Fourier Xform of Aperiodic Functions

Contents

Intro

This page will present the calculation of the forward and inverse Fourier Transform of a few functions, just to demonstrate the process using the analysis and synthesis functions. When at all possible, integration should be avoided. The section "Use of Tables" will describe how to use table lookups, and some straightforward manipulations, to calculate Fourier Transforms without integration.

Some Functions

This section briefly (re)introduces several functions that will be widely used in the ensuing discussion. If you are familiar with the functions, feel free to skip this section.

The Impulse Function, δ(x)

The impulse function (described in more detail elsewhere) is equal to zero everywhere but at x=0. The area of the impulse function is 1.

The Unit Pulse Function, Π(x)

The unit pulse function, Π(t), is a pulse of width=1, and height=1 centered horizontally about the origin.

It has a value of 0 if |t|>½ and a value of 1 if |t|≤½. Mathematically it can be defined as

$$\Pi (t) = \left\{ {\matrix{ {0,\quad \left| t \right| > {1 \over 2}} \cr
{1,\quad \left| t \right| \le {1 \over 2}} \cr } } \right.$$
Aside, an alternate definition of the pulse.

There are some mathematical subtleties that can make it preferable to define the value of the pulse function at the discontinuity to be at a point exactly halfway across the discontinuity,

$$\Pi (t) = \left\{ {\matrix{ {0,\quad \left| t \right| > {1 \over 2}} \cr
{{1 \over 2},\quad \left| t \right| = {1 \over 2}} \cr
{1,\quad |t| < {1 \over 2}} \cr } } \right.$$

but this is unimportant to these discussions.

The Scaled Pulse Function

We can easily scale the pulse function to have a height of A by multiplying by A. We can scale the width of the pulse to a value of Xp by dividing the value of x by Xp (so that the discontinuity occurs when x=±Xp/2).

The Unit Triangle Function, Λ(x)

The triangle function is zero except between ±1.

Later it will be useful to describe the unit triangle function as the convolution of two unit pulse functions, Λ(t)=Π(t)*Π(t).

The Scaled Triangle Function

As with the pulse function, we can scale the triangle pulse in width and height.

The Sinc Function, sinc(x)

The sinc() function (pronounced like "sink") comes up often in Fourier Analysis.

$${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( {\pi x} \right)} \over {\pi x}}$$

The sinc function has several important features:

The width and height of the sinc() function can be scaled in the same way as the pulse and triangle.

Some Fourier Transforms (by integration)

x(t)=δ(t), find X(ω)

$$\eqalign{
& x(t) = \delta (t) \cr
& X(\omega ) = \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt} = \int\limits_{ - \infty }^{ + \infty } {\delta (t){e^{ - j\omega t}}dt} = {e^{ - j\omega 0}} = 1 \cr} $$

Thus, the Fourier Transform of an impulse is a constant equal to 1, independent of frequency. Note that the derivation used the sifting property of the impulse to eliminate the integral.

x(t)=Π(t/Tp), find X(ω)

$$\eqalign{
x(t) = \Pi \left( {{t \over {{T_p}}}} \right) \cr
X(\omega ) = \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt } = \int\limits_{ - \infty }^{ + \infty } {\Pi \left( {{t \over {{T_p}}}} \right){e^{ - j\omega t}}dt } \cr} $$

but Π(t) is equal to 0 everywhere except between ±½Tp where it is equal to 1, so we change the limits of integrations and remove the pulse function.

$$X(\omega ) = \int\limits_{ - {1 \over 2}{T_p}}^{ + {1 \over 2}{T_p}} {{e^{ - j\omega t}}dt } = - \left. {{1 \over {j\omega }}{e^{ - j\omega t}}} \right|_{ - {1 \over 2}{T_p}}^{ + {1 \over 2}{T_p}} = {1 \over {j\omega }}\left( {{e^{ - j\omega {1 \over 2}{T_p}}} - {e^{ + j\omega {1 \over 2}{T_p}}}} \right)$$

This is simplified by using Euler's identity for the sine function (2jsin(x)=ejx-e-jx), and the fact that -sin(x)=sin(-x).

$$X(\omega ) = {1 \over {j\omega }}\left( {2j\sin \left( {{{\omega {T_p}} \over 2}} \right)} \right) = {2 \over \omega }\sin \left( {{{\omega {T_p}} \over 2}} \right)$$

This can be expressed in terms of the sinc() function by judiciously multiplying by π/π and Tp/Tp and rearranging:

$$X(\omega ) = {\pi \over \pi }{{{T_p}} \over {{T_p}}}{2 \over \omega }\sin \left( {{\pi \over \pi }{{\omega {T_p}} \over 2}} \right) = {T_p}{{\sin \left( {\pi {{\omega {T_p}} \over {2\pi }}} \right)} \over {\pi {{\omega {T_p}} \over {2\pi }}}} = {T_p}{\mathop{\rm sinc}\nolimits} \left( {{{\omega {T_p}} \over {2\pi }}} \right)$$

Note that the since function has amplitude Tp and has a zero at when ωTp/(2π) is an integer (i.e.,when ω is an integer multiple of 2π/Tp), except at ω=0.

