---
title: "Average power of sin(πn/3) via DTFS"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:dtfs-parseval-sine-power"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-parseval-sine-power
site: BCT Engineering
---

# Average power of sin(πn/3) via DTFS

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2075 Bhadra, 2+5 marks
- Video: [The Discrete-Time Fourier Series of a Signal by Inspection](https://www.youtube.com/watch?v=2_k3lUCrYBg) — Adam Panagos
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-parseval-sine-power
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

State Parseval's relation for discrete time Fourier series. Find the average power of discrete time periodic signal x[n] = sin(πn/3) using Fourier series coefficient.

## Short answer

Parseval's relation: (1/N) Σ_n |x[n]|^2 = Σ_k |a_k|^2. For x[n]=sin(πn/3) with period 6, coefficients are a_1 = -j/2, a_5 = j/2, others zero. Average power = |a_1|^2 + |a_5|^2 = 0.5 W.

## Answer

#### Parseval's relation for the DTFS
For a periodic $x[n]$ with period $N$ and coefficients $a_k$:

$$
\frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2
$$

The average power over one period equals the sum of the powers $|a_k|^2$ of the $N$ harmonic components.

#### Average power of $x[n] = \sin(\pi n/3)$
**Period:** $\omega_0 = \pi/3$, $\frac{\omega_0}{2\pi} = \frac{1}{6}$, so $N = 6$ and the fundamental is $2\pi/6 = \pi/3$.

**Coefficients:**

$$
x[n] = \frac{1}{2j}e^{j(2\pi/6)n} - \frac{1}{2j}e^{-j(2\pi/6)n}
$$

$$
a_1 = \frac{1}{2j} = -\frac{j}{2}, \qquad a_{-1} = a_5 = -\frac{1}{2j} = \frac{j}{2}
$$

and $a_0 = a_2 = a_3 = a_4 = 0$ in one period.

**Power by Parseval:**

$$
P = \sum_{k=0}^{5}|a_k|^2 = |a_1|^2 + |a_5|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{2}
$$

**Check in the time domain:** $x[n]$ for $n = 0 \dots 5$ is $0, \frac{\sqrt3}{2}, \frac{\sqrt3}{2}, 0, -\frac{\sqrt3}{2}, -\frac{\sqrt3}{2}$:

$$
P = \frac{1}{6}\left(4\times\frac{3}{4}\right) = \frac{3}{6} = 0.5
$$

**Answer:** average power $P = 0.5$ W.

## Check yourself

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

1. Parseval's relation for the DTFS says (1/N) sum |x[n]|^2 over a period equals:
   A. sum of |a_k|^2 over one period
   B. |sum of a_k|^2
   C. N sum of |a_k|^2
   D. sum of a_k
   Answer: A. Average power equals the sum of the power in the N harmonic components. (Key point: Parseval's relation)

2. What is the fundamental period of sin(pi n/3)?
   A. N = 3
   B. N = 6
   C. N = 2
   D. N = 12
   Answer: B. w0/(2 pi) = (pi/3)/(2 pi) = 1/6, so N = 6. (Key point: Fundamental period)

3. What are the nonzero coefficients of sin(2 pi n/6)?
   A. a_1 = a_5 = 1/2
   B. a_1 = 1/(2j), a_{-1} = a_5 = -1/(2j)
   C. a_0 = 1
   D. a_1 = a_{-1} = 1/2
   Answer: B. sin = (e^{j...} - e^{-j...})/(2j), so a_1 = 1/(2j) and a_{-1} = -1/(2j), which equals a_5. (Key point: Coefficients)

4. What is the average power of sin(pi n/3)?
   A. 1
   B. 0.25
   C. 0.75
   D. 0.5
   Answer: D. P = |a_1|^2 + |a_5|^2 = 1/4 + 1/4 = 0.5. (Key point: Average power)

## Related in Signals and Systems

- [Parseval's relation for DTFS](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtfs-parseval) — Signals and Systems, chapter 2, asked 8×
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- [DTFS coefficients are periodic](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtfs-coefficients-periodic) — Signals and Systems, chapter 2

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