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****************** Matlab Code ************
% For data sampled at 1000 Hz, design a lowpass filter with no more
than 1 dB of ripple in a passband from 0 to 10 Hz,
% and at least 40 dB of attenuation in the stopband. Find the
filter order and cutoff frequency.
Wp = 10/500; % 20pi corresponds to 10Hz
Ws = 50/500; % 100pi corresponds to 50Hz
clc
[n,Wn] = buttord(Wp,Ws,1,40)
%Specify the filter in terms of second-order sections and plot the
frequency response.
[z,p,k] = butter(n,Wn);
sys = zpk(z,p,k,1/500)
sos = zp2sos(z,p,k);
freqz(sos,512,1000)
title(sprintf('n = %d Butterworth Lowpass Filter',n))
%%%%% high pass filter
[z1,p1,k1] = butter(n,Wn,'high');
sys1 = zpk(z1,p1,k1,1/500)
sos1 = zp2sos(z1,p1,k1);
figure;
freqz(sos1,512,1000)
title(sprintf('n = %d Butterworth Highpass Filter',n))
************** End of Code ***************
Output:
Order = 4, Cutoff = 0.0319 rad/s

sys =
5.5214e-06 (z+1)^4
----------------------------------------------
(z^2 - 1.822z + 0.831) (z^2 - 1.917z + 0.9263)
Sample time: 0.002 seconds
Discrete-time zero/pole/gain model.
Transition band = 45Hz
High pass:

sys1 =
0.87736 (z-1)^4
----------------------------------------------
(z^2 - 1.822z + 0.831) (z^2 - 1.917z + 0.9263)
Sample time: 0.002 seconds
Discrete-time zero/pole/gain model.
*****************************************************************
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