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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>

Revision as of 12:05, 8 December 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.)

$ 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} $

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