# Enery and power when we have trigonometric functions (Fourier transform)

It is asked to evaluate the energy and power of the signal

$$x(t)=10\cos(100t+30°)-5\sin(220t-50°)$$

Since it is periodic, I need to find

$$\int _{-\infty}^\infty |x(t)|^2 dt \text{ and } \frac{1}{T}\int_{-T}^T |x(t)|^2 dt$$

Where $T$ is the period of $x(t)$ (which is $18$). By Parseval's theorem, we know the energy is conserved when we do a Fourier transform and I was trying to use it (but I couldn't). What is the best way of evaluating those integrals?

• seems like a maths question to me. Commented Mar 6, 2015 at 18:17
• The best way would be to open a table of integrals and use it. Commented Mar 6, 2015 at 18:19
• What sort of signal is it? Power, current, voltage? Commented Mar 6, 2015 at 18:23
• If $x(t)$ is always real (no imaginary part), you can drop the absolute value signs because $x^2(t)$ will always be positive. Commented Mar 9, 2015 at 14:36

For any periodic signal you get

$$\int_{-\infty}^{\infty}|x(t)|^2dt=\infty$$

i.e. the integral does not converge, and, consequently, the energy is infinite. The power is finite and can be computed from the following formula (which differs from yours by a factor of $\frac12$):

$$\overline{x^2(t)}=\frac{1}{2T}\int_{-T}^T|x(t)|^2dt$$

Note that $\cos^2 x=\frac12 (1+\cos(2x))$ and $\sin^2 x=\frac12 (1-\cos(2x))$. So after integrating over one period, the terms with double frequency and also the cross-term cancel out. So you finally get for the power

$$\overline{x^2(t)}=10^2\cdot \frac12 + 5^2\cdot\frac12=62.5$$

Note that the actual value of $T$ is irrelevant. This is a good thing because the value you got is wrong.

• +1 for pointing out that $T$ is irrelevant - the frequency of a sinusoidal function doesn't affect its power. Commented Mar 9, 2015 at 14:37