I want to show the Magnitude response of three filters, as one response

So, I am to design three butter filters, a low pass, band pass, and a high pass, with the biggest wrinkle being that they must each have a different order (other wise, I'd just create a band stop filter with two stops). (Technically 7, 8, and 10, but I'll get to that later.) I have no problem with that, but I want to get the freq response as a single line that shows what happens when all three filters are inline with one another. (Right now I can just show how each one campares, which is mostly useless, from my perspective. Below is where I am so far. Thank you for the help.

clear;
freqSample = 10000;
freqCutoffLow = 100;
bandPassLower = 500;
bandPassUpper = 1000;
freqCutoffHigh = 2000;
%as = [1/(freqCutoff*2*pi),1];
nLow = 8; % orderOfFilter;
nBandPass = 8;
nHighPass = 10;

wNLow= freqCutoffLow / (freqSample/2); %normalizedCutOffFreq

[lowA,lowB] = butter(nLow, wNLow);

[passA,passB] = butter(nBandPass,[bandPassLower bandPassUpper]/(freqSample/2));


[highA, highB] = butter(nHighPass, freqCutoffHigh/(freqSample/2),'high');

fvt = fvtool(lowA, lowB, passA, passB, highA, highB);
legend(fvt,'butterLow', 'butterHigh', 'butterPass')

Answer 1

[Max Murphy]

Could you use freqz and take the product of the frequency responses, assuming the filters will be cascaded in a linear system?

H_low = freqz(lowA,lowB,512);
H_pass = freqz(passA,passB,512);
H_high = freqz(highA,highB,512);

freqz will return the frequency response at a specified number of points, but since you have already determined your filter coefficients, the filter order will not be affected. Then, you can take the element-wise product of those responses.

Answer 2

[Star Strider]

It’s always worth experimenting.

It seems that convolving the numerators and denominators of the individual filters is appropriate:

freqSample = 10000;
freqCutoffLow = 100;
bandPassLower = 500;
bandPassUpper = 1000;
freqCutoffHigh = 2000;
%as = [1/(freqCutoff*2*pi),1];
nLow = 8; % orderOfFilter;
nBandPass = 8;
nHighPass = 10;
wNLow= freqCutoffLow / (freqSample/2); %normalizedCutOffFreq
[lowA,lowB] = butter(nLow, wNLow);
[passA,passB] = butter(nBandPass,[bandPassLower bandPassUpper]/(freqSample/2));
[highA, highB] = butter(nHighPass, freqCutoffHigh/(freqSample/2),'high');

Av = conv(conv(lowA,passA),highA);                              % Convolve
Bv = conv(conv(lowB,passB),highB);                              % Convolve

figure
freqz(Av, Bv, 2^14, freqSample)   
title('Convolved')

figure
freqz(lowA, lowB, 2^14, freqSample)
title('Low')

figure
freqz(passA, passB, 2^14, freqSample)
title('Bandpass')

figure
freqz(highA, highB, 2^14, freqSample)
title('High')

s = zeros(1,10001);
s(5000) = 1;
LowOut = filtfilt(lowA, lowB, s);                               % First Filter
BandOut = filtfilt(passA, passB, LowOut);                       % Cascade
AllOut = filtfilt(highA, highB, BandOut);                       % Cascade

FTs = fft(s);
FTA = fft(AllOut);
H = FTA ./ FTs;                                                 % Transfer Function
Fv = linspace(0, 1, fix(numel(s)/2)+1)*(freqSample/2);
Iv = 1:numel(Fv);

figure
plot(Fv, 20*log10(abs(H(Iv))/max(abs(H(Iv)))))
grid

The serial output of the three filters (the last figure) appears to give essentially the same result as the convolution (the first figure).

[Carl Eranio]

So, I did have a question, as I worked on the problem. The goal is to get what is below. Your answer seemed like it was on the best path, but I did end up with a question. When one convolutes, is that the same as creating multpile parralel paths for the signal to get form point a to point b, and therefore, allow you to get the output that blocks the two unwanted frequencies? Or would it have been better to add the three systems together? Just asking from curiosity, not from need. Thanks again for your help.

[Star Strider]

The convolution essentially connects them in series, which is what I assume you wanted, since you wrote ‘I want to get the freq response as a single line that shows what happens when all three filters are inline with one another’. (It would hgave removed all ambiguity had you stated ‘series’ or ‘parallel’ instead of ‘inline’.) My code does exactly that, and provides a numerical ‘proof’ of sorts by cascading the individual filters.

To connect them in series (i.e. cascade them), do as I did in my code.

To connect them in parallel, present the same signal to each filter, then add the results.

[Carl Eranio]

Thank you, that's my bad. And your answer makes sense. (I did learn alot about that from your code, and it really helps to have convolution means series in my head now.)

[Star Strider]

My pleasure.

No worries.

If my Answer helped you solve your problem, please Accept it!

Answer 3

[Carl Eranio]

So, my teammate worked out the answer we were looking for. It apears the answer, graphically speaking, was to only use the 'h' terms. I would like to thank the two of you for your help.

%% LAB2 EEET2447 Task 1

clear all
close all
clc; %clear

% 8th order Butterworth low-pass filter 
%reference*****************************
% http://mirlab.org/jang/books/audioSignalProcessing/filterDesign.asp?title=11-2%20Filter%20Design%20(%C2o%AAi%BE%B9%B3%5D%ADp)
%**************************************
fc1 = 100; % cutoff frequency
fs = 1e4; % sampling frequency
order1 = 8; 

[b,a] = butter(order1, 2*fc1/fs, 'low');
[sos,g] = tf2sos(b,a);
% === Plot frequency response
[hLow, wLow]=freqz(b, a);

%Get the frequencies in Hz
T = 1/fs;
Hertz = wLow/(2*pi*T);

plot(Hertz, abs(hLow)); 
title('Magnitude frequency response');
xlabel('Normalized Frequency');
xlim([0 2500]);
grid on
hold on

% 7th order band-pass butterworth filter

fmin = 500; %lower cut-off
fmax = 1000; %higher cut-off
order2 = 7;

[d,c] = butter(order2, [fmin/(fs/2), fmax/(fs/2)], 'bandpass');
[hPass, wPass]=freqz(d, c);
plot(wPass/pi*fs/2, abs(hPass)); 
grid on
hold on

% 10th order high-pass butterworth filter

fc3 = 2000; % cutoff frequency
order3 = 10; 

[f,e] = butter(order3, 2*fc3/fs, 'high');
[sos,g] = tf2sos(f,e);
% === Plot frequency response
[hHigh, wHigh]=freqz(f, e);
plot(wHigh/pi*fs/2, abs(hHigh)); 
grid on
hold on

[Max Murphy]

No problem. Looks like the filters were in parallel, apologies for the misleading answer! Star strider's comment below is the correct answer.