The Parseval’s theorem relates energy associated with a signal to its Fourier transform.
If \(p(t)\) is the power associated with the signal, the energy carried by the signal is
\(\displaystyle W = \int_{-\infty}^{\infty}p(t)\,dt\)
where \(\displaystyle p(t) = v^{2}(t)/R = i^{2}(t)R\)
In order to be able to 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 \(R = 1\:\Omega\) resistor, \(p(t) = i^{2}(t) = v^{2}(t) = f^{2}(t)\), where \(f(t)\) stands for either voltage or current.
\(\boxed{W_{1\,\Omega} = \int_{-\infty}^{\infty}f^{2}(t)\,dt}\)
Substituting the inverse Fourier transform \(f(t) = \mathcal{F}^{-1}\left[F(\omega)\right]\).
\(\displaystyle W_{1\,\Omega} = \int_{-\infty}^{\infty}f(t)\left[\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)\,e^{\displaystyle\,j\omega t}\,d\omega\right]\,dt\)
Reversing the order of integration,
\(\displaystyle W_{1\,\Omega} = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)\left[\int_{-\infty}^{\infty}f(t)\,e^{\displaystyle -j(-\omega)t}\,dt\right]\,d\omega\)
Applying the reversal property of Fourier transform,
\(\displaystyle W_{1\,\Omega} = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)F(-\omega)\,d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)F^{\ast}(\omega)\,d\omega\)
But if \(z = x + jy\), \(\displaystyle zz^{\ast} = (x + jy)(x - jy) = x^2 + y^2 = |z|^2\). Hence,
\(\boxed{W_{1\,\Omega} = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left|F(\omega)\right|^2\,d\omega}\)
Parseval’s theorem provides the physical significance of \(F(\omega)\), namely, that \(\left|F(\omega)\right|^2\) is a measure of the energy density (in joules per hertz) corresponding to \(f(t)\).
Parseval’s theorem states that the total energy delivered to a \(1\:\Omega\) resistor equals the total area under the square of \(f(t)\) or \(1/2\pi\) times the total area under the square of the magnitude of the Fourier transform of \(f(t)\).
\(\boxed{W_{1\,\Omega} = \int_{-\infty}^{\infty}f^{2}(t)\,dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left|F(\omega)\right|^2\,d\omega}\)
Since \(\left|F(\omega)\right|^2\) is an even function, it might be possible to integrate from \(0\) to \(\infty\) and double the result.
\(\boxed{W_{1\,\Omega} = \int_{-\infty}^{\infty}f^{2}(t)\,dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left|F(\omega)\right|^2\,d\omega = \frac{1}{\pi}\int_{0}^{\infty}\left|F(\omega)\right|^2\,d\omega}\)
Parseval’s theorem shows that the energy associated with a non-periodic signal is spread over the entire frequency spectrum, whereas the energy of a periodic signal is concentrated at the frequencies of its harmonic components.
Energy of some common signals.
\(\boxed{W_{1\,\Omega} = A^2 b}\quad\textrm{(rectangular pulse)}\)
\(\boxed{W_{1\,\Omega} = \frac{A^2 b}{2}}\quad\textrm{(half-cycle sinusoid)}\)
\(\boxed{W_{1\,\Omega} = \frac{A^2 b}{3}}\quad\textrm{(triangular pulse)}\)
where \(A\) and \(b\) are the height and width of the signal, respectively.
