--- title: "Fourier series of two-sided sawtooth with V" subject: "Electrical Circuit Theory" chapter: 6 key: "semester_3/electrical_circuit_theory:fourier-sawtooth-two-sided-v" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/fourier-sawtooth-two-sided-v site: BCT Engineering --- # Fourier series of two-sided sawtooth with V Electrical Circuit Theory · Chapter 6: Fourier Series and transform - Asked in: 2072 Chaitra, 8 marks - Video: [Electrical Engineering: Ch 18: Fourier Series (16 of 35) Find the Fourier Series: Saw-Tooth Wave***](https://www.youtube.com/watch?v=_s_P5XsaKdY) — Michel van Biezen - 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-two-sided-v - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-6 ## Question Find the trigonometric Fourier series for the waveform shown in figure below and plot the line spectrum. [Figure: v(t) with peak value V. For 0 < ωt < π the waveform falls linearly from 0 to −V at π; at ωt = π it jumps to 0 and rises linearly to V at 2π; at 2π it drops to 0, then falls again to −V at 3π, and so on (period 2π)] ## Short answer The waveform has half-wave symmetry and zero dc; only odd harmonics appear. The Fourier series includes both cosine and sine terms: v(t) = Σ_{n odd} [ (4V/(π² n²)) cos nωt − (2V/(nπ)) sin nωt ]. ## Answer **Reading the figure.** In 0 < θ < π (θ = ωt) the wave falls from 0 to −V; at θ = π it jumps to 0 and rises to +V at θ = 2π; then it drops to 0 and repeats. Period 2π: ```text v(θ) = −V θ/π 0 < θ < π v(θ) = V (θ − π)/π π < θ < 2π ``` Check: v(θ + π) = −v(θ), so the wave has half-wave symmetry; only odd harmonics are present and a0 = 0. #### Coefficients (odd n only) ```text an = (2/π)∫₀^π (−Vθ/π) cos nθ dθ = −(2V/π²)·(cos nπ − 1)/n² = 4V/(π² n²) bn = (2/π)∫₀^π (−Vθ/π) sin nθ dθ = −(2V/π²)·(−π cos nπ/n) = (2V/πn) cos nπ = −2V/(nπ) ``` using ∫₀^π θ cos nθ dθ = (cos nπ − 1)/n² and ∫₀^π θ sin nθ dθ = −π cos nπ/n. #### Fourier series ```text v(t) = Σ (n odd) [ (4V/π²n²) cos nωt − (2V/πn) sin nωt ] = V [ 0.405 cos ωt − 0.637 sin ωt + 0.045 cos 3ωt − 0.212 sin 3ωt + ... ] ``` Combined form: an cos nθ + bn sin nθ = cn cos(nθ + ψ), where cn = √(an² + bn²) = (2V/πn)√(1 + 4/(π²n²)) and ψ = arctan(−bn/an). #### Line spectrum (V = 1) | n | an | bn | cn | phase ψ (cos form) | |---|---|---|---|---| | 1 | 0.4053 | −0.6366 | 0.7547 | +57.5° | | 3 | 0.0450 | −0.2122 | 0.2169 | +78.0° | | 5 | 0.0162 | −0.1273 | 0.1284 | +82.7° | | 7 | 0.0083 | −0.0909 | 0.0913 | +84.8° | ```text n=1 |##########################| 0.7547 n=3 |####### | 0.2169 n=5 |#### | 0.1284 n=7 |### | 0.0913 n=9 |## | 0.0709 ``` Even harmonics (and the d.c. term) are absent. **Answer: v(t) = Σ (n = 1, 3, 5...) [ (4V/π²n²) cos nωt − (2V/nπ) sin nωt ].** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does v(θ + π) = -v(θ) imply? A. Only odd harmonics and a0 = 0 B. Only sine terms C. Only even harmonics D. Only cosines Answer: A. Half-wave symmetry removes the d.c. and even harmonics. (Key point: Half-wave symmetry) 2. Why can an and bn be computed with (2/π)∫ from 0 to π? A. Because the wave is even B. Because the wave is zero after π C. Because a0 = V D. Because of half-wave symmetry, for odd n Answer: D. Half-wave symmetric functions use half the period with factor 2/π. (Key point: Integral limits) 3. What are an and bn for odd n? A. an = 0, bn = -2V/(nπ) B. an = -4V/(π^2 n^2), bn = 2V/(nπ) C. an = 4V/(π^2 n^2), bn = -2V/(nπ) D. an = 2V/(nπ), bn = 4V/(π^2 n^2) Answer: C. Both sine and cosine terms occur since the wave is neither even nor odd. (Key point: Coefficients) 4. How is cn found? A. cn = an + bn B. cn = sqrt(an^2 + bn^2) C. cn = an - bn D. cn = an bn Answer: B. The combined cosine amplitude is the root sum of squares. (Key point: Combined amplitude) ## Related in Electrical Circuit Theory - [Fourier series of falling sawtooth V to 0](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-falling-sawtooth-v-to-0) — Electrical Circuit Theory, chapter 6 - [Fourier series of odd sawtooth, ±Vo](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-sawtooth-vo-odd) — Electrical Circuit Theory, chapter 6 - [Fourier series of half-period ramp pulses](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-ramp-half-period-vo) — Electrical Circuit Theory, chapter 6 - [Fourier series of rising sawtooth −V0 to V0](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-sawtooth-minus-v0-to-v0) — Electrical Circuit Theory, chapter 6 - [Fourier series of triangular wave, peak 20](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-triangular-wave-peak-20-minus) — Electrical Circuit Theory, chapter 6 - [Fourier series of triangular wave, peak 1](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Afourier-triangular-wave-peak-1-period) — Electrical Circuit Theory, chapter 6 ## Related in other subjects - [Average and RMS of sawtooth waveform](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Asawtooth-average-rms-value) — Fundamental of Electrical and Electronics Engineering, chapter 2 - [Spectrum of a sawtooth wave](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Asawtooth-fourier-spectrum) — Signals and Systems, chapter 2 - [Derive trigonometric Fourier series (CT)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-trigonometric-derivation) — Signals and Systems, chapter 2, asked 3× - [Average, rms, form and peak factor of sawtooth](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asawtooth-average-rms-form-peak) — Basic Electrical Engineering, chapter 5