--- title: "DTFS of sin(ω₀n) with N = 6" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-sine-period-six" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-sine-period-six site: BCT Engineering --- # DTFS of sin(ω₀n) with N = 6 Signals and Systems · Chapter 2: Fourier Series - Asked in: 2081 Asoj, 6 marks; 2082 Bhadra (new course), 4 marks (asked 2 times) - Video: [Determine DTFS of the signal and draw the spectrum | Numerical 1 on DTFS | EnggClasses](https://www.youtube.com/watch?v=HlEse_jD3ow) — EnggClasses - 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-sine-period-six - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Consider a signal x[n] = sin(ω₀n) with period N = 6. Find the Fourier series coefficient of the given signal x[n]. Plot spectra. Also find the average power. ## Short answer Using Euler's formula, a1 = -j/2, a5 = j/2, others zero. Parseval's relation gives average power P = |a1|^2 + |a5|^2 = 0.5 W. ## Answer Since $x[n] = \sin(\omega_0 n)$ has period $N = 6$, the fundamental frequency is $$ \omega_0 = \frac{2\pi}{N} = \frac{2\pi}{6} = \frac{\pi}{3} $$ #### Fourier series coefficients Using Euler's formula: $$ x[n] = \sin\frac{\pi n}{3} = \frac{1}{2j}e^{j(2\pi/6)n} - \frac{1}{2j}e^{-j(2\pi/6)n} $$ Comparing with $x[n] = \sum_{k=\langle 6\rangle}a_k e^{jk(2\pi/6)n}$: $$ a_1 = \frac{1}{2j} = -\frac{j}{2} = \frac{1}{2}e^{-j\pi/2}, \qquad a_{-1} = -\frac{1}{2j} = \frac{j}{2} = \frac{1}{2}e^{j\pi/2} $$ All other $a_k$ in one period are zero. The coefficients repeat every 6: $a_{-1} = a_5 = a_{11}$, $a_1 = a_7 = a_{-5}$, etc. | $k$ | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---| | $a_k$ | 0 | $-j/2$ | 0 | 0 | 0 | $j/2$ | | $\lvert a_k\rvert$ | 0 | 0.5 | 0 | 0 | 0 | 0.5 | | $\angle a_k$ | - | $-90^\circ$ | - | - | - | $+90^\circ$ | #### Spectra ```text |a_k| (repeats every 6) 0.5 | | | | -+--o--o--o--+--o--+--o--o--o--+--> k -5 -4 -3 -2 -1 0 1 2 3 4 5 angle a_k (deg) 90 | | -+--o--o--o--+--o--+--o--o--o--+--> k -90 | | -5 -4 -3 -2 -1 0 1 2 3 4 5 ``` Lines of height $1/2$ at $k = \pm1, \pm5, \pm7, \dots$; phase $-90^\circ$ at $k = 1, 7, -5, \dots$ and $+90^\circ$ at $k = -1, 5, \dots$ #### Average power By Parseval's relation: $$ P = \sum_{k=\langle 6\rangle}|a_k|^2 = |a_1|^2 + |a_5|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} $$ Check in time: samples over one period are $0, \frac{\sqrt3}{2}, \frac{\sqrt3}{2}, 0, -\frac{\sqrt3}{2}, -\frac{\sqrt3}{2}$, so $P = \frac{1}{6}\left(4\times\frac{3}{4}\right) = 0.5$. **Answer:** $a_1 = -j/2$, $a_{-1} = a_5 = j/2$, others 0; average power $P = 0.5$ W. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is w0 for period N = 6? A. pi/6 B. 2pi/3 C. pi/3 D. 6 pi Answer: C. w0 = 2pi/N = 2pi/6 = pi/3. (Key point: Fundamental frequency) 2. What are a_1 and a_-1 for sin(pi n/3)? A. a_1 = 1/(2j) = -j/2, a_-1 = j/2 B. a_1 = j/2, a_-1 = -j/2 C. a_1 = a_-1 = 1/2 D. a_1 = 1/2j, a_-1 = 1/2j Answer: A. sin x = (e^(jx) - e^(-jx))/(2j), so a_1 = 1/(2j) = -j/2 and a_-1 = -1/(2j) = j/2. (Key point: Coefficients) 3. Where else does the line a_-1 appear in one period k = 0..5? A. k = 1 B. k = 3 C. k = 4 D. k = 5 Answer: D. a_k repeats every 6, so a_-1 = a_5. (Key point: Periodicity) 4. What is the average power? A. 1/2 B. 1/4 C. 1 D. sqrt(3)/2 Answer: A. P = |a_1|^2 + |a_5|^2 = 1/4 + 1/4 = 1/2. 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