--- title: "PSD and average power of A cos(2πfct + θ)" subject: "Communication Systems" chapter: 1 key: "semester_6/communication_systems:esd-psd-acos-power" source: https://www.bctengineer.com/q/semester_6/communication_systems/esd-psd-acos-power site: BCT Engineering --- # PSD and average power of A cos(2πfct + θ) Communication Systems · Chapter 1: Introduction - Asked in: 2069 Bhadra (CS I), 4+6 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/esd-psd-acos-power - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1 ## Question Define energy and power spectral density functions. Find power spectral density and average power for the periodic signal defined by g(t) = A cos(2πfct + θ). ## Short answer Energy spectral density Ψ_g(f)=|G(f)|² for energy signals; power spectral density S_g(f)=lim |G_T(f)|²/T for power signals. For g(t)=A cos(2πfct+θ), autocorrelation is (A²/2)cos(2πfcτ), PSD is (A²/4)[δ(f-fc)+δ(f+fc)], average power is A²/2 watts into 1Ω. Phase θ does not affect PSD or power. ## Answer #### Energy spectral density For an **energy signal** $g(t)$ (finite energy, zero average power), the ESD is the squared magnitude of its Fourier transform: $$ \Psi_g(f)=\lvert G(f)\rvert^2\ \ (\text{J/Hz}),\qquad E=\int_{-\infty}^{\infty}\Psi_g(f)df $$ It is also the Fourier transform of the energy autocorrelation $R_g(\tau)=\int g(t)g^*(t-\tau)dt$. #### Power spectral density For a **power signal** (finite non-zero average power, infinite energy), the PSD is the power per unit bandwidth, defined with the truncated signal $g_T(t)$: $$ S_g(f)=\lim_{T\to\infty}\frac{\lvert G_T(f)\rvert^2}{T}\ \ (\text{W/Hz}),\qquad P=\int_{-\infty}^{\infty}S_g(f)df $$ By the Wiener–Khinchin theorem, $S_g(f)$ is the Fourier transform of the time-averaged autocorrelation $R_g(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}g(t)g^*(t-\tau)dt$. #### PSD and average power of $g(t)=A\cos(2\pi f_ct+\theta)$ The signal is periodic with period $T_0=1/f_c$, so the time average over one period is enough. **Step 1 – autocorrelation:** $$ \begin{aligned} R_g(\tau)&=\frac{1}{T_0}\int_{-T_0/2}^{T_0/2}A\cos(2\pi f_ct+\theta)\,A\cos(2\pi f_c(t-\tau)+\theta)\,dt\\ &=\frac{A^2}{2T_0}\int_{-T_0/2}^{T_0/2}\left[\cos(2\pi f_c\tau)+\cos(4\pi f_ct-2\pi f_c\tau+2\theta)\right]dt\\ &=\frac{A^2}{2}\cos(2\pi f_c\tau) \end{aligned} $$ The second cosine integrates to zero over a full period. Note that $R_g(\tau)$ does not depend on $\theta$. **Step 2 – PSD** (Fourier transform of $R_g(\tau)$, using $\cos 2\pi f_c\tau\leftrightarrow\frac{1}{2}[\delta(f-f_c)+\delta(f+f_c)]$): $$ S_g(f)=\frac{A^2}{4}\left[\delta(f-f_c)+\delta(f+f_c)\right] $$ So the PSD consists of two impulses, each of weight $A^2/4$, at $f=\pm f_c$. ```text S(f) A^2/4 ^ | ^ A^2/4 | | | -------+---+---+-------> f -fc 0 fc ``` **Step 3 – average power:** $$ \begin{aligned} P&=\int_{-\infty}^{\infty}S_g(f)\,df=\frac{A^2}{4}+\frac{A^2}{4}=\frac{A^2}{2}\\ P&=R_g(0)=\frac{A^2}{2}\ \ (\text{check}) \end{aligned} $$ **Answer:** $S_g(f)=\frac{A^2}{4}[\delta(f-f_c)+\delta(f+f_c)]$ and average power $P=\frac{A^2}{2}$ (watts into a 1 $\Omega$ load). The phase $\theta$ has no effect on PSD or power. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which type of signal is described by an energy spectral density? A. Finite energy, zero average power B. Finite non-zero power, infinite energy C. Any periodic signal D. Only random noise Answer: A. ESD |G(f)|^2 applies to energy signals (finite energy, zero average power); power signals use PSD. (Key point: ESD vs PSD signal types) 2. For g(t) = A cos(2πfct + θ), the time-averaged autocorrelation R_g(τ) is A. A^2 cos(2πfcτ + θ) B. (A^2/2) cos(2πfcτ) C. (A^2/2) cos(2πfcτ + 2θ) D. A^2/4 cos(2πfcτ) Answer: B. The double-frequency term integrates to zero over a period, leaving (A^2/2)cos(2πfcτ), independent of θ. (Key point: Autocorrelation of a cosine) 3. The PSD of A cos(2πfct + θ) consists of A. One impulse of weight A^2/2 at f = fc B. Two impulses of weight A^2/4 at f = ±fc C. Two impulses of weight A^2/2 at f = ±fc D. A continuous rectangular spectrum Answer: B. The Fourier transform of (A^2/2)cos(2πfcτ) gives (A^2/4)[δ(f-fc)+δ(f+fc)]. (Key point: PSD of a cosine) 4. What is the average power of A cos(2πfct + θ) and how does θ affect it? A. A^2, depends on θ B. A^2/2, depends on θ C. A^2/2, independent of θ D. A^2/4, independent of θ Answer: C. P = R_g(0) = sum of the two impulse weights = A^2/2; the phase θ does not appear. 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