---
title: "DT periodicity; energy/power of 5cosπt + sin5πt"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:dt-periodicity-energy-power-sinusoids"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-periodicity-energy-power-sinusoids
site: BCT Engineering
---

# DT periodicity; energy/power of 5cosπt + sin5πt

Signals and Systems · Chapter 1: Signals

- Asked in: 2075 Bhadra, 3+3 marks
- Video: [Find whether the following signals are periodic or not, If periodic find the fundamental period.](https://www.youtube.com/watch?v=9b-yKbDVNUo) — Engg-Course-Made-Easy
- 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-energy-power-sinusoids
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Derive necessary condition for a discrete time signal to be periodic. Determine whether the following signals are energy or power signal f(t) = 5 cos πt + sin 5πt

## Short answer

A discrete-time signal is periodic iff its frequency ratio ω₀/(2π) is rational. The signal f(t)=5cosπt+sin5πt is continuous-time; periods are 2 s and 0.4 s with rational ratio, making it periodic with T₀=2 s. Its average power is P=13 W, finite and nonzero, so f(t) is a power signal.

## Answer

#### Necessary condition for DT periodicity
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.

#### $f(t) = 5\cos\pi t + \sin 5\pi t$

$T_1 = 2\pi/\pi = 2$ s and $T_2 = 2\pi/5\pi = 0.4$ s; the ratio is 5 (rational), so $f(t)$ is periodic with $T_0 = 2$ s. A periodic signal has infinite energy, so check its power:

$$
\begin{aligned}
P &= \frac{1}{T_0}\int_0^{T_0}f^2(t)\,dt \\
&= \frac{1}{2}\int_0^{2}\left[25\cos^2\pi t + \sin^2 5\pi t + 10\cos\pi t\sin 5\pi t\right]dt \\
&= \frac{1}{2}\left[25 + 1 + 0\right] = 13
\end{aligned}
$$

(The cross term integrates to zero over a common period; each sinusoid of amplitude $A$ contributes $A^2/2$: $25/2 + 1/2 = 13$.)

$E = \infty$.

**Answer:** $f(t)$ is a **power signal** with $P = 13$ W (normalised).

## Check yourself

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

1. Why must ω0/2π be rational for a DT exponential to be periodic?
   A. Because ω0 must be an integer
   B. Because n is a real number
   C. Because e^(jω0 n) is always 1
   D. Because ω0 N = 2πm must hold with integers m and N
   Answer: D. x[n+N] = x[n] requires e^(jω0 N) = 1, so ω0/2π = m/N. (Key point: Periodicity rationale)

2. What is the fundamental period of f(t) = 5 cos πt + sin 5πt?
   A. 2 s
   B. 0.4 s
   C. 1 s
   D. 10 s
   Answer: A. T1 = 2 s and T2 = 0.4 s with ratio 5, so the common period is 2 s. (Key point: Period of f(t))

3. What is the power of f(t) = 5 cos πt + sin 5πt?
   A. 26
   B. 13
   C. 25
   D. 6
   Answer: B. Each sinusoid contributes A^2/2 and the cross term averages to 0: 25/2 + 1/2 = 13. (Key point: Power of sum of sinusoids)

4. How is f(t) classified?
   A. Energy signal, since P is finite
   B. Energy signal, since E is finite
   C. Power signal, since E is infinite and P is finite
   D. Neither
   Answer: C. A periodic signal has infinite energy but finite non-zero average power. (Key point: Classification)

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