Hi Mosharof,
Load the data
Plot the differences between time samples. We see that the data is not equally spaced in time, so using discrete-time methods, like fft, is not appropriate without further massaging the data into equally spaced samples that also catch all the transitions. Maybe that could be done by using a much smaller sampling period from the original source (which looks like it might be a Simulink model?). plot(diff(ts1.Time),'-o')
However, the signal is just the sum of rectangular pulses, and the rectangular pulse has a closed-form expression for its Continuous Time Fourier Transform (CTFT)
Here's one way we can construct the signal for analysis (this would be easier if the original source could save off the switch times and the associated signal levels at the switches).
Verify that the data has unique values:
Use diff to find when the signal level switches index = [1; find(diff(ts1.Data)~=0)];
The pattern is that the level changes only between 0 and 1, or 0 and -1. Construct the signal as the sum of rectangularPulse and build up its CTFT along the way. for ii = 2:2:numel(index)-1
if ts1.Time(index(ii)) < 10
s(t) = a*rectangularPulse(ts1.Time(index(ii)),ts1.Time(index(ii+1)),t);
Y(w) = Y(w) + fourier(s(t),t,w);
Verify that y(t) matches the original data
fplot(y(t),[0 20],'MeshDensity',200)
fplot was taking too long to make this plot Convert to an anonymous function
Yfunc = matlabFunction(Y(w));
and evaluate over some frequency range
plot(wval,abs(Yfunc(wval))),grid
plot(wval,180/pi*angle(Yfunc(wval))),grid
The Question stated that the data is actually one period of a periodic singal, which would be represented as a Continuous Fourier Series (CFS). The CFS coefficients are obtained by sampling the CTFT of a single period at multiples of the fundamental frequency (2*pi/T rad/sec, T = 20 sec) and scaling by 1/T. Here's a plot of the magnitude of the CFS coefficients over roughly the same frequency range used to plot the CTFT of the single period.
stem(2*pi*n/T,(1/T)*abs(Yfunc(2*pi*n/T)))
