# Filtering random Signal

My question is easy one actually.

First, I generate a random signal using randn() function of MATLAB like this:

Then, I design a FIR filter of order 200 of pass-band characteristics with the pass-band $[0.2\pi, 0.4\pi]$ using the MATLAB function fir2():

My questions are:

• What am I supposed to see when I filter a random signal using BPF?
• What change in the characteristics will be occur considering the theory?

The output signal will still be normally distributed, but its power spectrum, i.e. its frequency content, will obviously be different from the input signal. If $S_X(\omega)$ is the power spectrum of the input signal, which is approximately flat, then the power spectrum of the output signal is

$$S_Y(\omega)=|H(\omega)|^2S_X(\omega)$$

where $H(\omega)$ is the frequency response of the filter.

• Thanks for asnwer (: Actually I am wondering what am I supposed to see in band pass filtered random signal in time domain? I have some trouble about filtering the random signal I mean I may be applying the filter wrongly so I want to deduce or verify it. How to consider this problem? Commented Apr 27, 2015 at 18:48
• @mehmet: You should cross-check your filtering routine with some standard routine as provided by Matlab/Octave (filter.m) or some other package. You can't just look at the filtered time-domain signal and say, well that looks good, probably my filter works as it should. Commented Apr 27, 2015 at 18:51
• Well I had planned to filter() function of MATLAB but it takes num and den of the filter as arguements. I have the values of the band pass filter in bot time and frequency domain. I don't know how I can get the num and den from this point? Commented Apr 27, 2015 at 19:15
• @mehmet If h is the vector of filter coefficients returned from the function fir2(), and x is your random input signal, then you must use: y=filter(h,1,x); Commented Apr 27, 2015 at 19:21
• @mehmet: That's the same thing for an FIR filter, the filter coefficients are the same as the impulse response. Commented Apr 27, 2015 at 19:28

You will be creating a random band-pass signal.

What you are supposed to see for such a signal if you plot the time sequence, is a varying sinusoid of frequency $0.3 \pi$ (midpoint of the pass band). Amplitude and phase will vary randomly, depending on the signal bandwidth.

In your case, $0.2 \pi$ bandwidth is quite large compared to the carrier frequency of $0.3 \pi$, so the result should look quite erratic.

For a smaller bandwidth, amplitude and phase will vary more smoothly.

• This is very interesting! But I couldn't realize how I can deduce it would be a sinusoidal signal? Commented Apr 27, 2015 at 19:19
• @mehmet: I'm afraid you won't see anything that is similar to a sinusoid. The one but last sentence of Juancho's answer ("...quite erratic ...") is more accurate than the description of the output signal as a "varying sinusoid of frequency $0.3\pi$". The latter would only be visible if the bandwidth was much smaller. Commented Apr 27, 2015 at 19:25
• I see it is some kind of aliasing effect. If I used much smaller band width I would have almost a dirac in frequency domain and it would be a sinusoidal signal. Now I understand the idea. Thank to both of you guys! @MattL. Commented Apr 27, 2015 at 19:32