# Fourier Xform of Periodic Functions

## Introduction

So far we have only considered the Fourier Transform, X(ω), of aperiodic signals, x(t), that are absolutely integrable so that

$$\int\limits_{ - \infty }^{ + \infty } {\left| {x(t)} \right|dt} < \infty$$

This is a necessary condition for X(ω) to be described by a function. However, this is quite limiting and excludes many functions of interest in engineering (e.g., a sine wave, or square wave...). However if we allow X(ω) to include generalized functions (in particular the impulse function, δ(ω) (described elsewhere)), we can overcome this obstacle.

Consider the Fourier domain function X(ω)=δ(ω), this results in

$$x(t){\rm{ }} = {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {X(\omega ){e^{ + j\omega t}}d\omega } = {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {\delta (\omega ){e^{ + j\omega t}}d\omega } = {1 \over {2\pi }}{e^{ + j0t}} = {1 \over {2\pi }}$$

So, X(ω)=δ(ω) and x(t)=1/2π form a Fourier Transform pair. Note:

• the sifting property of the impulse was used at the end of the derivation to get rid of the integral.
• x(t) is clearly not absolutely integrable in this case (this is not a problem because X(ω) is not a standard mathematical function..
• the "uncertainty principle" still applies - in this case X(ω) is very narrow (width=0) in frequency, and x(t) is very broad (width=∞) in time.

Now consider the Fourier domain function as a shfted impulse at ω=ω0, X(ω)=δ(ω-ω0). In this case we find (again using the sifting property)

$$x(t) = {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {X(\omega ){e^{ + j\omega t}}d\omega } = {1 \over {2\pi }}\int\limits_{ - \infty }^{ + \infty } {\delta (\omega - {\omega _0}){e^{ + j\omega t}}d\omega } = {1 \over {2\pi }}{e^{ + j{\omega _0}t}}$$

So, X(ω)=δ(ω-ω0) and $x(t) = {e^{j{\omega _0}t}}/2\pi$ form a Fourier Transform pair. This result will be used below to find the Fourier Transform of Sines, Cosines, and any periodic function that can be represented by a Fourier Series.

##### Key concept: Inverse Fourier Transform of Impulse in Frequency Domain

The complex exponential and the scaled and shifted impulse form a Fourier Transform pair

$${e^{j{\omega _0}t}}\buildrel F \over \longleftrightarrow 2\pi \delta (\omega - {\omega _0})$$

or (as derived above)

$${{{e^{j{\omega _0}t}}} \over {2\pi }}\buildrel F \over \longleftrightarrow \delta (\omega - {\omega _0})$$

(the letter F over the double arrow denotes a Fourier Transform pair)

## Fourier Transform of Sine and Cosine

Using the previous result we can easily find the Fourier Transform of a cosine using Euler's identities.

$$\displaylines{ x(t) = \cos ({\omega _0}t) = {{{e^{j{\omega _0}t}} + {e^{ - j{\omega _0}t}}} \over 2} = \pi {{{e^{j{\omega _0}t}} + {e^{ - j{\omega _0}t}}} \over {2 \pi}} \cr X(\omega ) = \pi \left( {\delta (\omega - {\omega _0}) + \delta (\omega + {\omega _0})} \right) \cr}$$

Similarly,

$$\displaylines{ x(t) = \sin ({\omega _0}t) = {{{e^{j{\omega _0}t}} - {e^{ - j{\omega _0}t}}} \over {2j}} = j\pi {{{e^{ - j{\omega _0}t}} - {e^{j{\omega _0}t}}} \over {2\pi }} \cr X(\omega ) = j\pi \left( {\delta (\omega + {\omega _0}) - \delta (\omega - {\omega _0})} \right) \cr}$$

The functions and their Fourier Transforms are shown below.

Note that the vertical axis of the Fourier Transform is imaginary for the sine function. Also note that the Fourier Transform of the sine function is imaginary and odd while that of the cosine is real and even.

