Using obw to solve for the bandwidth of a signal

Hello everyone,

I have a bunch of symbols in the time domain and I am trying to do the fft of them and then calculate their bandwith. I have been trying to use the obw function to do this but the values do not seem to be right. My code is below:

clear
Fs = 1000;
Ts= 1/Fs;
symbol_duration= 1;
t= 0:Ts:symbol_duration;

%% number of functions is degree. Only one for testing
degree = 1;
p(1,:) = (2/symbol_duration)*cos(2*pi*15*t) + (2/symbol_duration)*cos(2*pi*50*t);

for i = 1:degree
    %%%%% https://www.mathworks.com/help/matlab/math/fourier-transforms.html
    xform=(1/Fs) .* fft( p(i,:) );
    n=length(p(i,:));
        
    fshift = (-n/2:n/2-1)*(Fs/n);
    yshift(i,:) = abs( fftshift(xform) );
    
    bw(i)= obw( yshift(i,:) ) 
    
    plot(fshift,yshift(i,:))
    xlim([0 Fs/2])
    hold on
end

bw = 3.0991

The plot of my fft appears to be correct, but bw always seems to be some very small number that seems incorrect. Any help would be appreciated!

Answer 1

[Ashutosh Singh Baghel]

Hi Russell,

I understand that 'obw' function is used for returning the 99% occupied bandwidth of the input signal. The function 'obw' uses input signal instead of 'fft' of that input signal. Also, it is good practise to give sampling frequency 'Fs' as second input to 'obw' function.

Please try in the following way -

Fs = 1000;
t= 0:1/Fs:1;
x = cos(2*pi*10*t) + cos(2*pi*60*t);
obw(x,Fs);

Please see MATLAB Documentation page on 'obw' for further references.

[Russell Geschrey]

How did you get that graph?

[Russell Geschrey]

I see why I was not getting a graph. I was setting the output to a variable. If I just call the function alone it gives a graph. This is indeed a powerful function. Thank you for your help.