Square Root MMSE

Could anyone please explain how to perform the derivation of the square-root MMSE equation using an extended channel matrix H (as it is said in the "A LOW COMPLEXITY SQUARE ROOT MMSE MIMO DECODER" article)? • I'm not 100% happy with the wording; a solution can be found, calculated, approximated… but not defined. However, this is a so-called ordinary least-squares problem, en.wikipedia.org/wiki/Ordinary_least_squares#Estimation , and the Moore-Penrose Pseudoinverse is the standard method for estimation of these. Wikipedia has an article "Proofs involving the M-P pseudoinverse" showing that in relative brevity. Nov 7 '21 at 12:19
• I suggest studying the classic equalizers, ZF and MMSE.
– MBaz
Nov 7 '21 at 15:38
• Thanks for the replies but the answers actually do not answer my question which was how to get equation (4) out of (2) using the concept of the extended channel matrix. I actually already understood the answer. I couldn't understand how to use ZF with extended H in order to solve the MMSE by using pseudo inverse
– dcs
Nov 7 '21 at 16:06

\begin{align} \underline{H}^H\underline{H}=\begin{bmatrix}H_{R\times T}^H&\sigma I_T\end{bmatrix}\begin{bmatrix}H_{R\times T}\\\sigma I_T\end{bmatrix}=H_{R\times T}^HH_{R\times T}+\sigma^2I_T \tag{a} \end{align}
\begin{align} \underline{H}^H\underline{y}=\begin{bmatrix}H_{R\times T}^H&\sigma I_T\end{bmatrix}\begin{bmatrix}y_{R\times 1}\\0_{T\times 1}\end{bmatrix}=H_{R\times T}^H y_{R\times 1}+0 \tag{b} \end{align}