(New page: == EXERCISE == Assume <math>|\alpha|<1</math>)
 
(SOLUTION)
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
 
== EXERCISE ==
 
== EXERCISE ==
Assume <math>|\alpha|<1</math>
+
Assume <math> |\alpha|<1 </math>
 +
 
 +
Compute the F.T. of <math>x_1[n]=\alpha^{n}u[n]</math>
 +
 
 +
 
 +
== SOLUTION ==
 +
<math>\,\mathcal{X}_1(\omega)=\mathcal{F}(x_1[n])=\sum_{n=-\infty}^{\infty}x_1[n]e^{-j\omega n}\,</math>
 +
 
 +
<math>\,=\sum_{n=-\infty}^{\infty}\alpha^{n}u[n]e^{-j\omega n}\,</math>
 +
 
 +
<math>\,=\sum_{n=0}^{\infty}\alpha^{n}e^{-j\omega n}\,</math>
 +
 
 +
<math>\,=\sum_{n=0}^{\infty}(\alpha e^{-j\omega })^{n}\,</math>
 +
 
 +
but <math>\,|\alpha e^{-j\omega }|<1\,</math>
 +
 
 +
<math>\,=\frac{1}{1-\alpha e^{-j\omega }}\,</math>

Latest revision as of 15:39, 24 October 2008

EXERCISE

Assume $ |\alpha|<1 $

Compute the F.T. of $ x_1[n]=\alpha^{n}u[n] $


SOLUTION

$ \,\mathcal{X}_1(\omega)=\mathcal{F}(x_1[n])=\sum_{n=-\infty}^{\infty}x_1[n]e^{-j\omega n}\, $

$ \,=\sum_{n=-\infty}^{\infty}\alpha^{n}u[n]e^{-j\omega n}\, $

$ \,=\sum_{n=0}^{\infty}\alpha^{n}e^{-j\omega n}\, $

$ \,=\sum_{n=0}^{\infty}(\alpha e^{-j\omega })^{n}\, $

but $ \,|\alpha e^{-j\omega }|<1\, $

$ \,=\frac{1}{1-\alpha e^{-j\omega }}\, $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood