Getting a filter error today when yesterday didn't?
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Is MATLAB a time varying system? Everything was working swimmingly for me yesterday, but I'm riddled with errors today, for code that has been unchanged since written yesterday.
For reference, I'm getting: Error using filter Initial conditions must be a vector of length max(length(a),length(b))-1, or an array with the leading dimension of size max(length(a),length(b))-1 and with remaining dimensions matching those of x.
I was able to get the stem plot associated with it from yesterday, and I know that this is what it's supposed to look like, indicating that it was written correctly:
Using the code:
tmin = 0;
tmax = 0.1;
dt = 1/2000;
t = tmin:dt:tmax;
x = 3*cos(2*pi*30*t) + cos(2*pi*200*t);
n = 79;
d = [1 zeros(1,n)];
a = [1 -1.294 0.64];
b = [1.812 -2.932 1.812];
q = filter(b,a,d,x);
stem(q)
0 commentaires
Réponses (1)
Chad Greene
le 28 Avr 2021
I suspect that yesterday you weren't including x in the filter. Yesterday you were filtering the d signal, which is just [1 0 0 0 0 ... 0]. When you filter d, all you're left with is the ringing from the impulse.
tmin = 0;
tmax = 0.1;
dt = 1/2000;
t = tmin:dt:tmax;
x = 3*cos(2*pi*30*t) + cos(2*pi*200*t);
n = 79;
d = [1 zeros(1,n)];
a = [1 -1.294 0.64];
b = [1.812 -2.932 1.812];
q = filter(b,a,d);
stem(q)
Were you intending instead to filter x?
q = filter(b,a,x);
stem(q)
0 commentaires
Voir également
Catégories
En savoir plus sur Filter Analysis dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!