DTFT, DFT, ZT, and FFT
555
>> step _ imp
****************************************************************
*********************** R E S U L T S *************************
****************************************************************
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
****************************************************************
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);
CRC_47760_CH005.indd 555
CRC_47760_CH005.indd 555
7/23/2008 1:47:43 PM
7/23/2008 1:47:43 PM