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');

 Akzeptierte Antwort

Walter Roberson
Walter Roberson am 9 Jun. 2018
Bearbeitet: Walter Roberson am 9 Jun. 2018

0 Stimmen

(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 Kommentare

Nikolay Megdanov
Nikolay Megdanov am 9 Jun. 2018
Bearbeitet: Nikolay Megdanov am 9 Jun. 2018
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 am 9 Jun. 2018
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 am 9 Jun. 2018
So what should I do, I just make the stopband boundaries to 1 and 1299 ?
Nikolay Megdanov
Nikolay Megdanov am 9 Jun. 2018
Nevermind, I found I messed up the frequencies I was thinking should be placed there, thanks for the answers!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by