--- title: "DTFS of 1 + sin + 3cos + cos(4πn/N + π/2)" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-coefficients-with-phase-term" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-coefficients-with-phase-term site: BCT Engineering --- # DTFS of 1 + sin + 3cos + cos(4πn/N + π/2) Signals and Systems · Chapter 2: Fourier Series - Asked in: 2082 Kartik, 5 marks; 2083 Bhadra (new course), 5 marks; 2083 Baisakh (new course), 4 marks (asked 3 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-with-phase-term - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Find and plot the Fourier series coefficients (magnitude and associated phase) for the following signal with period N. x[n] = 1 + sin((2π/N)n) + 3cos((2π/N)n) + cos((4π/N)n + π/2) ## Short answer Using Euler expansion: a0=1, a1=1.5−j0.5 (magnitude 1.581, phase −18.43°), a_{-1}=1.5+j0.5 (1.581, +18.43°), a2=j0.5 (0.5, +90°), a_{-2}=−j0.5 (0.5, −90°). Other coefficients zero within period; all repeat every N. ## Answer Fundamental frequency $\omega_0 = 2\pi/N$. Expand each term with Euler's formula and read off the coefficients of $e^{jk\omega_0 n}$: $$ \begin{aligned} x[n] &= 1 + \frac{1}{2j}\left[e^{j\omega_0 n} - e^{-j\omega_0 n}\right] + \frac{3}{2}\left[e^{j\omega_0 n} + e^{-j\omega_0 n}\right] \\ &\quad + \frac{1}{2}\left[e^{j(2\omega_0 n + \pi/2)} + e^{-j(2\omega_0 n + \pi/2)}\right] \\ &= 1 + \left(\frac{3}{2} + \frac{1}{2j}\right)e^{j\omega_0 n} + \left(\frac{3}{2} - \frac{1}{2j}\right)e^{-j\omega_0 n} \\ &\quad + \frac{1}{2}e^{j\pi/2}e^{j2\omega_0 n} + \frac{1}{2}e^{-j\pi/2}e^{-j2\omega_0 n} \end{aligned} $$ #### Fourier series coefficients (one period, $k = -2$ to $2$) 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 | $\frac{j}{2}$ | 0.5 | $+90^\circ$ | | $-2$ | $-\frac{j}{2}$ | 0.5 | $-90^\circ$ | | others in the period | 0 | 0 | - | Working: $|a_1| = \sqrt{1.5^2 + 0.5^2} = \sqrt{2.5} = 1.581$, $\angle a_1 = -\tan^{-1}(0.5/1.5) = -18.43^\circ$ ($-0.322$ rad). The coefficients repeat with period $N$: $a_{k+N} = a_k$ (e.g. $a_{N-1} = a_{-1}$, $a_{N-2} = a_{-2}$). Here $N \ge 5$ is assumed so that the harmonics do not overlap. #### Plots (one period, $k = -2 \dots 2$, repeating every $N$) ```text |a_k| (repeats every N) 1.581 | | 1 | | | 0.5 | | | | | -+--+--+--+--+--> k -2 -1 0 1 2 angle a_k (deg) 90 | 18.43 | | -+--+--o--+--+--> k -18.43 | | -90 | -2 -1 0 1 2 ``` Magnitude is even in $k$ and phase is odd in $k$, as expected for a real signal ($a_{-k} = a_k^*$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the coefficients found for this sum of sinusoids? A. Evaluate the analysis sum numerically only B. Expand each term with Euler's formula and read off coefficients of e^(jk w0 n) C. Take the derivative of x[n] D. Use Parseval's relation Answer: B. Writing sinusoids as complex exponentials gives a_k directly by comparison. (Key point: Method) 2. What is a_1? A. 3/2 + j/2 B. 3/2 - j/2 C. 3 - j D. 1/2 Answer: B. a_1 = 3/2 + 1/(2j) = 3/2 - j/2. (Key point: a_1 value) 3. What are |a_1| and the angle of a_1? A. 1.581 and -18.43 degrees B. 1.581 and +18.43 degrees C. 2.5 and -18.43 degrees D. 1.581 and -71.57 degrees Answer: A. |a_1| = sqrt(1.5^2 + 0.5^2) = 1.581; angle = -atan(0.5/1.5) = -18.43 degrees. (Key point: Magnitude and phase of a_1) 4. What is a_2 for the term cos(2 w0 n + pi/2)? A. 1/2 B. j/2 C. -j/2 D. 0 Answer: B. (1/2) e^(j pi/2) = j/2; a_-2 = -j/2. 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