How does the command: real(ifft(​fftshift(Y​))*N) operate?

1 view (last 30 days)
Hello, I am interested in knowing how this form of the inverse FFT command which is
real(ifft(fftshift(Y))*N)
works and its complexity for a vector Y of N sample points, because I have seen some basic ifft commands as seen here but this one seems to involve ifft and fft together.

Accepted Answer

Matt J
Matt J on 20 Apr 2021
Edited: Matt J on 20 Apr 2021
All of those oeprations are O(N) except for the IFFT which is O(Nlog(N)). So the chain of operations is O(Nlog(N)) overall.It would be slightly more efficient to re-implement it as,
N*real(ifft(fftshift(Y)))
since then the multiplication with N only needs to operate on real numbers.
  2 Comments
Matt J
Matt J on 20 Apr 2021
but this one seems to involve ifft and fft together.
No, it doesn't. But the FFT is also O(log(N)), so including an FFT step wouldn't increase the complexity either.
Tarek Hajj Shehadi
Tarek Hajj Shehadi on 20 Apr 2021
Thank you for the clarification, all is understood.

Sign in to comment.

More Answers (0)

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!