--- title: "DTFS of sin ω₀n, ω₀ = 2π/N" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-sine-period-five" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-sine-period-five site: BCT Engineering --- # DTFS of sin ω₀n, ω₀ = 2π/N Signals and Systems · Chapter 2: Fourier Series - Asked in: 2074 Bhadra, 4 marks - 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-five - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Find Fourier series coefficient of discrete time periodic signal x[n] = sin ω₀n where ω₀ = 2π/N. Plot the signal when N = 5. ## Short answer Using x[n] = (e^{jω₀n} - e^{-jω₀n})/(2j), coefficients are a1 = -j/2, a-1 = j/2, others zero per period N. For N=5, a1 = -j/2, a4 = j/2, a0=a2=a3=0. Signal values: 0, 0.951, 0.588, -0.588, -0.951 repeating. ## Answer Given $x[n] = \sin \omega_0 n$ with $\omega_0 = 2\pi/N$, the signal is periodic with fundamental period $N$. Its DTFS coefficients can be found by inspection, using Euler's formula. #### Fourier series coefficients $$ x[n] = \sin\left(\frac{2\pi}{N}n\right) = \frac{1}{2j}e^{j(2\pi/N)n} - \frac{1}{2j}e^{-j(2\pi/N)n} $$ Compare this with the synthesis equation $x[n] = \sum_{k=\langle N\rangle} a_k e^{jk(2\pi/N)n}$: $$ a_1 = \frac{1}{2j} = -\frac{j}{2}, \qquad a_{-1} = -\frac{1}{2j} = \frac{j}{2} $$ All other coefficients in one period are zero. DTFS coefficients repeat with period $N$, so $a_{k+N} = a_k$: - $a_1 = a_{1+N} = a_{1-N} = \dots = -j/2$ - $a_{-1} = a_{N-1} = a_{2N-1} = \dots = j/2$ Over $k = 0, 1, \dots, N-1$: $a_1 = -j/2$, $a_{N-1} = j/2$, and every other $a_k = 0$. **For N = 5:** $a_1 = -j/2$, $a_4 = j/2$, and $a_0 = a_2 = a_3 = 0$, repeating every 5 values of $k$. In magnitude form $|a_1| = |a_4| = 1/2$, with $\angle a_1 = -\pi/2$ and $\angle a_4 = +\pi/2$. #### Plot of the signal for N = 5 $x[n] = \sin(2\pi n/5)$: | n | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---| | x[n] | 0 | 0.951 | 0.588 | −0.588 | −0.951 | ```text x[n] 0.95 | * * 0.59 | | * | * | | | | | 0 --*-----+--+--+--+--*--+--+--+--+-- n 0 1 2 3 4 5 6 7 8 9 -0.59 | * | * | -0.95 | * * ``` The pattern repeats every 5 samples. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For x[n] = sin(2 pi n/N), what are the nonzero coefficients? A. a_1 = a_{-1} = 1/2 B. a_1 = 1/(2j) = -j/2 and a_{-1} = j/2 C. a_0 = 1 D. a_1 = j/2 and a_{-1} = -j/2 Answer: B. sin = (e^{j...} - e^{-j...})/(2j), so a_1 = 1/(2j) = -j/2 and a_{-1} = -1/(2j) = j/2. (Key point: Coefficients by inspection) 2. Over k = 0..4 for N = 5, which coefficients are nonzero? A. a_1 = -j/2 and a_4 = j/2 B. a_1 = j/2 and a_4 = -j/2 C. a_2 and a_3 D. Only a_0 Answer: A. a_{-1} is the same as a_{N-1} = a_4 by periodicity, so a_4 = j/2 and a_1 = -j/2. (Key point: Coefficients for N = 5) 3. What are the phases of a_1 and a_{-1}? A. +pi/2 and -pi/2 B. 0 and 0 C. -pi/2 and +pi/2 D. pi and -pi Answer: C. a_1 = -j/2 has angle -pi/2; a_{-1} = j/2 has angle +pi/2. (Key point: Phase) 4. For N = 5, x[1] = sin(2 pi/5) is about: A. 0.588 B. 0.951 C. 1 D. 0 Answer: B. sin(72 degrees) = 0.951; x[2] = sin(144 degrees) = 0.588. 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