--- title: "Fourier series of rising sawtooth −V0 to V0" subject: "Electrical Circuit Theory" chapter: 6 key: "semester_3/electrical_circuit_theory:fourier-sawtooth-minus-v0-to-v0" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-sawtooth-minus-v0-to-v0 site: BCT Engineering --- # Fourier series of rising sawtooth −V0 to V0 Electrical Circuit Theory · Chapter 6: Fourier Series and transform - Asked in: 2074 Chaitra, 8 marks - Video: [ECE202msu: Chapter 13 - Finding the Fourier Series of a Sawtooth](https://www.youtube.com/watch?v=N_PdVW_oim8) — ECE202msu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-sawtooth-minus-v0-to-v0 - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6 ## Question Obtain trigonometric Fourier series of the waveform shown in figure below and sketch the line spectra. [Figure: V(t) is a rising sawtooth: it starts at −V0 at ωt = 0, rises linearly through 0 at ωt = π to V0 at ωt = 2π, drops back to −V0 at ωt = 2π and repeats (rises to V0 at 4π), period 2π] ## Short answer The sawtooth wave rising from -V0 to V0 over 2π has zero DC and only sine terms: V(t) = -(2V0/π) Σ (1/n) sin nωt, n=1,2,3,… The amplitude spectrum decays as 1/n, all harmonics present, and phases are +90° in cosine form. ## Answer **Reading the figure.** V(t) rises linearly from −V0 at ωt = 0 to +V0 at ωt = 2π, then jumps back to −V0. The period is 2π (θ = ωt). ```text V(θ) = −V0 + (V0/π) θ = (V0/π)(θ − π) 0 < θ < 2π ``` The wave is odd about θ = π (the zero crossing), so the average is zero and the series has only sine terms. #### Coefficients ```text a0 = (1/2π)∫₀^2π V(θ) dθ = 0 an = (1/π)∫₀^2π (V0/π)(θ − π) cos nθ dθ = 0 (∫ θ cos nθ dθ = 0 over a full period) bn = (1/π)∫₀^2π (V0/π)(θ − π) sin nθ dθ = (V0/π²)·∫₀^2π θ sin nθ dθ (the π·sin term integrates to 0) ``` Using ∫₀^2π θ sin nθ dθ = [−θ cos nθ/n + sin nθ/n²]₀^2π = −2π/n: ```text bn = (V0/π²)(−2π/n) = −2V0/(nπ) ``` #### Fourier series ```text V(t) = −(2V0/π) [ sin ωt + (1/2) sin 2ωt + (1/3) sin 3ωt + ... ] ``` #### Line spectrum (amplitudes cn = |bn|, V0 = 1 shown) | n | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | cn (in units of V0) | 0.6366 | 0.3183 | 0.2122 | 0.1592 | 0.1273 | | phase (cosine form) | +90° | +90° | +90° | +90° | +90° | ```text n=1 |##########################| 0.6366 n=2 |############# | 0.3183 n=3 |######### | 0.2122 n=4 |###### | 0.1592 n=5 |##### | 0.1273 n=6 |#### | 0.1061 ``` All harmonics are present and the amplitude falls as 1/n. Writing −sin nθ = cos(nθ + 90°), each term has phase +90°. **Answer: V(t) = −(2V0/π) Σ (1/n) sin nωt, n = 1, 2, 3, ...** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does this sawtooth have only sine terms? A. It is odd about its zero crossing at θ = π, so a0 = an = 0 B. It is even about θ = 0 C. It has half-wave symmetry D. Its average is V0 Answer: A. Odd symmetry about the mean level means only sines. (Key point: Symmetry) 2. What is bn? A. 2V0/(nπ) B. -V0/(nπ) C. -4V0/(n^2 π^2) D. -2V0/(nπ) Answer: D. bn = (V0/π^2)(-2π/n) = -2V0/(nπ). (Key point: Coefficient) 3. Which harmonics are present and how do they decay? A. Only odd, falling as 1/n^2 B. Only odd, falling as 1/n C. All harmonics, amplitude falling as 1/n D. All, falling as 1/n^2 Answer: C. The jump discontinuity gives slow 1/n decay. (Key point: Harmonic content) 4. Why is each harmonic phase +90° in cosine form? A. Because sin nθ = cos(nθ + 90°) B. Because -sin nθ = cos(nθ + 90°) C. Because an is positive D. Because the wave starts at zero Answer: B. The negative bn flips the sign of sine, giving +90°. 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