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Lecture 2: Frequency Response

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Schedule :: Intro :: Inverse :: Properties :: Solution :: Transfer Functions :: System Realization :: Bilateral :: Frequency Response :: Rubber Sheet Geometry :: FIR Filters

Continuous time with transfer function $H(s)$ : $e^{j\omega t}
\rightarrow H(j\omega) e^{j\omega t}$ . An analogous result holds for discrete time systems.

Let the input to a discrete-time system be $f[k] = z^k$ (everlasting, so we don't have to worry about transients). Then

\begin{displaymath}y[k] = h[k] * z^k = z^k \sum_m h[m] z^{-m} = z^k H[z].
\end{displaymath}

More particularly, consider when $z = e^{\pm j\Omega}$ . We find that
\begin{displaymath}e^{j\Omega k} \rightarrow H(e^{j\Omega}) e^{j\Omega k}
\end{displaymath}


\begin{displaymath}e^{-j\Omega k} \rightarrow H(e^{-j\Omega}) e^{-j\Omega k}
\end{displaymath}

Adding:
\begin{displaymath}\cos \Omega k \rightarrow \Real(H(e^{j\Omega}) e^{j\Omega k})
\end{displaymath}

or, in polar form with $H(e^{j\Omega}) = \vert H(e^{j\Omega})\vert e^{j \arg
H(e^{j\Omega})}$ we find
\begin{displaymath}\cos \Omega k \rightarrow \vert H(e^{j\Omega})\vert \cos(\Omega k + \arg
H(e^{j\Omega k})).
\end{displaymath}

That is, the cosine is modified in amplitude and phase by the transfer function.
\begin{example}
For the system $y[k+1] - 0.8y[k] = f[k+1]$, determine the frequ...
...etermine the output. What about when
$f[k] = \cos(\pi/6 k - 0.2)$?
\end{example}
Note that $H(e^{j\Omega})$ is periodic.
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 18). Lecture 2: Frequency Response. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/node8.html. This work is licensed under a Creative Commons License Creative Commons License