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                              == '''Fundamentals of Laplace Transform''' ==
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[[Category:ECE 301]]
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[[Category:Fall 2008]]
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[[Category:mboutin]]
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[[Category:Homework]]
  
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=Homework 10, [[ECE301]] Fall 2008, Prof. [[user:mboutin|Boutin]]=
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----
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Click  [[Main_Page_ECE301Fall2008mboutin|here]] to return to the [[Main_Page_ECE301Fall2008mboutin|ECE301 Fall 2008 Course Page of Prof. Boutin]].
  
       <math>x(t) =\e^ (-at)\ \mathit{u}(t)</math>
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'''Quick Links to homework assignments'''
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[[Homework 1_ECE301Fall2008mboutin| 1]]
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|[[Homework 2_ECE301Fall2008mboutin| 2]]
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|[[Homework 3_ECE301Fall2008mboutin| 3]]
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|[[Homework 4_ECE301Fall2008mboutin| 4]]
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|[[Homework 5_ECE301Fall2008mboutin| 5]]
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|[[Homework 6_ECE301Fall2008mboutin| 6]]
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|[[Homework 7_ECE301Fall2008mboutin| 7]]
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|[[Homework 8_ECE301Fall2008mboutin| 8]]
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|[[Homework 9_ECE301Fall2008mboutin| 9]]
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|[[Homework 10_ECE301Fall2008mboutin| 10]]
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|[[homework 11_ECE301Fall2008mboutin| 11]]
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----
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                        == '''Fundamentals of Laplace Transform''' ==
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      Let the signal be:
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       <math>x(t) =e^ {-at} \mathit{u} (t).</math>
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      Here is how to compute the Laplace Transform of <math>x(t)</math>:
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      <math>
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\begin{align}
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X(s) &= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt, \\
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    &= \int_{-\infty}^{\infty}{e^{-at}}{e^{-st}}dt  ,\text{ since }\mathit{u} (t)=1,\text{ for }t>0, \text{ else }\mathit{u} (t)=0, \\
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    &=\frac{1}{s+a}. ~^*
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\end{align}
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</math>
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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 --[[User:Mboutin|Mboutin]] 11:58, 21 November 2008 (UTC)
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Correction of above:
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<math>
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\begin{align}
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X(s) &= \int_{-\infty}^{\infty}x(t){e^{-st}}\, dt, \\
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    &= \int_{0}^{\infty}{e^{-at}}{e^{-st}}dt  ,\text{ let  } s=b+j\omega,  \\
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    &=\int_{0}^{\infty}{e^{-(a+b+j\omega)t}}dt,  \\
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\end{align}
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</math>
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If <math>a+b\leq 0</math>, then the integral Diverges
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Else,
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<math>
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\begin{align}
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X(s) &=\frac{e^{-(a+b)t}e^{-j\omega t}}{-(a+b+j\omega)}|_0^\infty, \\
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    &=0-\frac{-1}{s+a},  \\
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    &=\frac{1}{s+a}
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\end{align}
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</math>
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* [[Homework _ECE301Fall2008mboutin#10 Daniel Morris: Properties of the Region of Convergence(ROC)]]
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* [[HW10 Jun Hyeong Park_ECE301Fall2008mboutin]]
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* [[HW10 Justin Kietzman- Properties of Laplace_ECE301Fall2008mboutin]]
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* [[HW10 Brian Thomas- More Properties of Laplace_ECE301Fall2008mboutin]]
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* [[HW10 Bavorndej Chanyasak_ECE301Fall2008mboutin]]
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* [[HW10 Sangwan Han HW_ECE301Fall2008mboutin#10]]
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* [[HW10 Sourabh Ranka_ECE301Fall2008mboutin]]
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* [[HW10 Emily Blount_ECE301Fall2008mboutin]]
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* [[HW10 Ananya Panja_ECE301Fall2008mboutin]]
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* [[HW10 Derek Hopper_ECE301Fall2008mboutin]]
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* [[HW10 Josh Long - Laplace Transform_ECE301Fall2008mboutin]]
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* [[HW10 Monil Goklani_ECE301Fall2008mboutin]]
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* [[HW10 David Record_ECE301Fall2008mboutin]]
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* [[HW10 Xujun Huang_ECE301Fall2008mboutin]]
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* [[HW10 Carlos Leon - Laplace transform table_ECE301Fall2008mboutin]]
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* [[HW10 Ben Moeller - Bigger Laplace Transform Table & Trig Identities_ECE301Fall2008mboutin]]
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* [[HW10 Zachary Curosh - Laplace Transform and Basic Properties_ECE301Fall2008mboutin]]
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* [[HW10 SangMo je_ECE301Fall2008mboutin]]
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* [[HW10 Seung Seob Lee_ECE301Fall2008mboutin]]
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* [[HW10 Thomas Wroblewski_ECE301Fall2008mboutin]]
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* [[HW10 Joseph Mazzei - Laplace Transform Pairs added to Table on Front Page_ECE301Fall2008mboutin]]
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* [[HW10 Daniel Barjum: Example illustration relationship between Fourier and Laplace transform_ECE301Fall2008mboutin]]
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*[[HW10 Vivek Ravi Region of Convergence for Laplace Transforms_ECE301Fall2008mboutin]]
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* [[HW10 William Schmidt Initial and Final Value Theorem_ECE301Fall2008mboutin]]

Latest revision as of 10:29, 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} $

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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