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[[Category: Properties]]
 
[[Category: Properties]]
 
[[Category: Convolution]]
 
[[Category: Convolution]]
 
+
=Framework for computing the CT Convolution of two unit step exponentials=
 
Let's take the convolution of the two most general unit-step exponentials in CT.
 
Let's take the convolution of the two most general unit-step exponentials in CT.
  
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(I know this is kinda long, but it is very detailed to show the process of how to get to the general simplified solution.)
 
(I know this is kinda long, but it is very detailed to show the process of how to get to the general simplified solution.)
  
: <math> x_1(t)=Ae^{Bt+C}u(Dt+E) \qquad x_2(t)=Fe^{Gt+H}u(It+J) </math>
+
<math> x_1(t)=Ae^{Bt+C}u(Dt+E) \qquad x_2(t)=Fe^{Gt+H}u(It+J) </math>
 +
 
 +
<math> \begin{align} x_1(t)*x_2(t) &= \int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau \\
 +
&=\int_{-\infty}^{\infty}Ae^{B\tau+C}u(D\tau+E)Fe^{G(t-\tau)+H}u(I(t-\tau)+J)d\tau \\
 +
&=AF\int_{-\infty}^{\infty}e^{B\tau+C+G(t-\tau)+H}u(D\tau+E)u(It-I\tau+J)d\tau; \;(u(D\tau+E)=0\;,for\;D\tau+E<0\;\rightarrow\;\tau<\frac{-E}{D}) \\
 +
&=AF\int_{\frac{-E}{D}}^{\infty}e^{\tau(B-G)+Gt+C+H}u(It-I\tau+J)d\tau;
 +
\;(u(It-I\tau+J)=0\;,for\;It-I\tau+J<0\;\rightarrow\;\tau>t+\frac{J}{I}) \\
 +
&=AF\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)+Gt+C+H}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\
 +
&=AFe^{Gt+C+H}\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\
 +
&=AFe^{Gt+C+H}\frac{1}{B-G}\left[e^{\tau(B-G)}\right]_{\frac{-E}{D}}^{t+\frac{J}{I}}\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\
 +
&=AFe^{Gt+C+H}\frac{1}{B-G}(e^{(t+\frac{J}{I})\cdot(B-G)}-e^{\frac{-E}{D}\cdot(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\
 +
&=\frac{AF}{B-G}(e^{Gt+CH+(t+\frac{J}{I})\cdot(B-G)}-e^{Gt+C+H-\frac{E}{D}(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\
 +
&=\frac{AF}{B-G}(e^{Bt+C+H+\frac{J}{I}(B-G)}-e^{Gt+C+H+\frac{E}{D}(G-B)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \end{align} </math>
  
: <math> x_1(t)*x_2(t)=\int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau </math>
 
: <math> \quad=\int_{-\infty}^{\infty}Ae^{B\tau+C}u(D\tau+E)Fe^{G(t-\tau)+H}u(I(t-\tau)+J)d\tau </math>
 
: <math> \quad=AF\int_{-\infty}^{\infty}e^{B\tau+C+G(t-\tau)+H}u(D\tau+E)u(It-I\tau+J)d\tau </math>
 
: <math> where\;u(D\tau+E)=0\;,for\;D\tau+E<0\;\rightarrow\;\tau<\frac{-E}{D} </math>
 
: <math> \quad=AF\int_{\frac{-E}{D}}^{\infty}e^{\tau(B-G)+Gt+C+H}u(It-I\tau+J)d\tau </math>
 
: <math> where\;u(It-I\tau+J)=0\;,for\;It-I\tau+J<0\;\rightarrow\;\tau>t+\frac{J}{I} </math>
 
: <math> \;\;\;=AF\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)+Gt+C+H}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) </math>
 
: <math> \;\;\;=AFe^{Gt+C+H}\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) </math>
 
: <math> \;\;\;=AFe^{Gt+C+H}\frac{1}{B-G}\left[e^{\tau(B-G)}\right]_{\frac{-E}{D}}^{t+\frac{J}{I}}\cdot u(t+\frac{J}{I}+\frac{E}{D}) </math>
 
: <math> \;\;\;=AFe^{Gt+C+H}\frac{1}{B-G}(e^{(t+\frac{J}{I})\cdot(B-G)}-e^{\frac{-E}{D}\cdot(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) </math>
 
: <math> \;\;\;=\frac{AF}{B-G}(e^{Gt+CH+(t+\frac{J}{I})\cdot(B-G)}-e^{Gt+C+H-\frac{E}{D}(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) </math>
 
: <math> \;\;\;=\frac{AF}{B-G}(e^{Bt+C+H+\frac{J}{I}(B-G)}-e^{Gt+C+H+\frac{E}{D}(G-B)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) </math>
 
  
 
Example: Problem 2 on Fall 06 Midterm 1:
 
Example: Problem 2 on Fall 06 Midterm 1:
  
