--- title: "DTFS of a DT pulse and e^(j4πn/7)x[n]" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-rectangular-modulated" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-rectangular-modulated site: BCT Engineering --- # DTFS of a DT pulse and e^(j4πn/7)x[n] Signals and Systems · Chapter 2: Fourier Series - Asked in: 2070 Bhadra, 5+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-rectangular-modulated - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Find out Fourier series coefficients of a periodic discrete time signal described over a period as x[n] = 2 for |n| ≤ 1; 0 for 1 < |n| ≤ 3. Using the Fourier series coefficients calculated above, find the Fourier series coefficients of the signal e^(j4πn/7) x[n]. ## Short answer x[n]=2 for |n|≤1 and 0 for 1<|n|≤3 with N=7. Its DTFS coefficients are a_k = (2/7) sin(3πk/7)/sin(πk/7) for k≠0, a0=6/7. For y[n]=e^{j4πn/7}x[n], frequency shifting gives b_k = a_{k-2}, moving the spectrum right by 2. ## Answer The signal is $x[n] = 2$ for $n = -1, 0, 1$ and $x[n] = 0$ for $n = \pm2, \pm3$. One period runs over $n = -3$ to $3$, so $N = 7$ and $\omega_0 = 2\pi/7$. ```text x[n] 2 * * * * * * | | | | | | --*--*--*--+--+--+--*--*--*--*--+--+--+--- n -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 ``` #### Fourier series coefficients of x[n] $$ a_k = \frac{1}{N}\sum_{n=-1}^{1} 2\,e^{-jk(2\pi/7)n} = \frac{2}{7}\left[1 + e^{-j2\pi k/7} + e^{j2\pi k/7}\right] = \frac{2}{7}\left[1 + 2\cos\frac{2\pi k}{7}\right] $$ In closed form (the standard result for a DT rectangular pulse with $N_1 = 1$): $$ a_k = \frac{2}{7}\cdot\frac{\sin(3\pi k/7)}{\sin(\pi k/7)}, \quad k \ne 0, \pm7, \dots; \qquad a_0 = \frac{2(2N_1+1)}{N} = \frac{6}{7} $$ | k | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---|---| | $a_k$ | 0.857 | 0.642 | 0.159 | −0.229 | −0.229 | 0.159 | 0.642 | The coefficients are real and even ($a_k = a_{-k} = a_{7-k}$), because $x[n]$ is real and even. They repeat with period 7. #### Coefficients of y[n] = e^{j4πn/7} x[n] $$ e^{j4\pi n/7} = e^{j2(2\pi/7)n} = e^{jM\omega_0 n} \quad \text{with } M = 2 $$ By the **frequency shifting property**, $e^{jM\omega_0 n}x[n] \xleftrightarrow{FS} a_{k-M}$, so $$ b_k = a_{k-2} = \frac{2}{7}\cdot\frac{\sin\left(3\pi(k-2)/7\right)}{\sin\left(\pi(k-2)/7\right)}, \qquad b_2 = \frac{6}{7} $$ | k | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---|---| | $b_k = a_{k-2}$ | 0.159 | 0.642 | 0.857 | 0.642 | 0.159 | −0.229 | −0.229 | **Answer:** $a_k = \frac{2}{7}\frac{\sin(3\pi k/7)}{\sin(\pi k/7)}$ with $a_0 = 6/7$, and $b_k = a_{k-2}$. The spectrum is the same shape, moved 2 places to the right (peak at $k = 2$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a_0 for this signal? A. 2/7 B. 6/7 C. 3/7 D. 2 Answer: B. a_0 is the average: (2+2+2)/7 = 6/7. (Key point: DC coefficient) 2. Which expression gives a_k for this signal? A. (2/7)[1 + 2cos(2πk/7)] B. (2/7)[1 + 2sin(2πk/7)] C. (1/7)[2 + cos(2πk/7)] D. 2[1 + 2cos(2πk/7)] Answer: A. a_k = (1/7) Σ 2 e^{-jk(2π/7)n} for n = -1,0,1 = (2/7)[1 + e^{-j2πk/7} + e^{j2πk/7}]. (Key point: Coefficient expression) 3. For y[n] = e^{j4πn/7} x[n], what are the coefficients b_k? A. a_{k+2} B. a_{k-2} C. a_{2k} D. 2 a_k Answer: B. e^{j4πn/7} = e^{j2ω0 n} with M = 2, and frequency shifting gives b_k = a_{k-M} = a_{k-2}. (Key point: Frequency shifting) 4. Where does the peak of the spectrum of y[n] lie? A. At k = 0 B. At k = 2 C. At k = -2 D. At k = 4 Answer: B. The spectrum moves 2 places to the right, so b_2 = a_0 = 6/7 is the peak. 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