Question

Let \( f(t) \) be a periodic signal with fundamental period \( T_0>0 \). Consider the signal \[ y(t) = f(\alpha t), \quad {where} \, \alpha>1. \] The Fourier series expansions of \( f(t) \) and \( y(t) \) are given by \[ f(t) = \sum_{k=-\infty}^{\infty} c_k e^{j \frac{2\pi}{T_0} k t}, \quad y(t) = \sum_{k=-\infty}^{\infty} d_k e^{j \frac{2\pi}{T_0 \alpha} k t}. \] Which of the following statements is/are TRUE?

• A. \( c_k = d_k \, {for all} \, k \)
• B. \( y(t) \) is periodic with a fundamental period \( \alpha T_0 \)
• C. \( c_k = \frac{d_k}{\alpha} \, {for all} \, k \)
• D. \( y(t) \) is periodic with a fundamental period \( \frac{T_0}{\alpha} \)

Hint:

When scaling the time variable in a periodic function, the period of the function is inversely proportional to the scaling factor. The Fourier coefficients remain unchanged.

Answer 1

The Correct Option is A, D

Step 1: Determine the relationship between \( c_k \) and \( d_k \)
The Fourier coefficients of \( y(t) = f(\alpha t) \) are related to the Fourier coefficients of \( f(t) \) by the substitution \( t \to \alpha t \). The scaling of the time variable by \( \alpha \) does not change the form of the coefficients; they are equal. Therefore: \[ c_k = d_k \, {for all} \, k \] Step 2: Determine the period of \( y(t) \)
If the period of \( f(t) \) is \( T_0 \), then the period of \( y(t) = f(\alpha t) \) will be scaled by \( \frac{1}{\alpha} \), so the period of \( y(t) \) is: \[ T_y = \frac{T_0}{\alpha} \] Thus, \( y(t) \) is periodic with a fundamental period \( \frac{T_0}{\alpha} \). Thus, the correct answers are (A) and (D).