Create a 4-pole, 300-6000HZ bandpass butterworth filter?

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Nick 2012-7-16

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评论： Mayank Lakhani 2015-6-8

Hi. I'm pretty new to filters. I want to design a bandpass 300-6000Hz 4-pole butterworth filter. Not sure about how to do this in matlab. I have managed to make progress in this, but I am not sure what I am doing anyway.

Would really appreciate if someone could show the code for this. Would be even more appreciative if you could explain how these filters work so I can do it on my own in the future :)

Thank you!

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Answer 1

Wayne King 2012-7-16

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编辑：Wayne King 2012-7-17

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If you want 4 poles and you have a bandpass filter, then you have to specify the order as 2. Also to convert from normalized frequency you divide by Fs/2 where Fs is the sampling frequency, not by Fs.

I'll assume the sampling rate is 20 kHz, but you have to specify it correctly.

Fs = 20000; 
[B,A] = butter(2,[300 6000]/(Fs/2));
% view magnitude response
fvtool(B,A,'Fs',Fs)
figure;
% view pole-zero plot
zplane(B,A)

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Mayank Lakhani 2015-6-8

hi there,

how to add signal in the fv tool. suppose my output signal is y = filter(B, A, x); than how to use fv tool?

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Answer 2

Matt Kindig 2012-7-16

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If you have the Signal Processing Toolbox, you can use the 'butter' function.

doc butter

Also, you mention that you have made some progress in this. Can you show us the code you've done so far?

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Nick 2012-7-16

Well I currently use [A,B] = butter(10,0.2)

Does this mean this is a 10-pole filter? And does what would that 0.2 mean?

Ryan 2012-7-16

编辑：Ryan 2012-7-16

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I have never used the toolbox and don't want to steer you in the wrong direction, but from the documentation I believe this:

[B,A] = butter(4,[300 6000],'stop');

acts as a 4th order bandstop filter between the cut off frequencies 300 and 6000 HZ. This does the opposite of what you're going for, but I don't see an option on 'ftype' to select otherwise. Possibly if you don't specify 'ftype' as 'stop','high' or 'low' it will act as a pass instead of stopband?

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Answer 3

Ryan G 2012-7-16

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编辑：Ryan G 2012-7-16

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If you look at the documentation it mentions you can define a 2 element frequency vector that you need to normalize to the sample rate of your data. Depending on the sample rate of your data you would want to implement something like this:

[b,a] = butter(4,[300 6000]/SampleRateOfData);

where the SampleRateOfData should be >6000. If you are designing a continuous time filter you could do

[b,a] = butter(4,[300 6000]/(2*pi),'s');

to design the filter in that domain.