0
$\begingroup$

As I know that in case of one tap OFDM equalizer, (ZF equalizer), we should have $N$ division complexity since we divide every subcarriers by its correspondent channel tap in frequency domain.

But I couldn't calculate how much exactly the computational complexity in case if we used MMSE equalizer for the same OFDM system.

Could you please help me how to calculate that?

Thank you

| improve this question | | | | |
$\endgroup$
1
$\begingroup$

For a SISO channel, assuming you have estimated channel coefficients for each sub-carrier $k$, the MMSE equalizer is $$ \hat{h}_k=\frac{h_k^*}{|h_k|^2 + \frac{\sigma_x^2}{N_0}} $$ So you can see already there is a multiplication in the numerator ($h_k^*$), and then there is a division by a term in the denominator. So for all $N$ subcarrier, this itself will be $2O(N) \rightarrow O(N)$ for large $N$. There is additional $|h_k|^2$ computation which is basically multiplication with conjugate. So if you have precomputed SNR ($\frac{\sigma_x^2}{N_0}$) and $h_k$ for all sub-carriers, complexity should still be linear on $N$. But this is a very simplified assumption because in most practical OFDM packets, there a pilot sequence periodically present so you may need to update $h_k$ using available pilot sequence.

| improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Note that this is only correct if the noise samples are uncorrelated. $\endgroup$ – AlexTP 2 days ago
  • $\begingroup$ Thank you for replying. You mean complexity is $N$ ? So why I usually notice the MMSE equalizer provides good results but it's more complex compared to ZF equalizer ?? $\endgroup$ – Fatima_Ali yesterday
  • $\begingroup$ @Fatima_Ali MMSE Equalizer is complex in the sense it has to compute the SNR along with additional multiplications compared to ZF Equalizer which is just a single division. So it is about $3N$ computations roughly if you have done precomputations. But if $N$ is large like $1024$, this is still linear in $N$. But instead of $3$ if it was $\log_2 (N) \times N = 10 \times N$, the difference in complexity would be apparent. Afraid to say, computational time order is another vast topic itself (en.wikipedia.org/wiki/Time_complexity). $\endgroup$ – jithin yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.