Question

A continuous time periodic signal \( x(t) \) is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If \( T \) is the period of \( x(t) \), then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad \text{(round off to the nearest integer).} \]

Hint:

To compute the average power of a periodic signal, use Parseval’s theorem: \[ \frac{1}{T} \int_0^T |x(t)|^2 dt = a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} a_n^2 \] where \( a_0 \) is the DC term and \( a_n \) are the amplitudes of harmonics.

Answer 1

This is the average power of a periodic signal. We apply Parseval’s theorem: \[ \frac{1}{T} \int_0^T |x(t)|^2 dt = a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} a_n^2 \] Given: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] Here, \( a_0 = 1 \), \( a_1 = a_2 = a_3 = 2 \) So, \[ \frac{1}{T} \int_0^T |x(t)|^2 dt = 1^2 + \frac{1}{2}[(2)^2 + (2)^2 + (2)^2] = 1 + \frac{1}{2}(4 + 4 + 4) = 1 + \frac{12}{2} = 1 + 6 = 7 \]