Making the MIMO virtual array size appear in the FFT and Capon angle resolution

The received signal at snapshot $l$ is: \begin{equation} X(:,l) = a_{r}(\theta)\beta(\theta)a_{t}^{T}(\theta)S(:,l)+ Z(:,l) \end{equation}

Where $X \in \mathbb{C}^{M\times L}$ is the receive signal matrix (M receive antennas, L snapshots), $a_r(\theta) \in \mathbb{C}^{M \times 1}$ is the receive steering vector, $a_t(\theta) \in \mathbb{C}^{N \times 1}$ is the transmit steering vector, $\beta(\theta)$ is the reflection coefficient (scalar), $S \in \mathbb{C}^{N\times L}$ is the transmit signal matrix, $Z \in \mathbb{C}^{M\times L}$ is the noise matrix.

As I said in my comment, in a MIMO system, if you transmit orthogonal waveforms, you can form a virtual array because you can have $N$ matched filters in each of your $M$ receive elements. The idea is that, since the waveforms are orthogonal, each of the $N$ matched filters on each element will produce a unique response since the cross correlation with all but one of the transmit waveforms will produce an inner product of 0, giving $N$ independent looks on each channel.

Mathematically, after matched filtering the signal model becomes: \begin{equation} Y = XS^{H} = a_{r}(\theta)\beta(\theta)a_{t}^{T}(\theta)R_{ss}+ZS^{H} \end{equation} where $R_{ss} = \frac{1}{L}SS^{H}$. Thus, $Y \in \mathbb{C}^{M\times N}$.

The virtual array steering vector is then the kronecker product of the transmit and receive steering vectors. Thus if we let $Y_{v} = \text{vec}(Y)$, we get \begin{equation} Y_{v} = \left(a_{t}(\theta)\otimes a_{r}(\theta)\right)\beta(\theta) + z_{v} \in\mathbb{C}^{MN\times 1} \end{equation}

Now, looking at the Capon equation they give \begin{equation} \hat{\beta}_{Capon}\left(\theta\right) = \frac{a_{r}^{H}(\theta)\hat{R}^{-1}XS^{H}a_{t}^{*}(\theta)}{L\left[a_{r}^{H}(\theta)\hat{R}^{-1}a_{r}(\theta)\right]\left[a_{t}^{T}(\theta)R_{ss}a_{t}^{*}(\theta)\right]} \end{equation} where $\hat{R} = \frac{1}{L}XX^{H}$, the virtual array becomes obvious by recognizing that $\hat{R}^{-1}XS^{H}$ = $R_{v}^{-1}Y_{v}$ where $R_{v} = E\left[Y_{v}Y_{v}^{H}\right]$. This is only available to us since we transmitted orthogonal waveforms, so from $Y = XS^{H}$ we can say \begin{equation} X = Y\left(S^{H}\right)^{+} = Y\left(SS^{H}\right)^{-1}S = Y(I)^{-1}S = YS \end{equation} where $I$ is the identity matrix, $\left(\cdot\right)^{+}$ is the pseudo-inverse, and we make use of the orthogonality condition for our transmit waveforms \begin{equation} \left\langle s_{n},s_{\bar{n}}\right\rangle = s_{n}s_{\bar{n}}^{H} = \sum_{l=0}^{L-1}s(n,l)s^{*}(\bar{n},l) = \delta_{n,\bar{n}} \end{equation} In other words, our transmit waveforms provide an orthonormal basis for the receive signal on each channel. Thus the Capon estimate can be rewritten as \begin{equation} \beta_{Capon}(\theta) = \frac{a_{v}^{H}(\theta)R_{v}^{-1}Y_{v}}{a_{v}^{H}(\theta)R_{v}^{-1}a_{v}(\theta)} \end{equation} where $a_{v}(\theta) = a_{t}(\theta)\otimes a_{r}(\theta)$, which is an $MN\times 1$ steering vector. Thus, the $MN$ virtual array was already in the equation, it was just difficult to see if you don't know the typical amplitude Capon formula. The typical formula for amplitude Capon doesn't have the second steering vector in the numerator. So, the fact that the second steering vector is there, which is necessary since $\hat{R}^{-1}XS^{H}$ is a matrix as opposed to a vector, shows the virtual array existence. In reality, however, it would be quite difficult I would think to estimate $R_{v}$, so it would be better left as the original form.