Line 52: | Line 52: | ||
--[[User:Cmcmican|Cmcmican]] 19:56, 28 January 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 19:56, 28 January 2011 (UTC) | ||
:<span style="color:green"> TA's comments: Excellent!</span> | :<span style="color:green"> TA's comments: Excellent!</span> | ||
− | --[[User:Ahmadi|Ahmadi]] 22:59, 29 January 2011 (UTC) | + | :--[[User:Ahmadi|Ahmadi]] 22:59, 29 January 2011 (UTC) |
+ | |||
+ | :Instructor's comments: The above answer shows a lot of detail, as I did when presenting the class example. However, once you are comfortable with all the steps, it is ok to skip a few details. In particular, the following answer would get full credit on the exam. | ||
+ | :<math> | ||
+ | \begin{align} | ||
+ | y(t)&=x(t)*h(t),\\ | ||
+ | & =\int_{-\infty}^{\infty} u(t)e^{\tau-t}u(t-\tau+3) d\tau,\\ | ||
+ | & =\int_{0}^{\infty} e^{-t} e^{\tau}u(t-\tau+3) d\tau,\\ | ||
+ | & =e^{-t} \int_{0}^{\infty} e^{\tau}u(t-\tau+3) d\tau, \\ | ||
+ | &=\begin{cases} | ||
+ | e^{-t} \int_{0}^{t+3} e^{\tau}d\tau, & \mbox{if }t > -3 \\ | ||
+ | 0, & \mbox{if }t \le -3 | ||
+ | \end{cases}, \\ | ||
+ | &= e^{-t}(e^{t+3}-1)u(t+3). | ||
+ | \end{align} | ||
+ | </math> | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. |
Revision as of 10:28, 30 January 2011
Contents
Practice Question on Computing the Output of an LTI system by Convolution
The unit impulse response h(t) of a CT LTI system is
$ h(t) = e^{-t} u(t+3). \ $
Use convolution to compute the system's response to the input
$ x(t)=u(t). \ $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
$ y(t)=x(t)*h(t)=\int_{-\infty}^{\infty} u(t)e^{\tau-t}u(t-\tau+3) d\tau $
but
$ u(\tau)= \begin{cases} 1, & \mbox{if }\tau \ge 0 \\ 0, & \mbox{if }\tau < 0 \end{cases} $
so
$ y(t)=\int_{0}^{\infty} e^{-t} e^{\tau}u(t-\tau+3) d\tau=e^{-t} \int_{0}^{\infty} e^{\tau}u(t-\tau+3) d\tau $
but
$ u(t-\tau+3)= \begin{cases} 1, & \mbox{if }(t-\tau+3) \ge 0 \\ 0, & \mbox{if }(t-\tau+3) < 0 \end{cases} = \begin{cases} 1, & \mbox{if }\tau \le t+3 \\ 0, & \mbox{if }\tau > t+3 \end{cases} $
so
$ y(t)=\begin{cases} e^{-t} \int_{0}^{t+3} e^{\tau}d\tau, & \mbox{if }t > -3 \\ 0, & \mbox{if }t \le -3 \end{cases} = e^{-t}(e^{t+3}-1)u(t+3) $
$ y(t)=(e^3-e^{-t})u(t+3)\, $
--Cmcmican 19:56, 28 January 2011 (UTC)
- TA's comments: Excellent!
- --Ahmadi 22:59, 29 January 2011 (UTC)
- Instructor's comments: The above answer shows a lot of detail, as I did when presenting the class example. However, once you are comfortable with all the steps, it is ok to skip a few details. In particular, the following answer would get full credit on the exam.
- $ \begin{align} y(t)&=x(t)*h(t),\\ & =\int_{-\infty}^{\infty} u(t)e^{\tau-t}u(t-\tau+3) d\tau,\\ & =\int_{0}^{\infty} e^{-t} e^{\tau}u(t-\tau+3) d\tau,\\ & =e^{-t} \int_{0}^{\infty} e^{\tau}u(t-\tau+3) d\tau, \\ &=\begin{cases} e^{-t} \int_{0}^{t+3} e^{\tau}d\tau, & \mbox{if }t > -3 \\ 0, & \mbox{if }t \le -3 \end{cases}, \\ &= e^{-t}(e^{t+3}-1)u(t+3). \end{align} $
Answer 2
Write it here.
Answer 3
Write it here.