--- title: "Random process; LTI output is WSS if input is WSS" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:random-process-lti-wssp" source: https://www.bctengineer.com/q/semester_6/communication_systems/random-process-lti-wssp site: BCT Engineering --- # Random process; LTI output is WSS if input is WSS Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2071 Shrawan (CS II), 2+6 marks - Video: [Transmission of a Random Process through a linear filter](https://www.youtube.com/watch?v=5f3L6k64Ywk) — Ali Muqaibel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/random-process-lti-wssp - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Define random process? Show that the output of LTI is WSSP if the input is also a WSSP. ## Short answer A random process is an ensemble of time functions, each associated with an outcome of a random experiment. For a WSS input, the output mean is constant (μ_X H(0)) and the autocorrelation depends only on time difference τ, so the output is also WSS. ## Answer #### 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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a random process X(t, s), what is X(t_k) at a fixed time t_k? A. A sample function B. A random variable C. A constant D. An autocorrelation function Answer: B. At a fixed time the process is a random variable; fixing the outcome instead gives a sample function. (Key point: Random process definition) 2. Which pair of conditions defines a wide-sense stationary process? A. Constant mean and autocorrelation depending only on time difference B. Zero mean and unit variance C. Constant PDF at every time instant D. Constant autocorrelation for all time differences Answer: A. WSS needs a constant mean and R_X(t1,t2) = R_X(t1 - t2). (Key point: WSS conditions) 3. If a WSS input with mean mu_X drives an LTI system with impulse response h(t), the output mean is A. mu_X times the integral of h^2(t) B. mu_X times H(0), a constant C. mu_X times H(f) at the signal frequency D. zero always Answer: B. Interchanging expectation and integration gives mu_X times the integral of h, which is mu_X H(0), independent of t. (Key point: Output mean) 4. Why does R_Y(t1,t2) depend only on tau = t1 - t2? A. Because h(t) is causal B. Because the input autocorrelation depends only on the difference (t1 - alpha) - (t2 - beta) C. Because the output mean is zero D. Because the system is non-linear Answer: B. E[X(t1-a)X(t2-b)] = R_X(t1 - t2 - a + b) by WSS, so the double integral is a function of tau alone. (Key point: Output autocorrelation depends on tau only) ## Related in Communication Systems - [LTI output of a WSS input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Alti-wssp-output) — Communication Systems, chapter 9 - [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× - [Ergodic process; WSS random signals through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aergodic-wss-through-lti) — Communication Systems, chapter 9, asked 2× - [NEB; mean and AC of WSS signal through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aneb-wssp-lti-mean-ac) — Communication Systems, chapter 9 - [Moments of random variable; NEB of RC-LPF; output noise power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amoments-neb-rc-lpf) — Communication Systems, chapter 9 - [Parseval's theorem; output PSD of LTI system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aparseval-power-lti-psd) — Communication Systems, chapter 1 ## Related in other subjects - [LTI output for 2δ(n) − δ(n−1) input](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-impulse-input) — Digital Signal Analysis and Processing, chapter 1 - [LTI output for h(t)=e^(-2t), rectangular input](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Alti-output-rect-input) — Data Communication, chapter 2, asked 3× - [LTI output, h[n]=(1/2)^n window](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-window-1) — Digital Signal Analysis and Processing, chapter 1, asked 4× - [LTI output, h[n]=(1/3)^n window](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-window-2) — Digital Signal Analysis and Processing, chapter 1, asked 4×