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:\(\begin{align*}\sin(\theta)&=\theta\,\mbox{sinc}\left(\frac{\theta}{\pi}\right)\end{align*}\)

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:

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