PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

546 Practical MATLAB® Applications for Engineers

Example 5.14

Let the impulse response of an LTI system be given by hn n

n

()
.

05

0

15

∑ and the system

input sequence be fn n

n

()
 cos( /)

15 5 3

0

15

∑ 

.

Create the script fi le syst_anal that returns the following plots:

a. f(n) versus n and h(n) versus n, using the stem command (Figure 5.60)

b. g(n) = {IDFT of the products of DFT [f(n)] times DFT [h(n)]} versus n (Figure 5.61)

c. gnn(n) = convolution of h(n) with g(n) versus n (Figure 5.61)

d. Estimate the error plot (in the time domain) of the two approaches (parts b

and c), given by (Figure 5.62)

error(n) = [g(n) − gnn(n)] versus n

MATLAB Solution

% Script file: syst _ anal

n = 0:15;

fn = cos(5*pi*n/3);

hn = 0.5.^n; % the length of convolution is 16*2-1

DFT _ F = fft(fn,31);

DFT _ H = fft(hn,31);

gn = ifft(DFT _ F.*DFT _ H);

figure(1)

subplot(2,1,1)

stem(n,fn)

ylabel(‘Amplitude’);axis([0 15 -1.2 1.2]);

title(‘ [Input sequence f(n)] vs n ‘)

subplot(2,1,2)

stem(n,hn)

ylabel(‘Amplitude ‘);axis([0 15 -0.5 1.2]);

xlabel(‘time index n’); title(‘[Impulse response h(n)] vs n’)

figure(2) % output plots

subplot(2,1,1)

nn=0:1:30;

plot(nn,gn);title(‘[IDFT of the products of the DFTs ] vs n’);

xlabel(‘time index n ’); ylabel(‘Amplitude’);axis([0 25 -1.5 2]);

subplot(2,1,2)

gnn=conv(fn,hn);

plot(nn,gnn);axis([0 25 -1.5 2]);

xlabel(‘time index n’), ylabel(‘Amplitude’)

title(‘[Direct convolution in the time domain] vs. n’);

figure(3)

error = gn-gnn;

bar(nn,error);title(‘ Error =[part(b)-part(c)] vs. n’ );

xlabel(‘time index n’), ylabel(‘Magnitude’);

The script fi le syst_anal is executed, and the results are as follows:

>> syst _ anal