Find power of a sum of sinusoids

Question

We got this question to solve:

Calculate the power of the signal: $$s(t) = 8\cos\left(20\pi t-\frac \pi4\right) + 4\sin(15\pi t)$$

Now, I thought of two approaches :

• Use Parseval theorem, so first find Fourier transform, then calculate the required integral, but then we get a impulse function in it and I don't know how to integrate square of $\delta$.

• Using the normal formula to calculate the power of a signal, and again in that, we get 3 terms, $\sin^2$ which is easy, then $\cos^2$ again simple, and then $\sin\cdot \cos$, which I know how to solve, but will be lengthy and I think there should be some another proper way to solve this question.

Any guidance would be appreciated.

Answer 1

An easy way to think about the power of a signal is to realize that it is the mean of the square of the signal. Now note that the cross-term $\cos(\omega_1t+\phi)\sin(\omega_2t+\phi)$ ($\omega_1\neq\omega_2$) doesn't have a (non-zero) mean, it's just equivalent to two sinusoids at the sum and difference frequencies. So you only get contributions from the squares of the individual terms. It's straightforward to show that the mean of a squared sinusoid with amplitude $A$ equals $A^2/2$. Now you should know everything to just write down the result without the need for any complex calculations.

Answer 2

HINT:

For real-valued periodic signal $x$ with period $T_0$ Parseval gives you $$ P=\frac 1T_0\int_{-T_0/2}^{T_0/2} x^2(t)\mathrm{d}t = \sum_{n=-\infty}^{\infty}\lvert c_n\rvert^2 $$
Where $c_n$'s are Fourier coefficients. You can then use the properties in Equations $(1)$ and $(2)$. $$ \mathcal F \big\{A\cos\left(2\pi f_0 t + \phi\right)\big\} = \frac A2\big[e^{j\phi}\delta\left(f - f_0\right) + e^{-j\phi}\delta\left(f + f_0\right)\big]\tag{1} $$

$$ \mathcal F \big\{A\sin\left(2\pi f_0 t + \phi\right)\big\} = j\frac A2\big[e^{-j\phi}\delta\left(f + f_0\right)-e^{j\phi}\delta\left(f - f_0\right)\big]\tag{2} $$

EDIT:

For Equation $(1)$ for instance, applying Parseval above you have the following:

\begin{align} P&= \sum_{n=-\infty}^{\infty}\lvert c_n\rvert^2\\ &= \lvert c_{-1}\rvert^2 + \lvert c_{1}\rvert^2\\ & = \frac{A^2}{4}\cdot 1 + \frac{A^2}{4}\cdot 1\tag{$\lvert e^{\pm j\phi} \rvert = 1$} \\ \end{align}