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The received signal of MIMO system is given as: $$ y= H \cdot x +n, (*)$$ where $H$ is a channel matrix.

$h(\tau_l)$ is a channel impulse response. It is determined as a matrix $\in \mathbb{C}^{Nr \times Nt}$, $\tau_l$ is delay.

My question: $H$ in (*) is a frequency response, fourier transform of $h(\tau_l)$, isnt't?

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No, $H_{rx,tx}$ is the channel matrix with $(r,t)^{th}$ element representing channels attenuation and phase rotation on transmit symbols $\vec{x}_{tx}$, from $t^{th}$ transmit antenna to $r^{th}$ receive antenna. The equation is neither in time-domain nor in frequency domain. It is on transmit symbols. For example, in OFDM systems, you get $y_{rx}$ (received symbols) only after FFT operation at the receiver, so you can understand the equation as in frequency domain. But for a Non-OFDM based system, like CDMA, you get received symbols $y_{rx}$ after demodulation with matched-filter and that would be in time-domain.

You can understand this equation as a relationship between $rx$ number of received symbols $y_{rx}$ and $tx$ number of transmitted symbols $x_{tx}$ in MIMO scenario where $n_{rx}$ is the AWGN for each of the $rx$ receive chain and $H_{rx,tx}$ is the channel matrix affecting the symbol transmitted from $t^{th}$ antenna and received at $r^{th}$ antenna. Since symbols are represented as vector points complex plane, hence each channel can only affect each symbol by changing its magnitude and giving a random phase rotation. It doesnot matter time-domain or frequency-domain. That depends on your Receive chain implementation of getting $\vec{y_{rx}}$.

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