# Why the transpose of wireless channel vector is taken in SIMO system?

I am getting confused about the received signal expression in case of SIMO wireless channel.

SIMO system=> Here we will have one antenna at Tx and multiple ($$N$$) antenna at Rx.

Therefore received signal will be $$\bar{y}(n) = \bar{h}s(n)+\bar{w}(n) \tag{1}$$

where bar indicates a vector, $$s(n)$$ is transmitted signal, $$\bar{w}(n)$$ is AWGN vector of dimension $$N \times 1$$, $$\bar{h}$$ is channel vector of dimension $$N \times 1$$.

I am confident in writing equation (1). But confusion starts because in many research papers, they take transpose of $$\bar{h}$$ i.e., $$\bar{h}^T$$ and hence the received signal expression becomes

$$\bar{y}(n) = \bar{h}^Ts(n)+\bar{w}(n) \tag{2}$$

My query is which of the equation among (1) and (2) is correct way of writing the expression for received signal.

I would really appreciate any help in overcoming this confusion.

• Can you give an example of the use of (2) in a paper that's accessible to us?
– Peter K.
Jun 1, 2022 at 15:48
• Thank you so much sir for your answer..... arxiv.org/pdf/2112.01336.pdf In this paper there is a base station (BS) with one antenna and intelligent reflecting surface (IRS) with $K$ elements. Therefore, the channel from BS to IRS is SIMO and they have mentioned conjugate transpose of it....
– paru
Jun 1, 2022 at 15:55
• Thanks for the link. I can't see anywhere where the signal model as in your (2) is used. I can see $h_{sn} + \mathbf{h}^H_{rn} \Theta_R \mathbf{h}_{sr}$ in equation (1) of the linked paper, but that's not the same as your equation (2). Which equation in the paper do you mean?
– Peter K.
Jun 1, 2022 at 17:19
• On page number 4 , above equation 2..... It starts with word "Denoting"
– paru
Jun 1, 2022 at 17:29
• That's just so they don't have to write $\mathbf{h}_{sr} = \left [ \begin{array}{c} h^1_{sr}\\ \vdots\\ h^K_{sr} \end{array} \right ]$. That is, they can write it as a row vector rather than a column vector. The row vector takes up less space on the page. The vectur $\mathbf{h}_{sr}$ is still a column vector.
– Peter K.
Jun 1, 2022 at 17:31

The notation before the paper's equation (2) is just so they don't have to write $$\mathbf{h}_{sr} = \left [ \begin{array}{c} h^1_{sr}\\ \vdots\\ h^K_{sr} \end{array} \right ].$$ That is, they can write it as a row vector rather than a column vector. The row vector takes up less space on the page. The vectur $$\mathbf{h}_{sr}$$ is still a column vector.