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_{0}^{T} {(x (t))}^{2} d t

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

P_{x} = \frac{1}{T} \int_{0}^{T} {(A \sin (t))}^{2} d t

Expanding the square:

P_{x} = \frac{A^{2}}{T} \int_{0}^{T} \sin^{2} (t) d t

Using the trigonometric identity:

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

We rewrite the integral:

P_{x} = \frac{A^{2}}{T} \int_{0}^{T} \frac{1 - \cos (2 t)}{2} d t = \frac{A^{2}}{2 T} \int_{0}^{T} 1 d t - \frac{A^{2}}{2 T} \int_{0}^{T} \cos (2 t) d t

Evaluating the integrals:

First integral:
$\frac{A^{2}}{2 T} \int_{0}^{T} 1 d t = \frac{A^{2}}{2 T} \cdot {[t]}_{0}^{T} = \frac{A^{2}}{2 T} \cdot T = \frac{A^{2}}{2}$
Second integral:
$\frac{A^{2}}{2 T} \int_{0}^{T} \cos (2 t) d t = \frac{A^{2}}{2 T} \cdot {[\frac{\sin (2 t)}{2}]}_{0}^{T} = \frac{A^{2}}{4 T} (\sin (2 T) - \sin (0)) = 0$
Since $\sin (2 T) = 0$ 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}

P_{x} = \frac{1}{T} \int_{0}^{T} A^{2} d t = \frac{A^{2}}{T} \int_{0}^{T} 1 d t = \frac{A^{2}}{T} \cdot {[t]}_{0}^{T} = \frac{A^{2}}{T} \cdot T = A^{2}

P_{x} = A^{2}