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[[Category:ECE301 S11 Exam 3 more practice]][[Category:ECE301 S11 Exam 3 more practice]][[Category:ECE301 S11 Exam 3 more practice]]
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[[Category:ECE301 S11 Exam 3 more practice]]
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[[Category:ECE301Spring2011Boutin]]
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[[Category:Problem_solving]]
  
=ECE301S11_more_practice_CT_conv_2=
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= Problem =
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Compute the convolution
  
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<math>z(t)=x(t)*y(t)  \ </math>
  
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between
  
Put your content here . . .
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<math> x(t) = e^{jwt}u(t+2)  \ </math>
 
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and
  
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<math>y(t) = e^{jwt}u(t-2)  \ </math>.
  
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= My Solution=
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<math>
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\begin{align}
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z(t) &= e^{jwt}u(t+2) * e^{jwt}u(t-2) \\
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&= \int_{-\infty}^{\infty} e^{jw\tau}u(\tau+2) e^{jw(t- \tau)}u(t - \tau -2)\mathrm{d}\tau \\
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&= \int_{-2}^{t-2} e^{jw\tau} e^{jw(t- \tau)}\mathrm{d}\tau \\
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&= e^{jwt} \int_{-2}^{t-2} 1 \mathrm{d}\tau \\
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&= \begin{cases}
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t e^{jwt} &, \text{when }t > 2 \\
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0 &, \text{else}\end{cases}
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\end{align}
  
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</math>
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==Comments==
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Write them here.
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----
 
[[ ECE301 S11 Exam 3 more practice|Back to ECE301 S11 Exam 3 more practice]]
 
[[ ECE301 S11 Exam 3 more practice|Back to ECE301 S11 Exam 3 more practice]]

Latest revision as of 05:04, 6 May 2011


Problem

Compute the convolution

$ z(t)=x(t)*y(t) \ $

between

$ x(t) = e^{jwt}u(t+2) \ $

and

$ y(t) = e^{jwt}u(t-2) \ $.

My Solution

$ \begin{align} z(t) &= e^{jwt}u(t+2) * e^{jwt}u(t-2) \\ &= \int_{-\infty}^{\infty} e^{jw\tau}u(\tau+2) e^{jw(t- \tau)}u(t - \tau -2)\mathrm{d}\tau \\ &= \int_{-2}^{t-2} e^{jw\tau} e^{jw(t- \tau)}\mathrm{d}\tau \\ &= e^{jwt} \int_{-2}^{t-2} 1 \mathrm{d}\tau \\ &= \begin{cases} t e^{jwt} &, \text{when }t > 2 \\ 0 &, \text{else}\end{cases} \end{align} $

Comments

Write them here.


Back to ECE301 S11 Exam 3 more practice

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva