()
 
Line 1: Line 1:
== HW10 ==
+
[[Category:ECE 301]]
'''Homework wiki:'''  
+
[[Category:Fall 2008]]
 +
[[Category:mboutin]]
 +
[[Category:Homework]]
 +
 
 +
=Homework 10, [[ECE301]] Fall 2008, Prof. [[user:mboutin|Boutin]]=
 +
----
 +
Click  [[Main_Page_ECE301Fall2008mboutin|here]] to return to the [[Main_Page_ECE301Fall2008mboutin|ECE301 Fall 2008 Course Page of Prof. Boutin]].
 +
 
 +
'''Quick Links to homework assignments'''  
 
[[Homework 1_ECE301Fall2008mboutin| 1]]
 
[[Homework 1_ECE301Fall2008mboutin| 1]]
 
|[[Homework 2_ECE301Fall2008mboutin| 2]]
 
|[[Homework 2_ECE301Fall2008mboutin| 2]]
Line 12: Line 20:
 
|[[Homework 10_ECE301Fall2008mboutin| 10]]
 
|[[Homework 10_ECE301Fall2008mboutin| 10]]
 
|[[homework 11_ECE301Fall2008mboutin| 11]]
 
|[[homework 11_ECE301Fall2008mboutin| 11]]
 +
----
 
        
 
        
  

Latest revision as of 11:29, 16 September 2013


Homework 10, ECE301 Fall 2008, Prof. Boutin


Click here to return to the ECE301 Fall 2008 Course Page of Prof. Boutin.

Quick Links to homework assignments 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11



                       == Fundamentals of Laplace Transform ==
     Let the signal be:
     $ x(t) =e^ {-at} \mathit{u} (t). $
     
     Here is how to compute the Laplace Transform of $ x(t) $:
     $  \begin{align} X(s) &= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt, \\      &= \int_{-\infty}^{\infty}{e^{-at}}{e^{-st}}dt   ,\text{ since }\mathit{u} (t)=1,\text{ for }t>0, \text{ else }\mathit{u} (t)=0, \\      &=\frac{1}{s+a}. ~^*  \end{align}  $

Note: the last equality (with a *) is untrue. Please do not write this on the test or you will get points marked off. I really appreciate this mistake being on Rhea, please do not erase it --Mboutin 11:58, 21 November 2008 (UTC)

Correction of above:

$ \begin{align} X(s) &= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt, \\ &= \int_{0}^{\infty}{e^{-at}}{e^{-st}}dt ,\text{ let } s=b+j\omega, \\ &=\int_{0}^{\infty}{e^{-(a+b+j\omega)t}}dt, \\ \end{align} $

If $ a+b\leq 0 $, then the integral Diverges

Else,

$ \begin{align} X(s) &=\frac{e^{-(a+b)t}e^{-j\omega t}}{-(a+b+j\omega)}|_0^\infty, \\ &=0-\frac{-1}{s+a}, \\ &=\frac{1}{s+a} \end{align} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn