--- title: "Autocorrelation of cos(200πt) at τ = π/4" subject: "Communication Systems" chapter: 1 key: "semester_6/communication_systems:psd-esd-autocorrelation-cos200pit" source: https://www.bctengineer.com/q/semester_6/communication_systems/psd-esd-autocorrelation-cos200pit site: BCT Engineering --- # Autocorrelation of cos(200πt) at τ = π/4 Communication Systems · Chapter 1: Introduction - Asked in: 2072 Asoj (CS I), 2+3 marks - Video: [Autocorrelation and Power Spectral Density (PSD) Examples in Digital Communications](https://www.youtube.com/watch?v=XWytSLZZP1A) — Iain Explains Signals, Systems, and Digital Comms - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/psd-esd-autocorrelation-cos200pit - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1 ## Question Define power and energy spectral density function. Given a signal x(t) = cos(200πt), find the auto correlation of x(t) at τ = π/4. ## Short answer The energy spectral density Ψ_x(f)=|X(f)|² (J/Hz) shows energy distribution for energy signals; power spectral density S_x(f)=lim_{T→∞}|X_T(f)|²/T (W/Hz) for power signals. For x(t)=cos(200πt), autocorrelation R_x(τ)=0.5 cos(200πτ). At τ=π/4, R_x≈−0.484 W. ## Answer #### Power and energy spectral density - **Energy spectral density (ESD):** for an energy signal, $\Psi_x(f) = |X(f)|^2$ (J/Hz); it shows how energy is distributed over frequency, and $E = \int\Psi_x(f)df$. - **Power spectral density (PSD):** for a power signal, $S_x(f) = \lim_{T\to\infty}\frac{|X_T(f)|^2}{T}$ (W/Hz); it shows how average power is distributed over frequency, and $P = \int S_x(f)df$. Each is the Fourier transform of the corresponding autocorrelation function. #### Autocorrelation of $x(t) = \cos(200\pi t)$ at $\tau = \pi/4$ $x(t)$ is periodic ($f_0 = 100$ Hz, $T_0 = 0.01$ s), so it is a power signal and we use the time-average autocorrelation: $$ \begin{aligned} R_x(\tau) &= \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}\cos(200\pi t)\cos(200\pi(t-\tau))\,dt \\ &= \lim_{T\to\infty}\frac{1}{2T}\int_{-T/2}^{T/2}\left[\cos(200\pi\tau) + \cos(400\pi t - 200\pi\tau)\right]dt \\ &= \frac12\cos(200\pi\tau) \end{aligned} $$ (the second term averages to zero). At $\tau = \pi/4$ s: $$ \begin{aligned} R_x\!\left(\frac{\pi}{4}\right) &= \frac12\cos\left(200\pi\cdot\frac{\pi}{4}\right) = \frac12\cos(50\pi^2) \\ &= \frac12\cos(493.48\ \text{rad}) = \frac12\times(-0.9689) \\ &\approx -0.484 \end{aligned} $$ ($493.48$ rad reduces to $493.48 - 78\times2\pi = 3.392$ rad, and $\cos 3.392 = -0.9689$.) **Answer:** $R_x(\tau) = \frac12\cos(200\pi\tau)$, so $R_x(\pi/4) = \frac12\cos(50\pi^2) \approx -0.484$ W (normalised). For comparison $R_x(0) = 0.5$ W = average power. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement is right? A. ESD is for power signals and PSD for energy signals B. ESD |X(f)|^2 has unit J/Hz; PSD has unit W/Hz C. Both have unit W D. ESD is the Fourier transform of x(t) Answer: B. ESD (J/Hz) is for energy signals; PSD (W/Hz) is for power signals. (Key point: ESD vs PSD) 2. Why is cos(200 pi t) treated with the time-average autocorrelation? A. It is an energy signal B. It is zero for t>0 C. It is periodic, so it is a power signal with infinite energy D. Its Fourier transform does not exist Answer: C. A sinusoid has finite power and infinite energy. (Key point: Signal type) 3. What is R_x(tau) for x(t)=cos(200 pi t)? A. cos(200 pi tau) B. (1/2) cos(200 pi tau) C. (1/2) cos(400 pi tau) D. 2 cos(200 pi tau) Answer: B. Using the product-to-sum identity, the double-frequency term averages to zero leaving (1/2)cos(200 pi tau). (Key point: Autocorrelation) 4. What does R_x(0) represent, and what is its value here? A. Average power, 0.5 W B. Total energy, infinity C. Peak amplitude, 1 D. Bandwidth, 100 Hz Answer: A. R_x(0)=P=1/2 W for a unit-amplitude cosine. 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