plot multiple PSD lines in single graph in color scale (imagesc)

Hi, I have 13 signals where I produce the PSD for each with the psd function using Welch's method object (spectrum.welch). But this only allows me to generate separate plots for each signal. Is there a way to express all 13 PSDs in pseudocolor (like imagesc) in a single figure where Y is each of the 13 signals, X is the frequencies, and the color is the power/frequency (dB/Hz)? So basically I want to generate a 13 x n matrix containing the PSD of each signal.

Example:

    Fs=10000; %sampling frequency in samples per second
    t=0:(1/Fs):1; %one second time vector, 10001 elements
    y=0.4*cos(2*pi*2000*t)+0.2*sin(2*pi*1000*t)+randn(size(t));
    h = spectrum.welch;
    Hpsd=psd(h,y,'Fs',Fs,'ConfLevel',0.95);
    figure, plot(Hpsd)

This only gives me 1 plot. If I do this 13 times, how can I create a 13 x n figure where the PSDs are expressed in color? I basically want a figure that looks like imagesc(rand(13,100)) but with PSDs and not random numbers.

Thanks in advance!

Answer 1

[Youssef Khmou]

Yes you can generate the 2D spectrum, here is an example with 13 signals as you said :

   Fs=80;
 f=10+rand(13,1)*10; % frequencies of the signals 
 t=0:1/Fs:10-1/Fs;
 y=zeros(13,length(t));
 for n=1:13
 y(n,:)=sin(2*pi*t*f(n));
 end
 % spectrum
 h=spectrum.welch;
 F=zeros(13,129);
 for n=1:13
     t=psd(h,y(n,:),'Fs',Fs);
      F(n,:)=t.Data;
 end
 figure, surface(F), shading interp

You add the x,y labels...

[S]

Thanks for the response. Although this certainly does give me a 2D spectrum, I'm a bit confused by the units in the vector of the data within dspdata. For example, in the above example for one of the iterations (say n=1), if you just plotted t ("plot(t)"), the units are in dB/Hz and range from -70 to 0. But the units are different from the nth vector in F where the values range ~e-9 to ~0.3. What are the units of F? Is there a way to plot the spectrum in dB/Hz, aka express F in dB/Hz?

[S]

I figured it out. I just need to multiply F by 10*log10 and the units become dB/Hz.

[Youssef Khmou]

ok.

Answer 2

[Image Analyst]

Perhaps the waterfall() function?