Question

Given \( y(t) = f(\alpha t) \), analyze how Fourier coefficients and time period are affected.

• A. \( C_k = d_k \ \forall k \)
• B. \( C_k = \alpha d_k \)
• C. If time period of \( f(t) \) is \( T_0 \), then the time period of \( f(\alpha t) \) is \( \frac{T_0}{\alpha} \)
• D. If time period of \( f(t) \) is \( T_0 \), then the time period of \( f(\alpha t) \) is \( \alpha T_0 \)

Hint:

In time-scaling \( y(t) = f(\alpha t) \), - The time period scales as \( T' = \frac{T_0}{\alpha} \). - The fundamental frequency scales as \( \omega_0' = \alpha \omega_0 \). - The Fourier coefficients remain unchanged.

Answer 1

The Correct Option is A

Effect of Time Scaling on Fourier Series.
Given \( y(t) = f(\alpha t) \), the Fourier series of \( f(t) \) is:
\[ f(t) = \sum_{n=-\infty}^{\infty} C_k e^{-jn\omega_0 t} \]
Applying time scaling:
\[ y(t) = f(\alpha t) = \sum_{n=-\infty}^{\infty} C_k e^{-jn\omega_0 (\alpha t)} \]
Rewriting:
\[ y(t) = \sum_{n=-\infty}^{\infty} C_k e^{-jn (\alpha \omega_0) t} \]
Comparing with the standard Fourier series form:
\[ y(t) = \sum_{n=-\infty}^{\infty} d_k e^{-jn\omega_0' t} \]
We get: - \( \omega_0' = \alpha \omega_0 \) (frequency scales by \( \alpha \)).
- Fourier coefficients remain the same: \( d_k = C_k \).
Checking the given options:
1. \( C_k = d_k \) (Correct)
2. \( C_k = \alpha d_k \) (Incorrect)
3. Time period transformation: \[ T' = \frac{T_0}{\alpha} \] % Option (Correct)
4. \( T' = \alpha T_0 \) (Incorrect).
Thus, the correct answers are (A) and (C).