Question

The exponential Fourier series representation of a continuous-time periodic signal \( x(t) \) is defined as \[ x(t) = \sum_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t} \] where \( \omega_0 \) is the fundamental angular frequency of \( x(t) \) and the coefficients of the series are \( a_k \). The following information is given about \( x(t) \) and \( a_k \): 
\( x(t) \) is real and even, having a fundamental period of 6. 
The average value of 

The average power of the signal \( x(t) \) (rounded off to one decimal place) is _________. 
 

Hint:

The average power of a signal can be calculated using the Fourier series coefficients and summing the squares of the non-zero coefficients.

Answer 1

Correct Answer: 31.9

The average power of a periodic signal \( x(t) \) is given by: \[ P_{\text{avg}} = \frac{1}{T} \int_0^T |x(t)|^2 dt \] where \( T = 6 \) is the period of \( x(t) \). The Fourier coefficients \( a_k \) are given as: 

Since \( x(t) \) is even, the power is given by: \[ P_{\text{avg}} = \sum_{k=-\infty}^{\infty} |a_k|^2. \] Using the given values of \( a_k \), we compute: \[ P_{\text{avg}} = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14. \] Thus, the average power of the signal is \( \boxed{31.9} \).