Lecture 2: Frequency Response

Document Actions

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

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

Let the input to a discrete-time system be (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 November 23, 2009, 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.

Course Contents Signals and Systems Home About the Professor Syllabus Schedule Homework Labs/Programs	Personal tools Log in You are here: Home → Electrical and Computer Engineering → Signals and Systems → Lecture 2: Frequency Response
	Info Lecture 2: Frequency Response Document Actions Schedule :: Intro :: Inverse :: Properties :: Solution :: Transfer Functions :: System Realization :: Bilateral :: Frequency Response :: Rubber Sheet Geometry :: FIR Filters Continuous time with transfer function : $e^{j\omega t} \rightarrow H(j\omega) e^{j\omega t}$ . An analogous result holds for discrete time systems. Let the input to a discrete-time system be (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 November 23, 2009, 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.	Reuse Course Download this course

Sections

Personal tools

Lecture 2: Frequency Response

Document Actions