--- title: "DTFS of 1 + 3cos(2πn/N + π/3)" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-cosine-with-phase" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-cosine-with-phase site: BCT Engineering --- # DTFS of 1 + 3cos(2πn/N + π/3) Signals and Systems · Chapter 2: Fourier Series - Asked in: 2077 Chaitra, 2+2+2 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-cosine-with-phase - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Obtain fourier series coefficients of the signal x[n] = 1 + 3cos(2πn/N + π/3), where N is the fundamental time period of x[n]. Also, plot magnitude and phase spectrum of the coefficients. ## Short answer Using Euler's formula, x[n] = 1 + 1.5 e^{jπ/3} e^{j(2π/N)n} + 1.5 e^{-jπ/3} e^{-j(2π/N)n}. Coefficients: a_0 = 1, a_1 = 1.5 e^{jπ/3}, a_{-1} = 1.5 e^{-jπ/3}, others zero. Magnitudes: 1, 1.5, 1.5; phases: 0°, 60°, -60°. Spectra repeat with period N. ## Answer With fundamental period $N$, $\omega_0 = 2\pi/N$. Expand the cosine using Euler's formula: $$ \begin{aligned} x[n] &= 1 + 3\cos\left(\frac{2\pi n}{N} + \frac{\pi}{3}\right) \\ &= 1 + \frac{3}{2}e^{j\pi/3}e^{j(2\pi/N)n} + \frac{3}{2}e^{-j\pi/3}e^{-j(2\pi/N)n} \end{aligned} $$ #### Fourier series coefficients Comparing with $x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}$: $$ a_0 = 1, \qquad a_1 = \frac{3}{2}e^{j\pi/3} = 0.75 + j1.299, \qquad a_{-1} = \frac{3}{2}e^{-j\pi/3} = 0.75 - j1.299 $$ All other coefficients in one period are zero, and $a_{k+N} = a_k$ (so $a_{N-1} = a_{-1}$, $a_{N+1} = a_1$). | $k$ | $a_k$ | $\lvert a_k\rvert$ | $\angle a_k$ | |---|---|---|---| | $-1$ | $1.5e^{-j\pi/3}$ | 1.5 | $-60^\circ$ | | 0 | 1 | 1 | $0^\circ$ | | 1 | $1.5e^{j\pi/3}$ | 1.5 | $+60^\circ$ | #### Magnitude spectrum ```text |a_k| (one period shown; repeats every N) 1.5 | | 1 | | | -o--+--+--+--o--> k -2 -1 0 1 2 ``` #### Phase spectrum ```text angle a_k (deg) 60 | -o--+--o--+--o--> k -60 | -2 -1 0 1 2 ``` The magnitude is even and the phase odd in $k$ (real signal), and both repeat with period $N$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Using Euler's formula, 3cos(2 pi n/N + pi/3) equals: A. 3 e^{j pi/3} e^{j(2 pi/N)n} B. (3/2) e^{j pi/3} e^{j(2 pi/N)n} + (3/2) e^{-j pi/3} e^{-j(2 pi/N)n} C. 3 e^{j pi/3} + 3 e^{-j pi/3} D. (3/2) e^{j(2 pi/N)n} + (3/2) e^{-j(2 pi/N)n} Answer: B. cos(theta) = (e^{j theta} + e^{-j theta})/2, and the phase pi/3 rides on each exponential with the sign of the exponent. (Key point: Euler expansion) 2. What are a_0, a_1 and a_{-1}? A. a_0 = 1, a_1 = 1.5 e^{j pi/3}, a_{-1} = 1.5 e^{-j pi/3} B. a_0 = 1, a_1 = 3 e^{j pi/3}, a_{-1} = 3 e^{-j pi/3} C. a_0 = 0, a_1 = 1.5, a_{-1} = 1.5 D. a_0 = 1, a_1 = 1.5 e^{-j pi/3}, a_{-1} = 1.5 e^{j pi/3} Answer: A. The constant 1 gives a_0 = 1, and the cosine of amplitude 3 splits into two coefficients of magnitude 1.5 with phases +60 and -60 degrees. (Key point: Coefficients) 3. With period N, how is a_{N-1} related to the other coefficients? A. a_{N-1} = a_1 B. a_{N-1} = 0 C. a_{N-1} = a_{-1} D. a_{N-1} = a_0 Answer: C. DTFS coefficients repeat with period N, so a_{N-1} = a_{-1}. (Key point: Periodicity of coefficients) 4. For this real signal, the magnitude and phase spectra are respectively: A. Odd and even in k B. Both even in k C. Both odd in k D. Even and odd in k Answer: D. For real signals a_{-k} = a_k*, so magnitude is even and phase is odd. 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