--- title: "DTFS of 1 + sin + 3cos + cos(4πn/N)" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-coefficients-no-phase-term" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-coefficients-no-phase-term site: BCT Engineering --- # DTFS of 1 + sin + 3cos + cos(4πn/N) Signals and Systems · Chapter 2: Fourier Series - Asked in: 2080 Asoj, 6 marks; 2070 Magh, 6 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-coefficients-no-phase-term - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Find the Fourier series coefficients of the signal x[n] = 1 + sin[(2π/N)n] + 3cos[(2π/N)n] + cos[(4π/N)n] ## Short answer Using Euler: a0 = 1, a1 = 1.5 - j0.5, a_{-1} = 1.5 + j0.5, a2 = a_{-2} = 0.5, others zero. |a0|=1, |a±1|≈1.581, |a±2|=0.5; phases: 0°, ∓18.43°, 0°. ## Answer Fundamental frequency $\omega_0 = 2\pi/N$. Rewrite each term with Euler's formula: $$ \begin{aligned} x[n] &= 1 + \frac{1}{2j}\left[e^{j\frac{2\pi}{N}n} - e^{-j\frac{2\pi}{N}n}\right] + \frac{3}{2}\left[e^{j\frac{2\pi}{N}n} + e^{-j\frac{2\pi}{N}n}\right] + \frac{1}{2}\left[e^{j\frac{4\pi}{N}n} + e^{-j\frac{4\pi}{N}n}\right] \\ &= 1 + \left(\frac{3}{2} + \frac{1}{2j}\right)e^{j(2\pi/N)n} + \left(\frac{3}{2} - \frac{1}{2j}\right)e^{-j(2\pi/N)n} + \frac{1}{2}e^{j2(2\pi/N)n} + \frac{1}{2}e^{-j2(2\pi/N)n} \end{aligned} $$ Comparing with $x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}$ (and using $\frac{1}{2j} = -\frac{j}{2}$): | $k$ | $a_k$ | $\lvert a_k\rvert$ | $\angle a_k$ | |---|---|---|---| | 0 | $1$ | 1 | $0$ | | 1 | $\frac{3}{2} - \frac{j}{2}$ | $\frac{\sqrt{10}}{2} = 1.581$ | $-18.43^\circ$ | | $-1$ | $\frac{3}{2} + \frac{j}{2}$ | 1.581 | $+18.43^\circ$ | | 2 | $\frac{1}{2}$ | 0.5 | $0$ | | $-2$ | $\frac{1}{2}$ | 0.5 | $0$ | | other $k$ in one period | 0 | 0 | - | Working: $|a_1| = \sqrt{1.5^2 + 0.5^2} = \sqrt{2.5} = 1.581$; $\angle a_1 = -\tan^{-1}(0.5/1.5) = -18.43^\circ$. The coefficients are periodic with period $N$: $a_{k+N} = a_k$, so for example $a_{N-1} = a_{-1}$ and $a_{N-2} = a_{-2}$ (assuming $N \ge 5$ so that the harmonics do not overlap). ```text |a_k| (repeats every N) 1.581 | | 1 | | | 0.5 | | | | | -+--+--+--+--+--> k -2 -1 0 1 2 ``` Since $x[n]$ is real, $a_{-k} = a_k^*$: the magnitude is even and the phase is odd. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are coefficients read off? A. Expand sines/cosines via Euler and compare with sum a_k e^(jk(2pi/N)n) B. Differentiate x[n] C. Evaluate a_k only at k = 0 D. Use convolution Answer: A. Euler's formula gives each complex exponential's coefficient directly. (Key point: Method) 2. What is a_1? A. 3/2 + j/2 B. 3/2 - j/2 C. 1/2 D. 3 Answer: B. a_1 = 3/2 + 1/(2j) = 3/2 - j/2. (Key point: a_1) 3. What is a_2 (from cos(4pi n/N))? A. 1 B. 1/2 C. j/2 D. 0 Answer: B. cos = (1/2)(e^(j...) + e^(-j...)), so a_2 = a_-2 = 1/2. (Key point: a_2) 4. Why is a_-1 = (3/2 + j/2)? A. Since x is real, a_-k = a_k* B. Since x is even C. Since x is periodic with N = 2 D. By coincidence Answer: A. Real signals have conjugate-symmetric coefficients: even magnitude, odd phase. 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