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6.8 Frequency Response 137

of an input signal. An exact value of the magnitude can also be obtained by
simple interpolation of the magnitude,

1-interp1(f,magnitude,1/20)

ans =
0.4260
which is the expected ca. 43% reduction of the amplitude of a sine wave
with period 20. The sine wave with period 15 experiences an amplitude
reduction of

1-interp1(f,magnitude,1/15)
ans =
0.6768

i.e., around 68% similar to the value observed at the beginning. The fre-
quency response can be calculated for all kinds of filters. It is a valuable

tool to predict the effects of a filter on signals in general. The phase re-

sponse can also be calculated from the complex frequency response of the

filter (Fig. 6.4):
phase = 180*angle(h)/pi;

plot(f,phase)
xlabel('Frequency'), ylabel('Phase in degrees')
title('Phase')
The phase angle is plotted in degrees. We observe frequent 180° jumps in
this plot that are an artifact of the arctangent function inside the function
angle. We can unwrap the phase response to eliminate the 180° jumps us-
ing the function unwrap.

plot(f,unwrap(phase))
xlabel('Frequency'),ylabel('Phase in degrees')
title('Phase')

Since the filter has a linear phase response, no shifts of the frequency com-
ponents of the signal occur relative to each other. Therefore we would not
expect any distortions of the signal in the frequency domain. The phase shift

of the filter can be computed using
interp1(f,unwrap(phase),1/20) * 20/360
ans =
-5.0000