Why are fft diagrams mirrored?

Why are the fft plot mirrored?

Answer 1

[Michelangelo Ricciulli]

Hi Dhyan, actually, it is not always the case, since it is true only for real signals. It can be proven mathematically that when you apply the FFT on a real signal (without imaginary elements) the abs() of FFT is perfectly symmetric. If you instead try to make the same thing on a signal that contains an imaginary elements, you will lose this symmetry.

If you're interested, here you can find more details.

[Michelangelo Ricciulli]

Hi Dhyan,

if you found this answer useful, could you please mark it as accepted? Thank you!

[Stephen23]

I accepted the answer as it seems to cover the OP's original question.

[Michelangelo Ricciulli]

Thank you very much, Stephen.

[John Buggeln]

Another way of thinking about it, (for real frequencies) is that after you reach the Nyquist limit (k = N/2) of the fourier transform matrix (which is filled with your component frequencies, indexed by k), your frequencies start to 'alias' to negative frequencies that are equivalent to the positive frequencies in magnitude. At the edges this is the easiest to see. The k =1 frequency will take you around the unit circle at a certain angular speed, and k= N-1 looks like k = 1, but in the negative direction. If you draw out the unit circle, and how far each waveform will 'travel around' in the each unit of time, this will make sense. Likewise, these pairings exist symmetrically around the nyquist limit as you approach it from the positive and negative direction. Therefore, you will have equal component frequencies, i.e. the 'mirroring' around that nyquist limit, N/2!