(New page: : <math> y[n] - y[n-1] = x[n]| </math> : <math> \Rightarrow H(\omega) = \frac {1}{1 - e^{-j \omega}} </math> From table in book: : <math> \begin{align} u[n] &\overset {\mathfrak{F}}{\lon...)
(No difference)

Revision as of 11:44, 12 December 2008

$ y[n] - y[n-1] = x[n]| $
$ \Rightarrow H(\omega) = \frac {1}{1 - e^{-j \omega}} $

From table in book:

$ \begin{align} u[n] &\overset {\mathfrak{F}}{\longleftrightarrow} \frac {1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) \\ 1 &\overset {\mathfrak{F}}{\longleftrightarrow} 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) \end{align} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood