Fourier Transforms

Introduction

This is a new page (October of 2021) to collect helpful information about Fourier Transforms

Syntax

• This page uses the following definition of the sinc function: \(\begin{align*}\mbox{sinc}(x)&=\frac{\sin(\pi x)}{\pi x}\end{align*}\) Note that other references (such as the zyBook) may omit the $$\pi$$. Given the version above, sinc(0)=1 and sinc(n$$\pi$$)=0 for all integers $$n\neq 0$$. Also note that:\(\begin{align*}\sin(\theta)&=\theta\,\mbox{sinc}\left(\frac{\theta}{\pi}\right)\end{align*}\)and finally\(\begin{align*}\mbox{sinc}(x)&=\frac{\sin(\pi x)}{\pi x}=\frac{e^{j\pi x}-e^{-j\pi x}}{2j\pi x}\end{align*}\)

Useful Fourier Transforms

On the Sakai page in Resources, there is a folder called "Ref: Tables" with two files in it. The AllTablesHVV.pdf version is the most relevant to this semester (HVV stands for Haykin and van Veen, who wrote a textbook we previously use and whose notation is most similar to the zyBook). Pages 8 and 9 are related to Continuous Fourier Transforms.

Periodic Square Wave

The periodic square wave of height 1 in the table has a width of $$W=2T_1$$; re-writing the Fourier Series coefficients in terms of the width $$W$$ and the period $$T=\frac{2\pi}{\omega_0}$$gives:

\(\begin{align*}X[k]&=\frac{\sin(k\omega_0T_1)}{k\pi}= \frac{\sin\left(k\left(\frac{2\pi}{T}\right)\left(\frac{W}{2}\right)\right)}{k\pi}= \frac{\color{blue}{\sin\left(\frac{k\pi W}{T}\right)}}{k\pi}= \frac{\color{blue}{\frac{k\pi W}{T}\mbox{sinc}\left(\frac{kW}{T}\right)}}{k\pi}= \frac{W}{T}\mbox{sinc}\left(k\frac{W}{T}\right) \end{align*}\)

using the definition of sinc from above. This means the Fourier Transform would be:

\(\begin{align*}X(j\omega)=\sum_{k=-\infty}^{\infty}\frac{2\sin(k\omega_0T_1)}{k}\delta(\omega-k\omega_0)= \sum_{k=-\infty}^{\infty}2\pi\frac{W}{T}\mbox{sinc}\left(k\frac{W}{T}\right)\delta(\omega-k\omega_0)\end{align*}\)

Centered Rectangular Pulse

The pulse of height 1 in the table has width of $$W=2T_1$$; re-writing the Fourier Transform in terms of the width $$W$$ gives:

\(\begin{align*}X(j\omega)&=\frac{2\,\sin(\omega T_1)}{\omega}=\frac{2\,\color{blue}{\sin\left(\frac{\omega W}{2}\right)}}{\omega}= \frac{2}{\omega}\color{blue}{\frac{\omega\,W}{2}\mbox{sinc}\left(\frac{\omega\,W}{2\pi}\right)}= W\mbox{sinc}\left(\frac{\omega W}{2\pi}\right)\end{align*}\)

using the definition of sinc from above.

General Rectangular Pulse

The general rectangular pulse in the table is given in terms of a shifted centered rectangular pulse. The width is $$2T_1=W=b-a$$ and the new center is $$t_0=\frac{a+b}{2}$$. Writing the Fourier Transform first as given in the table and then re-writing the Fourier Transform based on the width formula above gives:

\(\begin{align*}X(j\omega)&=\frac{2\,\sin\left(\omega \left(\frac{b-a}{2}\right)\right)}{\omega}\,\exp\left(-j\omega\left(\frac{a+b}{2}\right)\right)= W\mbox{sinc}\left(\frac{\omega \left(b-a\right)}{2\pi}\right)\,\exp\left(-j\omega\left(\frac{a+b}{2}\right)\right) \end{align*}\)

Sawtooth Pulse

A sawtooth pulse is of duration $$D$$ and height $$H$$ is a signal that is 0 before time 0 and after time $$D$$ and follows a straight line $$x(t)=Ht/D$$ for $$0<t<D$$. There are a few ways to get the Fourier Transform for this signal - two of them are given below:

Using Integral Property

The sawtooth pulse can be viewed as the integral of a signal made up of a rectangular pulse of width $$D$$ and height $$H/D$$ and an impulse of height $$-H$$ that fires off at $$D$$. That is to say, given that a sawtooth pulse $$x(t)=\frac{H}{D}t\left(u(t)-u(t-D)\right)$$, its derivative is $$y(t)=\frac{dx(t)}{dt}=\frac{H}{D}\left(u(t)-u(t-D)\right)+\frac{H}{D}t\left(\delta(t)-\delta(t-D)\right)=\frac{H}{D}\left(u(t)-u(t-D)\right)-H\delta(t-D)$$. It is possible to find the Fourier Transform of $$y(t)$$ using the tables:

\(\begin{align*}Y(j\omega)&=\frac{H}{D}D\,\mbox{sinc}\left(\frac{\omega D}{2\pi}\right)\,\exp\left(-j\omega\left(\frac{D}{2}\right)\right)-H\exp\left(-j\omega\left(\frac{D}{2}\right)\right)=H\,\mbox{sinc}\left(\frac{\omega D}{2\pi}\right)\,\exp\left(-j\omega\left(\frac{D}{2}\right)\right)-H\exp\left(-j\omega\left(\frac{D}{2}\right)\right)\end{align*}\)

From this we can see that:

\(\begin{align*}Y(j0)= H\mbox{sinc}\left(0\right)\,\exp\left(0\right)-H\exp\left(0\right)=0\end{align*}\)

and then we can use the integral property to get:

\(\begin{align*}X(j\omega)&=\frac{Y(j\omega)}{j\omega}+\pi Y(0)\delta(\omega)= \frac{1}{j\omega}\left( H\,\mbox{sinc}\left(\frac{\omega D}{2\pi}\right)\,\exp\left(-j\omega\left(\frac{D}{2}\right)\right)-H\exp\left(-j\omega\left(\frac{D}{2}\right)\right) \right)\end{align*}\)

This may look different from the version obtained above using the integral, but we can use the exponential form of the sinc function above to show both versions are the same:

\(\begin{align*} X(j\omega)&=\frac{1}{j\omega}\left( H\,\color{blue}{\mbox{sinc}\left(\frac{\omega D}{2\pi}\right)}\,\exp\left(-j\omega\left(\frac{D}{2}\right)\right)-H\exp\left(-j\omega\left(\frac{D}{2}\right)\right)\right) \\ ~&=\frac{1}{j\omega}\left( H\, \color{blue}{\left( \frac{\exp\left(j\omega\left(\frac{D}{2}\right)\right)-\exp\left(-j\omega\left(\frac{D}{2}\right)\right)}{j\omega D} \right)}\,\exp\left(-j\omega\left(\frac{D}{2}\right)\right)-H\exp\left(-j\omega\left(\frac{D}{2}\right)\right)\right)\\ ~&=\frac{H}{(j\omega)^2D}\left(1-e^{-j\omega D}-j\omega De^{-j\omega D}\right) \end{align*}\)