Line 1: Line 1:
 
[[Category:ECE301 S11 Exam 3 more practice]]
 
[[Category:ECE301 S11 Exam 3 more practice]]
  
=ECE301S11_more_practice_CT_conv_1=
+
= Problem =
 
+
<math>\begin{align} x(t) &= u(t) - u(t-1) \\ y(t) &= u(t+2) - u(t-2) \\ z(t) &= x(t) * y(t) \end{align}</math>
 
+
 
+
Put your content here . . .
+
  
 +
= Solution =
 +
<math>
 +
\begin{align}
 +
z(t) &= y(t) * x(t) \\
 +
&= \int_{-\infty}^{\infty}x(\tau) y(t-\tau)\mathrm{d}\tau \\
 +
&= \int_{-\infty}^{\infty}\left( u(\tau) - u(\tau-1) \right) \left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\
 +
&= \int_{0}^{1}\left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\
 +
&= \left[ (t-\tau+2)u(t-\tau+2)\right]\big|_0^1 - \left[ (t-\tau-2)u(t-\tau-2)\right]\big|_0^1 \\
 +
&= \left[ (t+1)u(t+1) - (t+2)u(t+2)\right] + \left[ -(t-3)u(t-3) + (t-2)u(t-2)\right]
 +
\end{align}
 +
</math>
  
  
  
 
[[ 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]]

Revision as of 06:17, 29 April 2011


Problem

$ \begin{align} x(t) &= u(t) - u(t-1) \\ y(t) &= u(t+2) - u(t-2) \\ z(t) &= x(t) * y(t) \end{align} $

Solution

$ \begin{align} z(t) &= y(t) * x(t) \\ &= \int_{-\infty}^{\infty}x(\tau) y(t-\tau)\mathrm{d}\tau \\ &= \int_{-\infty}^{\infty}\left( u(\tau) - u(\tau-1) \right) \left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\ &= \int_{0}^{1}\left( u(t-\tau+2) - u(t-\tau-2) \right)\mathrm{d}\tau \\ &= \left[ (t-\tau+2)u(t-\tau+2)\right]\big|_0^1 - \left[ (t-\tau-2)u(t-\tau-2)\right]\big|_0^1 \\ &= \left[ (t+1)u(t+1) - (t+2)u(t+2)\right] + \left[ -(t-3)u(t-3) + (t-2)u(t-2)\right] \end{align} $


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