Borrar filtros
Borrar filtros

MATLAB- ifft of simple lowpass filter returns NaN

22 visualizaciones (últimos 30 días)
Nathan Monroe
Nathan Monroe el 7 de Mzo. de 2017
Comentada: Nathan Monroe el 7 de Mzo. de 2017
Hello, I seem to be having issues using MATLAB's fft and ifft functions, when simulating a simple electrical circuit. Generally the strategy I've been taking is to fft the incoming time-domain signal, multiply the result by the frequency response of the circuit, and ifft the product to produce the time-domain output. However ifft seems to be producing NaN for the real part (and a small complex part). Any idea why this may be occurring and what to do about it? A simple example is below, using a first order RC butterworth filter. I'm really confused as to why this may be happening. I realize there are other ways to simulate such a simple system, however I'm applying this method to much more complex systems with existing frequency-domain models. Thanks!
clear all
Fs = 100;
n=1000;
t = (1/Fs:1/Fs:10); %time
test1 = cos(2*pi*5*t); %test funcion, 10Hz cosine
X = fft(test1,n); %1k pt fft
dF = Fs/n; %0.1Hz bin width
f = (dF*(0:n-1))-Fs/2; %frequency vector, centered around zero
c=1e-4;
r=1e3;
tf = (1./(j*2*pi*f*c))./(r+(1./(j*2*pi*f*c))); %filter frequency response
X_filtered = X.*tf;
out = ifft(X_filtered,n);

Respuesta aceptada

David Goodmanson
David Goodmanson el 7 de Mzo. de 2017
Hello Nathan, I believe the problem is because the frequency grid contains the point f=0, and your expression for tf comes up with nan at that one point. If you write tf as
tf = 1./(1+j*2*pi*f*r*c); %filter frequency response
you should get better results. (Or find the nan in the first expression and change it to 1).

Más respuestas (0)

Categorías

Más información sobre Spectral Measurements en Help Center y File Exchange.

Productos

Community Treasure Hunt

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

Start Hunting!

Translated by