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.

  • $\begingroup$ Another hint: $sin(x)cos(y) = 0.5(sin(x+y)+sin(x-y))$. Apply this, and take the mean. $\endgroup$
    – Hilmar
    Commented Jan 15, 2021 at 14:38
  • $\begingroup$ if the frequencies are different, the power of the sum of sinusoids is the sum of the powers of each sinusoid. $\endgroup$ Commented Jan 16, 2021 at 18:27

2 Answers 2


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.

  • $\begingroup$ Finally i ended up using the same method as mentioned in your answer. I would also like to know details about solving this problem using the second method which was elobrated in below answer by @Gillies $\endgroup$ Commented Jan 16, 2021 at 9:00
  • $\begingroup$ @HentaiOusama: You can't use the Fourier transform here because you're not interested in the energy (which is infinite for periodic signals) but in the power. What you can use are the Fourier series coefficients $c_n$, as shown in the first formula in Gilles' answer. $\endgroup$
    – Matt L.
    Commented Jan 16, 2021 at 13:26
  • $\begingroup$ Thx. This makes things more clear. $\endgroup$ Commented Jan 18, 2021 at 21:40


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 $(2)$. $$ \mathcal F \big\{A\cos\left(2\pi f_0 t + \phi\right)\big\} = \frac A2\big[e^{j\phi}\delta\left(f - f_0\right) + e^{-j\phi}\delta\left(f + f_0\right)\big]\tag{1} $$

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


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}

  • $\begingroup$ If you use Parseval's theorem for Fourier series, why would you need equations $(1)-(3)$? $\endgroup$
    – Matt L.
    Commented Jan 15, 2021 at 8:43
  • $\begingroup$ Instead of going through time-domain computation, you could actually simply use $(1)$ & $(2)$ only. $\endgroup$
    – Gilles
    Commented Jan 15, 2021 at 8:45
  • $\begingroup$ But how would you use them? You can't compute the power from the Fourier transform of the signal. $\endgroup$
    – Matt L.
    Commented Jan 15, 2021 at 8:46
  • $\begingroup$ But there you have your $c_n$'s, using $(1)$ and $(2)$ in Parseval gives you $A^2/2$ in the end. $\endgroup$
    – Gilles
    Commented Jan 15, 2021 at 8:50
  • 1
    $\begingroup$ Yes, you're totally right, my understanding from OP's first bullet point is that OP got to the transform with no problem. Equation $(3)$ could be skipped actually, I'll fix that. $\endgroup$
    – Gilles
    Commented Jan 15, 2021 at 10:11

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