PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

DTFT, DFT, ZT, and FFT 555

>> step _ imp

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*********************** R E S U L T S *************************

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The first 11 coefficients of the step response

using the simulation of the difference equation are :

a =

2.0000 3.0000 3.5000 3.7500 3.8750 3.9375 3.9688

3.9844 3.9922 3.9961 3.9980

using the command dstep are :

b =

2.0000 3.0000 3.5000 3.7500 3.8750 3.9375 3.9688

3.9844 3.9922 3.9961 3.9980

The first 11 coefficients of the impulse response

using the simulation of the difference equation are :

c =

2.0000 1.0000 0.5000 0.2500 0.1250 0.0625 0.0313

0.0156 0.0078 0.0039 0.0020

using the command impz are:

d =

2.0000 1.0000 0.5000 0.2500 0.1250 0.0625 0.0313

0.0156 0.0078 0.0039 0.0020

The cumulative step error [a-b] is :

ans =

0

The cumulative impulse error [c-d] is :

ans =

0

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Example 5.19

The objective of this example is to illustrate the computational effi ciency of evaluating

the fft when the sequence length is a power of 2.

Create the script fi les time_analysis_ 512 and time_analysis_ 513, that returns the time

statistics by evaluating the fft of the signal f(t) = cos(2π850t), twice using 512 and 513

samples, respectively, with a sampling rate of Ts = 1/8000 in both.

MATLAB Solution

% Script file: time _ analysis _ 512

% statistic analysis using 512 points; a power of 2

clear;

tstart = clock; tic;

ta = cputime;

fs = 8000;Ts=1/fs;N=512;

t = linspace(0,Ts*511,512);signal _ 512 = cos(2*pi*850.*t);

DFT _ 512 = fft(signal _ 512,512);

estimea = etime(clock,tstart);

tictoca = toc;

t1= cputime-ta;

disp(‘********************* R E S U L T S **********************’)

disp(‘********** T I M E S T A T I S T I C S **********’)

disp(‘ f(n) consist of 512 samples ‘)

fprintf(‘est.time = %6.6f\n’,estimea);