New answers tagged ifft
1
Kevin's Solution solved it for me, but the solution in the time domain is shifted in y by a value that depends on the max x value of the signal and the form of the signal itself.
Edit: I just found the solution myself. It is shifted in y by the mean(y), so if I subtract mean(y)/2 of my ifft output I get the signal
2
[Pictures to follow] Let us start with a thought experiment (which can be simulated): imagine a constant signal with value $c$. Add a full period of a pure sine with non-zero frequency. If you can remove this harmonic contribution by zeroing out its frequency bin in the Fourier domain, then the resulting inverse Fourier signal will still have mean $c$. So ...
1
The magnitude of the FFT will likely scale by the number of samples N depending on the specific algorithm you use. So the IFFF will be 1/N. Just multiply your FFT bins by N to normalize it. If you use any windowing this will change the result accordingly
-1
For your sub-carrier formula, you forgot the $j$ in front of the $b_n$; without that, you lose all your data, because your real and imaginary parts get summed. Can't do that! The whole idea of complex equivalent baseband is that the real anand imaginary parts are independent; this does not only apply to OFDM, but to any baseband technique; for example, QPSK ...
Top 50 recent answers are included
