---
title: "Energy/power signals; periodicity of complex exponential"
subject: "Digital Signal Analysis and Processing"
chapter: 1
key: "semester_7/digital_signal_analysis_and_processing:energy-power-periodic-1"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/energy-power-periodic-1
site: BCT Engineering
---

# Energy/power signals; periodicity of complex exponential

Digital Signal Analysis and Processing · Chapter 1: Discrete Time Signals and Systems

- Asked in: 2074 Asoj, 2+2 marks
- Video: [ss15_A03-4 (DT Complex Exponential Signals)](https://www.youtube.com/watch?v=F4eLVAVnz1M) — Feng-Li Lian
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/energy-power-periodic-1
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-1

## Question

Define energy and power type discrete time signal. Check whether signal x[n] = e^(j(πn/3 + π/4)) is periodic or not. If it is periodic, state its periodic time.

## Short answer

Energy signal has finite energy, zero power; power signal has finite nonzero power, infinite energy. x[n]=e^(j(πn/3+π/4)) is periodic because ω₀/2π = 1/6 is rational. Fundamental period N=6 samples.

## Answer

#### Energy and power signals
- **Energy signal:** a signal with finite energy, 0 < E < ∞, where E = Σ |x[n]|² (sum over all n). Its average power is zero. Example: x[n] = (½)ⁿ u[n], E = 1/(1 − ¼) = 4/3.
- **Power signal:** a signal with finite non-zero average power, 0 < P < ∞, where P = lim(N→∞) 1/(2N+1) Σ(n=−N to N) |x[n]|². Its energy is infinite. Example: unit step, and every periodic signal.

A signal cannot be both; some signals (e.g. x[n] = n) are neither.

#### Periodicity of x[n] = e^(j(πn/3 + π/4))
A discrete exponential e^(jω₀n) is periodic only if ω₀/2π is a rational number, i.e. x[n + N] = x[n] for some integer N.

Here ω₀ = π/3, so

f₀ = ω₀/2π = (π/3)/(2π) = 1/6, which is rational.

So the signal **is periodic**. Check: x[n + 6] = e^(j(πn/3 + 2π + π/4)) = x[n].

The fundamental period is N = 1/f₀ = **6 samples** (the phase π/4 does not affect the period).

This signal has |x[n]| = 1, so it is a power signal with P = 1.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which describes an energy signal?
   A. Finite non-zero average power and infinite energy
   B. Finite energy 0 < E < infinity, hence zero average power
   C. Infinite energy and infinite power
   D. Finite energy and non-zero average power
   Answer: B. An energy signal has finite energy, so its average power over infinite time goes to zero. (Key point: Energy signal definition)

2. A discrete exponential e^(jw0 n) is periodic only if
   A. w0/2pi is a rational number
   B. w0 is any real number
   C. w0 is a multiple of pi/4
   D. w0 < pi
   Answer: A. Periodicity needs an integer N with w0 N = 2 pi k, which requires w0/2pi to be rational. (Key point: Periodicity condition)

3. For x[n] = e^(j(pi n/3 + pi/4)), what is the fundamental period?
   A. 3
   B. 12
   C. It is not periodic
   D. 6
   Answer: D. f0 = (pi/3)/(2pi) = 1/6, so N = 6. The phase pi/4 does not change the period. (Key point: Period of the given signal)

4. Since |x[n]| = 1 for all n, this signal is
   A. an energy signal with E = 1
   B. neither energy nor power
   C. a power signal with P = 1
   D. an energy signal with P = 1
   Answer: C. Energy is infinite (sum of 1s), while average power is 1, so it is a power signal. (Key point: Power of periodic signal)

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