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Signals and Systems – Rayleigh’s Energy Theorem
Energy of a Signal
The energy of a signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is defined as the area under the curve of square of magnitude of that signal, i.e.,
$$\mathrm{\mathit{E}\:\mathrm{=}\:\int_{-\infty}^{\infty}\left|\mathit{x}\mathrm{\left(\mathit{t}\right)} \right|^{\mathrm{2}}\:\mathit{dt}}$$
The energy signal exists only of the energy (E) of the signal is finite, i.e., only if 0 < E < $\infty$.
Rayleigh’s Energy Theorem
Statement - The Rayleigh’s energy theorem states that the integral of the square of magnitude of a function (i.e., energy of the function) is equal to the integral of the square of magnitude of its Fourier transform, i.e.,
$$\mathrm{\mathit{E}\:\mathrm{=}\:\int_{-\infty}^{\infty}\left|\mathit{x}\mathrm{\left(\mathit{t}\right)} \right|^{\mathrm{2}}\:\mathit{dt}\:\mathrm{=}\:\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\mathit{X}\mathrm{\left(\mathit{\omega }\right)} \right|^{\mathrm{2}}\:\mathit{d\omega }}$$
Proof
Consider a function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ such that its Fourier transform pair is,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$
Assume $\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}$ is the conjugate of the function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ and its Fourier transform pair is,
$$\mathrm{\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{X}^{*}\mathrm{\left(\mathit{-\omega }\right)}}$$
Then, the energy of the signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is given by,
$$\mathrm{\mathit{E}\:\mathrm{=}\:\int_{-\infty}^{\infty}\left|\mathit{x}\mathrm{\left(\mathit{t}\right)} \right|^{\mathrm{2}}\:\mathit{dt}}$$
$$\mathrm{\because \left|\mathit{x}\mathrm{\left(\mathit{t}\right)} \right|^{\mathrm{2}}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}}$$
$$\mathrm{\therefore \mathit{E}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}}$$
Now, by replacing the function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ by its inverse Fourier transform, we get,
$$\mathrm{\mathit{E}\:\mathrm{=}\:\int_{-\infty}^{\infty}\:\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\mathrm{\left[\frac{1}{2\pi} \int_{-\infty}^{\infty }\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\mathit{e^{j\omega t}d\omega}\right]}\:\mathit{dt}}$$
By interchanging the order of integration in above equation, we get,
$$\mathrm{\mathit{E}\:\mathrm{=}\:\frac{1}{2 \pi}\int_{-\infty}^{\infty }\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\left[\int_{-\infty}^{\infty }\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\mathit{e^{j\omega t}\:dt} \right]\:\mathit{d\omega }}$$
$$\mathrm{\Rightarrow \mathit{E}\:\mathrm{=}\:\frac{1}{2 \pi}\int_{-\infty}^{\infty }\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathit{X}^{*}\mathrm{\left(\mathit{-\omega}\right)}\:\mathit{d\omega}}$$
$$\mathrm{\because \mathit{X}\mathrm{\left(\mathit{\omega}\right)}\mathit{X}^{*}\mathrm{\left(\mathit{-\omega}\right)}\:\mathrm{=}\:\left| \mathit{X}\mathrm{\left(\mathit{\omega}\right)}\right|^{\mathrm{2}}}$$
Therefore, we have,
$$\mathrm{\mathit{E}\:\mathrm{=}\:\frac{1}{2 \pi}\int_{-\infty}^{\infty }\left| \mathit{X}\mathrm{\left(\mathit{\omega}\right)}\right|^{\mathrm{2}}\:\mathit{d\omega}\:\mathrm{=}\:\int_{\infty}^{\infty}\left| \mathit{x}\mathrm{\left(\mathit{t}\right)}\right|^{\mathrm{2}}\:\mathit{dt}}$$
This is called the Rayleigh’s energy theorem. The Rayleigh’s energy theorem is called the Parseval’s theorem for energy signals.
The above expression proves that the integral of the square of a signal is equal to the integral of square of its Fourier transform.
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