Question:medium
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?
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?
Show 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.
Updated On: Feb 4, 2026
- \( c_k = d_k \, {for all} \, k \)
- \( y(t) \) is periodic with a fundamental period \( \alpha T_0 \)
- \( c_k = \frac{d_k}{\alpha} \, {for all} \, k \)
- \( y(t) \) is periodic with a fundamental period \( \frac{T_0}{\alpha} \)
Show Solution
The Correct Option is A, D
Solution and Explanation
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).
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).
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