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Lecture 5: Representation of Nonperiodic Functions

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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

We have seen how we can represent periodic functions in terms of sines and cosines. This is a natural thing to contemplate (at least retrospectively). What if we want to represent nonperiodic functions in terms of periodic functions? This seems kind of strange at first. Why would we want to? Part of the answer comes because we know how LTI systems respond to sines and cosines. It turns out to be closely related to sinusoidal steady-state analysis of circuits and provide some useful tools for design of signal processing and communication systems.

To represent a nonperiodic function in terms of periodic functions, one thing we will need to to use periodic functions that might have infinite period. It will also turn out to add up a whole lot of them: more than a countable infinite. Here is one approach to the problem.

Suppose we have a function $f(t)$ . We construct a periodic extension of $f(t)$ called $f_{T_0}(t)$ , where $f(t)$ is repeated every $T_0$ seconds. Since $f_{T_0}(t)$ is periodic, it has a Fourier series

\begin{displaymath}f_{T_0}(t) = \sum_{n} D_n e^{jn\omega_0 t} \end{displaymath}


\begin{displaymath}D_n = \frac{1}{T_0}\int_{-T_0/2}^{T_0/2} f_{T_0}(t)e^{-jn\omega_0
t} dt

and $\omega_0 = \frac{2\pi}{T_0}$ . The integral is the same (since $f(t)$ is zero outside of $-T_0,T_0$ ) as

\begin{displaymath}D_n = \frac{1}{T_0}\int_{-\infty}^{\infty} f(t)e^{-jn\omega_0 t} dt

Now let's define the function

\begin{displaymath}F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt \end{displaymath}

Comparing these, we can see that

\begin{displaymath}D_n = \frac{1}{T_0}F(n\omega_0) \end{displaymath}

So we get the F.S. coefficients by sampling the function $F(\omega)$ . In the limit as $T_0 \rightarrow \infty$ , we sample infinitely often.

Now the F.S. for $f_{T_0}(t)$ is

\begin{displaymath}f_{T_0}(t) = \sum_n \frac{F(n\omega_0)}{T_0}e^{jn\omega_0 t}

Let $\Delta\omega = 2\pi/T_0$ and write

\begin{displaymath}f_{T_0}(t) = \sum_n \frac{F(n\Delta\omega_0)\Delta\omega}{2\pi}
e^{jn\Delta\omega t}

Now in the limit as $T_0 \rightarrow \infty$ we get

\begin{displaymath}f(t) = \lim_{T_0 \rightarrow \infty}f_{T_0}(t) = \frac{1}{2\pi}
\int_{-\infty}^\infty F(\omega)e^{j\omega t} d\omega

So we have been able to reconstruct our nonperiodic function from $F(\omega)$ . These are important enough they bear repeating:

\begin{displaymath}\boxed{F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt} \end{displaymath}

This is called the Fourier Transform of $f(t)$ . The inverse Fourier transform is

\begin{displaymath}\boxed{f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty
F(\omega)e^{j\omega t} d\omega}

We also write

\begin{displaymath}F(\omega) = \Fc[f(t)]\quad\quad f(t) = \Fc^{-1}[F(\omega)]


\begin{displaymath}f(t) \Leftrightarrow F(\omega) \end{displaymath}

This is notation similar to what we used for Laplace transforms.

Some things to notice:

  1. Notice that to get $f(t)$ we are doing the same kind of thing we did for F.S. except that we add up a continuum of frequencies.
  2. Suppose that $f(t)$ is causal. In the Laplace transform, set $s=e^{j\omega}.$ Then we get the Laplace transform back! But notice that the integral goes from $-\infty$ to $\infty$ . So we explicitly don't assume that $f(t)$ is causal. We thus do not typically use the F.T. to examine things like transient response: use the Laplace transform for that.

    As will be mentioned below, however, the F.T. cannot be used for unstable signals, since the integral does not exist in this case. The F.T. is not usually used to examine stability, as the L.T. can.

  3. Notice again that in the forward transform, a negative is used in the exponent.
  4. The frequency variable in the F.T. is $\omega$ , which is frequency in radians/sec. If we make the change of variable $\omega
= 2\pi f$ and find the transform in terms of $f$ we get the following transform pair:

    \begin{displaymath}F(2\pi f) = \int_{-\infty}^{\infty} f(t) e^{-j2\pi f t} dt \end{displaymath}

    This is sometimes written simply as $F(f)$ (which, strictly speaking, is not correct notation, so you have to be careful). In this notation, we would have

    \begin{displaymath}\boxed{F(f) = \int_{-\infty}^\infty f(t) e^{-j2\pi f t}\,dt.}

    The inverse transform is

    \begin{displaymath}f(t) = \int_{-\infty}^\infty F(2\pi f)e^{j2\pi ft} d\omega

    This has the convenience of not having the factor of $1/2\pi$ in front of it. This is usually written (again, by an abuse of notation)

    \begin{displaymath}\boxed{f(t) = \int_{-\infty}^\infty F(f)e^{j2\pi ft} df }

As we did for F.S., we can plot the magnitude and phase of the F.T. We can write

\begin{displaymath}F(\omega) = \vert F(\omega)\vert e^{j\angle F(\omega)} \end{displaymath}

Notice that if $f(t)$ is real, then $F(-\omega) =
\overline{F(\omega)}$ (they are complex conjugates) so that

\begin{displaymath}\vert F(-\omega)\vert = \vert F(\omega)\vert

\begin{displaymath}\angle F(-\omega) = -\angle F(\omega) \end{displaymath}

or, in other words, if $f(t)$ is a real function, then the magnitude function is even and the phase function is odd.

Find the F.T. of $f(t) = e^{-at}u(t)$
...a) \end{displaymath}Plot and discuss. What kind of filter is this?

Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 26). Lecture 5: Representation of Nonperiodic Functions. 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