# channel impulse and frequency response: H?

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?

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 $$\vec{y_{rx}}$$.