This question is regarding applying minimum mean square operation at the receiver side of communication system.

I have read in paper that

Given an output signal from communication system $$R(k)= H(k)S(k) + N(k)$$

an MMSE filter $$G(k)= \frac{H^*(k)}{|H(k)|^2+\frac{\sigma_n^2}{\sigma_s^2}}$$

My incomplete solution

I think we need to find G(K) such that

$$\mathbb{E}[G(k)H(k)S(k)+G(k)N(k)- S(k)]^2$$

Does anyone know how should I continue, thanks alot.

  • $\begingroup$ I think you should start from the MMSE optimization: Find G(k) that minimizes $\mathbb{E} \left[( S(k)-G(k))^2 \right]$ $\endgroup$
    – Oliver
    Commented May 25, 2015 at 7:52
  • $\begingroup$ @Oliver: The criterion you suggested doesn't make much sense. Why should the equalizer $G$ approximate the data $S$? $\endgroup$
    – Matt L.
    Commented May 25, 2015 at 10:34
  • $\begingroup$ @MattL You are right Dr. Matt L. The proper minimization should be find G(k) such that $$\mathbb{E}[|S(k)-G(k)R(k)|^2]$$ is minimized. $\endgroup$
    – Oliver
    Commented May 25, 2015 at 11:10
  • $\begingroup$ @Oliver: OK, now there's just one thing that remains confusing: in your comment, $G$ is the equalizer (just as in the OP), but in your answer, $G$ is the filtered received signal, and $W$ is the equalizer. $\endgroup$
    – Matt L.
    Commented May 25, 2015 at 11:15
  • $\begingroup$ Yes..That is right Dr. Matt.. $\endgroup$
    – Oliver
    Commented May 25, 2015 at 11:21

2 Answers 2


For simplicity of notation I'll leave out the index $k$. You've already noted that the error is

$$E=R\cdot G-S=S(HG-1)+NG\tag{1}$$

The MSE can be computed by integrating the power spectrum of the error. Since the power spectrum is real-valued and non-negative, its integral (i.e., the MSE) can be minimized by making the power spectrum as small as possible. Assuming white data and white noise with powers $\sigma_s^2$ and $\sigma_n^2$, respectively, which are independent of each other, we can compute the error power spectrum from (1) as


The expression for the equalizer response $G$ that minimizes (2) can be found in an elegant way by completing the square in (2):


The expression (3) can be minimized by choosing $G$ such that the left-hand term vanishes:


which is the optimal MMSE solution.

It is not immediately obvious that (2) and (3) are identical, but proving their equivalence is simple (yet tedious) by multiplying both expressions out and comparing them.


Assume that your $G(k)=W(k)R(k)$, that is, a constant multiple of $R(k)$. Then, according to the MMSE theory, the error $S(k)-G(k)$ is orthogonal to $R(k)$. In analytical terms, this means that \begin{equation} \mathbb{E}[(S(k)-G(k))R^*(k)]=0 \end{equation} On expanding the terms in the above equation we get \begin{equation} \mathbb{E}[S(k)R^*(k)]=\mathbb{E}[G(k)R^*(k)]\\ \end{equation}

Now, \begin{equation} \mathbb{E}[S(k)R^*(k)]=H^*(k)\sigma^2_s\\ \end{equation} and

\begin{equation} \begin{split} \mathbb{E}[G(k)R^*(k)]&=\mathbb{E}[W(k)R(k) (H^*(k)S^*(k)+N^*(k))]\\ &=W(k)\mathbb{E}[(H(k)S(k)+N(k)) (H^*(k)S^*(k)+N^*(k))]\\ &=W(k)(|H(k)|^{2}\sigma^2_s+ \sigma_n^2)\\ \end{split} \end{equation} And therefore from second equation, \begin{equation} H^*(k)\sigma^2_s = W(k)(|H(k)|^{2}\sigma^2_s+ \sigma_n^2) \end{equation} from which the filter $W(k)$ can be readily derived.

  • $\begingroup$ thank you, clear and nice. You say that the error is $S-G$, shouldn''t it be $RG-S$ instead?@oliver $\endgroup$
    – Tyrone
    Commented May 26, 2015 at 18:12
  • $\begingroup$ @Tyrone, It is a notation error. The error is $S-G=S-WR$. The $W$ in my derivation is the $G$ in your case. So the error is $S-GR$. $\endgroup$
    – Oliver
    Commented May 27, 2015 at 0:56

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