--- title: "Even/odd signals; energy or power check" subject: "Signals and Systems" chapter: 1 key: "semester_4/signals_and_systems:even-odd-energy-power-check" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/even-odd-energy-power-check site: BCT Engineering --- # Even/odd signals; energy or power check Signals and Systems · Chapter 1: Signals - Asked in: 2082 Kartik, 3+4 marks - Video: [Even and Odd Signals](https://www.youtube.com/watch?v=c_YLkvk_ZXI) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/even-odd-energy-power-check - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1 ## Question Define even and odd signal with necessary diagram. Check whether the following signal is energy or power signal. (i) x(t) = 5A cos(ωt + φ) (ii) x[n] = sin(n/4) ## Short answer Even: x(-t)=x(t), symmetric about vertical axis; odd: x(-t)=-x(t), antisymmetric. (i) 5A cos(ωt+φ) is periodic, average power P = 25A^2/2, so it is a power signal. (ii) sin(n/4) is aperiodic but bounded, its average power is 1/2; hence also a power signal. ## Answer #### Even and odd signals - **Even:** $x(-t) = x(t)$, symmetric about the vertical axis. Example: $\cos t$. - **Odd:** $x(-t) = -x(t)$, antisymmetric about the origin, $x(0) = 0$. Example: $\sin t$. ```text Even: x(t) = |t| Odd: x(t) = t \ | / | / \ | / | / \ | / | / --------+--------> t ---------+---------> t 0 / | / | mirror image about 180 deg symmetry vertical axis about origin ``` Any signal can be split as $x_e(t) = \frac{1}{2}[x(t)+x(-t)]$, $x_o(t) = \frac{1}{2}[x(t)-x(-t)]$. #### (i) $x(t) = 5A\cos(\omega t + \phi)$ It is periodic with $T = 2\pi/\omega$, so compute power over one period: $$ \begin{aligned} P &= \frac{1}{T}\int_0^{T} 25A^2\cos^2(\omega t+\phi)\,dt \\ &= \frac{25A^2}{T}\int_0^{T}\frac{1+\cos(2\omega t+2\phi)}{2}\,dt = \frac{25A^2}{2} \end{aligned} $$ $E = \infty$ (infinite duration). **Power signal**, $P = 12.5A^2$. #### (ii) $x[n] = \sin(n/4)$ Here $\omega_0 = 1/4$, $\omega_0/2\pi = 1/(8\pi)$ is irrational, so the sequence is **not periodic**, but it is bounded and lasts forever. $$ \begin{aligned} P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\sin^2\frac{n}{4} \\ &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\frac{1-\cos(n/2)}{2} = \frac{1}{2} \end{aligned} $$ because the sum of $\cos(n/2)$ stays bounded while $2N+1 \to \infty$. Energy $E = \sum\sin^2(n/4) = \infty$. **Answer:** (i) power signal, $P = 25A^2/2$; (ii) power signal (aperiodic), $P = 1/2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The average power of 5A cos(wt + phi) is: A. 5A^2 B. 25A^2/2 C. 25A^2 D. 5A/2 Answer: B. Amplitude 5A gives P = (5A)^2/2 = 12.5 A^2. (Key point: Power of 5A cos(wt+phi)) 2. Why is a periodic cosine a power signal, not an energy signal? A. Its energy is zero B. It lasts forever, so E is infinite while P is finite C. Its power is infinite D. It is odd Answer: B. Each period adds equal energy forever, giving E = infinity but finite average power. (Key point: Periodic implies power) 3. Is x[n] = sin(n/4) periodic? A. Yes, N = 8 B. Yes, N = 4 C. No, since w0/2pi = 1/(8 pi) is irrational D. No, because it is odd Answer: C. No integer N satisfies N/4 = 2 pi m, so it is aperiodic. (Key point: sin(n/4) periodicity) 4. The signal sin(n/4) is: A. An energy signal B. A power signal with P = 1/2 C. Neither D. A power signal with P = 1 Answer: B. sin^2 = (1 - cos(n/2))/2 averages to 1/2 because the cosine sum stays bounded; energy is infinite. 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