I'm a little confused about how 'rich' the scattering environment should be for MIMO to work (i.e. for the channel matrix to be full rank). For instance, let's say we have a Tx and Rx with two antennas each, and under the two-ray model (or the flat/plane earth model), one path is direct from the Tx to the Rx and another path follows a reflection from the ground.

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Can MIMO be used here? In other words, can two separate spatial streams be obtained?


The Channel Impulse Response (CIR) for a beamforming vector $\mathbf{w} \in \mathbb{C}^{N_T \times 1}$ (which corresponds to your spatial streams) is the following

$$\mathbf{c}(n) = \sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \mathbf{w} \delta(n - \tau_k) \tag{1}$$ where $q$ is the number of paths (is $2$ due to the two-ray propagation model), $\alpha_k$ is the complex gain of the $k^{th}$ path, $ \mathbf{a}_{R,k}$ and $\mathbf{a}_{T,k}$ are the receive and transmit steering vectors of the $k^{th}$ path (depending on the antenna placement and the Angles-of-Arrival and Angles-of-Departures of the path) and finally $\tau_k$ is the propagation delay of the $k^{th}$ path.

I can see that in your case $\mathbf{a}_{T,k} \in \mathbb{C}^{2 \times 1},\mathbf{a}_{R,k} \in \mathbb{C}^{2 \times 1},\mathbf{w} \in \mathbb{C}^{2 \times 1}$. Note that equation $1$ could be written as $$\mathbf{c}(n) = \Big(\sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \delta(n - \tau_k) \Big) \mathbf{w}$$ So in order to "equalize" to get $\mathbf{w}$, you need to know the matrix $\mathbf{H} = \sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \delta(n - \tau_k) $.

  • $\begingroup$ @Ahmad Perhaps two spatial streams can be obtained based on the nature of H? In other words, if its rank is 2, then 2 streams can be resolved separately? $\endgroup$ – V-Red Aug 1 '19 at 12:21
  • $\begingroup$ @V-Red indeed. You could see the spatial streams in the vector $\mathbf{w}$ as in a precoding matrix $\mathbf{P}$ multiplied by vector of streams $\mathbf{s}$, i.e.$\mathbf{w} = \mathbf{P}\mathbf{s}$. Hence $$\mathbf{c}(n) = \Big(\sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top\mathbf{P} \delta(n - \tau_k) \Big) \mathbf{s}$$ Then your $\mathbf{H} = \sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \mathbf{P} \delta(n - \tau_k)$ and now you can start arguing on the rank of $\mathbf{H}$. $\endgroup$ – Ahmad Bazzi Aug 1 '19 at 12:24

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