--- title: "NEB; mean and AC of WSS signal through LTI" subject: "Communication Systems" chapter: 9 key: "semester_6/communication_systems:neb-wssp-lti-mean-ac" source: https://www.bctengineer.com/q/semester_6/communication_systems/neb-wssp-lti-mean-ac site: BCT Engineering --- # NEB; mean and AC of WSS signal through LTI Communication Systems · Chapter 9: Noise in Communication Systems - Asked in: 2070 Asar (CS II), 3+5 marks - Video: [DCLec09:Transmission of Random Process through an LTI System](https://www.youtube.com/watch?v=K2Eed98UhIo) — Zareena Nishanth - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/neb-wssp-lti-mean-ac - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-9 ## Question Define noise equivalent bandwidth. Find mean and AC function at the output when a WSSP signal is passed through the LTI system. ## Short answer Noise equivalent bandwidth is B_N = ∫|H(f)|^2 df / |H(0)|^2. For a WSS input, output mean is constant μ_Y = μ_X H(0), and output autocorrelation R_Y(τ) = h(τ)*h(-τ)*R_X(τ) depends only on τ, so output is WSS. ## 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 (or zero-frequency) gain, that passes the **same noise power** as the actual filter when both are driven by the same white noise. If white noise of PSD $N_0/2$ is applied, actual output power $= \frac{N_0}{2}\int_{-\infty}^{\infty}|H(f)|^2df$ and ideal output power $= \frac{N_0}{2}|H(0)|^2 \cdot 2B_N$. Equating: $$ B_N = \frac{\int_0^{\infty}|H(f)|^2\,df}{|H(0)|^2} $$ ```text |H(f)|^2 |H(0)|^2 +-------+ area of rectangle |\ | = area under |H(f)|^2 | \ | | \__ | | ~~|~~~___ +-------+------------> f 0 BN ``` Example: RC-LPF has $B_N = 1/(4RC) = 1.57 f_{3dB}$. Then output noise power is simply $N_0 B_N |H(0)|^2$. #### Mean and autocorrelation at the output Let a WSS process $X(t)$ (mean $\mu_X$, autocorrelation $R_X(\tau)$) drive a stable LTI system with impulse response $h(t)$: $$ Y(t) = \int_{-\infty}^{\infty} h(\alpha)X(t-\alpha)\,d\alpha $$ **Mean** $$ \mu_Y = E[Y(t)] = \int_{-\infty}^{\infty} h(\alpha)E[X(t-\alpha)]\,d\alpha = \mu_X\int_{-\infty}^{\infty}h(\alpha)\,d\alpha = \mu_X H(0) $$ The output mean is the input mean times the DC gain, a constant. **Autocorrelation** $$ \begin{aligned} R_Y(t, t+\tau) &= E[Y(t)Y(t+\tau)] \\ &= \int\!\!\int h(\alpha)h(\beta)\,E[X(t-\alpha)X(t+\tau-\beta)]\,d\alpha\,d\beta \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} h(\alpha)h(\beta)\,R_X(\tau+\alpha-\beta)\,d\alpha\,d\beta \end{aligned} $$ This depends only on $\tau$, so $Y(t)$ is also WSS. In compact form: $$ R_Y(\tau) = h(\tau) * h(-\tau) * R_X(\tau) $$ **Power spectral density** (Fourier transform, using $h(-\tau) \leftrightarrow H^*(f)$): $$ S_Y(f) = H(f)H^*(f)S_X(f) = |H(f)|^2 S_X(f) $$ **Output power** $$ E[Y^2(t)] = R_Y(0) = \int_{-\infty}^{\infty}|H(f)|^2S_X(f)\,df $$ For white input, $S_X = N_0/2$, this gives $N_0|H(0)|^2B_N$, which links back to the NEB. ## 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 an ideal rectangular filter that A. has the same 3 dB point B. has the same gain and passes the same noise power for white noise input C. passes the same signal power D. has the same phase response Answer: B. NEB is defined by equal output noise power from the same white noise with the same reference gain. (Key point: Definition of NEB) 2. The NEB B_N equals A. the integral of |H(f)| over f, divided by |H(0)| B. the integral of |H(f)|^2 from 0 to infinity, divided by |H(0)|^2 C. |H(0)|^2 times the integral of |H(f)|^2 D. the integral of H(f)^2 from 0 to infinity Answer: B. Equating actual and ideal output power gives B_N = (integral from 0 to infinity of |H|^2 df) / |H(0)|^2. (Key point: NEB formula) 3. A WSS process with mean mu_X passes through an LTI system. The output mean is A. mu_X B. mu_X H(0) C. mu_X |H(0)|^2 D. 0 Answer: B. The mean is mu_X times the integral of h(t), which is the DC gain H(0). (Key point: Output mean) 4. In compact form, R_Y(tau) and S_Y(f) are A. R_X(tau) * h(tau) and H(f) S_X(f) B. h(tau) * h(-tau) * R_X(tau) and |H(f)|^2 S_X(f) C. h(tau) h(-tau) R_X(tau) and |H(f)| S_X(f) D. h(tau) * R_X(tau) and |H(f)|^2 S_X(f) Answer: B. Convolving with h(tau) and h(-tau) corresponds to multiplying the PSD by H(f)H*(f) = |H(f)|^2. (Key point: Output autocorrelation and PSD) ## Related in Communication Systems - [Stochastic process; WSS random signal through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Astochastic-wss-through-lti) — Communication Systems, chapter 9, asked 2× - [LTI output of a WSS input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Alti-wssp-output) — Communication Systems, chapter 9 - [Moments of random variable; NEB of RC-LPF; output noise power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amoments-neb-rc-lpf) — Communication Systems, chapter 9 - [Random process; LTI output is WSS if input is WSS](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Arandom-process-lti-wssp) — Communication Systems, chapter 9 - [Noise equivalent bandwidth; matched filter is time reversed](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aneb-matched-filter-reverse) — Communication Systems, chapter 9 - [Ergodic process; WSS random signals through LTI](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aergodic-wss-through-lti) — Communication Systems, chapter 9, asked 2× ## Related in other subjects - [LTI output, h[n]=(1/2)^n window](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-window-1) — Digital Signal Analysis and Processing, chapter 1, asked 4× - [LTI output, h[n]=(1/3)^n window](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-window-2) — Digital Signal Analysis and Processing, chapter 1, asked 4× - [LTI output with sparse samples, verify](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-sparse-samples) — Digital Signal Analysis and Processing, chapter 1, asked 4× - [LTI output for h=2u[n]−2u[n−4]](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-step-difference) — Digital Signal Analysis and Processing, chapter 1, asked 4×