Question

Which one of the following options is the correct Fourier series of the periodic function \( f(x) \) described below: \[ f(x) = \begin{cases} 0, & \text{if } -2 < x < -1, \\ 2k, & \text{if } -1 \leq x < 1, \quad \text{period} = 4, \\ 0, & \text{if } 1 \leq x < 2. \end{cases} \]

• A. \( f(x) = \frac{k}{2} + \frac{2k}{\pi} \left( \cos \frac{\pi x}{2} - \cos \frac{3\pi x}{2} + \frac{5}{5} \cos \frac{5\pi x}{2} + \dots \right) \)
• B. \( f(x) = \frac{k}{2} + \frac{2k}{\pi} \left( \sin \frac{\pi x}{2} - \sin \frac{3\pi x}{2} + \frac{5}{5} \sin \frac{5\pi x}{2} + \dots \right) \)
• C. \( f(x) = k + \frac{4k}{\pi} \left( \cos \frac{\pi x}{2} - \cos \frac{3\pi x}{2} + \frac{5}{5} \cos \frac{5\pi x}{2} + \dots \right) \)
• D. \( f(x) = k + \frac{4k}{\pi} \left( \sin \frac{\pi x}{2} - \sin \frac{3\pi x}{2} + \frac{5}{5} \sin \frac{5\pi x}{2} + \dots \right) \)

Hint:

For periodic functions with a given period, use the Fourier series expansion to express the function as a sum of sine and cosine terms. For even functions, only cosine terms will remain.

Answer 1

The Correct Option is C

Given that \( f(x) \) is a periodic function with a period of 4, we need to find the Fourier series for this function. Since the function is even, we know that all the sine terms in the Fourier series will be zero. We will only have cosine terms in the Fourier series expansion.

Fourier Coefficients:
The Fourier series is given by: \[ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos \frac{n \pi x}{L}. \] The period \( L = 4 \), so the cosine terms will be of the form \( \cos \frac{n \pi x}{2} \).

We first calculate \( a_0 \), the average or DC component: \[ a_0 = \frac{1}{L} \int_{-L/2}^{L/2} f(x) \, dx = \frac{1}{4} \int_{-2}^{2} f(x) \, dx. \] From the given function, we know that: \[ a_0 = 2k \quad \text{(as the function is constant and equal to \( 2k \) for \( -1 \leq x < 1 \))}. \]

Next, we calculate the \( a_n \) coefficients: \[ a_n = \frac{2}{L} \int_{-L/2}^{L/2} f(x) \cos \frac{n \pi x}{L} \, dx. \] For \( n = 1, 2, 3, \dots \), the integrals for these coefficients will give us the terms in the series. After performing the integration (which involves calculating the integrals for cosine terms), we get: \[ a_n = \frac{4k}{n\pi} \sin \left( \frac{n \pi}{2} \right). \]

Thus, the Fourier series for this function is: \[ f(x) = k + \frac{4k}{\pi} \left( \cos \frac{\pi x}{2} - \cos \frac{3\pi x}{2} + \frac{1}{5} \cos \frac{5\pi x}{2} + \dots \right). \] Therefore, the correct answer is option (C).

- This Fourier series consists of only cosine terms because \( f(x) \) is an even function. The sine terms are eliminated due to the symmetry of the function.