Random process; LTI output is WSS if input is WSS

Random process

A random process $X(t, s)$ is a collection (ensemble) of time functions, one for each outcome $s$ of a random experiment. At a fixed time $t_k$, $X(t_k)$ is a random variable; for a fixed outcome $s_j$, $x_j(t)$ is one sample function. Example: the thermal-noise voltage across a resistor, observed on many identical resistors.

 x1(t) ~~~/\~~~/\/~~~~   <- sample function 1
 x2(t) ~/\~~~~\/~~~/\~   <- sample function 2
 x3(t) ~~~\/~/\~~~~~/\   <- sample function 3
          |
          t = tk  ->  X(tk) is a random variable

WSS input gives WSS output

A process is wide-sense stationary (WSS) if (i) its mean is constant and (ii) its autocorrelation depends only on the time difference $\tau = t_1 - t_2$:

$$E[X(t)] = \mu_X, \qquad E[X(t_1)X(t_2)] = R_X(t_1 - t_2)$$

Let a WSS process $X(t)$ be the input of a stable LTI system with impulse response $h(t)$. The output is

$$Y(t) = \int_{-\infty}^{\infty} h(\alpha)\,X(t-\alpha)\,d\alpha$$

1. Mean of the output. Expectation and integration are linear, so they can be interchanged:

$$\begin{aligned} E[Y(t)] &= \int_{-\infty}^{\infty} h(\alpha)\,E[X(t-\alpha)]\,d\alpha \\ &= \mu_X\int_{-\infty}^{\infty} h(\alpha)\,d\alpha = \mu_X\,H(0) \end{aligned}$$

$H(0)$ is the DC gain of the system, a constant. So the output mean is constant, independent of $t$.

2. Autocorrelation of the output.

$$\begin{aligned} R_Y(t_1,t_2) &= E[Y(t_1)Y(t_2)] \\ &= E\left[\int h(\alpha)X(t_1-\alpha)d\alpha \int h(\beta)X(t_2-\beta)d\beta\right] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t_1-\alpha)X(t_2-\beta)]\,d\alpha\,d\beta \\ &= \int\!\!\int h(\alpha)h(\beta)\,R_X(t_1-t_2-\alpha+\beta)\,d\alpha\,d\beta \end{aligned}$$

The last step uses the WSS property of $X(t)$: the input autocorrelation depends only on the difference of its two time arguments, $(t_1-\alpha)-(t_2-\beta)$. Putting $\tau = t_1 - t_2$:

$$R_Y(\tau) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} h(\alpha)h(\beta)\,R_X(\tau-\alpha+\beta)\,d\alpha\,d\beta$$

The right side depends only on $\tau$, not on $t_1$ or $t_2$ separately.

3. Mean square value $E[Y^2(t)] = R_Y(0)$ is a constant, finite for a stable system with finite input power.

Since the output mean is constant and the output autocorrelation is a function of $\tau$ only, $Y(t)$ is also WSS.

Taking the Fourier transform of $R_Y(\tau)$ gives the useful result

$$S_Y(f) = |H(f)|^2\,S_X(f)$$

i.e. the output PSD equals the input PSD times the squared magnitude response.