Exponential Fourier series of sawtooth wave

Reading the figure. f(t) = t for −1/2 < t < 1, i.e. it rises from −1/2 to 1, then jumps back to −1/2 and repeats. Period T = 3/2 s, so ω0 = 2π/T = 4π/3 rad/s.

The exponential Fourier series is f(t) = Σ cn e^(jnω0t), cn = (1/T)∫ f(t) e^(−jnω0t) dt.

d.c. term

c0 = (1/T)∫_{−1/2}^{1} t dt = (2/3)·(1/2)(1 − 1/4) = 1/4

cn for n ≠ 0

Let a = nω0 = 4πn/3. Use ∫ t e^(−jat) dt = e^(−jat)(jt/a + 1/a²):

cn = (2/3) [ e^(−jat)(jt/a + 1/a²) ] from t = −1/2 to t = 1
   = (2/3) [ e^(−ja)(j/a + 1/a²) − e^(ja/2)(−j/(2a) + 1/a²) ]

For a = 4πn/3: e^(−ja) = e^(−j4πn/3) = e^(j2πn/3) = e^(ja/2), so the 1/a² terms cancel:

cn = (2/3) e^(j2πn/3) · (3j/2a) = j e^(j2πn/3)/a = 3j e^(j2πn/3) / (4πn)
|cn| = 3/(4π|n|),    arg cn = 90° + 120°·n   (for n > 0);  c(−n) = cn*

Series

f(t) = 1/4 + Σ (n ≠ 0) [ 3j e^(j2πn/3) / (4πn) ] e^(j4πnt/3)

Real form: f(t) = 1/4 − Σ (n = 1 to ∞) [3/(2πn)] sin(4πnt/3 + 2πn/3).

Spectrum (checked numerically in Python)

  n=0 |##########################| 0.25
 n=±1 |######################### | 0.2387
 n=±2 |############              | 0.1194
 n=±3 |########                  | 0.0796
 n=±4 |######                    | 0.0597
 n=±5 |#####                     | 0.0477

Answer: f(t) = 1/4 + Σ (n ≠ 0) [3j e^(j2πn/3)/(4πn)] e^(j4πnt/3), with ω0 = 4π/3 rad/s.