--- title: "Fourier series coefficients and plot" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:fourier-series-coefficients-plot" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/fourier-series-coefficients-plot site: BCT Engineering --- # Fourier series coefficients and plot Signals and Systems · Chapter 2: Fourier Series - Asked in: 2073 Magh, 5 marks - Video: [Fourier series| Fourier series coefficient Examples- Signals and systems](https://www.youtube.com/watch?v=_pmvLPYYowI) — Dr.Santosh Tondare Engineering Tutorials - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/fourier-series-coefficients-plot - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Find Fourier series coefficients of signal and plot the coefficients. [The signal is not printed on the paper.] ## Short answer For a periodic square wave of width 2T₁: a0 = 2T₁/T, ak = sin(kω₀T₁)/(kπ) for k≠0. With 50% duty cycle T=4T₁: a0=0.5, a1=1/π≈0.318, a3=-1/(3π)≈-0.106, even harmonics zero. Coefficients are real and even. ## Answer The paper does not print the signal. This answer takes the standard textbook case: a **periodic square wave** of period $T$ with $x(t) = 1$ for $|t| < T_1$ and $x(t) = 0$ for $T_1 < |t| < T/2$. ```text x(t) 1 ___ ___ ___ | | | | | | ______| |_____| |_____| |___ t -T1 T1 T-T1 T+T1 (pulse centred at t = 0) ``` #### Coefficients With $\omega_0 = 2\pi/T$, the DC term is $$ a_0 = \frac{1}{T}\int_{-T_1}^{T_1} 1\,dt = \frac{2T_1}{T} $$ For $k \ne 0$: $$ \begin{aligned} a_k &= \frac{1}{T}\int_{-T_1}^{T_1} e^{-jk\omega_0 t}dt = \frac{1}{T}\left[\frac{e^{-jk\omega_0 t}}{-jk\omega_0}\right]_{-T_1}^{T_1} \\ &= \frac{2}{k\omega_0 T}\cdot\frac{e^{jk\omega_0 T_1} - e^{-jk\omega_0 T_1}}{2j} = \frac{\sin(k\omega_0 T_1)}{k\pi} \end{aligned} $$ The coefficients are real, because $x(t)$ is real and even. #### Example values for T = 4T₁ (50% duty cycle) Here $\omega_0 T_1 = \pi/2$, so $a_k = \frac{\sin(k\pi/2)}{k\pi}$: | k | 0 | ±1 | ±2 | ±3 | ±4 | ±5 | |---|---|---|---|---|---|---| | $a_k$ | 1/2 | $1/\pi$ ≈ 0.318 | 0 | $-1/(3\pi)$ ≈ −0.106 | 0 | $1/(5\pi)$ ≈ 0.064 | #### Plot of the coefficients ```text a_k 0.50 | 0.32 | | | 0.06 | | | | | 0 -----+--+--+--+--+--+--+--+--+--+--+--- k -5 -4 -3 -2 -1 0 1 2 3 4 5 -0.11 | | ``` The coefficients are samples, at $\omega = k\omega_0$, of the sinc-shaped envelope $\frac{\omega_0}{\pi}\cdot\frac{\sin(\omega T_1)}{\omega}$. Even harmonics vanish for a 50% duty cycle, and the odd ones decrease as $1/k$ with alternating sign. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a unit pulse train of width 2T1 and period T, a_0 is: A. T1/T B. 2T1/T C. T/(2T1) D. 1/2 always Answer: B. a_0 is the average value, 2T1/T. (Key point: DC coefficient) 2. The nonzero-k coefficient is: A. cos(k w0 T1)/(k pi) B. sin(k w0 T1)/k C. sin(k w0 T1)/(k pi) D. 2 sin(k w0 T1)/pi Answer: C. Integration gives a_k = sin(k w0 T1)/(k pi). (Key point: General coefficient) 3. For T = 4T1, which values are correct? A. a_{+-2} = 1/(2 pi) and a_{+-3} = 1/(3 pi) B. a_0 = 1/2, a_{+-1} = 1/pi, a_{+-2} = 0, a_{+-3} = -1/(3 pi) C. a_0 = 1, a_{+-1} = 1/pi D. a_{+-1} = 1/2 and a_0 = 1/pi Answer: B. a_k = sin(k pi/2)/(k pi): a_0 = 1/2, a_1 = 1/pi, a_2 = 0, a_3 = -1/(3 pi). (Key point: Values for T = 4T1) 4. The coefficients are samples of an envelope of what shape? A. Sinc-shaped, sin(w T1)/w type B. Exponential decay C. Gaussian D. Constant Answer: A. They are samples at w = k w0 of (w0/pi) sin(w T1)/w. 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