# Getting the power of a signal from its Fourier transform?

I have a non-periodic signal that contains the sinc function in the time domain and so it is a bit difficult to calculate its power (because of the integral) through: $${P_x} = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |x(t)|^2 dt$$

This is my signal: $$x(t)=A\cos(\omega_1t)+B\,\textrm{sinc}(\omega_2t)$$. I wonder whether one can calculate the power in the frequency domain, since (Rayleigh's Theorem): $$\int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-\infty}^{\infty} |X(f)|^2 df$$

Note that due to the periodic component in the signal ($$A\cos(\omega_1t))$$, the integral

$$\int_{-\infty}^{\infty}|x(t)|^2dt=\int_{-\infty}^{\infty}|X(f)|^2df\tag{1}$$

doesn't exist, i.e., the energy of the given signal is infinite. The signal is a so-called power signal, having finite power and infinite energy.

What you can do is compute its power using the equation given in your question:

$$P_x=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt\tag{2}$$

but you can't use Parseval's theorem to compute that limit.

However, you can use Parseval's theorem indirectly to show that the contributions to the limit $$(2)$$ from all components of $$|x(t)|^2$$ containing the $$\textrm{sinc}$$ function go to zero as $$T\to\infty$$, because all those components have finite energy, hence zero power. So the power of the given signal just equals the power of $$A\cos(\omega_1t)$$, and I'm sure you know how to compute that power.

• "You can use Parseval's theorem indirectly". This is the observation that led me in asking this question. I am afraid that there might not be a direct way (a formula) for calculating the power from fourier transform. However, getting rid of the sinc function from the power expression is good progress! Jul 5 '20 at 13:23