Questions: A periodic signal with fundamental period T0=2 have complex exponential Fourier Series coefficients D0=1, D1=1, D2=0.5 j and D-2=0.5 j, Find the average power of the periodic signal x(t)

Solution Steps

To find the average power of a periodic signal using its Fourier Series coefficients, we use the formula for average power, which is the sum of the squares of the magnitudes of the coefficients. Specifically, the average power \( P \) is given by:

\[ P = \sum_{n=-\infty}^{\infty} |D_n|^2 \]

In this case, we only have a few non-zero coefficients, so we will sum the squares of their magnitudes.

The given Fourier Series coefficients are \( D_0 = 1 \), \( D_1 = 1 \), \( D_2 = 0.5j \), and \( D_{-2} = 0.5j \). These are the non-zero coefficients that will contribute to the average power of the signal.

The magnitudes of the coefficients are calculated as follows:

• \( |D_0| = |1| = 1 \)
• \( |D_1| = |1| = 1 \)
• \( |D_2| = |0.5j| = 0.5 \)
• \( |D_{-2}| = |0.5j| = 0.5 \)

The average power \( P \) of the signal is the sum of the squares of the magnitudes of the non-zero coefficients: \[ P = |D_0|^2 + |D_1|^2 + |D_2|^2 + |D_{-2}|^2 \] Substituting the magnitudes: \[ P = 1^2 + 1^2 + 0.5^2 + 0.5^2 = 1 + 1 + 0.25 + 0.25 = 2.5 \]

Final Answer