Fequency Domain Oversampling - decimating when going back to time domain

Question

I'm currently trying to simulate Frequency Domain Equalization, and the effect that over sampling in the frequency domain has on the affect of the equalization. If I understand things correctly this is the process that's supposed to happen using a single Carrier when transmitting

1. Modulate and transmit the symbols to create $x(n)$, each symbol being $N$ samples.
2. add noise, $z(n)$, and channel effects, $h(n)$, for the receiver, $ y(n) = h(n)\star x(n) + z(n)$
3. Zero pad, where $ZP$ is the number of added zeros, in the time domain to over sample in the frequency domain. $FFT(y(n),N+ZP)$. Note that increasing the order of the FFT should be the same as zero padding in the time domain
4. Do Equalization in the frequency domain
5. Go back to the time domain
6. Demodulate the symbols

The part I'm not sure about is step 5, where we go back to the time domain. In order for my number of samples per symbol to be correct, I need N samples, but I have $N+ZP$ samples if I take the same size $IFFT$. Doing the Zero padding and taking the $FFT$ should be interpolation... but i'm not sure how to do the decimation so that I have the correct number of samples. Does anyone have any idea about this? Thanks!

Answer 1

Your time domain waveform after 6 will be zero-padded as in step 3 with the additional "smearing" due to your equalization to remove the channel effects that will go into the zero padded area. The entire waveform in time will be delayed by the group delay of your frequency domain equalization, which may not necessarily match the group delay of the channel distortion.

If you applied no equalization (by multiplying in frequency by your frequency domain equalization); the IFFT would recover your time domain waveform in its original form with the noise and channel distortion with the additional zeros at the end that were introduced by the zero-padding when taking the FFT. (Along with boundary conditions at the transition such as Gibbs phenomena due to the effective rectangular window applied). Multiplying by your equalizer in frequency is identical to convolving in time the original time domain waveform with the Fourier transform of your equalizer fucntion, and therefore will delay this waveform based on the group delay of this filter (where group delay is $\frac{d\phi}{d\omega}$). Equalizers are typically not linear phase (unless your channel distortion was which is unlikely) so the group delay is not constant versus frequency; but the point is the resultant delay of the recovered original waveform will be completely dependent on the net delay of the channel and equalizer. With knowledge of the channel and specific equalizer function, the delay can be determined.