--- title: "Noise equivalent bandwidth; optimum detector impulse response" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:neb-optimum-detector" source: https://www.bctengineer.com/q/semester_6/communication_systems/neb-optimum-detector site: BCT Engineering --- # Noise equivalent bandwidth; optimum detector impulse response Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2081 Bhadra (CS II), 4+6 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-optimum-detector - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Explain what Noise Equivalent Bandwidth represents of a filter. Derive the impulse response of the optimum detector in the presence of additive white noise. ## Short answer Noise equivalent bandwidth B_N of a filter is the bandwidth of an ideal rectangular filter with the same peak gain that passes identical noise power under white noise input. The optimum detector for a known pulse in additive white Gausian noise is the matched filter; maximising output SNR via the Schwarz inequality yields impulse response hopt(t) = k s(T−t), a time-reversed and delayed pulse. ## 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: an RC low-pass filter has $B_N=\frac{1}{4RC}=\frac\pi2f_{3dB}$. #### Impulse response of the optimum detector An **optimum detector** is a receiver (filter followed by sampler and threshold device) that makes the decision with the **minimum probability of error**. For a known pulse in additive white Gaussian noise, this is achieved by the filter that **maximises the output signal-to-noise ratio at the sampling instant**; this optimum filter is 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 is noise equivalent bandwidth? A. Bandwidth of an ideal rectangular filter with the same gain that passes the same noise power B. The 3 dB bandwidth C. The bandwidth where gain is zero D. Twice the signal bandwidth Answer: A. It equates the noise power of the real and the ideal filters. (Key point: NEB meaning) 2. Which formula gives B_N? A. Integral of |H| df B. Integral from 0 to infinity of |H|^2 df divided by |H(0)|^2 C. |H(0)|^2 divided by the integral of |H|^2 D. f_3dB / 2 Answer: B. Equating N0 times that integral with N0 B_N |H(0)|^2 gives it. (Key point: NEB formula) 3. What is B_N for an RC low-pass filter? A. 1/(2 pi RC) B. 1/RC C. 1/(4RC) = (pi/2) f_3dB D. f_3dB Answer: C. It is larger than the 3 dB bandwidth by pi/2. (Key point: RC example) 4. What is the optimum detector's impulse response? A. h(t) = k s(t) B. h(t) = k s(t + T) C. h(t) = k / s(t) D. h(t) = k s(T - t) Answer: D. Time reversal and a delay of T. 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