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- <math>\displaystyle Let:\;x_1(t)=x(t)=e^{-2t}u(t)</math> | - <math>\displaystyle Let:\;x_1(t)=x(t)=e^{-2t}u(t)</math> | ||
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- <math>\displaystyle \;\;\;=\frac{1}{2}(1-e^{-2t})\cdot u(t)</math> | - <math>\displaystyle \;\;\;=\frac{1}{2}(1-e^{-2t})\cdot u(t)</math> | ||
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Latest revision as of 20:33, 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) $
- $ \displaystyle \;\;\;\;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) $
