LTI and Deterministic Channel

Can someone explain me the statistical relationship of random signals with a linear time-invariant and deterministic channel to the output of the channel when having random signal as an input?

If your input "random signal" is modeled as a zero-mean wide-sense-stationary random process $$\{X(t)\}$$ with autocorrelation function $$R_X(t)$$, then the output of the channel (which is basically an LTI system with impulse response $$h(t)$$) is also a zero-mean wide-sense-stationary process $$\{Y(t)\}$$ with autocorrelation function $$R_Y(t)$$ given by $$R_Y = R_h \star R_X$$ where $$R_h$$ is the autocorrelation function of the impulse response $$h(t)$$ $$\big($$that is, $$R_h(t) = \int_{-\infty}^\infty h(t)h(t+\tau) \, \mathrm d\tau$$ or $$\int_{-\infty}^\infty h(t)h(t-\tau) \, \mathrm d\tau$$ for left-handed folks$$\big)$$ and $$\star$$ notes convolution. Indeed, since both $$R_h$$ and $$R_X$$ are even functions, that convolution can be written as a cross-correlation if you like, which makes for a nice mantra to murmur to impress your boss during your presentation: "The output autocorrelation function is found by cross-correlating the input autocorrelation with the channel autocorrelation".

If you prefer power spectral densities, then $$S_Y(f) = |H(f)|^2S_X(f)$$ where $$H(f)$$ is the channel transfer function.

• Does this change in case i have a limited bandwidth? – Shkodrani Jan 9 at 8:28
• No ${}{}{}{}{}$ – Dilip Sarwate Jan 10 at 2:39
• Bless you mate! – Shkodrani Jan 10 at 6:41

If the channel (LTI filter) is $$h(t)$$ and the input signal is $$X(t)$$, then the output of the channel is $$Y(t) = X(t)\star h(t) = \int_{-\infty}^{\infty}X(\tau)h(t-\tau)\,d\tau = \int_{-\infty}^{\infty}X(t-\tau)h(\tau)\,d\tau$$ where $$\star$$ is the convolution.

EDIT: Statistical relationships between the input and output signals:

1- The mean of the output signal:

$$E[Y(t)] = \int_{-\infty}^{\infty}h(t-\tau)E[X(\tau)]\,d\tau$$

2- The cross-correlation between the input and output signals:

$$R_{YX}(t_1, t_2) = E[Y(t_1)X(t_2)] = \int_{-\infty}^{\infty} h(t_1-\tau)\underbrace{E[X(\tau)X(t_2)]}_{R_{XX}(\tau,\, t_2)}\,d\tau$$

where $$R_{XX}(t_1,\,t_2)$$ is the auto-correlation function of the random signal $$X(t)$$.

3- The auto-correlation function of the output signal

$$R_{YY}(t_1, t_2) = E[Y(t_1)Y(t_2)] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}h(t_1-\tau_1)h(t_2-\tau_2)\underbrace{E[X(\tau_1)X(\tau_2)]}_{R_{XX}(\tau_1,\,\tau_2)}\,d\tau_1\,d\tau_2$$

• The question is about "random signals" $x(t)$ and so perhaps a little more explanation is needed? – Dilip Sarwate Jan 9 at 0:21
• @DilipSarwate My mistake. I thought the question was about the instantaneous output of the filter. I edited my answer for the statistical relationships. Thanks – BlackMath Jan 9 at 8:18
• Does this explanation change in case i have a limited bandwidth, or not? – Shkodrani Jan 9 at 8:27