Question

The equation of average power of a periodic signal x(t) is given as ___________

(a) \(\sum_{k=0}^{\infty}|c_k|^2\)

(b) \(\sum_{k=-\infty}^{\infty}|c_k|\)

(c) \(\sum_{k=-\infty}^0|c_k|^2\)

(d) \(\sum_{k=-\infty}^{\infty}|c_k|^2\)

I have been asked this question in examination.

My enquiry is from Frequency Analysis of Continuous Time Signal topic in chapter Frequency Analysis of Signals and Systems of Digital Signal Processing

Answer 1

Right answer is (d) \(\sum_{k=-\infty}^{\infty}|c_k|^2\)

The explanation: The average power of a periodic signal x(t) is given as

The average power of a periodic signal x(t) is given as \(\frac{1}{T_p} \int_{t_0}^{t_0+T_p}|x(t)|^2 dt\)

=\(\frac{1}{T_p} \int_{t_0}^{t_0+T_p}x(t).x^* (t) dt\)

=\(\frac{1}{T_p} \int_{t_0}^{t_0+T_p}x(t).\sum_{k=-\infty}^{\infty} c_k * e^{-j2πkF_0 t} dt\)

By interchanging the positions of integral and summation and by applying the integration, we get

=\(\sum_{k=-\infty}^{\infty}|c_k |^2\)