--- title: "Ergodic process; matched filter impulse response" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:ergodic-matched-filter" source: https://www.bctengineer.com/q/semester_6/communication_systems/ergodic-matched-filter site: BCT Engineering --- # Ergodic process; matched filter impulse response Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2081 Baisakh (CS II), 2+6 marks - Video: [What is a Matched Filter?](https://www.youtube.com/watch?v=Ci-EjiMJo3I) — Iain Explains Signals, Systems, and Digital Comms - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/ergodic-matched-filter - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question What is Ergodic Stochastic Process? Derive the expression of impulse response of the matched filter. ## Short answer An ergodic stochastic process is a stationary process whose time averages equal its ensemble averages, allowing measurement of mean and correlation from a single sample function. The matched filter impulse response for a known pulse s(t) in white noise is obtained by maximising SNR with the Schwarz inequality, giving hopt(t) = k s(T−t), a reversed and delayed pulse. ## Answer #### Ergodic stochastic process A stationary random process is **ergodic** if its **time averages** (computed from a single sample function over a long time) are equal to its **ensemble averages** (computed across all sample functions at one time): $$ \langle x(t)\rangle=\lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)dt=E[X(t)]=m_X $$ and similarly $\langle x(t)x(t+\tau)\rangle=R_X(\tau)$. Ergodicity lets us measure mean, power and autocorrelation of noise from one long record (e.g. with a DC meter or power meter). Every ergodic process is stationary, but not every stationary process is ergodic. #### Impulse response of the matched filter **Set-up.** The received signal is $x(t)=s(t)+w(t)$, $0\le t\le T$, where $s(t)$ is a known pulse and $w(t)$ is white noise of two-sided PSD $N_0/2$. It passes through an LTI filter $h(t)\leftrightarrow H(f)$ and is sampled at $t=T$: ```text s(t)+w(t) +--------+ y(t) sample y(T) ------------>| h(t) |---------->--/ ----> decision +--------+ at t=T ``` Output $y(t)=s_o(t)+n_o(t)$ with $$ s_o(T)=\int_{-\infty}^{\infty}H(f)S(f)e^{j2\pi fT}df,\qquad E[n_o^2]=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df $$ We want $H(f)$ that maximises the output peak signal-to-noise ratio $$ \eta=\frac{\lvert s_o(T)\rvert^2}{E[n_o^2]}=\frac{\left\lvert\int H(f)S(f)e^{j2\pi fT}df\right\rvert^2}{\frac{N_0}{2}\int\lvert H(f)\rvert^2df} $$ **Schwarz inequality:** $\left\lvert\int\phi_1\phi_2\,df\right\rvert^2\le\int\lvert\phi_1\rvert^2df\int\lvert\phi_2\rvert^2df$, with equality only when $\phi_1(f)=k\,\phi_2^*(f)$. Put $\phi_1=H(f)$ and $\phi_2=S(f)e^{j2\pi fT}$: $$ \eta\le\frac{\int\lvert H\rvert^2df\int\lvert S\rvert^2df}{\frac{N_0}{2}\int\lvert H\rvert^2df}=\frac{2}{N_0}\int_{-\infty}^{\infty}\lvert S(f)\rvert^2df=\frac{2E}{N_0} $$ (using Rayleigh's theorem, $E$ = pulse energy). The maximum is reached when $$ H_{opt}(f)=k\,S^*(f)\,e^{-j2\pi fT} $$ **Impulse response.** Take the inverse transform. For real $s(t)$, $S^*(f)=S(-f)$, which is the transform of $s(-t)$; the factor $e^{-j2\pi fT}$ is a delay of $T$: $$ \begin{aligned} h_{opt}(t) &= k\int_{-\infty}^{\infty}S^*(f)e^{-j2\pi fT}e^{j2\pi ft}df \\ &= k\int_{-\infty}^{\infty}S(-f)e^{-j2\pi f(T-t)}df \\ &= k\,s(T-t) \end{aligned} $$ So the optimum (matched) filter impulse response is the input pulse **reversed in time and delayed by $T$**. Choosing the sampling instant $T$ equal to the pulse duration makes $h(t)$ causal. ```text s(t) h(t) = s(T - t) |\ /| | \ / | | \ / | ---+---+---> t ---+---+---> t 0 T 0 T ``` **Results:** maximum SNR $=2E/N_0$ depends only on pulse energy, not on pulse shape; the filter's frequency response is the conjugate of the pulse spectrum, so it passes strongly the frequencies where the signal is strong. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What makes a process ergodic? A. Time averages of one sample function equal ensemble averages B. Its mean is zero C. Its PSD is flat D. Its samples are independent Answer: A. Ergodicity lets one long record give mean, power and ACF. (Key point: Ergodic definition) 2. Which statement is correct? A. Every stationary process is ergodic B. Every ergodic process is stationary, but not every stationary process is ergodic C. Ergodic processes need not be stationary D. The two are equivalent Answer: B. Ergodicity is the stronger property. (Key point: Ergodic vs stationary) 3. What transfer function maximises the output SNR? A. H(f) = k S*(f) e^(-j2 pi f T) B. H(f) = k S(f) e^(j2 pi f T) C. H(f) = k / S(f) D. H(f) = k |S(f)|^2 Answer: A. Equality in the Schwarz inequality needs H to be proportional to the conjugate of S(f)e^(j2 pi f T), i.e. k S*(f) e^(-j2 pi f T). (Key point: Optimum transfer function) 4. What is the impulse response of the optimum detector? A. h(t) = k s(t - T), the pulse simply delayed by T B. h(t) = k s(t), the pulse itself C. h(t) = k s(-t) with no delay D. h(t) = k s(T - t), the pulse reversed in time and delayed by T Answer: D. The conjugate gives time reversal s(-t) and the exponential gives a delay T, so h(t) = k s(T - t). 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