: <math> Let:\;x_1(t)=x(t)=e^{-2t}u(t) </math>
+
<math> Let:\;x_1(t)=x(t)=e^{-2t}u(t) \qquad x_2(t)=h(t)=u(t) </math>
: <math> \;\;\;\;x_2(t)=h(t)=u(t)\ </math>
+
 
: <math> Thus: </math>
+
<math> Thus:\;A=1,\;B=-2,\;C=0,\;D=1,\;E=0,\;F=1,\;G=0,\;H=0,\;I=1,\;J=0 </math>
: <math> A=1,\;B=-2,\;C=0,\;D=1,\;E=0,\;F=1,\;G=0,\;H=0,\;I=1,\;J=0 </math>
+
 
: <math> x(t)<em>h(t)=x_1(t)</em>x_2(t)\ </math>
+
<math> \begin{align} x(t)*h(t)&=x_1(t)*x_2(t) \\
: <math> \;\;\;=\frac{1\cdot1}{-2-0}(e^{-2t+0+0+\frac{0}{1}(-2-0)}-e^{0t+0+0+\frac{0}{1}(0--2)})\cdot u(t+\frac{0}{1}+\frac{0}{1}) </math>
+
&=\frac{1\cdot1}{-2-0}(e^{-2t+0+0+\frac{0}{1}(-2-0)}-e^{0t+0+0+\frac{0}{1}(0--2)})\cdot u(t+\frac{0}{1}+\frac{0}{1}) \\
: <math> \;\;\;=\frac{-1}{2}(e^{-2t}-1)\cdot u(t) </math>
+
&=\frac{-1}{2}(e^{-2t}-1)\cdot u(t) \\
: <math> \;\;\;=\frac{1}{2}(1-e^{-2t})\cdot u(t) </math>
+
&=\frac{1}{2}(1-e^{-2t})\cdot u(t) \end{align} </math>

Latest revision as of 10:04, 30 January 2011

Framework for computing the CT Convolution of two unit step exponentials

Let's take the convolution of the two most general unit-step exponentials in CT.

This solution can be very helpful in checking your work for convolutions of this form. Just plug in your numbers for the capital letters.

(I know this is kinda long, but it is very detailed to show the process of how to get to the general simplified solution.)

$ x_1(t)=Ae^{Bt+C}u(Dt+E) \qquad x_2(t)=Fe^{Gt+H}u(It+J) $

$ \begin{align} x_1(t)*x_2(t) &= \int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau \\ &=\int_{-\infty}^{\infty}Ae^{B\tau+C}u(D\tau+E)Fe^{G(t-\tau)+H}u(I(t-\tau)+J)d\tau \\ &=AF\int_{-\infty}^{\infty}e^{B\tau+C+G(t-\tau)+H}u(D\tau+E)u(It-I\tau+J)d\tau; \;(u(D\tau+E)=0\;,for\;D\tau+E<0\;\rightarrow\;\tau<\frac{-E}{D}) \\ &=AF\int_{\frac{-E}{D}}^{\infty}e^{\tau(B-G)+Gt+C+H}u(It-I\tau+J)d\tau; \;(u(It-I\tau+J)=0\;,for\;It-I\tau+J<0\;\rightarrow\;\tau>t+\frac{J}{I}) \\ &=AF\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)+Gt+C+H}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=AFe^{Gt+C+H}\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=AFe^{Gt+C+H}\frac{1}{B-G}\left[e^{\tau(B-G)}\right]_{\frac{-E}{D}}^{t+\frac{J}{I}}\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=AFe^{Gt+C+H}\frac{1}{B-G}(e^{(t+\frac{J}{I})\cdot(B-G)}-e^{\frac{-E}{D}\cdot(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=\frac{AF}{B-G}(e^{Gt+CH+(t+\frac{J}{I})\cdot(B-G)}-e^{Gt+C+H-\frac{E}{D}(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=\frac{AF}{B-G}(e^{Bt+C+H+\frac{J}{I}(B-G)}-e^{Gt+C+H+\frac{E}{D}(G-B)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \end{align} $


Example: Problem 2 on Fall 06 Midterm 1:

$ Let:\;x_1(t)=x(t)=e^{-2t}u(t) \qquad x_2(t)=h(t)=u(t) $

$ Thus:\;A=1,\;B=-2,\;C=0,\;D=1,\;E=0,\;F=1,\;G=0,\;H=0,\;I=1,\;J=0 $

$ \begin{align} x(t)*h(t)&=x_1(t)*x_2(t) \\ &=\frac{1\cdot1}{-2-0}(e^{-2t+0+0+\frac{0}{1}(-2-0)}-e^{0t+0+0+\frac{0}{1}(0--2)})\cdot u(t+\frac{0}{1}+\frac{0}{1}) \\ &=\frac{-1}{2}(e^{-2t}-1)\cdot u(t) \\ &=\frac{1}{2}(1-e^{-2t})\cdot u(t) \end{align} $

Alumni Liaison

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

Francisco Blanco-Silva