0
$\begingroup$

I have a signal vector x=[a b c..] having 10 data values. I replicated this signal and created another vector y=[a b c.. a b c..] totaling to 20 data values.

But the Magnitude FFT spectrum results of signal x and y are different i.e. abs(fft(x)) and abs(fft(y)) are different. Any reasons what is happening.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

This is due to Parsevals Theorem. The energy before and after DFT is equal (or depending on the normalization factor, at least proportional).

So, if you double the input energy by duplicating the input, you also get a different output energy. In your special case, you have a two-times repetition of the input, this corresponds to a discrete spectrum, where only every second bin is non-zero. Hence, the value on these bins needs to increase to fulfill the energy constraint.

Edit - More explanation on the discrete spectrum: For the continuous-time Fourier Transform, every periodic time domain function of period $T$ has a discrete spectrum, where the spectrum consists only of discrete lines (Dirac impulses) with distance $1/T$.

A similar property holds for the DFT: if you input a periodic time function, you get out a discrete spectrum (i.e. it only every $n$th frequency bin is non-zero). Namely, if your time-domain input function has period of $N/L$, your spectrum is discrete, i.e. every $L$th bin is non-zero ($N$ is the DFT length).

In your case, your input function has period $N/2$, i.e. only every 2nd frequency bin is non-zero. Now, the double input energy (you have input 2 periods instead of one period) needs to be represented on these bins, leading to higher amplitudes compared to a single period.

$\endgroup$
1
  • $\begingroup$ I would appreciate if you could elaborate on your statement " In your special case.......this corresponds to a discrete spectrum, where only every second bin is non-zero. " The Magnitude FFT result of my replicated signal (i.e.) abs(fft(y)) produced zero for every alternate data points. I could not understand why this happened. $\endgroup$
    – Arun Raj
    Commented Jan 28, 2017 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.