---
title: "Ergodic process; WSS random signals through LTI"
subject: "Communication Systems"
chapter: 9
key: "semester_6/communication_systems:ergodic-wss-through-lti"
source: https://www.bctengineer.com/q/semester_6/communication_systems/ergodic-wss-through-lti
site: BCT Engineering
---

# Ergodic process; WSS random signals through LTI

Communication Systems · Chapter 9: Noise in Communication Systems

- Asked in: 2074 Chaitra (CS II), 2+10 marks
- Video: [DCLec09:Transmission of Random Process through an LTI System](https://www.youtube.com/watch?v=K2Eed98UhIo) — Zareena Nishanth
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/ergodic-wss-through-lti
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9

## Question

What do you mean by Ergodic Stochastic Process? Explain with necessary derivation passage of wide-sense random signals through a LTI.

## Short answer

An ergodic process has time averages equal to ensemble averages, so mean and autocorrelation can be measured from a single record. Passing WSS X(t) through LTI h(t) gives output Y with mean m_X H(0) and autocorrelation R_Y(τ)=h(τ)*h(-τ)*R_X(τ), keeping Y WSS; output PSD is |H(f)|²S_X(f).

## Answer

#### Ergodic stochastic process

A stationary random process is **ergodic** if its **time averages**, taken over one long sample function, equal its **ensemble averages**, taken across all sample functions at one instant:

$$
\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)\,dt=E[X(t)]=m_X,\qquad \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)x(t+\tau)\,dt=R_X(\tau)
$$

Ergodicity allows the mean (DC value), power and autocorrelation of noise to be measured from a single long record. An ergodic process must be stationary, but a stationary process need not be ergodic.

#### Passage of WSS random signals through an LTI system

Let a WSS process $X(t)$ (mean $m_X$, autocorrelation $R_X(\tau)$, PSD $S_X(f)$) be applied to a stable LTI system with impulse response $h(t)$ and $H(f)$:

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

```text
  X(t) (WSS)   +--------------+   Y(t)
 ------------->| h(t) , H(f)  |------------->
  R_X, S_X     +--------------+  R_Y, S_Y
```

**1. Mean of output.** Expectation and integration can be interchanged:

$$
\begin{aligned}
E[Y(t)] &= \int_{-\infty}^{\infty}h(\alpha)E[X(t-\alpha)]\,d\alpha = m_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha \\
&= m_X\,H(0) \quad(\text{a constant, independent of } t)
\end{aligned}
$$

**2. Autocorrelation of output.**

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

because $X$ is WSS, its autocorrelation depends only on the time difference $(t+\tau-\beta)-(t-\alpha)=\tau+\alpha-\beta$. The result depends only on $\tau$, not on $t$:

$$
R_Y(\tau)=h(\tau)*h(-\tau)*R_X(\tau)
$$

**3. Mean-square value** $E[Y^2]=R_Y(0)=\int\!\!\int h(\alpha)h(\beta)R_X(\alpha-\beta)d\alpha d\beta$ is finite and constant.

Since the mean is constant and the autocorrelation depends only on $\tau$, **$Y(t)$ is also wide-sense stationary**.

**4. Output PSD.** Taking the Fourier transform of $R_Y(\tau)$ ($h(\tau)\to H(f)$, $h(-\tau)\to H^*(f)$):

$$
S_Y(f)=H(f)H^*(f)S_X(f)=\lvert H(f)\rvert^2S_X(f)
$$

and the output power is $P_Y=R_Y(0)=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2S_X(f)\,df$. The phase of $H(f)$ has no effect on the output PSD.

**Example (white noise through RC low-pass filter).** $S_Y(f)=\frac{N_0/2}{1+(2\pi fRC)^2}$, $R_Y(\tau)=\frac{N_0}{4RC}e^{-\lvert\tau\rvert/RC}$, output power $\frac{N_0}{4RC}$.

**Cross-correlation:** $R_{XY}(\tau)=h(\tau)*R_X(\tau)$; for white-noise input, $R_{XY}(\tau)=\frac{N_0}{2}h(\tau)$, so correlating output with input measures $h(t)$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does it mean for a stationary random process to be ergodic?
   A. Its time averages over one long sample function equal its ensemble averages
   B. Its mean is always zero
   C. Its autocorrelation is a delta function
   D. Every stationary process is automatically ergodic
   Answer: A. Ergodicity means one long sample function gives the same mean and autocorrelation as averaging over the ensemble. A stationary process need not be ergodic. (Key point: Meaning of ergodic)

2. If a WSS process with mean m_X passes through an LTI system with transfer function H(f), the output mean is:
   A. m_X H(f)
   B. m_X times the integral of |h(t)|^2
   C. m_X H(0)
   D. zero for any input
   Answer: C. E[Y] = m_X times the integral of h(α) dα, which is m_X H(0), a constant. (Key point: Output mean of LTI)

3. For a WSS input, which relations hold for the output?
   A. R_Y(τ) = h(τ)*R_X(τ) and S_Y = H S_X
   B. R_Y(τ) = h(τ)*h(-τ)*R_X(τ) and S_Y(f) = |H(f)|^2 S_X(f)
   C. R_Y(τ) = R_X(τ) and S_Y(f) = S_X(f)
   D. R_Y(τ) = h(τ) R_X(τ) and S_Y(f) = |H(f)| S_X(f)
   Answer: B. Double convolution with h(τ) and h(-τ) transforms to H(f)H*(f) = |H(f)|^2 multiplying S_X(f). (Key point: Output autocorrelation and PSD)

4. Why is the output Y(t) also wide-sense stationary?
   A. Because h(t) is real
   B. Because the mean is constant and R_Y depends only on τ, since R_X depends on the time difference
   C. Because Y(t) is ergodic
   D. Because the system has no memory
   Answer: B. Y has constant mean m_X H(0) and its autocorrelation depends only on τ, which are exactly the WSS conditions. (Key point: Output is WSS)

## Close variants set in other exams

- [Stochastic process; WSS random signal through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Astochastic-wss-through-lti) — Communication Systems, chapter 9, asked 2×

## Related in Communication Systems

- [Random process; LTI output is WSS if input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arandom-process-lti-wssp) — Communication Systems, chapter 9