part of a hologram can reproduce the entire image. $$

(1,2,3,?) \longrightarrow (a,b,c,d) + \text{noise}$$

An incomplete reference wave can regenerate a complete object wave with additional noise. This is a possible model for biological memory.