# Parsevals Theorem

#### PARSEVAL'S THEOREM

Parseval's theorem demonstrates one practical use of the Fourier transform. It relates the energy carried by a signal to the Fourier transform of the signal.
If $p(t)$ is the power associated with the signal, the energy carried by the signal is
$$W=\int_{-\infty}^{\infty} p(t) d t \tag{1}$$
In order to be able compare the energy content of current and voltage signals, it is convenient to use a $1-\Omega$ resistor as the base for energy calculation. For a 1- $\Omega$ resistor,
$$p(t)=v^{2}(t)=i^{2}(t)=f^{2}(t)$$
where $f(t)$ stands for either voltage or current. The energy delivered to the $1-\Omega$ resistor is
$$W_{1 \Omega}=\int_{-\infty}^{\infty} f^{2}(t) d t \tag{2}$$
Parseval's theorem states that this same energy can be calculated in the frequency domain as
$$W_{1 \Omega}=\int_{-\infty}^{\infty} f^{2}(t) d t=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|F(\omega)|^{2} d \omega \tag{3}$$
Parseval’s theorem states that the total energy delivered to a 1-0 resistor equals the total area under the square of $f (t)$ or $1/2π$ times the total area under the square of the magnitude of the Fourier transform of $f (t)$.
Parseval's theorem relates energy associated with a signal to its Fourier transform. It provides the physical significance of $F(\omega)$, namely, that $|F(\omega)|^{2}$ is a measure of the energy density (in joules per hertz) corresponding to $f(t)$.
To derive Eq. (3), we begin with Eq. (2) and substitute Eq.(A) for one of the $f(t)$ 's. We obtain
$$W_{1 \Omega}=\int_{-\infty}^{\infty} f^{2}(t) d t=\int_{-\infty}^{\infty} f(t)\left[\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega\right] d t \tag{4}$$
The function $f(t)$ can be moved inside the integral within the brackets, since the integral does not involve time:
$$W_{1 \Omega}=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) F(\omega) e^{j \omega t} d \omega d t \tag{5}$$
Reversing the order of integration,
\begin{aligned}W_{1 \Omega} &=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega)\left[\int_{-\infty}^{\infty} f(t) e^{-j(-\omega) t} d t\right] d \omega \\&=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) F(-\omega) d \omega=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) F^{*}(\omega) d \omega\end{aligned} \tag{6}
But if $z=x+j y, z z^{*}=(x+j y)(x-j y)=x^{2}+y^{2}=|z|^{2}$. Hence,
$$W_{1 \Omega}=\int_{-\infty}^{\infty} f^{2}(t) d t=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|F(\omega)|^{2} d \omega \tag{7}$$
as expected. Equation (7) indicates that the energy carried by a signal can be found by integrating either the square of $f(t)$ in the time domain or $1 / 2 \pi$ times the square of $F(\omega)$ in the frequency domain.
Since $|F(\omega)|^{2}$ is an even function, we may integrate from 0 to $\infty$ and double the result, that is,
$$W_{1 \Omega}=\int_{-\infty}^{\infty} f^{2}(t) d t=\frac{1}{\pi} \int_{0}^{\infty}|F(\omega)|^{2} d \omega$$
We may also calculate the energy in any frequency band $\omega_{1}<\omega<\omega_{2}$ as
$$W_{1 \Omega}=\frac{1}{\pi} \int_{\omega_{1}}^{\omega_{2}}|F(\omega)|^{2} d \omega$$
Notice that Parseval's theorem as stated here applies to nonperiodic functions. As evident in Eq. (7), Parseval's theorem shows that the energy associated with a nonperiodic signal is spread over the entire frequency spectrum, whereas the energy of the periodic signal is concentrated at the frequencies of its harmonic components.
Example 1: The voltage across a $10-\Omega$ resistor is $v(t)=5 e^{-3 t} u(t) \mathrm{V}$. Find the total energy dissipated in the resistor.
Solution:
We can find the energy using either $f(t)=v(t)$ or $F(\omega)=V(\omega)$. In the time domain,
\begin{aligned}W_{10 \Omega} &=10 \int_{-\infty}^{\infty} f^{2}(t) d t=10 \int_{0}^{\infty} 25 e^{-6 t} d t \\&=\left.250 \frac{e^{-6 t}}{-6}\right|_{0} ^{\infty}=\frac{250}{6}=41.67 \mathrm{~J}\end{aligned}
In the frequency domain,
$$F(\omega)=V(\omega)=\frac{5}{3+j \omega}$$
so that
$$|F(\omega)|^{2}=F(\omega) F^{*}(\omega)=\frac{25}{9+\omega^{2}}$$
Hence, the energy dissipated is
\begin{aligned}W_{10 \Omega} &=\frac{10}{2 \pi} \int_{-\infty}^{\infty}|F(\omega)|^{2} d \omega=\frac{10}{\pi} \int_{0}^{\infty} \frac{25}{9+\omega^{2}} d \omega \\&=\left.\frac{250}{\pi}\left(\frac{1}{3} \tan ^{-1} \frac{\omega}{3}\right)\right|_{0} ^{\infty}=\frac{250}{\pi}\left(\frac{1}{3}\right)\left(\frac{\pi}{2}\right)=\frac{250}{6}=41.67 \mathrm{~J}\end{aligned}
Example 2: Calculate the fraction of the total energy dissipated by a $1-\Omega$ resistor in the frequency band $0 < \omega < 10 \mathrm{rad} / \mathrm{s}$ when the voltage across it is $v(t)=e^{-2 t} u(t)$. Solution:
Given that $f(t)=v(t)=e^{-2 t} u(t)$, then
$$F(\omega)=\frac{1}{2+j \omega} \quad \Longrightarrow \quad|F(\omega)|^{2}=\frac{1}{4+\omega^{2}}$$
The total energy dissipated by the resistor is
\begin{aligned}W_{1 \Omega} &=\frac{1}{\pi} \int_{0}^{\infty}|F(\omega)|^{2} d \omega=\frac{1}{\pi} \int_{0}^{\infty} \frac{d \omega}{4+\omega^{2}} \\&=\frac{1}{\pi}\left(\left.\frac{1}{2} \tan ^{-1} \frac{\omega}{2}\right|_{0} ^{\infty}\right)=\frac{1}{\pi}\left(\frac{1}{2}\right) \frac{\pi}{2}=0.25 \mathrm{~J}\end{aligned}
The energy in the frequencies $0<\omega<10$ is
\begin{aligned}W &=\frac{1}{\pi} \int_{0}^{10}|F(\omega)|^{2} d \omega=\frac{1}{\pi} \int_{0}^{10} \frac{d \omega}{4+\omega^{2}}=\frac{1}{\pi}\left(\left.\frac{1}{2} \tan ^{1} \frac{\omega}{2}\right|_{0} ^{10}\right) \\&=\frac{1}{2 \pi} \tan ^{-1} 5=\frac{1}{2 \pi}\left(\frac{78.69^{\circ}}{180^{\circ}} \pi\right)=0.218 \mathrm{~J}\end{aligned}
Its percentage of the total energy is
$$\frac{W}{W_{1 \Omega}}=\frac{0.218}{0.25}=87.4 \%$$