--- title: "Periodicity condition for e^(jωn)" subject: "Signals and Systems" chapter: 1 key: "semester_4/signals_and_systems:dt-periodicity-condition" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-periodicity-condition site: BCT Engineering --- # Periodicity condition for e^(jωn) Signals and Systems · Chapter 1: Signals - Asked in: 2079 Jestha, 4 marks; 2074 Bhadra, 3 marks; 2073 Magh, 4 marks (asked 3 times) - Video: [Periodic and Non-Periodic Signals](https://www.youtube.com/watch?v=Gj9pDC_OBQU) — 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/dt-periodicity-condition - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1 ## Question Derive the necessary condition for the discrete time signal x[n] = e^(jωn) to be periodic. ## Short answer A discrete-time complex exponential e^{jωn} is periodic if and only if ω/(2π) is a rational number m/N. Then the fundamental period is N = 2πm/ω, where m is the smallest integer making N an integer. ## Answer A discrete-time signal is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$. For $x[n] = e^{j\omega_0 n}$: $$ \begin{aligned} x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N} \end{aligned} $$ For $x[n+N] = x[n]$ we need $$ e^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer} $$ $$ \frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number} $$ So **$e^{j\omega_0 n}$ is periodic only if $\omega_0/2\pi$ is a rational number.** The fundamental period is the smallest positive integer $N = 2\pi m/\omega_0$, taking $m$ as the smallest integer that makes $N$ an integer (with $m$ and $N$ having no common factor). The same condition holds for $\cos(\omega_0 n+\theta)$ and $\sin(\omega_0 n+\theta)$, because they are sums of such exponentials. This differs from continuous time, where $e^{j\omega_0 t}$ is periodic for **every** $\omega_0 \neq 0$ (period $2\pi/|\omega_0|$). In discrete time $n$ takes only integer values, so $N$ must be an integer. **Example:** $e^{j(3\pi/4)n}$: $\omega_0/2\pi = 3/8$ (rational), so periodic with $N = 8$. But $e^{j2n}$: $\omega_0/2\pi = 1/\pi$ (irrational), so it is **not** periodic. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For x[n] = e^(j w0 n) to have period N, what must hold? A. e^(j w0 N) = 1 B. e^(j w0 N) = -1 C. e^(j w0 N) = j D. w0 N = pi Answer: A. x[n+N] = x[n] e^(j w0 N), so equality for all n requires e^(j w0 N) = 1. (Key point: Periodicity requirement) 2. The condition w0 N = 2 pi m leads to which necessary condition for periodicity? A. w0 must be an integer B. w0 must be a multiple of pi C. w0/(2 pi) must be a rational number D. w0 must be less than pi Answer: C. w0/(2 pi) = m/N, a ratio of integers, so it must be rational. (Key point: Rational frequency condition) 3. Why is e^(j w0 t) periodic for every w0 != 0 in continuous time but not in discrete time? A. Discrete signals have no frequency B. The DT period N must be an integer because n takes only integer values C. CT signals are always real D. DT signals are always aperiodic Answer: B. In CT the period 2 pi/|w0| can be any real number, but in DT the period must be an integer, forcing rational w0/2 pi. (Key point: CT vs DT difference) 4. Which signal is NOT periodic? A. e^(j(3 pi/4)n) B. e^(j2n) C. e^(j(2 pi/5)n) D. e^(j pi n) Answer: B. For e^(j2n), w0/2 pi = 1/pi is irrational, so no integer period exists. 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