Question

In the Fourier series expansion of two functions \( f_1(t) = 4t^2 + 3 \) and \( f_2(t) = 6t^3 + 7t \) in the interval \( -\frac{T}{2} \) to \( +\frac{T}{2} \), the Fourier coefficients \( a_n \) and \( b_n \) (\( a_n \) and \( b_n \) are coefficients of \( \cos(n\omega t) \) and \( \sin(n\omega t) \), respectively) satisfy:

• A. \( a_n = 0 \) and \( b_n \neq 0 \) for \( f_1(t) \); \( a_n \neq 0 \) and \( b_n = 0 \) for \( f_2(t) \)
• B. \( a_n \neq 0 \) and \( b_n = 0 \) for \( f_1(t) \); \( a_n = 0 \) and \( b_n \neq 0 \) for \( f_2(t) \)
• C. \( a_n \neq 0 \) and \( b_n \neq 0 \) for \( f_1(t) \); \( a_n = 0 \) and \( b_n \neq 0 \) for \( f_2(t) \)
• D. \( a_n = 0 \) and \( b_n \neq 0 \) for \( f_1(t) \); \( a_n \neq 0 \) and \( b_n \neq 0 \) for \( f_2(t) \)

Hint:

In Fourier analysis, even functions contain only cosine terms (\( a_n \)) and odd functions only sine terms (\( b_n \)).

Answer 1

The Correct Option is B

Step 1: Identify the symmetry of the functions.
The function \( f_1(t) = 4t^2 + 3 \) is an even function because \( f_1(-t) = f_1(t) \). The function \( f_2(t) = 6t^3 + 7t \) is an odd function because \( f_2(-t) = -f_2(t) \).
Step 2: Recall Fourier series properties.
- For even functions: only cosine terms (\( a_n \)) are present. - For odd functions: only sine terms (\( b_n \)) are present.
Step 3: Apply to given functions.
\[ f_1(t) \Rightarrow a_n \neq 0, \ b_n = 0, \quad f_2(t) \Rightarrow a_n = 0, \ b_n \neq 0. \] Step 4: Final Answer.
Hence, the correct relation is option (B).