relation between the fft of full and segmented signal
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Hello,
Suppose I the following signal of 10s:
fs=1e4;
t_duration=10; % duration of signal (seconds)
t = 0:1/fs:t_duration-1/fs; % time samples
s=sin(2*pi*1000*t); %narrowband signal from the source
N=length(s);
I compute its FFT and plot 2-sided spectrum as:
S=fft(s);
f=fs*[0:N-1]/N;
stem(abs(S)); %peaks at 10001 and 90001
Now, if i break the signal into 10 segments of 1s duration each and take its fft() and plot, the bins are somehow jumbled.
s1=s(1:10000);
N1=length(s1);
S1=fft(s1);
f1=fs*[0:N1-1]/N1;
stem(abs(S1)); %peaks at 1001 and 9001
Ofcourse, if I plot the magnitude spectrum with 'f' and 'f1' vectors, I get the peaks at right frequencies.
But without 'f' and 'f1', how can I relate/interpret this 1s segment result to my original 10s long signal? please clarify.
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Wayne King
il 21 Set 2012
Modificato: Wayne King
il 21 Set 2012
That's because the spacing between the DFT bins depends not only on the sampling frequency, but ALSO on the length of the input signal.
The spacing is Fs/N, in the second case you have reduced the length of the signal by a factor of 10 so your DFT bins are spaced much more 10 times farther apart.
In the first case, your spacing is 0.1 Hz. In the second case, your spacing is 1 Hz.
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