--- title: "FT of periodic signals; pulse example" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ft-periodic-pulse-example" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-periodic-pulse-example site: BCT Engineering --- # FT of periodic signals; pulse example Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2072 Asoj, 4+4 marks - Video: [7.6: The Continuous-Time Fourier Transform of Periodic Signals](https://www.youtube.com/watch?v=chfVJux_7tk) — ECE Done Better - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-periodic-pulse-example - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Derive the expression for Fourier transform of continuous-time periodic signals. Using this expression obtain the Fourier transform of periodic signal x(t) = 1 for 0 < t < T, 0 otherwise. ## Short answer FT of a periodic signal: X(jω)=∑ 2π a_k δ(ω−kω0). For one period of width T, a_k = sin(kω0T/2)/(kπ) e^{-jkω0T/2} (k≠0), a0=T/T0. The spectrum is an impulse train with envelope 2 sin(ωT/2)/ω ω0 and linear phase. ## Answer #### FT of a continuous-time periodic signal Let $x(t)$ be periodic with period $T_0$ and $\omega_0 = 2\pi/T_0$. Its exponential Fourier series is $$ x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T_0}\int_{T_0}x(t)e^{-jk\omega_0 t}dt $$ The inverse FT of $2\pi\delta(\omega - k\omega_0)$ is $\frac{1}{2\pi}\int 2\pi\delta(\omega-k\omega_0)e^{j\omega t}d\omega = e^{jk\omega_0 t}$, so $e^{jk\omega_0 t} \leftrightarrow 2\pi\delta(\omega - k\omega_0)$. By linearity, $$ X(j\omega) = \sum_{k=-\infty}^{\infty} 2\pi\,a_k\,\delta(\omega - k\omega_0) $$ #### Example: periodic pulse **Assumption:** In one period $T_0$, $x(t) = 1$ for $0 < t < T$ and $0$ for $T < t < T_0$ (with $T < T_0$), repeated every $T_0$. **Fourier coefficients:** $$ \begin{aligned} a_k &= \frac{1}{T_0}\int_0^{T} e^{-jk\omega_0 t}dt = \frac{1}{T_0}\cdot\frac{1 - e^{-jk\omega_0 T}}{jk\omega_0} \\ &= \frac{1}{T_0}\cdot\frac{e^{-jk\omega_0 T/2}\left(2j\sin(k\omega_0 T/2)\right)}{jk\omega_0} \\ &= \frac{\sin(k\omega_0 T/2)}{k\pi}\,e^{-jk\omega_0 T/2}, \quad k \ne 0 \end{aligned} $$ (using $T_0\omega_0 = 2\pi$), and $a_0 = \frac{T}{T_0}$ (the average value, also the limit of the formula as $k\to0$). **Fourier transform:** $$ X(j\omega) = \frac{2\pi T}{T_0}\delta(\omega) + \sum_{k\ne0}\frac{2\sin(k\omega_0 T/2)}{k}\,e^{-jk\omega_0 T/2}\,\delta(\omega - k\omega_0) $$ **Interpretation:** - The spectrum is a set of impulses at $\omega = k\omega_0$. Their strengths follow a sinc-shaped envelope $\frac{2\sin(\omega T/2)}{\omega}\cdot\omega_0$, which is $\omega_0$ times the FT of one pulse. - The factor $e^{-jk\omega_0 T/2}$ is a linear phase because the pulse is centred at $t = T/2$ rather than at $t=0$. - Example: duty cycle $T/T_0 = 1/2$ gives $a_0 = 1/2$ and $a_k = 0$ for even $k \ne 0$ (only odd harmonics). ```text x(t) 1 +---+ +---+ +---+ | | | | | | ---+---+-----+---+-----+---+---> t 0 T T0 T0+T 2T0 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is X(jω) for a periodic signal? A. Σ a_k δ(ω - kω0) B. Σ 2π a_k δ(ω - k) C. 2π a_0 δ(ω) D. Σ 2π a_k δ(ω - kω0) Answer: D. Each harmonic maps to an impulse of area 2π a_k. (Key point: General expression) 2. What is a_0 for the pulse of width T with period T0? A. T/T0 B. T0/T C. T/(2T0) D. 1 Answer: A. a_0 is the average value, (1/T0)∫0^T 1 dt = T/T0. (Key point: DC term) 3. What is a_k for k ≠ 0? A. sin(kω0T/2)/(kπ) B. sin(kω0T/2)/(kπ) · e^(-jkω0T/2) C. sin(kω0T)/(kπ) · e^(-jkω0T) D. 1/(jkπ) Answer: B. Evaluating the integral and using T0ω0 = 2π; the phase appears because the pulse is centred at T/2. (Key point: Coefficients) 4. If T/T0 = 1/2, which harmonics are present? A. Only even harmonics B. All harmonics C. Only odd harmonics (plus DC) D. Only DC Answer: C. sin(kπ/2) = 0 for even k. 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