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I have some problem when i learning MIMO,first thing is that i know designing beamforming and MRC can ensure better receive signals,but they are not the same,so i have some question about them.

1.I know the MRC is the one of method to ensure better received signal,so if I have already found the beamforming direction now,do i still do MRC when calculating the SNR?,like the example below

2.If the transmitter use $N_T$ antenna to transmit one signal to the receiver which is with $N_R$ antenna,now i do SVD to the MIMO channel $\mathbf H$ ($N_R \times N_T $ matrix ),and find the best beamfoming direction $f_A,(N_T \times 1$ column vector), from it.Now i have beamforming now ,so i don't need to do MRC?

3.Now the received signal is $y_t=\sqrt{P}\mathbf H \vec f_Ax+\vec n$,so now the $SNR=\frac{P||H\vec f_A||^2}{\sigma^2 _n}$?

Because i heard my classmate said that $SNR$ is not $\frac{P||H\vec f_A||^2}{\sigma ^2 _n}$,but $\frac{P||\vec u^HH\vec f_A||^2}{\sigma ^2_n}$,and this $\vec u^H$ is calculate the SVD too.The first column of $U$,SVD=$U\Sigma V^H$.Because $H\vec f_Ax$ is still a vetor,and we can't not calculate SNR from a vector,it must be calculate from a value,and $\vec u^HH\vec f_A$ is a value

Is my SNR formula right or my classmate's is right?Does anyone know the answer about my question?It confused me for lots of months

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You are doing transmit beamforming and choosing your vector $\mathbf{f}_A$ so that you beamforming in the "best" direction. So you transmit the signal: $\mathbf{z}=\sqrt{P} \mathbf{f}_A x$, where $\mathbf{z}$ is a length $N_T$ vector. Now you receive the signal: $\mathbf{y} = \sqrt{P}\mathbf{H}\mathbf{f}_A x + \mathbf{n}$, which is a length $N_R$ vector.

The post-MRC SNR should still be a vector. You will have a SNR for each transmit stream, so $N_T$ SNR values. To compute the SNR of the $k^{\text{th}}$ transmit stream after MRC you take the channel gains from transmit antenna $k$ to each of the $N_R$ receive antennas, $\mathbf{h}_k$, and compute its output power: $\gamma_k = P ||\mathbf{h}_k^H \mathbf{H} \mathbf{f}_A||^2$. Or, you can also equivalently do it all at once by: $||\mathbf{H}^H \mathbf{H} \mathbf{f}_A||^2$.

Your formula is actually computing the pre-MRC SNR, whereas your friends was showing how to get the post-MRC SNR. Hope this helps!

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  • $\begingroup$ So can i say post-MRC SNR is more precise than pre-MRC SNR,so it is better to caluculate the post -MRC SNR?I also want to ask that where have i done "MRC" already?because i think i just do SNR like usual $\endgroup$ – care we Nov 13 at 23:56

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