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You are here: Home Electrical and Computer Engineering Signals and Systems Lecture 5: System Response Using Fourier Transforms

Lecture 5: System Response Using Fourier Transforms

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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

Let $f(t)$ be the input to an LTIC system with transfer function $H(s)$ and output $y(t)$ . The response to the everlasting signal $f(t) =
e^{j\omega t}$ is $y(t) = H(\omega) e^{j\omega t}$ . Thus $f(t) = e^{j n
\Delta \omega t} \rightarrow H(n \Delta \omega) e^{j n \Delta \omega
t}$ . Now let the input be

\begin{displaymath}\frac{1}{2\pi} \lim_{\Delta \omega \rightarrow 0} \sum_n
F(n \Delta \omega) e^{j n \Delta \omega t}.
\end{displaymath}

By the definition of the FT, this is $f(t)$ . By linearity, the output is

\begin{displaymath}y(t) = \frac{1}{2\pi} \lim_{\Delta \omega \rightarrow 0} \sum_n
F(n \Delta \omega) H(n \Delta \omega) e^{j n \Delta \omega t}.
\end{displaymath}

That is,

\begin{displaymath}y(t) = \frac{1}{2\pi} \int_{\infty}^\infty F(\omega)H(\omega)...
...frac{1}{2\pi} \int_{\infty}^\infty Y(\omega) e^{j\omega t}\,dt
\end{displaymath}

so we can say

\begin{displaymath}Y(\omega) = F(\omega) H(\omega).
\end{displaymath}

In a sense, this is the whole point of Fourier analysis!
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 26). Lecture 5: System Response Using Fourier Transforms. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/5_5node5.html. This work is licensed under a Creative Commons License Creative Commons License