--- title: "Spectrum of a sawtooth wave" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:sawtooth-fourier-spectrum" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/sawtooth-fourier-spectrum site: BCT Engineering --- # Spectrum of a sawtooth wave Signals and Systems · Chapter 2: Fourier Series - Asked in: 2079 Jestha, 8 marks - Video: [Signal Representation #5. How to Build a Sawtooth Wave Using Fourier Series in Frequemcy Domain.](https://www.youtube.com/watch?v=hNgzYx0OCBs) — Technologies Discussion - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/sawtooth-fourier-spectrum - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Calculate and plot magnitude and phase spectrum of the following periodic signal. [Figure: sawtooth wave of period 1: x(t) rises linearly from 0 at t = 0 to 1 at t = 1 and then drops back to 0, i.e. x(t) = t for 0 ≤ t < 1, repeating every 1 s; shown from t = −3 to t = 4] ## Short answer Integrate x(t)=t over period T=1. DC a0=0.5; for k≠0, a_k = j/(2πk). Magnitude |a_k|=1/(2π|k|), phase 90° for k>0, –90° for k<0. ## Answer One period: $x(t) = t$ for $0 \le t < 1$, with period $T = 1$ s, so $\omega_0 = 2\pi/T = 2\pi$ rad/s. $$ a_k = \frac{1}{T}\int_0^T x(t)e^{-jk\omega_0 t}dt = \int_0^1 t\,e^{-j2\pi kt}dt $$ #### DC term ($k = 0$) $$ a_0 = \int_0^1 t\,dt = \frac{1}{2} $$ #### Coefficients for $k \ne 0$ Integrate by parts with $u = t$, $dv = e^{-j2\pi kt}dt$, $v = \frac{e^{-j2\pi kt}}{-j2\pi k}$: $$ \begin{aligned} a_k &= \left[\frac{t\,e^{-j2\pi kt}}{-j2\pi k}\right]_0^1 - \int_0^1\frac{e^{-j2\pi kt}}{-j2\pi k}dt \\ &= \frac{e^{-j2\pi k}}{-j2\pi k} + \frac{1}{j2\pi k}\left[\frac{e^{-j2\pi kt}}{-j2\pi k}\right]_0^1 \\ &= \frac{1}{-j2\pi k} + \frac{1}{j2\pi k}\cdot\frac{e^{-j2\pi k} - 1}{-j2\pi k} \end{aligned} $$ Since $e^{-j2\pi k} = 1$, the second term is zero: $$ a_k = \frac{-1}{j2\pi k} = \frac{j}{2\pi k}, \qquad k \ne 0 $$ #### Magnitude and phase $$ |a_k| = \frac{1}{2\pi|k|}, \qquad \angle a_k = \begin{cases} +90^\circ\ (\pi/2), & k > 0 \\ -90^\circ\ (-\pi/2), & k < 0 \\ 0, & k = 0 \end{cases} $$ | $k$ | 0 | $\pm1$ | $\pm2$ | $\pm3$ | $\pm4$ | |---|---|---|---|---|---| | $\lvert a_k\rvert$ | 0.5 | 0.159 | 0.0796 | 0.0531 | 0.0398 | | $\angle a_k$ ($k>0$) | $0^\circ$ | $90^\circ$ | $90^\circ$ | $90^\circ$ | $90^\circ$ | | $\angle a_k$ ($k<0$) | - | $-90^\circ$ | $-90^\circ$ | $-90^\circ$ | $-90^\circ$ | So the Fourier series is $$ x(t) = \frac{1}{2} + \sum_{k\ne0}\frac{j}{2\pi k}e^{j2\pi kt} = \frac{1}{2} - \sum_{k=1}^{\infty}\frac{1}{\pi k}\sin(2\pi kt) $$ #### Spectra (lines at $\omega = 2\pi k$) ```text |a_k| 0.5 | 0.159 | | | 0.0796 | | | | | 0.0531 | | | | | | | -+--+--+--+--+--+--+--> k -3 -2 -1 0 1 2 3 angle a_k (deg) 90 | | | -+--+--+--o--+--+--+--> k -90 | | | -3 -2 -1 0 1 2 3 ``` The magnitude falls as $1/k$ (because of the jump discontinuity each period), is even in $k$, and the phase is odd: $+90^\circ$ for positive $k$ and $-90^\circ$ for negative $k$. **Check:** at $t = 0.25$, the series $\frac{1}{2} - \frac{1}{\pi}\left(1 - \frac{1}{3} + \frac{1}{5} - \dots\right) = \frac{1}{2} - \frac{1}{\pi}\cdot\frac{\pi}{4} = 0.25$, which equals $x(0.25)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are T and w0 for the sawtooth? A. T = 1, w0 = 2pi B. T = 1, w0 = 1 C. T = 2pi, w0 = 1 D. T = 1, w0 = pi Answer: A. w0 = 2pi/T = 2pi rad/s. (Key point: Fundamental frequency) 2. What is a_0? A. 1 B. 0 C. 1/2 D. 1/pi Answer: C. Integral of t from 0 to 1 is 1/2, the average. (Key point: DC term) 3. What is a_k for k not equal to 0? A. j/(2 pi k) B. 1/(2 pi k) C. -j/(2 pi k) D. 1/(pi k) Answer: A. By parts, a_k = -1/(j 2 pi k) = j/(2 pi k). (Key point: Coefficients) 4. What are |a_k| and the phase? A. 1/(pi|k|), phase 0 B. 1/(2 pi|k|), +90 for k>0 and -90 for k<0 C. 1/(2 pi|k|), phase 90 for all k D. 1/(2 pi k^2), phase 90 Answer: B. a_k = j/(2 pi k) is imaginary with sign of k. 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