I have an OFDM system with Number of sub-carriers $N= 1024$, modulated data using $QAM$ modulation is transmitted via those subcarriers as follows:

modulation --> ifft --> adding CP --> upsampling --> conv channel --> adding noise

Then at the receiving side, the received signal is processed as follows, discarding the channel

delay --> downsampling --> CP removal --> fft --> MMSE equalizer --> demodulation

The issue which I am facing in that system is in the equzlizer step, When using the original channel, I can not get the performance back. Howenver, when I estimate that channel after the step of fft and then use the estimated channel, The performance is OK !!

Why does that issue happens? I think because of the upsampling step, because when I don't upsample the signal, that becomes fine but when I use the upsampline and use then equalize using the original channel, I can't get the performance back.

  • $\begingroup$ Can you clarify what exactly you are doing when you use the original channel? What do you do with the channel to correct for that channel distortion? $\endgroup$ – Dan Boschen Jun 16 '20 at 13:03
  • $\begingroup$ with the original channel $h$, I just use $H = fft(h,N)$; and then use MMSE equalizer using the frequency-domain channel $H$. where in the second case where I can get the performance well, I replace $h$ by the estimate channel gotten from LS estimation. $\endgroup$ – Fatima_Ali Jun 16 '20 at 13:12
  • $\begingroup$ But your correcting for the channel, not applying it again. If the channel was minimum phase you would use the inverse of the channel and not the channel itself (for example)--but given most channels are NOT minimum phase we can't simply invert it. So I am trying to see how you are using the channel directly but still don't quite follow. $\endgroup$ – Dan Boschen Jun 16 '20 at 13:16
  • $\begingroup$ @DanBoschen After obtaining $H$ I calculate the equalizer coefficients $Gz = conj(H)/(H.*conj(H) + 1./Beta)$ where * means multiplication. Then I multipy the received signal after the fft by the equalizer coefficient $Gz$, is that right? When I calculate $H$ based on the estimated channel, It's ok, but when I calculate it using the original channel $h$, I can't get the performance back. $\endgroup$ – Fatima_Ali Jun 17 '20 at 0:44
  • $\begingroup$ I am sorry I am not completely following, I think I would need to see the actual data/processing in detail. Hopefully someone else recognizes this more immediately to help you! $\endgroup$ – Dan Boschen Jun 17 '20 at 2:45

In simulation, we do upsample to simulate the "analog" signal which is a continuous signal in reality but we represent it as a high-time-resolution in our digital/discrete simulation. When we upsample our data before going through the channel, we should also upsample the "impulse response" (IR) of the channel. It means that the timing-base of both signal and IR of channel must be same. For example, without upsampling, take for example of a channel with an IR of {4,0,0,3,0,1.5,0,0,0}. First, We may apply our signal (without upsampling) to this channel. Then, we upsample our signal with a factor of 2 and want to apply that to this channel again. this time we have to upsample our channel's IR with a factor of 2 just like the signal. So, the upsampled channel'IR would be: {4,0,0,0,0,3,0,0,1.5,0,0 ....}. So, the time-base or time-resolution of both signal and channel should be equal.

  • $\begingroup$ Thank you for your reply, but when should I estimate the channel after that ? should it be estimated before the down-sampling ? If I estimated it after the fft, it means I will get estimated channel of length L while the channel used during the convolution is of length Fs*L, where Fs is the upsampling factor. $\endgroup$ – Fatima_Ali Jun 17 '20 at 14:38
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    $\begingroup$ We perform upsampling just for simulating the analog corresponding signal. So, our main processing should be done after downsampling. We apply the upsampled version of channel for convolution (on analog/air signal) but, when doing channel estimation, we will estimate the downsampled version of channel. $\endgroup$ – Hamid R. Tanhaei Jun 17 '20 at 14:54
  • $\begingroup$ In real test, the channel will not be upsampled, right ? Is that normal ? $\endgroup$ – Fatima_Ali Jun 20 '20 at 12:33

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