# FFT of time-compressed signal not outputting correct amplitude spectrum

Given a signal g(t) <--> G(f), a time-compression g(at) <--> 1/(abs(a)) * G(f/abs(a)), but not with the FFT as I've found out. Performing the FFT on g(a*t) doesn't scale the amplitude of the G(f/abs(a)) spectrum at all. What gives?

Fs=100;
t=-5:1/Fs:5;
y=sin(2*pi*t);

a=4;
g=sin(2*pi*a*t);

m = length(y);
n = pow2(nextpow2(m));
Y = fftshift(fft(y,n));
G = fftshift(fft(g,n));

f0 = (-n/2:n/2-1)*(Fs/n);

figure;
plot(f0, abs(Y)/Fs);

figure;
plot(f0, abs(G)/Fs);


Here's y: Here's g. It's at 4 Hz, but the amplitude is basically the same as y when it should be 25%. Also, why does the y-axis go up to 5 and not 1? I scaled the abs(fft(y)) and abs(fft(g)) by Fs so shouldn't the max amplitude be 1? – AJN
Oct 4 '20 at 14:21

You have 4 times as many cycles in the time -5:ts:5. If your y and g signals had the same number of cycles and different frequencies, then the expected result would have appeared I believe. In short, g is not time compressed version of y as we can verify by simply counting the number of cycles in each signal.

## Time domain comparing time compressed signal and higher frequency signal higher frequency signal has more cycles that time compressed version.

## Code comparing time compression and just a higher frequency


fs = 100;

time = -5:1/fs:5;

y = sin(2*pi*time);

time2 = -5/4 : 1/fs : 5/4;
g = sin(2*pi*4*time2);

h = sin(2*pi*4*time);

f1 = linspace(0, fs, length(time));
f2 = linspace(0, fs, length(time2));

plot(f1, abs( fft(y) )/fs, '.-');
hold on;
plot(f2, abs( fft(g) )/fs, '.-');
plot(f1, abs( fft(h) )/fs, '.-');

legend('original', 'time compressed', 'higher frequency');

grid on;
xlim([0 10]);


## Result • Thanks for your reply, but I'm not sure I follow. What should I do differently in the MATLAB to produce the expected result? Oct 4 '20 at 14:29
• @user50420 I wanted to show that the code in your question doesn't time compress the original signal. It creates a new signal having more cycles than the original. It is not the time compressed version of the original signal. I have also added a code which I think is the correct way for time compression. It seems to give expected (?) results.
– AJN
Oct 4 '20 at 14:40
• Ok, I figured that since they were both plotted over the same time interval that they were the same signal, but I suppose they should have the same number of cycles instead. Oct 4 '20 at 14:46
• Yes. Imagine if it was a non periodic signal which was being time compressed. We would have expected time compression to preserve the number of peaks and valleys and not introduce new ones.
– AJN
Oct 4 '20 at 14:47