--- title: "Matched filter impulse response is reverse delayed input" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:matched-filter-reverse-delayed" source: https://www.bctengineer.com/q/semester_6/communication_systems/matched-filter-reverse-delayed site: BCT Engineering --- # Matched filter impulse response is reverse delayed input Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2073 Shrawan (CS II), 8 marks; 2071 Shrawan (CS II), 6 marks (asked 2 times) - Video: [Matched Filter - Need, Properties, Decision Threshold, & Impulse Response Derivation.](https://www.youtube.com/watch?v=XpGiDG0Bj2Y) — PASSIONATE PROFESSOR - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/matched-filter-reverse-delayed - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Prove that the impulse response of the matched filter is reverse delayed version of the input signal. ## Short answer Using Schwarz inequality, the optimum frequency response is H(f)=k S*(f) e^{-j2πfT}. Inverse Fourier transform gives h(t)=k s(T-t), confirming the impulse response is the time-reversed and delayed copy of the input pulse, sampled at T. ## Answer A **matched filter** is a linear filter whose impulse response is matched to (a time-reversed, delayed copy of) a known input pulse $s(t)$, $h(t)=k\,s(T-t)$, so that it gives the **maximum possible peak signal-to-noise ratio** at the sampling instant $t=T$ when the pulse is received in additive white noise. It is the optimum detector for digital signals in AWGN; in practice it is built as a correlator. **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. **Example:** for a rectangular pulse $s(t)=A$, $0\le t\le T$, $h(t)=kA$ for $0\le t\le T$, which is the same rectangle; it behaves as an **integrate-and-dump** circuit. The output is a triangle with peak $kA^2T$ at $t=T$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the matched filter achieve? A. Minimum bandwidth of the output pulse B. Maximum peak output SNR at t = T for a known pulse in white noise C. Exact recovery of the pulse shape at the output D. Removal of all noise from the signal Answer: B. It maximises the peak SNR at the sampling instant; it does not preserve pulse shape or eliminate noise. (Key point: Matched filter purpose) 2. What is the maximum output SNR of the matched filter? A. 2E/N0 B. E/N0 C. E/(2 N0) D. 2E N0 Answer: A. The Schwarz bound gives (2/N0) times the integral of |S|^2, which is 2E/N0 by Rayleigh's theorem. (Key point: Maximum SNR) 3. When does the Schwarz inequality hold with equality here? A. When H(f) is constant B. When H(f) = S(f) C. When H(f) = k S*(f) e^(-j2 pi f T) D. When the noise PSD is zero Answer: C. Equality needs phi1 = k phi2*, giving H as the conjugate spectrum with a delay factor. (Key point: Equality condition) 4. For a rectangular pulse of duration T, what does the matched filter reduce to? A. A differentiator B. An integrate-and-dump circuit with a triangular output peaking at t = T C. A high-pass filter with a zero at T D. An ideal low-pass filter with output peak at T/2 Answer: B. The reversed delayed rectangle is the same rectangle, so h(t) is rectangular and the output is a triangle peaking at T. 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