Module 4 - Summary

Summary Notes Module 4

• Any amplitude distribution can be expressed as a sum of plane waves, i.e., $\exp[j 2\pi (u x + v y)].$ Compared to the mathematical expression of a plane wave in Module 2, $\exp[j(k_x x + k_y y)],$ then $k_x = 2\pi u = \frac{2\pi}{\lambda} \sin \theta_x$ and $k_y = 2\pi v = \frac{2\pi}{\lambda} \sin \theta_y$. This means that the angle of the plane waves along the x- and y-direction can be estimated by $\theta_x = \sin^{-1}(\lambda u) \quad \text{and} \quad \theta_y = \sin^{-1}(\lambda v).$

• An imaging system must be a linear shift-invariant system. Consider that an imaging system is represented by a $\mathbf{S}$ operator.

A system is linear if the output is equal to the sum of the individual outputs.

A system is shift-invariant if the output of a displaced input signal is the same function but displaced by the same amount.

Only when the optical system is linear space-invariant (LSI), the output signal, $g(x, y)$, can be expressed as the two-dimensional (2D) convolution ($\otimes_2$) between the input signal, $f(x, y)$, and the impulse response of the system, $h(x, y)$:

$$g(x, y) = f(x, y) \otimes_2 h(x, y).$$

In the Fourier domain, the spectrum of the output signal, $G(u, v)$, is the product between the spectrum of the input signal, $F(u, v)$, and the optical transfer function, $H(u, v)$:

$$G(u, v) = F(u, v) H(u, v).$$ This means that the transfer function $H(u, v)$ is the 2D Fourier transform of the impulse response, $h(x, y)$: $$H(u, v) = \text{FT}[h(x, y)].$$

• Free-space propagator under the Fresnel approximation

$$h(x, y) = \frac{e^{j k z}}{j \lambda z} \exp\left[ j \frac{k}{2z}(x^2 + y^2) \right],$$ and $$H(u, v) = \frac{e^{j k z}}{j \lambda z} \exp\left[-j \pi \lambda z (u^2 + v^2)\right].$$

Condition of the Fresnel region:

$$N_F \frac{\theta_m^2}{4} \ll 1,$$ where $N_F = \frac{a^2}{\lambda z}$ is the Fresnel number and $\theta_m = \frac{a}{z}$ is the maximum aperture angle.

• Free-space propagator under the Fraunhofer approximation ($z \gg$)

$$g(x, y) = f(x, y) \otimes_2 h(x, y) = F \left( \frac{x}{\lambda z}, \frac{y}{\lambda z} \right) = F \left( \frac{x}{M}, \frac{y}{M} \right),$$ where $M$ is the magnification factor.

For $z \gg$, the amplitude distribution (also known as the diffraction pattern) of an object transmittance $f(x, y)$ is a scaled version of its Fourier Transform under Fraunhofer approximation.

Condition of the Fraunhofer region:

$$z \gg \frac{b_{\text{input}}^2}{\lambda} \quad \text{and} \quad z \gg \frac{a_{\text{output}}^2}{\lambda},$$ where $b_{\text{input}}$ is the maximum lateral extent in the input plane and $a_{\text{output}}$ is the maximum lateral extent in the output plane.

• Converging and diverging (i.e., optical) lenses make Fourier transforms.

Consider a converging lens. If an object with amplitude transmittance $t(x, y)$ is placed at the front focal plane of a lens (i.e., the distance between the object/input and lens planes is equal to the lens’ focal length, $f$), a scaled replica of the Fourier transform of the object transmittance is found at its back focal plane:

$$T \left( \frac{x}{\lambda f}, \frac{y}{\lambda f} \right).$$

• Calculation of the complex amplitude distribution of an arbitrary optical system with an input transmittance $t(x, y)$. The arbitrary optical system is represented by its ABCD matrix.

If $B = 0$, the complex amplitude distribution at the output plane is the product of a quadratic phase wavefront and a scaled replica of the object transmittance:

$$u(x, y) = e^{j k L_0 / A} \exp \left[ j \frac{k C}{2A} (x^2 + y^2) \right] t \left( \frac{x}{A}, \frac{y}{A} \right),$$ where $L_0$ is the geometrical distance between the input and output planes and $k = \frac{2\pi}{\lambda}$ is the wave number.

Else, if $B \neq 0$, the complex amplitude distribution at the output plane becomes:

$$u(x, y) = \frac{e^{j k L_0}}{j \lambda B} \exp \left[ j \frac{k D}{2B} (x^2 + y^2) \right] \int_{-\infty}^{\infty} t(x_0, y_0) \exp \left[ j \frac{k A}{2B} (x_0^2 + y_0^2) \right] \exp \left[ -j \frac{2\pi}{\lambda B} (x x_0 + y y_0) \right] \, dx_0 \, dy_0.$$

• 4f imaging system if both lenses have the same focal length

• 4f imaging system if both lenses have different focal lengths

• Spatial filtering in a 4f imaging system by inserting a pupil with transmittance $p(x, y)$ at the Fourier plane of a 4f system. The pupil filters out some object frequencies.

Thus, the complex amplitude at the image plane (i.e., back focal plane of the $L_2$ lens) is:

$$u(x, y) = \frac{1}{M^2} t \left( \frac{x}{M}, \frac{y}{M} \right) \otimes_2 P \left( \frac{x}{\lambda f_2}, \frac{y}{\lambda f_2} \right) = \frac{1}{M^2} t \left( \frac{x}{M}, \frac{y}{M} \right) \otimes_2 h(x, y),$$ where $M = - \frac{f_2}{f_1}$ is the lateral magnification, $\otimes_2$ is the 2D convolution operator, and $P(u, v)$ is the 2D Fourier transform of the pupil transmittance. Based on the equation, the impulse response of the 4f imaging system is a scaled replica of the pupil’s Fourier transform, $h(x, y) = P \left( \frac{x}{\lambda f_2}, \frac{y}{\lambda f_2} \right)$. Therefore, the transfer function is a scaled replica of the pupil’s transmittance: $$H(u, v) = \text{FT}[h(x, y)] = p \left( -\frac{\lambda f_2 u}{f_1}, -\frac{\lambda f_2 v}{f_1} \right).$$