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===Answer 1=== | ===Answer 1=== | ||
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+ | <math>y(t)=x(t)*h(t)=\int_{-\infty}^{\infty} u(t)e^{\tau-t}u(t-\tau+3) d\tau</math> | ||
+ | |||
+ | but | ||
+ | |||
+ | <math>u(\tau)= | ||
+ | \begin{cases} | ||
+ | 1, & \mbox{if }\tau \ge 0 \\ | ||
+ | 0, & \mbox{if }\tau < 0 | ||
+ | \end{cases}</math> | ||
+ | |||
+ | so | ||
+ | |||
+ | <math>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</math> | ||
+ | |||
+ | but | ||
+ | |||
+ | <math>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}</math> | ||
+ | |||
+ | so | ||
+ | |||
+ | <math>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)</math> | ||
+ | |||
+ | |||
+ | <math>y(t)=(e^3-e^{-t})u(t+3)\,</math> | ||
+ | |||
+ | --[[User:Cmcmican|Cmcmican]] 19:56, 28 January 2011 (UTC) | ||
+ | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. |
Revision as of 15:56, 28 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)
Answer 2
Write it here.
Answer 3
Write it here.