I read this paper: https://arxiv.org/pdf/1305.2460.pdf

And there is a formula about MMSE:$\mathbf W^*_{MMSE}= E[\mathbf s \mathbf y^*]E[\mathbf y \mathbf y^*]^{-1}$

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$\mathbf W _{BB}$:Baseband combiner

$\mathbf W _{RF}$:RF combiner

$\mathbf W $:The combination of $\mathbf W_{RF}$ and $\mathbf W _{BB}$

$\mathbf W^*$ : the hermitian of $\mathbf W$

$\mathbf s$:the signal on the left

$\mathbf y$:the signal when $\mathbf s$ goes through the channel

Why can the :$\mathbf W^*_{MMSE} $ be written as $E[\mathbf s \mathbf y^*]E[\mathbf y \mathbf y^*]^{-1}$.why should we multiply the these two signal ,$\mathbf s$ and $ \mathbf y$,and average them ?Does anyone have information about this formula ?

  • $\begingroup$ There is a reference in the paper, adressing this formula: T. Kailath, A. H. Sayed, and B. Hassibi, Linear estimation. Prentice Hall New Jersey, 2000, vol. 1. $\endgroup$ Mar 28, 2019 at 14:26

1 Answer 1


The mean squared error equalizer is one that minimizes the mean squred error


which is the equalizer that minimizes the Euclidean distance between the transmitted signal $\mathbf{s}$ and the equalized received signal $\mathbf{Wy}$. To find the value of $\mathbf{W}$ the minimizes $J(\mathbf{W})$, take the derivative with respect to $\mathbf{W}$ and put it equal to $\mathbf{0}$, which will result in

$$-2\mathbb{E}\left[\left(\mathbf{s}-\mathbf{Wy}\right)\mathbf{y}^H\right]=\mathbf{0}$$ Solve the above equation with respect to $\mathbf{W}$ you get the MMSE equalizer.

Note: $(.)^H$ is the conjugate transpose, which is equivalent to $(.)^*$ in the paper.


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