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Lecture 5: Signal Energy and Parseval's Formula

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Schedule :: Fourier Series Formulas :: Nonperiodic Functions :: Useful Functions :: Simple Functions :: Fourier Transforms :: Properties :: System Analysis :: Signal Distortion :: Filters :: Parseval's Formulas

If we assume for definiteness a unit load (a 1 $\Omega$ resistor), then if $f(t)$ is the voltage across the resistor or the current through the resistor, then the energy dissipated is

\begin{displaymath}E_f = \int_{-\infty}^\infty f^2(t) dt \end{displaymath}

If this quantity is finite, then $f(t)$ is said to be an energy signal . Not every signal that we consider analytically is an energy signal: for example, $f(t) =\cos(\omega_o t)$ is not an energy signal.

Substituting in for the $f(t)$ in terms of the inverse F.T.,

\begin{displaymath}E_f = \int_{-\infty}^\infty f(t)\left[\frac{1}{2\pi}
...fty F(\omega)\int_{-\infty}^\infty
f(t)e^{j\omega t}dt d\omega \end{displaymath}

So we can write
\begin{displaymath}E_f = \frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)F(-\omega)
d\omega \end{displaymath}

For a real $f(t)$ , the F.T. satisfies $F(\omega) = F^*(-\omega)$ (they are conjugates), so that
\begin{displaymath}E_f = \frac{1}{2\pi}\int_{-\infty}^\infty \vert F(\omega)\vert^2 d\omega

This is Parseval's theorem for Fourier Transforms. It can be written using inner product notation as
\begin{displaymath}\la f(t),f(t) \ra = \frac{1}{2\pi}\la F(\omega),F(\omega) \ra. \end{displaymath}

It can also be generalized for products of different functions as
\begin{displaymath}\la f(t),g(t) \ra = \frac{1}{2\pi} \la F(\omega),G(\omega) \ra.

We can think of an increment of energy lying in an increment of bandwidth:
\begin{displaymath}\Delta E_f = \frac{1}{2\pi} \vert F(\omega)\vert^2 \Delta \omega \end{displaymath}

To get the total energy, add up all of the pieces. The function $\vert F(\omega)\vert^2$ is therefore referred to as the energy spectral density of the function: it tells where the energy is distributed in frequency.

The energy which is passed by a bandpass filter with cutoff frequencies $\omega_1$ and $\omega_2$ is

\begin{displaymath}E_f(\omega_1,\omega_2) = \frac{1}{2\pi}\int_{\omega_1}^{\omega_2}
\vert F(\omega)\vert^2 d\omega \end{displaymath}

Find the energy in the signal $f(t) = e^{-at}u(t)$, and find
...}{2} = \tan^{-1}\frac{W}{a} \end{displaymath}so $W = 12.706$\ rad.

Some observations:

  1. We can use the Parseval's theorem to integrate some messy functions by simply applying the theorem. For example, we can compute
    \begin{displaymath}I = \int_0^\infty \frac{\sin^2(x)}{x^2} dx

    by writing
    \begin{displaymath}I = \frac{1}{2} \int_{-\infty}^\infty \frac{\sin^2(x)}{x^2} d...
...2 d\omega =
\frac{\pi}{4} \int_{-1}^1 1d\omega = \frac{\pi}{2}

    The previous example also demonstrated how we can sometimes do easier integrals in one domain than the other.
  2. We can also use the idea of the last example to introduce the idea of an essential bandwidth : include enough bandwidth that most of the energy of the signal can get through. Often the 95% or 99% level is sufficient.
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 26). Lecture 5: Signal Energy and Parseval\'s Formula. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: This work is licensed under a Creative Commons License Creative Commons License