In a MIMO system we have: $y= \mathbf{H} \star \mathbf{x} +n$ with $x \in \mathbb{C}^{N_t \times 1}$ ($N_t$ the number of transmission antennas) and $\star$ the convolution product.

For the matrix, $\mathbf{H}$ we have: $\mathbf {H}=(h_{i,j}(t,\tau))$ with $i \in \{1,...,N_r \}$, $j \in \{1,...,N_t \}$, $t$ the time and $\tau$ the delay for the impulse response

Here is my question: What is the dimension of $h_{i,j}(t,\tau)$?It's a number or a vector because in many book, there is $size(\mathbf{H})=N_r \times N_t$. But if $h_{i,j} (t,\tau)$ is a vector, it's false.

An example is article Survey of channel and radio propagation models for wireless MIMO systems.

  • $\begingroup$ I suppose $N_r$ is the number of receive antennas. Could you please cite the paragraph/book that said "$y= \mathbf{H} \star \mathbf{x} +n$ where $\star$ is the convolution product" and "$\mathbf {H}=(h_{i,j}(t,\tau))$ with $i \in \{1,...,N_r \}$, $j \in \{1,...,N_t \}$, $t$ the time and $\tau$ the delay for the impulse response"? $\endgroup$ – AlexTP Apr 28 '18 at 8:19
  • $\begingroup$ If you are talking about Eq (8) in your article, in two dimensional space defined by TX-RX antenna, $h_{ij}(t,\tau)$ can be thought as a scalar. In time dimension, $h_{ij}(t,\tau)$ itself is a continuous function. For $N_t=1$, you have the classic function of LTI systems. $\endgroup$ – AlexTP Apr 28 '18 at 14:57
  • $\begingroup$ For example, if I have $h_{ij}(t,\tau)=h_{i,j}(\tau)$ then $h_{i,j}(\tau)$ is a scalar. And if I have $N_t=1$ then $\mathbf{H} \in \mathbb{C}^{N_r \times 1}$ Is that correct or not? $\endgroup$ – user35419 Apr 28 '18 at 18:57
  • $\begingroup$ @user35419 No, only $\mathbf{H}(\tau) \in \mathbb{C}^{N_r \times 1}$. If the channel is flat, you come up to Eq (9) in your article. Please read! The article is quite clear. $\endgroup$ – AlexTP Apr 29 '18 at 8:35

Sorry, I'm starting again. $\mathbf{H(t,\tau)}=(h_{ij}(t,\tau))$.

If I have $\mathbf{H}(t,\tau)=\mathbf{H(\tau)}$ and $N_t=1$ then $\mathbf{H}(\tau) \in \mathbb{C}^{N_r \times 1}$

  • 2
    $\begingroup$ Your revised question no longer contains a question. It also seems to contain less information than the original. If you would like assistance, then you will need to provide clear information and clarify what your question is exactly. $\endgroup$ – hops Apr 29 '18 at 17:06
  • $\begingroup$ You should edit your original question, rather than put any modifications as an "answer". "Answers" are supposed to be answers to your question. $\endgroup$ – auspicious99 Aug 16 '20 at 17:04

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