Question

A continuous-time periodic signal \( x(t) = 1 + 2 \cos(4\pi t) + 2 \cos(6\pi t) \). If \( T \) is the time period of \( x(t) \), then \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt = \ \_\_\_\_\_? \]

• A. 7
• B. 4
• C. 5
• D. 9

Hint:

For periodic signals, \( \frac{1}{T} \int_0^T |x(t)|^2 \, dt \) represents the average power over one period.

Answer 1

The Correct Option is A

Step 1: The signal \( x(t) \) is given by: \[ x(t) = 1 + 2 \cos(4\pi t) + 2 \cos(6\pi t) \] The time period \( T \) of the signal is the least common multiple of the periods of the cosine terms. The periods of \( \cos(4\pi t) \) and \( \cos(6\pi t) \) are \( \frac{1}{2} \) and \( \frac{1}{3} \) seconds, respectively. Thus, the fundamental period \( T \) is 1 second.
Step 2: We need to calculate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \] First, find \( |x(t)|^2 \): \[ |x(t)|^2 = \left( 1 + 2 \cos(4\pi t) + 2 \cos(6\pi t) \right)^2 \]
Expanding this expression: \[ |x(t)|^2 = 1 + 4 \cos(4\pi t) + 4 \cos(6\pi t) + 4 \cos^2(4\pi t) + 8 \cos(4\pi t) \cos(6\pi t) + 4 \cos^2(6\pi t) \]
Now, use the trigonometric identity for \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \) and the product-to-sum identities for the cross terms: \[ \int_0^T |x(t)|^2 \, dt = T \left( 1 + 4 \cdot 0 + 4 \cdot 0 + 4 \cdot \frac{T}{2} + 0 + 4 \cdot \frac{T}{2} \right) \] After performing the integration and simplifying, the result is: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt = 7 \]
Thus, the value is 7.