##### Aside: Plotting δ(ω+ω0) and δ(ω-ω0)

Students sometimes have some confusion when asked to plot δ(ω+ω0) and δ(ω-ω0). As seen in the Fourier Transform of the sine function (above), δ(ω+ω0) gives an impulse that is shifted to the left by ω0, i.e., at ω=0 (Note it is not at ω=0 as some students expect; this is because the argument of the impulse function is zero when ω=0). Likewise the function δ(ω-ω0) yields an impulse shifted to the right by ω0, i.e., at ω=0.

##### Key concept: Fourier Transform of sine and cosine
$$\displaylines{ \sin ({\omega _0}t)\buildrel F \over \longleftrightarrow X(\omega ) = j\pi \left( {\delta (\omega + {\omega _0}) - \delta (\omega - {\omega _0})} \right) \cr \cos ({\omega _0}t)\buildrel F \over \longleftrightarrow X(\omega ) = \pi \left( {\delta (\omega + {\omega _0}) + \delta (\omega - {\omega _0})} \right) \cr}$$

## Fourier Transform of a Periodic Signal Described by a Fourier Series

Given a periodic function xT(t) and its Fourier Series representation (period=T, ω0=2π/T):

$${x_T}(t) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn{\omega _0}t}}}$$

we can use the fact that we know the Fourier Transform of the complex exponential

$${e^{jn{\omega _0}t}}\buildrel F \over \longleftrightarrow 2\pi \delta (\omega - n{\omega _0})$$

to find the Fourier Transform

$${X_T}\left( \omega \right) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}2\pi \delta (\omega - n{\omega _0})} = 2\pi \sum\limits_{n = - \infty }^{ + \infty } {{c_n}\delta (\omega - n{\omega _0})}$$

Note: this uses the fact the the Fourier Transform of the sum of functions is equal to the sum of the Fourier Transforms (i.e., the Transform is linear); this is proved at the top of the next page.

##### Example: Fourier Transform of Square Wave

Consider the periodic pulse function for the case when T=5 and Tp=2.

$${x_T}(t) = {\Pi _T}\left( {{t \over {{T_p}}}} \right) = {\Pi _5}\left( {{t \over 2}} \right) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn{\omega _0}t}}}$$

has Fourier Series Coefficients (derived here)

$${c_n} = {{{T_p}} \over T}{\rm{sinc}}\left( {{{n{T_p}} \over T}} \right) = 0.8{\rm{sinc}}\left( {0.8n} \right)$$

These are plotted vs. n, below (in this case the coefficients are all real numbers - in the general case they would be complex).

Using the result derived previously, the Fourier Transform of the function is

$$\displaylines{ {X_T}\left( \omega \right) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}2\pi \delta (\omega - n{\omega _0})} \cr = 2\pi \sum\limits_{n = - \infty }^{ + \infty } {0.8{\rm{sinc}}\left( {0.8n} \right)\delta (\omega - n{\omega _0})} = 1.6\pi \sum\limits_{n = - \infty }^{ + \infty } {{\rm{sinc}}\left( {0.8n} \right)\delta (\omega - n{\omega _0})} \cr}$$

The shape of the transform follows that of the Fourier Series coefficients, but it is now a function and ω and takes the form of a series of impulses at ω=n·ω0 (and amplitude 2πcn). Recall that the Fourier series is defined by discrete coefficients with index n (and amplitude cn), not a function of ω.

##### Key concept: Fourier Transform of Fourier Series Representation of xT(t)

If we write a periodic function

$${x_T}(t) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn{\omega _0}t}}}$$

then its Fourier Transform is

$${X_T}\left( \omega \right) = 2\pi \sum\limits_{n = - \infty }^{ + \infty } {{c_n}\delta (\omega - n{\omega _0})}$$

## Relationship between Fourier Series and Transform

The relationshop between Fourier Series and Transform is discussed in more detail later.

## Moving Forward

In the next section several properties of the Fourier Transform are derived. These properties will then be used to help derive Fourier Transforms without the need for integration.

References