--- title: "FS coefficients of periodic square wave" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:square-wave-fourier-coefficients" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/square-wave-fourier-coefficients site: BCT Engineering --- # FS coefficients of periodic square wave Signals and Systems · Chapter 2: Fourier Series - Asked in: 2076 Bhadra, 5 marks - Video: [(English)S&S || Example 3.5 || Fourier Series || Periodic Square Wave](https://www.youtube.com/watch?v=PF1xnvSjjOg) — Electrical Engineering Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/square-wave-fourier-coefficients - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question A periodic square wave as shown in figure below is defined over one period as: x(t) = A for |t| < T₁; 0 for T₁ < |t| < T/2. Determine the Fourier series coefficients for x(t). [Figure: periodic rectangular pulse train of height A and period T; each pulse is centred on t = 0, ±T, … and extends from −T₁ to T₁] ## Short answer With period T and ω0=2π/T, a_0 = 2A T1/T. For k≠0, a_k = (A sin(k ω0 T1))/(k π). For a 50% duty cycle (T=4T1), a_0 = A/2 and a_k = A sin(kπ/2)/(kπ), with even harmonics zero. ## Answer The signal has period $T$, so $\omega_0 = 2\pi/T$. Over one period ($-T/2$ to $T/2$) it equals $A$ only for $|t| < T_1$. $$ a_k = \frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_{-T_1}^{T_1}A\,e^{-jk\omega_0 t}dt $$ #### DC term ($k = 0$) $$ a_0 = \frac{1}{T}\int_{-T_1}^{T_1}A\,dt = \frac{2AT_1}{T} $$ This is the average value (height times duty cycle). #### For $k \ne 0$ $$ \begin{aligned} a_k &= \frac{A}{T}\left[\frac{e^{-jk\omega_0 t}}{-jk\omega_0}\right]_{-T_1}^{T_1} = \frac{A}{T}\cdot\frac{e^{jk\omega_0 T_1} - e^{-jk\omega_0 T_1}}{jk\omega_0} \\ &= \frac{2A}{k\omega_0 T}\sin(k\omega_0 T_1) \\ &= \frac{A\sin(k\omega_0 T_1)}{k\pi} \qquad (\omega_0 T = 2\pi) \end{aligned} $$ #### Result $$ a_k = \begin{cases} \dfrac{2AT_1}{T}, & k = 0 \\[2mm] \dfrac{A\sin(k\omega_0 T_1)}{k\pi} = \dfrac{2AT_1}{T}\,\text{sinc}\left(\dfrac{2kT_1}{T}\right), & k \ne 0 \end{cases} $$ (with $\text{sinc}(x) = \sin(\pi x)/(\pi x)$). The coefficients are real and even because $x(t)$ is real and even. **Special case, 50% duty cycle ($T = 4T_1$):** $\omega_0 T_1 = \pi/2$, so $a_0 = A/2$ and $a_k = \frac{A\sin(k\pi/2)}{k\pi}$: $a_{\pm1} = A/\pi$, $a_{\pm2} = 0$, $a_{\pm3} = -A/3\pi$, $a_{\pm5} = A/5\pi$, i.e. even harmonics are zero. ```text a_k / A (T = 4T1) 0.5 | 0.318 | | | 0.0637 | | | | | -+--o--+--o--+--+--+--o--+--o--+--> k -0.106 | | -5 -4 -3 -2 -1 0 1 2 3 4 5 ``` The envelope of $a_k$ follows a sinc shape; making the pulse narrower (smaller $T_1/T$) spreads the spectrum wider. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a pulse train of height A, width 2T1 and period T, a_0 equals: A. A T1/T B. A T/(2 T1) C. 2 A T1/T D. A/2 always Answer: C. a_0 is the average: height times duty cycle = A(2T1)/T. (Key point: DC term) 2. For k not equal to 0, a_k is: A. A cos(k w0 T1)/(k pi) B. A sin(k w0 T1)/(k pi) C. A sin(k w0 T1)/k D. 2 A sin(k w0 T1)/(k w0) Answer: B. Integrating gives (2A/(k w0 T)) sin(k w0 T1), and w0 T = 2 pi gives A sin(k w0 T1)/(k pi). (Key point: General coefficient) 3. Why are the coefficients real for this signal? A. Because the period is real B. Because A is positive C. Because the signal is periodic D. Because x(t) is real and even Answer: D. A real, even signal has real, even coefficients. (Key point: Real coefficients) 4. For T = 4 T1, which coefficients vanish and what is a_{+-1}? A. Odd harmonics vanish; a_{+-1} = A/2 B. Even harmonics (k not 0) vanish; a_{+-1} = A/pi C. No harmonics vanish; a_{+-1} = A/pi D. Even harmonics vanish; a_{+-1} = A/(2 pi) Answer: B. sin(k pi/2) is zero for even k, and for k = 1 gives A/pi. (Key point: 50% duty cycle) ## Related in Signals and Systems - [FS of ±1.5 square wave, ω₀ = π](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asquare-wave-fs-fundamental-pi) — Signals and Systems, chapter 2 - [Convergence; coefficients of ±1.5 square wave](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asquare-wave-convergence-coefficients) — Signals and Systems, chapter 2 - [FS of unit-amplitude rectangular pulses](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Arectangular-pulse-train-fs) — Signals and Systems, chapter 2 - [Periodic signal whose Fourier series fails](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afourier-series-existence-examples) — Signals and Systems, chapter 2 - [Fourier series coefficients and plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afourier-series-coefficients-plot) — Signals and Systems, chapter 2 - [FT of periodic signals; rect pulse and constant](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-periodic-rect-constant) — Signals and Systems, chapter 3 ## Related in other subjects - [Fourier series coefficients of a square wave](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afs-coefficients-square-wave) — Data Communication, chapter 3 - [Fourier series of square wave and DFT](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Afourier-series-square-wave-dft) — Instrumentation and Measurement, chapter 2 - [Fourier series of triangle train, Vo and T](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-train-of-isosceles-triangles-vo) — Electrical Circuit Theory, chapter 6, asked 2× - [Fourier series of triangle train, peak 10](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-train-of-isosceles-triangles-10v) — Electrical Circuit Theory, chapter 6, asked 2×