Exercise 1 1a) Try help kaiserord to get started. Beware that the ...

Johannes J. Struijk, Signalbehandling, 2004, Lecture 8
Exercise 1
1a)
Try help kaiserord to get started.
Beware that the Kaiser window method only gives an approximate solution. You may
have to increase N by 1 or 2 if the specifications are not perfectly met.
The MatLab code:
F = [20 30]; % First frequency band (pass) is 0-20, second band (stop) is 30-50
A = [1 0]; % defining the bands as pass and stop respectively
DEV = [0.11 0.01] % -1 dB and –40 dB, the FIR filter will use the smallest of the
two for both bands! This means that the pass band will be over dimensioned.
Fs = 100; % sample frequency
[N,wn,beta,type] = kaiserord(F,A,DEV,Fs);
b = fir1(N,wn,type,kaiser(N+1,beta),’noscale’);
1b)
freqz(b,1);
See figure below.
Zooming in at ws=30/50=0.6 shows that the attenuation is a little more than 40 dB.
Zooming in at wp=20/50=0.4 shows that the attenuation is much less than the allowed
0.11.
The filter thus easily meets the requirements.
50
Magnitude (dB)
0
-50
-100
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency (×π rad/sample)
0
Phase (degrees)
-500
-1000
-1500
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency (×π rad/sample)
You can use the filter to filter the signal ekg by:
y = filter(b,1,ekg);
1c)
The type of filter can easily be obtained by looking at the impulse response:
impz(b,1);

Johannes J. Struijk, Signalbehandling, 2004, Lecture 8
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Johannes J. Struijk, Signalbehandling, 2004, Lecture 8
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The impulse response has even symmetry with 0

Johannes J. Struijk, Signalbehandling, 2004, Lecture 8
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3b)
Parks-McClellan
Kaiser
Filter order M = 14 M = 24
Stopband ripple equiripple not equiripple
Passband ripple equiripple not equiripple, much better
than the specifications
Remez (Parks-McClellan) thus meets the specifications with a much lower filter order
than Kaiser.