Problem when graphing digital filter chebyshev

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

Answer 1

[Paul]

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.

[Juan Chehin]

That code again returns an incorrect graph. For example at 250 Hz it shows 0.9441

[Paul]

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]

n and Rp are fixed. That is, n = 3 and Rp = 0.5 are the requirements

[Paul]

In that case, you can't get the graph in the figure, at least not using cheby1.

[Juan Chehin]

Sorry, the filter order n is 6

[Paul]

Did you try the code at the top of this answer with n=6? Does it yield the desired result?

[Juan Chehin]

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]

Do you now have the desired result or is there still an open question? Did you intentionally reduce Fs from 44000 to 4400?