Question

Let a continuous-time signal be \( x(t) = e^{j9t} + e^{j5t} \), where \( j = \sqrt{-1} \) and \( t \) is in seconds. The fundamental period of the magnitude of \( x(t) \), in seconds, is:

• A. \( \pi \)
• B. \( 2\pi \)
• C. \( \frac{5\pi}{2} \)
• D. \( 9\pi \)

Hint:

To find the period of a sum of exponentials \( e^{j\omega t} \), use the LCM of their individual periods. For magnitude, the fundamental period remains the same.

Answer 1

The Correct Option is B

Step 1: Analyze the signal components.
The signal is:
\[ x(t) = e^{j9t} + e^{j5t} \] Let’s denote this as the sum of two complex exponentials.

Step 2: Compute individual periods.
The period \( T \) of a signal \( e^{j\omega t} \) is given by: \[ T = \frac{2\pi}{\omega} \] So:
- \( T_1 = \frac{2\pi}{9} \)
- \( T_2 = \frac{2\pi}{5} \)

Step 3: Fundamental period of the sum.
To find the fundamental period of the sum \( x(t) \), we take the LCM of the individual periods:
\[ \text{LCM} \left( \frac{2\pi}{9}, \frac{2\pi}{5} \right) = 2\pi \cdot \text{LCM} \left( \frac{1}{9}, \frac{1}{5} \right) = 2\pi \]
Step 4: Period of the magnitude.
The magnitude:
\[ |x(t)| = |e^{j9t} + e^{j5t}| \] Since both components are periodic with \( 2\pi \), the magnitude is also periodic with:
\[ \boxed{2\pi} \]