Introduction To Fourier Optics Third Edition Problem Solutions (TRENDING • FULL REVIEW)

This level of detail turns a simple answer into a pedagogical tool.

Problem Statement: Calculate the Fourier transform of the function $f(x) = \textrect(x/a)$ where $a > 0$.

Solution: Recall the definition of the rectangular function: $$ \textrect\left(\fracxa\right) = \begincases 1 & |x| < a/2 \ 0 & \textotherwise \endcases $$

The Fourier transform $\mathcalFf(x)$ is defined as $F(f_x) = \int_-\infty^\infty f(x) e^-j 2\pi f_x x dx$. This level of detail turns a simple answer

$$ F(f_x) = \int_-a/2^a/2 (1) e^-j 2\pi f_x x dx $$

Integrating: $$ F(f_x) = \left[ \frace^-j 2\pi f_x x-j 2\pi f_x \right]_-a/2^a/2 $$ $$ F(f_x) = \frac1-j 2\pi f_x \left( e^-j \pi f_x a - e^j \pi f_x a \right) $$

Using Euler's formula, $e^j\theta - e^-j\theta = 2j\sin(\theta)$: $$ F(f_x) = \frac2j \sin(\pi f_x a)j 2\pi f_x = \frac\sin(\pi f_x a)\pi f_x $$ The third edition includes new problems on sampled

Using the definition of the sinc function, $\textsinc(z) = \frac\sin(\pi z)\pi z$: $$ F(f_x) = a \cdot \textsinc(a f_x) $$

Key Insight: The width of the function in the space domain ($a$) is inversely proportional to the width of the spectrum in the frequency domain.

The third edition includes new problems on sampled apertures and digital Fourier transforms (Chapter 9). A common problem asks: “Using a 2D FFT, compute the Fraunhofer pattern of a hexagonal aperture. Compare with the analytical Fourier transform.” Without a carefully explained solution, a student might

A robust solution must address:

Without a carefully explained solution, a student might simply run fft2 in MATLAB and misinterpret the output.