# What's the computational complexity of MMSE equalizer in OFDM system

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.

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.
• 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 ?? – Fatima_Ali May 24 '20 at 1:15
• @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). – jithin May 24 '20 at 4:42