Let $ f(t) $ be a periodic signal with fundamental period $ T_00 $. Consider the signal $$ y(t) = f(\alpha t), \quad {where} \, \alpha1. $$ 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?

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?

cdquestions Admin · Accepted Answer

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).

Answer

$ c_k = d_k \, {for all} \, k $

Answer

$ y(t) $ is periodic with a fundamental period $ \alpha T_0 $

Answer

$ c_k = \frac{d_k}{\alpha} \, {for all} \, k $

Answer

$ y(t) $ is periodic with a fundamental period $ \frac{T_0}{\alpha} $

Show Hint

The Correct Option is A, D

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