0.6 Lecture 7:continuous time fourier series for periodic signals  (Page 2/3)

The area of each period of sN(t) is simply

$\text{Area}={\int }_{-T/2}^{T/2}{S}_N(t)\text{dt}={\int }_{-T/2}^{T/2}\frac{1}{T}\sum _{n=-N}^N{e}^{\mathrm{j2\pi }\text{nt}/T}\text{dt}=\sum _{n=-N}^N\frac{1}{T}{\int }_{-T/2}^{T/2}{e}^{\mathrm{j2\pi }\text{nt}/T}\text{dt}$

The integral is zero except when n = 0 when it equals T so that Area = 1. Thus, each lobe of sN(t) becomes: tall, height is sN(nT) = (2N + 1)/T ; narrow, width is 2T/(2N + 1); and its area is 1. Thus, the partial sum approaches an infinite impulse train of unit area,

$s(t)=\sum _{n=-\infty }^{\infty }\delta (t-\text{nT})=\frac{1}{T}\sum _{n=-\infty }^{\infty }{e}^{\mathrm{j2\pi }\text{nt}/T}$

2/ Fourier transform of a periodic impulse train

We have two expressions for a periodic impulse train,

$s(t)=\sum _{n=-\infty }^{\infty }\delta (t-\text{nT})=\frac{1}{T}\sum _{n=-\infty }^{\infty }{e}^{\mathrm{j2\pi }\text{nt}/T}$

The Fourier transform of each expression is

$S(f)=\sum _{n=-\infty }^{\infty }{e}^{-\mathrm{j2\pi }\text{nTf}}=\frac{1}{T}\sum _{n=-\infty }^{\infty }\delta (f-\frac{n}{T})$

Therefore, the Fourier transform of a periodic impulse train in time is a periodic impulse train in frequency.

3/ Relation of Fourier series spectrum to Fourier transform of a periodic impulse train

Therefore, the Fourier transform of the periodic impulse train has an impulse at the frequency of each Fourier series component and the area of the impulse equals the Fourier series coefficient.

III. FOURIER TRANSFORM OF AN ARBITRARY PERIODIC FUNCTION

1/ Representation of a periodic function

An arbitrary periodic function can be generated by convolving a pulse, xT (t), that represents one period of the periodic function with a periodic impulse train, $s(t)={\sum }_n\delta (t-\text{nT})$

$x(t)=x_T(t)\ast s(t)=x_T(t)\ast \sum _{n=-\infty }^{\infty }\delta (t-\text{nT})=\sum _{n=-\infty }^{\infty }{x}_T(t-\text{nT})$

2/ Fourier transform of a periodic function

The Fourier transform of the periodic function is

$x(t)=x_T(t)\ast s(t)\stackrel{F}{\iff }X(f)=X_T(f)\times S(f)$

$X(f)=X_T(f)\times \frac{1}{T}\sum _{n=-\infty }^{\infty }\delta (f-\frac{n}{T})=\sum _{n=-\infty }^{\infty }(\frac{X_T(\frac{n}{T})}{T})\delta (f-\frac{n}{T})$

An important conclusion is that the Fourier transform of a periodic function consists of impulses in frequency at multiples of the fundamental frequency. Thus, periodic continuous time functions can be represented by a countably infinite number of complex exponentials.

3/ Fourier series coefficients

The Fourier transform of the periodic function is

$Xf=\sum _{n=-\infty }^{\infty }(\frac{X{}_T(\frac{n}{T})}{T})\delta (f-\frac{n}{T})=\sum _{n=-\infty }^{\infty }\frac{1}{2}(\frac{\text{sin}\mathrm{n\pi }/2}{\mathrm{n\pi }/2})\delta (f-\frac{n}{T})$

Recall that

$X{}_T(f)={\int }_{-T/2}^{T/2}{X}_T(t)e^{-\mathrm{j2\pi }\text{ft}}\text{dt}$

Therefore,

$\frac{X_T(\frac{n}{T})}{T}=\frac{1}{T}{\int }_{-T/2}^{T/2}{X}_T(t)e^{-\mathrm{j2\pi }\text{ft}}\text{dt}=X[n]$

Therefore, for an arbitrary periodic continuous time function, the Fourier transform consists of impulses (located at the harmonic frequencies) whose areas are the Fourier series coefficients.

4/ Fourier series of a square wave — generation of the square wave

We will find the Fourier series of a square wave by finding the Fourier transform of one period.

$x(t)=x_T(t)\ast s(t)$

5/ Fourier series of a square wave — Fourier transform of one period of the square wave

$x_T(t)\stackrel{F}{\iff }{X}_T(f)$

6/ Fourier series of a square wave — Fourier transform of square wave

To obtain the Fourier transform of the square wave, we take the Fourier transform of one period of the square wave and multiply it by the Fourier transform of the periodic impulse train.

$X(f)=X_T(f)\times S(f)=\frac{T}{2}(\frac{\text{sin}\pi \text{fT}/2}{\pi \text{fT}/2})\times \frac{1}{T}\sum _{n=-\infty }^{\infty }\delta (f-\frac{n}{T})=\sum _{n=-\infty }^{\infty }\frac{1}{2}(\frac{\text{sin}\mathrm{n\pi }/2}{\mathrm{n\pi }/2})\delta (f-\frac{n}{T})$

Two-minute miniquiz problem

Problem 18-1 — Fourier series of the square wave

a) Determine the Fourier series coefficients of the square wave.

b) From the Fourier series coefficients determine the average value of the square wave.