Fourier Series Solved Examples

Introduction

The best way to master Fourier series is through practice. In this section, we work through complete examples of computing Fourier series coefficients for common engineering waveforms. Each example demonstrates a systematic approach: identify the period, exploit symmetry to simplify calculations, compute the integrals, and verify results using Parseval's theorem. These are exactly the types of problems you will encounter in university exams and GATE.

Example 1: Rectangular Pulse Train

Problem: Find the exponential Fourier series of a periodic rectangular pulse train with period $T_0 = 4$, pulse width $\tau = 1$, and amplitude $A = 2$, centered at the origin.

Solution:

Step 1: Identify parameters. $T_0 = 4$, $\omega_0 = 2\pi/4 = \pi/2$, pulse from $t = -0.5$ to $t = 0.5$.

Step 2: Compute $c_n$:

$$c_n = \frac{1}{T_0}\int_{-\tau/2}^{\tau/2} A e^{-jn\omega_0 t}dt = \frac{1}{4}\int_{-0.5}^{0.5} 2e^{-jn(\pi/2)t}dt$$

$$= \frac{1}{2}\left[\frac{e^{-jn\pi t/2}}{-jn\pi/2}\right]_{-0.5}^{0.5} = \frac{1}{2} \cdot \frac{2\sin(n\pi/4)}{n\pi/2}$$

$$= \frac{2\sin(n\pi/4)}{n\pi} = \frac{1}{2}\text{sinc}(n/4)$$

Step 3: DC component: $c_0 = A\tau/T_0 = 2 \times 1/4 = 0.5$

Step 4: Verify with Parseval's theorem: $$P_{time} = \frac{1}{4}\int_{-0.5}^{0.5} 4\,dt = \frac{4 \times 1}{4} = 1$$

The sum $\sum|c_n|^2$ should also equal 1. ✓

Example 2: Sawtooth Wave

Problem: Find the Fourier series of a sawtooth wave: $x(t) = t/\pi$ for $-\pi < t < \pi$, period $2\pi$.

Solution:

The function is odd ($x(-t) = -x(t)$), so $a_n = 0$ for all $n$, and $c_n$ is purely imaginary.

$$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{t}{\pi} e^{-jnt}dt$$

Using integration by parts with $u = t/\pi$, $dv = e^{-jnt}dt$:

$$c_n = \frac{1}{2\pi}\left[\frac{t}{\pi} \cdot \frac{e^{-jnt}}{-jn}\right]_{-\pi}^{\pi} - \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{\pi}\cdot\frac{e^{-jnt}}{-jn}dt$$

The second integral evaluates to zero (integral of $e^{-jnt}$ over a full period). The first term:

$$c_n = \frac{1}{2\pi^2}\left[\frac{\pi e^{-jn\pi}}{-jn} - \frac{(-\pi)e^{jn\pi}}{-jn}\right] = \frac{1}{2\pi}\cdot\frac{-e^{-jn\pi} - e^{jn\pi}}{jn}$$

Wait, let me redo more carefully. Since $e^{-jn\pi} = e^{jn\pi} = (-1)^n$:

$$c_n = \frac{(-1)^n}{-jn\pi} \cdot \frac{1}{1} = \frac{(-1)^{n+1}}{jn\pi} = \frac{j(-1)^n}{n\pi}$$

Trigonometric form: $b_n = \frac{2(-1)^{n+1}}{n\pi}$

$$x(t) = \frac{2}{\pi}\left[\sin t - \frac{\sin 2t}{2} + \frac{\sin 3t}{3} - \cdots\right]$$

Example 3: Half-Wave Rectified Sine

Problem: Find the Fourier series of $x(t) = \sin(t)$ for $0 < t < \pi$ and $x(t) = 0$ for $\pi < t < 2\pi$, period $2\pi$.

