--- title: "Noise equivalent bandwidth; matched filter is time reversed" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:neb-matched-filter-reverse" source: https://www.bctengineer.com/q/semester_6/communication_systems/neb-matched-filter-reverse site: BCT Engineering --- # Noise equivalent bandwidth; matched filter is time reversed Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2073 Chaitra (CS II), 10+2 marks - Video: [Matched Filter- Communication Systems](https://www.youtube.com/watch?v=DKHOfYFwK6Q) — sam bytes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/neb-matched-filter-reverse - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Explain noise equivalent bandwidth. Prove that the impulse response of matched filter is a time reversed delayed version of input signal Si(t). ## Short answer Noise equivalent bandwidth is the bandwidth of an ideal rectangular filter with the same peak gain that passes the same noise power from white noise. The matched filter impulse response is h(t)=k s(T-t), obtained by maximizing output SNR via Schwarz inequality, yielding H_opt(f)=k S*(f)e^{-j2πfT}. ## Answer #### Noise equivalent bandwidth The **noise equivalent bandwidth (NEB)** $B_N$ of a filter is the bandwidth of an **ideal rectangular filter**, with the same peak (usually zero-frequency or centre) gain $\lvert H(0)\rvert$, that passes the **same total noise power** as the actual filter when both are fed with white noise. White noise $N_0/2$ into $H(f)$ gives output power $$ P_{actual}=\frac{N_0}{2}\int_{-\infty}^{\infty}\lvert H(f)\rvert^2df=N_0\int_0^{\infty}\lvert H(f)\rvert^2df $$ The ideal filter of gain $\lvert H(0)\rvert$ and bandwidth $B_N$ gives $P_{ideal}=N_0B_N\lvert H(0)\rvert^2$. Equating: $$ B_N=\frac{\int_0^{\infty}\lvert H(f)\rvert^2df}{\lvert H(0)\rvert^2} $$ ```text |H(f)|^2 |H(0)|^2 +---------+ equal areas | ... | | .. | | .|.. | | ...... --+---------+----------> f 0 B_N ``` NEB lets us write the output noise power simply as $N=N_0B_N\lvert H(0)\rvert^2$ (or $kTB_N$ per unit gain), which is used in noise figure and SNR calculations. For a band-pass filter $\lvert H(0)\rvert$ is replaced by the centre-frequency gain $\lvert H(f_0)\rvert$. **Example: RC low-pass filter.** For the RC low-pass filter (series $R$, shunt $C$): $$ H(f)=\frac{1}{1+j2\pi fRC},\qquad \lvert H(f)\rvert^2=\frac{1}{1+(2\pi fRC)^2},\qquad \lvert H(0)\rvert=1 $$ $$ \begin{aligned} B_N &= \int_0^{\infty}\frac{df}{1+(2\pi fRC)^2}\qquad (\text{let } u=2\pi fRC,\ df=\tfrac{du}{2\pi RC}) \\ &= \frac{1}{2\pi RC}\int_0^{\infty}\frac{du}{1+u^2}=\frac{1}{2\pi RC}\cdot\frac{\pi}{2} \\ &= \frac{1}{4RC} \end{aligned} $$ Since the 3 dB bandwidth is $f_{3dB}=\frac{1}{2\pi RC}$, $$ B_N=\frac{\pi}{2}f_{3dB}\approx1.57\,f_{3dB} $$ The NEB is larger than the 3 dB bandwidth because the RC response rolls off slowly (only $-20$ dB/decade) and passes noise well beyond $f_{3dB}$. Output noise power $=N_0B_N=\frac{N_0}{4RC}$. **Example: ideal filter.** For an ideal LPF of bandwidth $B$, $B_N=B$ exactly; for practical filters with steep skirts (e.g. high-order Butterworth), $B_N$ approaches the 3 dB bandwidth. #### Impulse response of matched filter is time-reversed, delayed input **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 gives the most weight to the frequencies where the signal is strongest. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The noise equivalent bandwidth of a filter is the bandwidth of: A. The ideal rectangular filter with the same peak gain that passes the same white noise power B. The band between the half-power points C. The band where the filter phase is linear D. The band carrying 99% of the signal energy Answer: A. NEB equates the noise power of the real filter with an ideal rectangular filter of equal peak gain. (Key point: Definition of NEB) 2. For a single RC low-pass filter, the noise equivalent bandwidth is: A. 1/(2πRC) B. 1/(4RC), which is (π/2) f_3dB C. 1/(RC) D. π/(4RC) Answer: B. The integral of 1/(1+(2πfRC)^2) from 0 to infinity is 1/(4RC) = 1.57 f_3dB. (Key point: NEB of RC low-pass) 3. By the Schwarz inequality, the maximum output peak SNR of the matched filter is: A. E/N0 B. 2E/N0, depending only on pulse energy C. 2E/N0 times the pulse duration D. Dependent on pulse shape, highest for rectangular pulses Answer: B. eta ≤ (2/N0) times the integral of |S(f)|^2 df = 2E/N0, independent of the pulse shape. (Key point: Maximum SNR of matched filter) 4. Why is h_opt(t) = k s(T - t) a time-reversed, delayed copy of s(t)? A. S*(f) gives s(-t) and the factor e^(-j2πfT) delays it by T B. S*(f) gives s(t) and e^(-j2πfT) advances it by T C. S*(f) is the derivative of s(t) D. The Schwarz inequality forces h to be zero before t = T Answer: A. For real s(t), S*(f) = S(-f) is the transform of s(-t), and e^(-j2πfT) adds the delay T. 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