I am performing FFT on a real odd function and the resultant transform has zero amplitude in the last bin. Essentially if Y= rfft(X), then Y[-1] is always zero. I stumbled on this answer which says

An anti-symmetric waveform has to be zero at the center of an even length window.

Now, this might be very easy to prove but I cannot think of why this is true. If I use the formula then to the first term should be zero, not $N/2$ th. It's clear to me why it should be real but why should the amplitude be zero for real-valued, odd function. Any help will be appreciated.


A real symmetric function has the following property $$f(t) = f(-t)$$

whereas a real anti-symmetric function has the following property $$f(t) = -f(-t)$$

So at $t=0$ you have $$f(0) = -f(0)$$ which implies that $$f(0) = 0$$

A complex conjugate-symmetric function has the following property $$f(t) = f^*(-t)$$

whereas a complex conjugate-antisymmetric function has the following property $$f(t) = -f^*(-t)$$

So at $t=0$ you have $$f(0) = -f^*(0)$$ which implies that $f(0)$ is real and further $$f(0) = 0$$

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  • $\begingroup$ Thanks for the answer. You are working in the time domain. My question was about the FFT of a discrete anti-symmetric function, so even if f(0) =0 that doesn't imply that the N/2 th bin of Fourier transformed data will be zero. $\endgroup$ – Prav001 Nov 2 '19 at 6:16
  • $\begingroup$ I'm not working on time-domain as $t$ refers to any dummy variable. At times when it is hard to make clear communications, it's best if you can post one of your discrete-tim anti-symmetric sequences of length less than 10. So that I can also see why its last bin of FFT is zero? $\endgroup$ – Fat32 Nov 2 '19 at 10:25

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