--- title: "Realization of matched filter" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:matched-filter-realization" source: https://www.bctengineer.com/q/semester_6/communication_systems/matched-filter-realization site: BCT Engineering --- # Realization of matched filter Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2070 Asar (CS II), 4 marks - Video: [DC#24 Matched Filter Receiver in digital communication || EC Academy](https://www.youtube.com/watch?v=LxCrF2fPc7E) — EC Academy - 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-realization - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Realize the matched filter with relevant mathematical support. ## Short answer The matched filter impulse response is h(t) = k s(T-t), derived from Schwarz's inequality, giving maximum SNR = 2E/N_0. Realizations include a correlator, an integrate-and-dump circuit, or a tapped delay line. ## Answer A **matched filter** is the linear filter that maximises the output peak-signal-to-noise ratio at the sampling instant $T$ for a known pulse $s(t)$ in white noise. Its impulse response is $h(t) = k\,s(T-t)$. **Derivation (brief).** Input $s(t) + n(t)$, noise PSD $N_0/2$. At $t = T$: $$ \left(\frac{S}{N}\right)_o = \frac{\left|\int H(f)S(f)e^{j2\pi fT}df\right|^2}{\frac{N_0}{2}\int|H(f)|^2df} \le \frac{\int|H|^2df\int|S|^2df}{\frac{N_0}{2}\int|H|^2df} = \frac{2E}{N_0} $$ by Schwarz's inequality. Equality holds when $H(f) = kS^*(f)e^{-j2\pi fT}$, i.e. $h(t) = k\,s(T-t)$, giving $(S/N)_{max} = 2E/N_0$. **Realization 1: Integrate-and-dump (rectangular pulse).** For $s(t) = A$, $0 \le t \le T$: $h(t) = kA$ for $0 \le t \le T$, a rectangular impulse response, which is an integrator over the last $T$ seconds: $$ y(T) = \int_0^{T} r(t)\,dt $$ ```text R S1 (dump) r(t)--/\/\--+--(-A+)--+-------+---> y(T) | op-amp| | | +---||----+ +-/-+ short C C at end of T ``` An op-amp integrator (or RC with $RC \gg T$) integrates over one bit; the output is sampled at $T$, then the capacitor is shorted (dumped) so the next bit starts from zero. **Realization 2: Correlator.** Since $$ y(T) = \int_0^T r(\tau)\,h(T-\tau)\,d\tau = k\int_0^T r(\tau)\,s(\tau)\,d\tau $$ the matched filter output at $t=T$ equals the correlation of $r(t)$ with a stored replica of $s(t)$: ```text r(t) --->(X)--->[ integrate 0..T ]--> sample at T --> y(T) ^ s(t) (local replica) ``` **Realization 3: Tapped delay line** (for general pulses): delays, tap weights equal to samples of $s(T-t)$, and a summer, giving a digital (FIR) matched filter. Note: correlator and matched filter give the same value only at $t = T$; their outputs differ at other times. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The impulse response of the matched filter for a pulse s(t) sampled at t = T is A. k s(t) B. k s(T - t) C. k s(-t) D. k ds/dt Answer: B. Equality in Schwarz's inequality needs H(f) = k S*(f) e^(-j2 pi f T), i.e. h(t) = k s(T - t). (Key point: Matched filter impulse response) 2. The maximum peak output SNR of a matched filter in white noise of PSD N_0/2 is A. E/N_0 B. 2E/N_0 C. E/(2N_0) D. 4E/N_0 Answer: B. Schwarz's inequality bounds the SNR by 2E/N_0, reached by the matched filter. (Key point: Maximum output SNR) 3. Why is the capacitor dumped (shorted) at the end of each bit in an integrate-and-dump matched filter? A. To amplify the output B. So that the next bit's integration starts from zero C. To remove the carrier D. To reduce the noise variance Answer: B. The output is sampled at T and then the integrator is reset, so charge from one bit does not carry into the next. 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