Solution:

$$a_0 = \frac{1}{\pi}\int_0^{\pi}\sin t\,dt = \frac{1}{\pi}[-\cos t]_0^{\pi} = \frac{1}{\pi}(1+1) = \frac{2}{\pi}$$

$$a_n = \frac{1}{\pi}\int_0^{\pi}\sin t\cos(nt)\,dt = \frac{1}{\pi}\int_0^{\pi}\frac{1}{2}[\sin((1+n)t) + \sin((1-n)t)]dt$$

For $n \neq 1$: $$a_n = \frac{1}{2\pi}\left[\frac{-\cos((1+n)t)}{1+n} + \frac{-\cos((1-n)t)}{1-n}\right]_0^{\pi}$$

After evaluation (using $\cos(k\pi) = (-1)^k$):

$$a_n = \frac{-(1+(-1)^n)}{\pi(1-n^2)} = \begin{cases} \frac{-2}{\pi(1-n^2)} & n \text{ even} \\ 0 & n \text{ odd}, n \neq 1 \end{cases}$$

For $n = 1$: $b_1 = \frac{1}{\pi}\int_0^{\pi}\sin^2 t\,dt = \frac{1}{2}$

Final result: $$x(t) = \frac{1}{\pi} + \frac{1}{2}\sin t - \frac{2}{\pi}\left(\frac{\cos 2t}{1 \cdot 3} + \frac{\cos 4t}{3 \cdot 5} + \frac{\cos 6t}{5 \cdot 7} + \cdots\right)$$

Example 4: Triangular Wave

Problem: Find the Fourier series of a symmetric triangular wave with amplitude $A$ and period $T_0$.

$$x(t) = \frac{4A}{T_0}|t| - A, \quad -T_0/2 < t < T_0/2$$

Actually, let's use the standard symmetric triangle: $x(t) = A(1 - 4|t|/T_0)$ for $|t| < T_0/2$.

Solution: The signal is even, so $b_n = 0$.

$$a_n = \frac{2}{T_0}\int_{-T_0/2}^{T_0/2}x(t)\cos(n\omega_0 t)dt$$

By symmetry (even function): $= \frac{4}{T_0}\int_0^{T_0/2}A(1-2t/T_0)\cdot 2\cos(n\omega_0 t)dt$

After integration by parts (details omitted for brevity):

$$a_n = \begin{cases} \frac{8A}{n^2\pi^2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$$

$$x(t) = \frac{8A}{\pi^2}\left[\cos(\omega_0 t) + \frac{\cos(3\omega_0 t)}{9} + \frac{\cos(5\omega_0 t)}{25} + \cdots\right]$$

Note: Coefficients decay as $1/n^2$ (faster than square wave's $1/n$) because the triangle is continuous.

Example 5: Power Calculation Using Parseval's

Problem: A periodic signal has $c_0 = 1$, $c_1 = c_{-1} = 0.5$, $c_2 = c_{-2} = 0.3$, $c_3 = c_{-3} = 0.1$, and all other $c_n = 0$. Find the total power and the percentage in the first three harmonics.

Solution:

Total power: $P = |c_0|^2 + 2|c_1|^2 + 2|c_2|^2 + 2|c_3|^2$

$= 1 + 2(0.25) + 2(0.09) + 2(0.01) = 1 + 0.5 + 0.18 + 0.02 = 1.7$

Power in DC + first harmonic: $1 + 0.5 = 1.5$ → $1.5/1.7 = 88.2\%$

Power in first two harmonics: $1.5 + 0.18 = 1.68$ → $1.68/1.7 = 98.8\%$

This illustrates how most power is typically concentrated in the lower harmonics.

Example 6: Applying Time-Shift Property

Problem: The Fourier series of $x(t) = \sum c_n e^{jn\omega_0 t}$ is known. Find the coefficients of $y(t) = x(t - T_0/6)$.

Solution: By time-shifting property:

$$d_n = c_n e^{-jn\omega_0 T_0/6} = c_n e^{-jn\pi/3}$$

The magnitudes $|d_n| = |c_n|$ are unchanged — only phases shift by $-n\pi/3$ radians.

Tips for Computing Fourier Series

1. Always check symmetry first — it can eliminate half the computation
2. Use integration by parts for polynomial × exponential products
3. Remember: $\int_0^{2\pi} e^{jnt}dt = 0$ for $n \neq 0$ (orthogonality)
4. Verify with Parseval's — compute power both ways as a check
5. Known results: Memorize the square wave, triangle wave, and pulse train results — many problems reduce to these through properties

Key Takeaways

• Systematic approach: identify period → check symmetry → compute integrals → verify
• Square wave: $\sim 1/n$ (odd harmonics only, discontinuous)
• Triangle wave: $\sim 1/n^2$ (odd harmonics only, continuous)
• Pulse train: sinc envelope in frequency (bandwidth inversely proportional to pulse width)
• Parseval's theorem provides a verification tool and enables power calculations without time-domain integration