Question

For a WSS random process, the autocorrelation $R_x(\tau)$ as $\tau \rightarrow \infty$ approaches:

• A. Zero always
• B. Mean square value
• C. Square of the mean
• D. Signal power

Hint:

Remember these two critical reference points for the autocorrelation function of a WSS process: 1. At $\tau = 0$: $R_x(0) = E[X^2(t)] \rightarrow$ Mean square value (Total Power). 2. At $\tau \rightarrow \infty$: $R_x(\infty) = (E[X(t)])^2 \rightarrow$ Square of the mean (DC Power component).

Answer 1

The Correct Option is C

Concept: For a Wide-Sense Stationary (WSS) random process $X(t)$, the autocorrelation function is defined as: $$R_x(\tau) = E[X(t)X(t+\tau)]$$ When the time separation $\tau$ grows infinitely large ($\tau \rightarrow \infty$), the two random variables $X(t)$ and $X(t+\tau)$ become completely independent of one another for any physical, non-periodic ergodic random process. From fundamental probability theory, if two random variables $A$ and $B$ are statistically independent, the expectation of their product equals the product of their individual mathematical expectations: $$E[AB] = E[A] \cdot E[B]$$ Step-by-step Proof:
• Apply the independence condition as the time separation $\tau$ approaches infinity: $$\lim_{\tau \rightarrow \infty} R_x(\tau) = \lim_{\tau \rightarrow \infty} E[X(t)X(t+\tau)] = E[X(t)] \cdot E[\lim_{\tau \rightarrow \infty} X(t+\tau)]$$
• Since the process is Wide-Sense Stationary, its mean value is fully constant over all time steps, meaning $E[X(t)] = \mu_x$ and $E[X(t+\tau)] = \mu_x$.
• Substituting this value back into our limit statement: $$\lim_{\tau \rightarrow \infty} R_x(\tau) = \mu_x \cdot \mu_x = \mu_x^2$$ The term $\mu_x^2$ is mathematically designated as the square of the mean.