400 C H A P T E R 6: Application to Control and Communications
the filter is analog by including an ’s’ as one of the arguments. Once the coefficients of the filter
are obtained, we could then either solve the differential equation from these coefficients or use the
Fourier transform, which we choose to do. Symbolic MATLAB is thus used to compute the Fourier
transform of the inputX(), and after generating the frequency response functionH(j)from the
filter coefficients, we multiply these two to getY(), which is inversely transformed to obtainy(t).
To obtainH(j)symbolically we multiply the coefficients of the numerator and denominator
obtained frombutterby variables(j)nwherencorresponds to the order of the coefficient in
the numerator or the denominator, and then add them. The poles of the designed filter and its
magnitude response are shown in Figure 6.23, as well as the inputx(t)and the outputy(t). The
following script was used for the filter design and the filtering of the given signal.− 6 − 4 − (^202)
− 5
− 4
− 3
− 2
− 1
0
1
2
3
4
5
jΩ
0 10 20 30
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
|H
(j
Ω
)|
σ Ω
0 5 10 15 20
− 5
0
5
t
x(
t)
0 5 10 15 20
− 5
0
5
t
y(
t)
cos(10t)−2cos(5t)+4sin(20t)
FIGURE 6.23
Filtering of an analog signalx(t)using a low-pass Butterworth filter. Notice that the output of the filter is
approximately the sinusoid of 5 rad/sec inx(t), as the other two components have been attenuated.
%%%%%%%%%%%%%%%%%%%
% Example 6.8 -- Filtering with Butterworth filter
%%%%%%%%%%%%%%%%%%%
clear all; clf
syms t w
x = cos(10∗t)− 2 ∗cos(5∗t) + 4∗sin(20∗t); % input signal
X = fourier(x);
N = 3; Whp = 5; % filter parameters
[b, a] = butter(N, Whp, ’s’); % filter design
W = 0:0.01:30; Hm = abs(freqs(b, a, W)); % magnitude response in W
% filter output
n = N:−1:0; U = ( j∗w).ˆn
num = b−conj(U’); den = a−conj(U’);