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For a flat fading channel (SISO system) we have: $y=h \star s+n$

But for a mimo system, we have: $y=Hx+n$ with $H \in \mathbb{C}^{N_r \times N_t}$

Why is there not the product of convolution for a mimo system?

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    $\begingroup$ what are $H$ and $h$? Their meanings answer your question. $\endgroup$ – AlexTP Apr 15 '18 at 13:40
  • $\begingroup$ can you explain alextp? I know that $H=(h_{ij})$ $\endgroup$ – user35104 Apr 15 '18 at 15:35
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I think it should be:

$$ \boldsymbol{y}_{i} = {H}_{i} \boldsymbol{x} + \boldsymbol{n}_{i} $$

For the $ i $ -th antenna and $ {H}_{i} $ being the convolution matrix of this specific channel.
You could write this in a Matrix Form where $ H $ becomes a tensor.
In case $ \forall i, j, \; {H}_{i} = {H}_{j} $ then it can be made simpler.

You could also work on the samples:

$$ \boldsymbol{y} \left[ m \right] = H \left[ m \right] \boldsymbol{x} \left[ m \right] + n $$

Where in this case $ {y}_{i} \left[ n \right] $ is the $ i $ -th antenna at time sample $ m $ and the $ i $ -th row of $ H \left[ m \right] $ is the response of the channel $ i $ at time $ m $.
This model allows time variant channels.

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Convolution or simple product are used according to the channel's frequency selectivity, regardless transmitter and receiver antenna configurations (SISO or MIMO system).

If the channel is frequency selective (within source signal bandwidth, the channel is a filter like), then you have to use convolution between channel's impulse response and the source signal. Else (frequency flat fading channel), you have to use the simple product. So your first equation (SISO system) didn't describe a frequency flat fading channel.

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