I have a sound signal x(n). i want to make a random signal r(n) of the same length but half the magnitude of x(n). How do i make r(n) have half the magnitude of x(n)?

I have a sound signal x(n). i want to make a random signal r(n) of the same length but half the magnitude of x(n). How do i make r(n) have half the magnitude of x(n)?

Answer 1

[Star Strider]

Random signals are of course random, so its being exactly half the amplitude is probably not going to be the situation everywhere. You can scale the randn output by multiplying it by the standard deviation you want, so:

signal = ...;                                       % Signal Vector
noise = std(signal)*0.5*randn(size(signal));        % Noise Vector

This scales the noise vector to be 0.5 times the standard deviation of your signal vector. That is probably as close as you can get to what you want to do.

[Saiankita Anand]

I did what u said. but after I plotted the frequency spectrum and time waveform of both x(n) and the random signal rs(n) (which u named as noise in your statement), rs(n) is not half of x(n). It can be seen from both the plots. how do I correct this?

[Star Strider]

I don’t know how you created those signals. The noise signal will have broadband frequency content. It will also not be modulated such as in the time-domain plot.

Random signals are random by definition, and Gaussian random noise simulates real-world random noise most accurately. Do this to see that the standard deviation of both the ‘signal’ and ‘noise’ vectors are as desired:

t = linspace(0,2000*pi, 1E+5);
signal = sin(t/10);                                 % Signal Vector
noise = std(signal)*0.5*randn(size(signal));        % Noise Vector

sd_sig = std(signal)
sd_rdn = std(noise)

sd_sig =
   707.1068e-003

sd_rdn =
   353.3841e-003

I don’t know what there is to ‘correct’. From a statistical perspective, it works.

EDIT — In electrical engineering as I learned it, the ‘amplitude’ of a periodic signal is the root-mean-square (RMS) of the signal, this being equivalent of the standard deviation of the signal for all practical purposes. The peak-to-peak amplitude (P-P) is RMS/sqrt(2).

I maintain that convention in my Answer here.

[Saiankita Anand]

Thank you, Star Strider for your help. I am fairly new to MATLAB and hence the confusion. I kinda did understand what you are saying here. Allow me to ask you if I have further doubts in this.

[Star Strider]

My pleasure, as always.

I will answer as many of your questions as I have the knowledge to provide.