(New page: 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 ...)
 
Line 5: Line 5:
 
(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.)
  
−
- <img alt="tex:x_1(t)=Ae^{Bt+C}u(Dt+E)" />
+
- <math>\displaystyle x_1(t)=Ae^{Bt+C}u(Dt+E)</math>
  
−
- <img alt="tex:x_2(t)=Fe^{Gt+H}u(It+J)" />
+
- <math>\displaystyle x_2(t)=Fe^{Gt+H}u(It+J)</math>
  
−
- <img alt="tex:x_1(t)*x_2(t)=\int_{\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau" />
+
- <math>\displaystyle x_1(t)*x_2(t)=\int_{\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau</math>
  
−
- <img alt="tex:\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>\displaystyle \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>
  
−
- <img alt="tex:\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>\displaystyle \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>
  
−
- <img alt="tex:where\;u(D\tau+E)=0\;,for\;D\tau+E<0\;\rightarrow\;\tau<\frac{-E}{D}" />
+
- <math>\displaystyle where\;u(D\tau+E)=0\;,for\;D\tau+E<0\;\rightarrow\;\tau<\frac{-E}{D}</math>
  
−
- <img alt="tex:\quad=AF\int_{\frac{-E}{D}}^{\infty}e^{\tau(B-G)+Gt+C+H}u(It-I\tau+J)d\tau" />
+
- <math>\displaystyle \quad=AF\int_{\frac{-E}{D}}^{\infty}e^{\tau(B-G)+Gt+C+H}u(It-I\tau+J)d\tau</math>
  
−
- <img alt="tex:where\;u(It-I\tau+J)=0\;,for\;It-I\tau+J<0\;\rightarrow\;\tau>t+\frac{J}{I}" />
+
- <math>\displaystyle where\;u(It-I\tau+J)=0\;,for\;It-I\tau+J<0\;\rightarrow\;\tau>t+\frac{J}{I}</math>
  
−
- <img alt="tex:\;\;\;=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>\displaystyle \;\;\;=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>
  
−
- <img alt="tex:\;\;\;=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>\displaystyle \;\;\;=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>
  
−
- <img alt="tex:\;\;\;=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>\displaystyle \;\;\;=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>
  
−
- <img alt="tex:\;\;\;=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>\displaystyle \;\;\;=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>
  
−
- <img alt="tex:\;\;\;=\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>\displaystyle \;\;\;=\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>
  
−
- <img alt="tex:\;\;\;=\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>\displaystyle \;\;\;=\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:
−
 
 
−
  &nbsp;
 
  
−
- <img alt="tex:Let:\;x_1(t)=x(t)=e^{-2t}u(t)" />
+
&nbsp;
  
−
- <img alt="tex:\;\;\;\;x_2(t)=h(t)=u(t)" />
+
- <math>\displaystyle Let:\;x_1(t)=x(t)=e^{-2t}u(t)</math>
  
−
- <img alt="tex:Thus:" />
+
- <math>\displaysytle \;\;\;\;x_2(t)=h(t)=u(t)</math>
  
−
- <img alt="tex:A=1,\;B=-2,\;C=0,\;D=1,\;E=0,\;F=1,\;G=0,\;H=0,\;I=1,\;J=0" />
+
- <math>\displaystyle Thus:</math>
  
−
- <img alt="tex:x(t)*h(t)=x_1(t)*x_2(t)" />
+
- <math>\displaystyle A=1,\;B=-2,\;C=0,\;D=1,\;E=0,\;F=1,\;G=0,\;H=0,\;I=1,\;J=0</math>
  
−
- <img alt="tex:\;\;\;=\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>\displaystyle x(t)*h(t)=x_1(t)*x_2(t)</math>
  
−
- <img alt="tex:\;\;\;=\frac{-1}{2}(e^{-2t}-1)\cdot u(t)" />
+
- <math>\displaystyle \;\;\;=\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>
  
−
- <img alt="tex:\;\;\;=\frac{1}{2}(1-e^{-2t})\cdot u(t)" />
+
- <math>\displaystyle \;\;\;=\frac{-1}{2}(e^{-2t}-1)\cdot u(t)</math>
 +
 
 +
- <math>\displaystyle \;\;\;=\frac{1}{2}(1-e^{-2t})\cdot u(t)</math>
  
 
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Revision as of 20:32, 16 March 2008

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.)

- $ \displaystyle x_1(t)=Ae^{Bt+C}u(Dt+E) $

- $ \displaystyle x_2(t)=Fe^{Gt+H}u(It+J) $

- $ \displaystyle x_1(t)*x_2(t)=\int_{\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau $

- $ \displaystyle \quad=\int_{\infty}^{\infty}Ae^{B\tau+C}u(D\tau+E)Fe^{G(t-\tau)+H}u(I(t-\tau)+J)d\tau $

- $ \displaystyle \quad=AF\int_{\infty}^{\infty}e^{B\tau+C+G(t-\tau)+H}u(D\tau+E)u(It-I\tau+J)d\tau $

- $ \displaystyle where\;u(D\tau+E)=0\;,for\;D\tau+E<0\;\rightarrow\;\tau<\frac{-E}{D} $

- $ \displaystyle \quad=AF\int_{\frac{-E}{D}}^{\infty}e^{\tau(B-G)+Gt+C+H}u(It-I\tau+J)d\tau $

- $ \displaystyle where\;u(It-I\tau+J)=0\;,for\;It-I\tau+J<0\;\rightarrow\;\tau>t+\frac{J}{I} $

- $ \displaystyle \;\;\;=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}) $

- $ \displaystyle \;\;\;=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}) $

- $ \displaystyle \;\;\;=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}) $

- $ \displaystyle \;\;\;=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}) $

- $ \displaystyle \;\;\;=\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}) $

- $ \displaystyle \;\;\;=\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}) $

Example: Problem 2 on Fall 06 Midterm 1:

 

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

- $ \displaysytle \;\;\;\;x_2(t)=h(t)=u(t) $

- $ \displaystyle Thus: $

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

- $ \displaystyle x(t)*h(t)=x_1(t)*x_2(t) $

- $ \displaystyle \;\;\;=\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}) $

- $ \displaystyle \;\;\;=\frac{-1}{2}(e^{-2t}-1)\cdot u(t) $

- $ \displaystyle \;\;\;=\frac{1}{2}(1-e^{-2t})\cdot u(t) $

<style> ul { list-style: none; } li { margin-bottom: 1.5em; } </style>

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Alumni Liaison

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

Francisco Blanco-Silva