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== HW10 == | == HW10 == | ||
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+ | '''Homework wiki:''' | ||
+ | [[Homework 1_ECE301Fall2008mboutin| 1]] | ||
+ | |[[Homework 2_ECE301Fall2008mboutin| 2]] | ||
+ | |[[Homework 3_ECE301Fall2008mboutin| 3]] | ||
+ | |[[Homework 4_ECE301Fall2008mboutin| 4]] | ||
+ | |[[Homework 5_ECE301Fall2008mboutin| 5]] | ||
+ | |[[Homework 6_ECE301Fall2008mboutin| 6]] | ||
+ | |[[Homework 7_ECE301Fall2008mboutin| 7]] | ||
+ | |[[Homework 8_ECE301Fall2008mboutin| 8]] | ||
+ | |[[Homework 9_ECE301Fall2008mboutin| 9]] | ||
+ | |[[Homework 10_ECE301Fall2008mboutin| 10]] | ||
+ | |[[homework 11_ECE301Fall2008mboutin| 11]] | ||
Revision as of 01:22, 5 December 2008
HW10
Homework wiki: 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} $
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