If we vary the width of the pulse function, note that the width (and height) of the Fourier Transform also varies. As the pulse function becomes narrower (red→blue→yellow) the width of the Fourier Transform (sinc()) becomes broader and lower.

Aside: Uncertainty Principle (Π/sinc)

Take the width of the rectangular pulse in time to be ΔT=Tp, and the width of the sinc() function to be the distance between zero crossings near the origin, Δω=4π/Tp. The product of the widths is constant.

ΔT·Δω=4π

This is a form of the "uncertainty principle" of the Fourier Transform. It tells us that as a function becomes more localized in time (ΔT decreases) it becomes less localized in frequency (Δω increases), and vice-versa.

An extreme example of this is the impulse, δ(t), that is extremely localized (it is non-zero at only one instant of time). Its Fourier Transform is equal to 1; i.e., it is spread out uniformly in frequency.

Aside: Uncertainty Principle (Gaussian)

Though not proven here, it is well known that the Fourier Transform of a Gaussian function in time

$$x(t) = {1 \over {\sigma \sqrt {2\pi } }}{e^{ - {{{t^2}} \over {2{\sigma ^2}}}}}$$

is a Gaussian function in frequency

$$X(\omega ) = {e^{ - {{{\sigma ^2}{\omega ^2}} \over 2}}}$$

If we take the width of x(t) to be the variance, ΔT=σ2, then the width of X(ω) is Δω=1/σ2. The product of the widths in this case is simply equal to ΔT·Δω=1. As one Gaussian gets broader, the other gets narrower. As σ→∞, x(t) becomes infinitely narrow (i.e., an impulse) and X(ω) becomes infintiely broad (a constant).

X(ω)=Π(ω/Ωp), find x(t)

This derivation (and result) is similar to the previous one. This is not surprising since the equations for forward and inverse transform are so similar. This similarity is formalized as a property later.

$$\eqalign{
X(\omega ) &= \Pi \left( {{\omega \over {{\Omega_p}}}} \right) \cr
x(t) &= {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {X(\omega ){e^{ + j\omega t}}d\omega } = {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {\Pi \left( {{\omega \over {{\Omega_p}}}} \right){e^{ + j\omega t}}d\omega } \cr
&= {1 \over {2\pi }}\int\limits_{ - {1 \over 2}{\Omega_p}}^{ + {1 \over 2}{\Omega_p}} {{e^{ + j\omega t}}d\omega } = \left. {{1 \over {2\pi jt}}{e^{ + j\omega t}}} \right|_{ - {1 \over 2}{\Omega_p}}^{ + {1 \over 2}{\Omega_p}} = {1 \over {2\pi jt}}\left( {{e^{ + {{j{\Omega_p}t} \over 2}}} - {e^{ - {{j{\Omega_p}t} \over 2}}}} \right) \cr
&= {1 \over {2\pi jt}}\left( {2j\sin \left( {{{{\Omega_p}t} \over 2}} \right)} \right) = {2 \over {2\pi t}}\sin \left( {{{{\Omega_p}t} \over 2}} \right) \cr
&= {\pi \over \pi }{{{\Omega_p}} \over {{\Omega_p}}}{2 \over {2\pi t}}\sin \left( {{\pi \over \pi }{{{\Omega_p}t} \over 2}} \right) = {{{\Omega_p}} \over {2\pi }}{{\sin \left( {\pi {{{\Omega_p}t} \over {2\pi }}} \right)} \over {\pi {{{\Omega_p}t} \over {2\pi }}}} = {{{\Omega_p}} \over {2\pi }}{\mathop{\rm sinc}\nolimits} \left( {{{{\Omega_p}t} \over {2\pi }}} \right) \cr} $$

If you compare these results with the previous you'll notice that the shapes of the signal and its transform have been switched (along with some scaling of the horizontal and vertical axes).

x(t)=Λ(t), find X(ω)

This derivation is harder than the previous ones because the integral is harder (integration by parts, or a table lookup is necessary) and two integrals are required, since the Λ(t) function is defined as two piecewise linear functions. We will find later that this Fourier Transform can be easily calculated using a table of properties.

$$X(\omega ) = \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}dt} = \int\limits_{ - \infty }^{ + \infty } {\Lambda (t){e^{ - j\omega t}}dt} $$

We start by define Λ(t) as a piecewise function and rewrite the integral accordingly (note the < and ≤ signs are interchangeable for our purposes because the function is continuous). The process is quite lengthy, it is presented here for completeness, and to show (later) that other methods using Fourier Transform properties are better (i.e., easier). Feel free to skip to the final result.

$$\eqalign{
\Lambda (t) &= \left\{ {\matrix{
{1 + t,\quad - 1 \le t < 0} \cr
{1 - t,\quad 0 \le t \le + 1} \cr
{0,\quad \quad otherwise} \cr } } \right. \cr
X(\omega ) &= \int\limits_{ - 1}^{ + 1} {\Lambda (t){e^{ - j\omega t}}dt} = \int\limits_{ - 1}^0 {\left( {1 + t} \right){e^{ - j\omega t}}dt} + \int\limits_0^{ + 1} {\left( {1 - t} \right){e^{ - j\omega t}}dt} \cr
&= \int\limits_{ - 1}^{ + 1} {{e^{ - j\omega t}}dt} + \int\limits_{ - 1}^0 {t{e^{ - j\omega t}}dt} - \int\limits_0^{ + 1} {t{e^{ - j\omega t}}dt} \quad \quad {\rm{(use\ table\ lookup\ for\ last\ two\ integrals)}} \cr
&= \left. { - {1 \over {j\omega }}{e^{ - j\omega t}}} \right|_{ - 1}^{ + 1} + \left. {{{{{\rm{e}}^{ - j\omega t{\kern 1pt} }}{\kern 1pt} \left( {1 + j{\mkern 1mu} {\kern 1pt} \omega {\mkern 1mu} {\kern 1pt} {\rm{t}}} \right)} \over {{\omega ^2}}}} \right|_{ - 1}^0 - \left. {{{{{\rm{e}}^{ - j\omega t{\kern 1pt} }}{\kern 1pt} \left( {1 + j{\mkern 1mu} {\kern 1pt} \omega {\mkern 1mu} {\kern 1pt} {\rm{t}}} \right)} \over {{\omega ^2}}}} \right|_0^{ + 1} \cr
&= - {1 \over {j\omega }}\left( {{e^{ - j\omega ( + 1)}} - {e^{ - j\omega ( - 1)}}} \right) + {{{\kern 1pt} {{\rm{e}}^{ - j\omega 0{\kern 1pt} }}{\kern 1pt} \left( {1 + j{\mkern 1mu} {\kern 1pt} \omega {\mkern 1mu} {\kern 1pt} {\rm{0}}} \right) - \left( {{{\rm{e}}^{ - j\omega ( - 1){\kern 1pt} }}{\kern 1pt} \left( {1 + j{\mkern 1mu} {\kern 1pt} \omega ( - 1){\mkern 1mu} {\kern 1pt} } \right)} \right)} \over {{\omega ^2}}} - \cdots\cr
&\quad \quad \quad \quad \quad \quad {{{\kern 1pt} \left( {{{\rm{e}}^{ - j\omega ( + 1){\kern 1pt} }}{\kern 1pt} \left( {1 + j{\mkern 1mu} {\kern 1pt} \omega ( + 1){\mkern 1mu} {\kern 1pt} } \right)} \right) - {{\rm{e}}^{ - j\omega t{\kern 1pt} }}{\kern 1pt} \left( {1 + j{\mkern 1mu} {\kern 1pt} \omega {\mkern 1mu} {\kern 1pt} {\rm{t}}} \right)} \over {{\omega ^2}}} \cr
&= - {1 \over {j\omega }}\left( {{e^{ - j\omega }} - {e^{ + j\omega }}} \right) + {{{\kern 1pt} 1 - \left( {{{\rm{e}}^{ + j\omega {\kern 1pt} }}{\kern 1pt} \left( {1 - j{\mkern 1mu} {\kern 1pt} \omega {\mkern 1mu} {\kern 1pt} } \right)} \right)} \over {{\omega ^2}}} - {{{\kern 1pt} \left( {{{\rm{e}}^{ - j\omega {\kern 1pt} }}{\kern 1pt} \left( {1 + j{\mkern 1mu} {\kern 1pt} \omega {\mkern 1mu} {\kern 1pt} } \right)} \right) - 1} \over {{\omega ^2}}} \cr} $$

We can simplify this by expanding all the terms and bringing j to the numerator. All of the terms with ω in the denominator cancel, leaving only those with ω2.

