Find the Fourier series of the “sawtooth waveform” below using analytical methods

Find the Fourier series of the “sawtooth waveform” below using analytical methods.

The Correct Answer and Explanation is :

To find the Fourier series representation of the given sawtooth waveform analytically, let’s follow the standard Fourier series approach.

Step 1: Define the Sawtooth Function

A sawtooth wave is a periodic function, typically defined over one period ( T ). For this problem, assume the function is:

[
f(t) = \frac{t}{T}, \quad -\frac{T}{2} \leq t \leq \frac{T}{2}
]

Since the function is odd and has a linear increase, we expect only sine terms in the Fourier series (no cosine terms or constant term).

Step 2: Fourier Series General Form

For a general periodic function with period ( T ), the Fourier series is:

[
f(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\omega_0 t) + \sum_{n=1}^{\infty} b_n \sin(n\omega_0 t)
]

where the fundamental frequency is ( \omega_0 = \frac{2\pi}{T} ), and the Fourier coefficients are:

[
a_0 = \frac{2}{T} \int_{-T/2}^{T/2} f(t) dt
]

[
a_n = \frac{2}{T} \int_{-T/2}^{T/2} f(t) \cos(n\omega_0 t) dt
]

[
b_n = \frac{2}{T} \int_{-T/2}^{T/2} f(t) \sin(n\omega_0 t) dt
]

Step 3: Compute the Fourier Coefficients

• Compute ( a_0 ): Since ( f(t) = t/T ) is odd, the integral over a symmetric period cancels out, so: [
  a_0 = 0
  ]
• Compute ( a_n ) (Cosine Coefficients): [
  a_n = \frac{2}{T} \int_{-T/2}^{T/2} \frac{t}{T} \cos(n\omega_0 t) dt
  ] Since ( f(t) \cos(n\omega_0 t) ) is an odd function, the integral evaluates to zero, meaning ( a_n = 0 ).
• Compute ( b_n ) (Sine Coefficients): [
  b_n = \frac{2}{T} \int_{-T/2}^{T/2} \frac{t}{T} \sin(n\omega_0 t) dt
  ] Using integration by parts, [
  b_n = \frac{2}{T^2} \int_{-T/2}^{T/2} t \sin(n\omega_0 t) dt
  ] Solving this integral results in: [
  b_n = \frac{2}{n\pi} (-1)^n
  ]

Step 4: Write the Fourier Series

Since ( a_0 = 0 ) and ( a_n = 0 ), the Fourier series reduces to:

[
f(t) = \sum_{n=1}^{\infty} \frac{2 (-1)^n}{n\pi} \sin(n\omega_0 t)
]

where ( \omega_0 = \frac{2\pi}{T} ).

Explanation

1. The function ( f(t) = t/T ) is an odd function, so it only has sine terms in its Fourier series.
2. The Fourier series represents the sawtooth wave as an infinite sum of sine functions.
3. The coefficients ( b_n = \frac{2 (-1)^n}{n\pi} ) show the amplitude of each sine component, which decreases as ( 1/n ), indicating the harmonic nature of the wave.
4. The alternating sign ( (-1)^n ) ensures the correct waveform shape.

This result is crucial in signal processing, as sawtooth waves are used in music synthesis and communication systems.