# What is impulse response in simple AWGN channel?

Assume a sender and receiver communicate through an AWGN channel. Let $$x(t)$$ be a transmitted signal and $$y(t)$$ be a received signal and $$z(t)$$ be the Gaussian noise. It is known that $$y(t) = kx(t-\Delta t) + z(t)$$ where $$k$$ is the channel gain (say due to attenuation) and $$\Delta t$$ is the delay. I was wondering what is the channel impulse response? I assume it should be $$h(t) = k\delta(t - \Delta t) + z(t)$$? Is it correct?

A linear system cannot add anything new to its input signal, so additive noise is modeled separately from any linear distortion of the signal. In your case with just scaling and delay, the channel impulse response is $$h(t)=k\delta(t-\Delta t)$$, and the complete channel is modeled by first filtering the input signal with an LTI system with impulse response $$h(t)$$ and then adding the noise.
• I know what you mean but saying a linear system cannot add anything to its input signal is misleading I think. You could add another time shifted and scaled version of the input to itself, like a system whose impulse response is $h(t)=k_1\delta(t - \Delta t_1) + k_2\delta(t - \Delta t_2)$. Jan 17 at 17:25