$$\eqalign{
X(\omega ) &= j{{{e^{ - j\omega }} - {e^{ + j\omega }}} \over \omega } + {{{\kern 1pt} 2 - {{\rm{e}}^{ + j\omega {\kern 1pt} }} + j\omega {{\rm{e}}^{ + j\omega {\kern 1pt} }} - {{\rm{e}}^{ - j\omega {\kern 1pt} }} - j{\mkern 1mu} {\kern 1pt} \omega {{\rm{e}}^{ - j\omega {\kern 1pt} }}} \over {{\omega ^2}}}\quad \quad {\rm{(collect\ terms\ with\ the\ same\ denominator)}} \cr
&= {{j{e^{ - j\omega }} - j{e^{ + j\omega }} + j\omega {{\rm{e}}^{ + j\omega {\kern 1pt} }} - j{\mkern 1mu} {\kern 1pt} \omega {{\rm{e}}^{ - j\omega {\kern 1pt} }}} \over \omega } + {{2 - {{\rm{e}}^{j\omega {\kern 1pt} }} - {{\rm{e}}^{ - j\omega {\kern 1pt} }}} \over {{\omega ^2}}}\quad \quad {\rm{(first\ term\ goes\ to\ zero)}} \cr
&= {{2 - {{\rm{e}}^{j\omega {\kern 1pt} }} - {{\rm{e}}^{ - j\omega {\kern 1pt} }}} \over {{\omega ^2}}} = {{2 - \left( {{{\rm{e}}^{j\omega {\kern 1pt} }} + {{\rm{e}}^{ - j\omega {\kern 1pt} }}} \right)} \over {{\omega ^2}}} = {{2 - 2\cos \left( \omega \right)} \over {{\omega ^2}}}\quad \quad {\rm{(Euler's\ Identity\ for\ cosine)}} \cr
&= 2{{1 - \cos \left( \omega \right)} \over {{\omega ^2}}} = 4 {{{\rm{si}}{{\rm{n}}^2}\left( {{\omega \over 2}} \right)} \over {{\omega ^2}}}\quad \quad \left( {{\rm{trig\ identity, 1 - cos(}}\omega {\rm{) = 2si}}{{\rm{n}}^2}\left( {{\omega \over 2}} \right)} \right) \cr} $$

Finally, this can be expressed as a quite simple result in terms of sinc() functions by rearranging the result.

$$X(\omega ) = 4{{{\rm{si}}{{\rm{n}}^2}\left( {{\omega \over 2}} \right)} \over {{\omega ^2}}} = 4{{{\rm{si}}{{\rm{n}}^2}\left( {\pi \left( {{\omega \over {2\pi }}} \right)} \right)} \over {4{{\left( {\pi \left( {{\omega \over {2\pi }}} \right)} \right)}^2}}} = {{\mathop{\rm sinc}\nolimits} ^2}\left( {{\omega \over {2\pi }}} \right)$$

The transform is always positive since it goes as sinc2() and goes to zero when ω is an integer multiple of 2π, except at the origin. Also, because of the sinc2(), the transform decays more rapidly to the left and right of the origin than did that of the rectangular pulse. Though not shown, as the triangle pulse becomes narrower in time, the transform grows broader in frequency, and lower in amplitude.

x(t)=A Single Sawtooth, find X(ω)

This derivation requires either integration by parts or a table lookup to perform the integral, these details are not given in the following.

$$\eqalign{
x(t) &= sawtooth = \left\{ {\matrix{
{0,\quad t < 0,\;t > 1} \cr
{t,\;\;\quad 0 \le t \le 1} \cr} } \right. \cr
X(\omega ) &= \int\limits_{ - \infty }^{ + \infty } {x(t){e^{ - j\omega t}}d\omega } = \int\limits_0^1 {t{e^{ - j\omega t}}dt} = (by\;table\;lookup) = \left. {{{{{\rm{e}}^{ - j\omega {\kern 1pt} t}}{\mkern 1mu} {\kern 1pt} \left( {j\omega {\mkern 1mu} {\kern 1pt} t + 1} \right)} \over {{\omega ^2}}}} \right|_0^1 \cr
&= {{{{\rm{e}}^{ - j\omega {\kern 1pt} }}{\mkern 1mu} {\kern 1pt} \left( {j\omega {\mkern 1mu} {\kern 1pt} + 1} \right) - 1} \over {{\omega ^2}}} \cr} $$

In this case no plot of the function and its transform is given. It is more difficult in this case because the transform is complex so two plots would be needed (typically magnitude and phase, or, less frequently, the real and imaginary parts).

Moving Ahead

So far all of the functions we have considered have been aperiodic. With a little cleverness we can also find the Fourier Transform of periodic functions.


References

© Copyright 2005 to 2015 Erik Cheever    This page may be freely used for educational purposes.

Erik Cheever       Department of Engineering         Swarthmore College