0 个投票

Can someone tell me why do I get an error: Error using cheb2ord (line 69) The cutoff frequencies must be within the interval of (0,1).
I have normalized the frequencies by fn so what's the deal?
clear all;
close all;
clc;
fs=2600;
fn=fs/2;
ws=[0 1300]/fn;
wp=[710 839]/fn;
rp=0.1;
rs=25;
[n,wn]=cheb2ord(wp,ws,rp,rs);
[nz,dz] = cheby2(n,rs,wn,'pass');

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Walter Roberson
Walter Roberson 2018-6-9
编辑：Walter Roberson 2018-6-9

0 个投票

(0,1) indicates 0 to 1 exclusive. Your ws is exactly 0 and exactly 1.
Also, you probably need 'bandpass' on your cheby2 instead of 'pass'

4 个评论
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Nikolay Megdanov
Nikolay Megdanov 2018-6-9
编辑：Nikolay Megdanov 2018-6-9
ok but what if my assignment requires the stopband's boundaries to be at 0 and 1300 ?
Yeah I fixed the bandpass on the cheby2.
Walter Roberson
Walter Roberson 2018-6-9
1/0 and infinity divided by infinity would be involved in the calculation. If you were to try to pursue the mathematical formula you would end up with point theoretically undefined (except possibly in limit.)
Nikolay Megdanov
Nikolay Megdanov 2018-6-9
So what should I do, I just make the stopband boundaries to 1 and 1299 ?
Nikolay Megdanov
Nikolay Megdanov 2018-6-9
Nevermind, I found I messed up the frequencies I was thinking should be placed there, thanks for the answers!

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