Appendix D

This appendix presents the derivation of power expressions for both sinusoidal and direct current (DC) signals.

Power of sinusoid#

The power of a periodic function is defined as the integral of its squared value over one period, normalized by the period duration:

\[ P_x = \frac{1}{T} \int\limits_{0}^{T} \left( x(t) \right)^2 dt \]

For the sine function \( x(t) = A\sin(t) \):

\[ P_x = \frac{1}{T} \int\limits_{0}^{T} \left( A\sin(t) \right)^2 dt \]

Expanding the square:

\[ P_x = \frac{A^2}{T} \int\limits_{0}^{T} \sin^2(t) \, dt \]

Using the trigonometric identity:

\[ \sin^2(t) = \frac{1 - \cos(2t)}{2} \]

We rewrite the integral:

\[ P_x = \frac{A^2}{T} \int\limits_{0}^{T} \frac{1 - \cos(2t)}{2} \, dt = \color{red}{\frac{A^2}{2T} \int\limits_{0}^{T} 1 \, dt} \color{#222832} {\ -\ } \color{green}{\frac{A^2}{2T} \int\limits_{0}^{T} \cos(2t) \, dt} \]

Evaluating the integrals:

First integral:
\[ \color{red}{\frac{A^2}{2T} \int\limits_{0}^{T} 1 \, dt = \frac{A^2}{2T} \cdot \left[ t \right]_0^T = \frac{A^2}{2T} \cdot T = \frac{A^2}{2}} \]
Second integral:
\[ \color{green}{\frac{A^2}{2T} \int\limits_{0}^{T} \cos(2t) \, dt = \frac{A^2}{2T} \cdot \left[ \frac{\sin(2t)}{2} \right]_0^T = \frac{A^2}{4T} \left( \sin(2T) - \sin(0) \right) = 0} \]
Since \( \sin(2T) = 0 \) for a full period \( T \).

Final Result:

\[ P_x = \frac{A^2}{2} - 0 = \frac{A^2}{2} \]

Thus, the power of a sine function over one period is:

                                                \[
                                                P_x = \frac{A^2}{2}
                                                \]
                                            

The power of the DC component#

\[ P_x = \frac{1}{T} \int\limits_{0}^{T} A^2 \, dt = \frac{A^2}{T} \int\limits_{0}^{T} 1 \, dt = \frac{A^2}{T} \cdot \left[ t \right]_0^T = \frac{A^2}{T} \cdot T = A^2 \]