Problem when graphing digital filter chebyshev
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Hello, I need to design a chebyshev digital filter, with cutoff frequencies of 250 and 2000 Hz, it should cut at 0.7 at these two frequencies. The code I am using does not meet the objectives, it is the following:
Fs = 44000;
Fn = Fs/2;
n = 3; Rp = 0.5;
Wn = [250 2000]/Fn;
[b a]=cheby1(n,Rp,Wn,'bandpass')
[h,w]=freqz(b,a)
plot(w,abs(h))
What I need to graph is what is shown in the image. The order is 3 and the ripple is 0.5
回答（1 个）
Need some some addition arguments to freqz(). Check its doc page for more information:
Fs = 44000;
Fn = Fs/2;
n = 3; Rp = 0.5;
Wn = [250 2000]/Fn;
[b a]=cheby1(n,Rp,Wn,'bandpass');
[h,f]=freqz(b,a,1024,Fs);
semilogx(f,abs(h))
Adjust n and Rp for the desired result.
8 个评论
Juan Chehin
2021-8-30
That code again returns an incorrect graph. For example at 250 Hz it shows 0.9441
Paul
2021-8-30
As I said, you'll need to adjust the n and Rp inputs to cheby1 to match that plot. Also, if you increase n, consider chaning the 1024 in the call to freqz to something bigger, like 4096.
Juan Chehin
2021-9-1
n and Rp are fixed. That is, n = 3 and Rp = 0.5 are the requirements
Paul
2021-9-1
In that case, you can't get the graph in the figure, at least not using cheby1.
Juan Chehin
2021-9-1
Sorry, the filter order n is 6
Paul
2021-9-1
Did you try the code at the top of this answer with n=6? Does it yield the desired result?
Juan Chehin
2021-9-1
Yes, but with some modifications
Fs=4400;
Fn = Fs/2;
n = 6;Rp = 0.5;
[b,a]=cheby1(n,Rp,[250/Fn,2000/Fn]);
[h,f]=freqz(b,a,1024,Fs);
plot(f,abs(h));
xlabel('Frequency (Hz)')
Paul
2021-9-1
Do you now have the desired result or is there still an open question? Did you intentionally reduce Fs from 44000 to 4400?
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