# Extreme gain for low pass filter

3 Ansichten (letzte 30 Tage)
Sebastian Holmqvist am 21 Nov. 2012
% Filter constants
Wp = 5e3*2*pi;
Ws = 9.5e3*2*pi;
Rp = 2;
Rs = 45;
% Butterworth lowest order
[n,Wn] = buttord(Wp,Ws,Rp,Rs,'s');
disp([10 'Butterworth order: ' num2str(n) ' @ ' num2str(Wn/(2*pi)) ' [Hz]']);
[z,p,k] = butter(n,Wn,'s');
disp(['Zeros: ' num2str(size(z,1)) 10 ...
'Poles: ' num2str(size(p,1)) 10 ...
'Gain: ' num2str(k)]);
> Butterworth order: 9 @ 5342.252 [Hz]
> Zeros: 0
> Poles: 9
> Gain: 54092634016296818598907515083157976121344
How is this gain possible? Is this a weird bug?
##### 0 Kommentare-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden

Melden Sie sich an, um zu kommentieren.

### Akzeptierte Antwort

Teja Muppirala am 21 Nov. 2012
Bearbeitet: Teja Muppirala am 21 Nov. 2012
I don't think that is a weird bug. That large value looks correct to me.
The DC gain should = 1. In other words, if you plug in s = 0, you should get 1. Since there are no zeros, this implies that the product of the poles is the gain:
G(s) = K/[ (s-p(1)) * (s-p(2)) * (s-p(3)) * ... (s-p(9)) ]
G(0) = K/[ (-p(1)) * (-p(2)) * (-p(3)) * ... (-p(9)) ] = 1
--> Therefore K = prod(-p)
Actually plugging it in
prod(-p)
ans =
5.4093e+40
Which is precisely what K is.
##### 1 KommentarKeine anzeigenKeine ausblenden
Sebastian Holmqvist am 21 Nov. 2012
Excellent explanation! We are to implement this in a typical Sallen-Key circuit and were a bit thrown by the large numbers. But we see now that the gain K relates to the pole, not directly to any physical components in our amplifier.

Melden Sie sich an, um zu kommentieren.

### Kategorien

Mehr zu Butterworth finden Sie in Help Center und File Exchange

### Community Treasure Hunt

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

Start Hunting!

Translated by