0.3 Energy and power

Energy and power

The energy of a continuous-time signal $x\left(t\right)$ is given as

Plancherel's theorem says (for more information see  Comment 2 ): If the signal $x\left(t\right)$ has finite energy then its Fourier transform $X\left(f\right)$ has the same energy:

Comment 2 Plancherel theorem is a result in harmonic analysis, first proved byMichel Plancherel. In its simplest form it states that if a function f is in both $L_1\left(\mathrm{I}\mathrm{R}\right)$ and $L_2\left(\mathrm{I}\mathrm{R}\right)$ , then its Fourier transform is in $L_2\left(\mathrm{I}\mathrm{R}\right)$ ; moreover the Fourier transform map is isometric. This implies that the Fourier transform map restricted to $L_1\left(\mathrm{I}\mathrm{R}\right)\cap L_2\left(\mathrm{I}\mathrm{R}\right)$ has a unique extension to a linear isometric map $L_2\left(\mathrm{I}\mathrm{R}\right)\to L_2\left(\mathrm{I}\mathrm{R}\right)$ . This isometry is actually a unitary map.

Periodic signals have of course infinite energy; therefore, one introduces the power $P_x$ of the signal $x\left(t\right)$ , which is the average energy over one period. Energy is measured in Joule,power is measured in Watt=Joule/Sec.

The analog of Plancherel's theorem is Parseval's theorem which applies to $T$ -periodic signals and says

We may derive Parseval's theorem as follows, using [link] and ${\left|a\right|}^2=a·a^{*}$ :

Here, we used that ${\int }_{-T/2}^{T/2}{e}^{j2\pi (k-n)t/T}$ equals $T$ when $k=n$ (since $e^0=1$ ), but equals 0 when $k\ne n$ since (since $e^{js}=cos\left(s\right)+jsin\left(s\right)$ , which are integrated over several periods).

A similar computation can be carried out for Plancherel's equation. However, some difficulties arise due to the integrals over infiniteintervals (see  Comment 3 below). Also, a justification of Plancherel could be given by performinga limit of infinite period in Parseval's equation (see  Comment 4 below).

For finite discrete signals the analog is simply the fact, that DFT is unitary up to a stretching factor. More precisely, the matrix $DF{T}_K$ leaves angles intact and stretches length by $\sqrt{K}$ . Intuitively, one may think of the DFT as a rotation and a stretching. In other words, to perform a DFT simplymeans to change the coordinate system into a new one, and to change length measurement by a factor $\sqrt{K}$ . Thus:

Note that the DFT Fourier coefficients are complex numbers; thus, the absolute value has to be taken (for a complex number $a$ we have ${\left|a\right|}^2=a·a^{*}$ , which is usually different from $a^2$ —unless $a$ is by chance real valued).

Comment 3 A “hand-waving” argument for Plancherel's theorem runs as follows, using [link] and ${\left|a\right|}^2=a·a^{*}$ :

Comment 4 With Parseval's equation established one may provide a derivation of Plancherel's formula which is more convincing, tho more difficult to makerigorous. Assume that the signal $x\left(t\right)$ is time-limited, say defined on $-T/2<t<T/2$ . Its Fourier-series coefficients are $X_k=1/T{\int }_{-T/2}^{T/2}x\left(t\right)e^{-j2\pi kt/T}dt$ . Comparing to the Fourier transform we findthus $T·X_k=X(k/T)$ . We may interpret $x$ as being periodically extended and use Parseval's equation which says $P_x=\sum {\left|X_k\right|}^2$ ; this allows the following computation