It should be obvious that if there is no interference but only zero-mean random noise, as it is discussed in many textbooks. When there is known deterministic interference, what is the capacity now?

In details, the discrete channel model is as follows: $$\mathbf{y}(n) = \mathbf{H}\mathbf{x}(n) + \mathbf{i}(n) +\mathbf{n}(n),$$ where $\mathbf{y}(n)$ is the received signal vector at time index $n$, $\mathbf{x}(n)$ $\mathbf{i}(n)$ $\mathbf{n}(n)$ are the transmitted waveform, interference signal vector and the noise for time $n$, $\mathbf{H}$ is the channel which is known and does not change. I believe the answer should be $\log|\mathbf{I}+\mathbf{R}_n^{-1}\mathbf{H}\mathbf{R}_{x}\mathbf{H}^H|$, where $\mathbf{R}_n$ is the covariance of the noise. The effect of interference is directly removed from the receiver. Is this correct?

Another problem is what if the values of the interference are unknown to the receiver and transmitter but they only know the value for any time index is deterministic? What can we say about the capacity? What will the receiver and transmitter do in practice? will the receiver try to estimate the interference?


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