--- title: "DT unit step; periodicity of e^(jωn)" subject: "Signals and Systems" chapter: 1 key: "semester_4/signals_and_systems:dt-unit-step-periodicity" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-unit-step-periodicity site: BCT Engineering --- # DT unit step; periodicity of e^(jωn) Signals and Systems · Chapter 1: Signals - Asked in: 2072 Asoj, 1+3 marks - Video: [Discrete Time Unit Impulse and Step Signal](https://www.youtube.com/watch?v=DQAeINlf_i8) — PMRF Teachings - 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-unit-step-periodicity - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1 ## Question Define discrete time unit step signal. Derive the necessary condition for the signal x[n] = e^(jωn) to be periodic. ## Short answer DT unit step: u[n]=1 for n≥0, 0 for n<0. For x[n]=e^(jωn) to be periodic with integer period N, we need e^(jωN)=1 → ωN=2πm → ω/(2π)=m/N rational. The fundamental period is the smallest such N. ## Answer #### Discrete-time unit step signal The discrete-time unit step $u[n]$ is a sequence that is zero for negative $n$ and one for $n \ge 0$: $$ u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases} $$ It can be written as a sum of shifted impulses, $u[n] = \sum_{k=0}^{\infty} \delta[n-k]$, and it is used to make signals causal (switch them on at $n = 0$). ```text u[n] 1 o o o o o ... | | | | | 0 --o--o--+--+--+--+--+---> n -2 -1 0 1 2 3 4 ``` #### Condition for $x[n] = e^{j\omega n}$ to be periodic $x[n]$ is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$: $$ \begin{aligned} e^{j\omega (n+N)} &= e^{j\omega n} \\ e^{j\omega n}\, e^{j\omega N} &= e^{j\omega n} \\ \Rightarrow e^{j\omega N} &= 1 \end{aligned} $$ $e^{j\theta} = 1$ only when $\theta$ is an integer multiple of $2\pi$, so $$ \omega N = 2\pi m \quad \Rightarrow \quad \frac{\omega}{2\pi} = \frac{m}{N}, \quad m, N \text{ integers} $$ **Condition:** $e^{j\omega n}$ is periodic only if $\omega/2\pi$ is a **rational number**. The fundamental period is the smallest such $N$: write $\omega/2\pi = m/N$ in lowest terms, then $N$ is the period. Example: $\omega = 3\pi/4$ gives $\omega/2\pi = 3/8$, so $N = 8$. But $\omega = 1/6$ gives $\omega/2\pi = 1/(12\pi)$, which is irrational, so $e^{jn/6}$ is **not** periodic (unlike the continuous-time $e^{j\omega t}$, which is periodic for every $\omega$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the DT unit step u[n]? A. 1 for n ≥ 0 and 0 for n < 0 B. 1 only at n = 0 C. 1 for n > 0 and 0 otherwise D. 1 for all n Answer: A. u[n] switches on at n = 0 and stays 1. (Key point: Unit step definition) 2. How can u[n] be written using impulses? A. u[n] = δ[n] only B. u[n] = Σ δ[n-k], k = 0 to ∞ C. u[n] = Σ δ[n+k], k = 0 to ∞ D. u[n] = n δ[n] Answer: B. The step is a train of unit impulses at n = 0, 1, 2, ... (Key point: Step from impulses) 3. Why does e^(jωN) = 1 require ωN = 2πm? A. Because ω is always 2π B. Because e^(jθ) = 0 C. e^(jθ) = 1 only when θ is a multiple of 2π D. Because N = 1 Answer: C. This leads to ω/2π = m/N, which must be rational. (Key point: Periodicity condition) 4. Is e^(jn/6) periodic? A. Yes, with N = 6 B. Yes, with N = 12 C. Yes, with N = 12π D. No, since ω/2π = 1/(12π) is irrational Answer: D. No integer N satisfies N/6 = 2πm. 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