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I have a real orthogonal design for codeword...

$$ \begin{bmatrix} s1 & s2 \\ -s2 & s1 \end{bmatrix} $$

$$ s=\begin{bmatrix} s1\\ s2 \end{bmatrix} $$

$$ Y^T=\sqrt{\frac{P}{Mt}} h^T X + n^T $$

...and receive signal:

$$ Y1=\begin{bmatrix} y1\\ y2 \end{bmatrix} $$

So for:

$$ X = \begin{bmatrix} x1 & x2 \\ -x2 & x1 \end{bmatrix} $$

and $$ Y = \begin{bmatrix} y1 & y2 \end{bmatrix} $$

...i have:

$$ Y=\sqrt{\frac{P}{Mt}} \cdot \begin{bmatrix}h1 & h2\\ h2 & -h1\end{bmatrix}\cdot \begin{bmatrix}x1 & x2\end{bmatrix}^T+N $$

...right?

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    $\begingroup$ Can you please check that the formulas have been transferred as you intended them to be? $\endgroup$ – A_A Mar 14 '18 at 8:50
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Your analysis is correct. I would write it as follows: if you define the transmitted matrix as \begin{equation*} X = \begin{bmatrix} x_1 & -x_2 \\ x_2 & x_1 \end{bmatrix} \end{equation*} with one column transmitted at each symbol interval (that is, first you transmit $[x_1, x_2]$ and then $[-x_2, x_1]$), then the system can be written as $$Y=hX$$ with $h = [h_1, h_2]$, or equivalently it can be written as $$Y = Hx$$ with \begin{equation*} H = \begin{bmatrix} h_1 & h_2 \\ -h_2 & h_1 \end{bmatrix} \end{equation*} and $x=[x_1, x_2]^T$.

Of course, you can also transpose all vectors and matrices and arrive at equivalent results.

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  • $\begingroup$ what is the number of a time snapshots I need to transmit the 2 symbols? Is it 2? how can I provide an explanation of it? $\endgroup$ – LenaPark Oct 29 '18 at 11:08
  • $\begingroup$ @LenaPark You need two "symbol periods" to transmit two symbols. One column of $X$ is transmitted every period. Over two periods, you transmit a total of four symbols (the four elements of $X$), but two are repeated, so the effective rate is two symbols over two periods. $\endgroup$ – MBaz Oct 29 '18 at 13:43
  • $\begingroup$ to determine how many time snapshots I need I should look at a number of symbols. If I have 3 symbols, then I need 3 snapshots, right? $\endgroup$ – LenaPark Oct 30 '18 at 12:42
  • $\begingroup$ Not exactly -- you look at the number of columns of the code matrix $X$. Each column is transmitted in a different symbol period. Each element of the column is transmitted by a different antenna. $\endgroup$ – MBaz Oct 30 '18 at 12:55
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If you transmit $[x_1\,\,\,x_2]$ in the first time slot, and $[-x_2\,\,\,x_1]$ in the second time slot, the received signals will be

$$y_1=h_1x_1+h_2x_2+n_1$$

and

$$y_2=-h_1x_2+h_2x_1+n_2$$

This can be written in matrix-vector form as

$$\mathbf{y}=\mathbf{H}\mathbf{x}+\mathbf{n}$$.

where

$$\mathbf{H}=\begin{pmatrix}h_1&h_2\\h_2&-h_1\end{pmatrix}$$

and $$\mathbf{x}=\begin{pmatrix}x_1\\x_2\end{pmatrix}$$

Note that although the channel coefficients maybe complex-valued, the only part that will affect the transmission is the real part since the signals are real. In this case

$$\mathbf{H}^T\mathbf{H}=(h_1^2+h_2^2)\mathbf{I}_{2\times 2}$$

which means that the two signals can be completely decoupled, and detected optimally separately.

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