Fourier Transforms

Introduction

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

Syntax

• This page uses the following synthesis and analysis equations for the Fourier transform:

\( \begin{align*} x(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}dt & X(j\omega)&=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\end{align*}\)

• 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 Integration

The sawtooth pulse can be written as $$x(t)=\frac{H}{D}t\left(u(t)-u(t-D)\right)$$ and therefore its Fourier Transform may be found directly with:

\(\begin{align*} X(j\omega)&=\int_{-\infty}^{\infty}x(t)\,e^{-j\omega t}\,dt\\ ~&=\int_{-\infty}^{\infty}\frac{H}{D}t\left(u(t)-u(t-D)\right)\,e^{-j\omega t}\,dt\\ ~&=\int_{0}^{D}\frac{H}{D}t\,e^{-j\omega t}\,dt\\ ~&=\frac{H}{D}\left[-\frac{(j\omega t+1)e^{-j\omega t}}{(j\omega)^2}\right]_{0}^{D}\\ ~&=\frac{H}{D}\left( -\frac{(j\omega D+1)e^{-j\omega D}}{(j\omega)^2} +\frac{1}{(j\omega)^2}\right)\,\mbox{which can be rearranged in reverse order as}\\ ~&=\frac{H}{(j\omega)^2D}\left(1-e^{-j\omega D}-j\omega De^{-j\omega D}\right) \end{align*}\)

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

\(\begin{align*}y(t)&=\frac{dx(t)}{dt}\\~&=\frac{H}{D}\left(u(t)-u(t-D)\right)+\frac{H}{D}t\left(\delta(t)-\delta(t-D)\right)\\~&=\frac{H}{D}\left(u(t)-u(t-D)\right)-H\delta(t-D)\end{align*}\)

It is possible to find the Fourier Transform of $$y(t)$$ using the tables:

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

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:

\(\begin{align*} x(t)&=\int_{-\infty}^{t}y(\tau)\,d\tau\\X(j\omega)&=\frac{Y(j\omega)}{j\omega}+\pi Y(0)\delta(\omega)\\~&= \frac{1}{j\omega}\left( H\,\mbox{sinc}\left(\frac{\omega D}{2\pi}\right)\,\exp\left(-j\omega\left(\frac{D}{2}\right)\right)-H\exp\left(-j\omega D\right) \right)+0\end{align*}\)

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:

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

Finite Duration Accumulated Singularity Functions

If you have a signal of finite duration (that is, a finite stretch where it is nonzero) made up of singularity functions, there is a shortcut for finding the Fourier Transform. They key thing is that the function needs to always be 0 to the left of some time and to the right of some later time. For the singularity functions in the non-zero zone, you can use the integral property ignoring the $$\pi\,X(j0)\,\delta(\omega)$$ part. That is to say you can replace unit steps, ramps, quadratics, and cubics with $$1/(j\omega)$$, $$1/(j\omega)^2$$, $$1/(j\omega)^3$$, and $$1/(j\omega)^3$$, multiplied by the appropriate exponential for the time shift. But again, this only works if the signal is nonzero for a finite amount of time!

Centered Rectangular Pulse revisited

If $$x(t)=u\left(t+\frac{W}{2}\right)-u\left(t-\frac{W}{2}\right)$$, we can find the Fourier Transform as:

\(\begin{align*} x(t)&=u\left(t+\frac{W}{2}\right)-u\left(t-\frac{W}{2}\right)\\ X(j\omega)&=\frac{1}{j\omega}e^{j\omega W/2}-\frac{1}{j\omega}e^{-j\omega W/2},\mbox{ and using Euler's representation,}\\ ~&=\frac{1}{j\omega}(2j\sin(\omega W/2))=\frac{2\sin(\omega W/2)}{\omega}=W\mbox{sinc}\left(\frac{\omega W}{2\pi}\right)\end{align*}\)

General Rectangular Pulse revisited

If $$x(t)=u\left(t-a\right)-u\left(t-b\right)$$ with $$b>a$$, we can find the Fourier Transform as:

\(\begin{align*} x(t)&=u\left(t-a\right)-u\left(t-b\right)\\ X(j\omega)&=\frac{1}{j\omega}e^{-j\omega a}-\frac{1}{j\omega}e^{-j\omega b}=\frac{1}{j\omega}\left(e^{-j\omega a}-e^{-j\omega b}\right)\end{align*}\)

This version is perfectly acceptable for ECE 280; however, often you will see differences between exponentials with imaginary exponents written with trig functions. They way it works is to pull out an exponential with the average of the exponents:

\(\begin{align*} e^{-j\omega a}-e^{-j\omega b}&=e^{-j\omega\left(\frac{b+a}{2}\right)}\left(e^{j\omega\left(\frac{b-a}{2}\right)}-e^{-j\omega\left(\frac{b-a}{2}\right)}\right)\\ &=e^{-j\omega\left(\frac{b+a}{2}\right)}\left(2j\sin\left(\omega\left(\frac{b-a}{2}\right)\right)\right)\end{align*}\)

which means

\(\begin{align*} X(j\omega)&=\frac{1}{j\omega}\left(e^{-j\omega\left(\frac{b+a}{2}\right)}\left(2j\color{blue}{\sin\left(\omega\left(\frac{b-a}{2}\right)\right)}\right)\right)\\ ~&=\frac{1}{j\omega}\left(e^{-j\omega\left(\frac{b+a}{2}\right)}\left(2j\color{blue}{\omega\left(\frac{b-a}{2}\right)\mbox{sinc}\left(\omega\left(\frac{b-a}{2\pi}\right)\right)}\right)\right)\\ ~&=(b-a)\mbox{sinc}\left(\omega\left(\frac{b-a}{2\pi}\right)\right)\left(e^{-j\omega\left(\frac{b+a}{2}\right)}\right)\end{align*}\)

Given that $$W=b-a$$, this is exactly what we found above!

Sawtooth revisited

If $$x(t)=\frac{H}{D}t\left(u(t)-u(t-D)\right)$$, we can re-write it as an accumulation of value and slope changes with:

\(\begin{align*} x(t)&=\frac{H}{D}r(t)-\frac{H}{D}r(t-D)-Hu(t-D)\end{align*}\)

and then find

\(\begin{align*} X(j\omega)&=\frac{H}{D}\frac{1}{(j\omega)^2}-\frac{H}{D}\frac{1}{(j\omega)^2}e^{-j\omega D}-H\frac{1}{j\omega}e^{-j\omega D}\\ ~&=\frac{H}{(j\omega)^2D}\left(1-e^{-j\omega D}-j\omega De^{-j\omega D}\right)\end{align*}\)