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

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood
The text passage provides a detailed derivation of the convolution of two unit-step exponentials in continuous time (CT), followed by an example problem with a specific set of parameters and its solution.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood
user
The text passage provides a detailed derivation of the convolution of two unit-step exponentials in continuous time (CT), followed by an example problem with a specific set of parameters and its solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER