phase plot fft (strange)values
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
KE BR
le 9 Août 2020
Réponse apportée : Sulaymon Eshkabilov
le 9 Août 2020
i Have the following code;
Fs = 500;
t = 0:1/Fs:10-1/Fs;
s = cos(2*pi*15*t - 30) + cos(2*pi*40*t +60);
NFFT=length(s);
ft =(1/NFFT)*fft(s);
freq = Fs*(0:(NFFT/2))/NFFT;
Am=abs(ft);
tol = 1e-6;
ft(abs(ft) < tol) = 0;
Ph = angle(ft);
Am = Am*2;
Am = Am';
Am = Am(1:end/2,:);
freq = freq';
freq = freq(1:end-1,:);
Ph=Ph';
Ph = Ph(1:end/2,:);
plot(freq,Ph*(180/pi));
How do i interpretend the phase of this signal? shouldnt the phase plot give me -30° at 15Hz and 60° at 40Hz? How does it come that the fft gives 81.13° and -162.3°?
Kind regards
0 commentaires
Réponse acceptée
Sulaymon Eshkabilov
le 9 Août 2020
Hi,
There was a small but crucial mistake while defining 's' signal. Here is the corrected one:
s = cos(2*pi*15*t -30*pi/180) + cos(2*pi*40*t+60*pi/180);
Good luck.
0 commentaires
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Fourier Analysis and Filtering 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!