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[[Category: asan]]
 
[[Category: asan]]
 
[[Category: Exams]]
 
[[Category: Exams]]
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=Problem=
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The unit impulse response of an LTI system is the CT signal
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<math> h(t)=e^{-t}u(t). \ </math>
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What is the system's response to the input
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<math>x(t)= u(t-1) ? \ </math>
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=Solution =
 
We are given the input to an LTI system along with the system's impulse response and told to find the output y(t).  Since the input and impulse response are given, we simply use convolution on x(t) and h(t) to find the system's output.
 
We are given the input to an LTI system along with the system's impulse response and told to find the output y(t).  Since the input and impulse response are given, we simply use convolution on x(t) and h(t) to find the system's output.
  

Latest revision as of 10:57, 30 January 2011

Problem

The unit impulse response of an LTI system is the CT signal

$ h(t)=e^{-t}u(t). \ $

What is the system's response to the input

$ x(t)= u(t-1) ? \ $

Solution

We are given the input to an LTI system along with the system's impulse response and told to find the output y(t). Since the input and impulse response are given, we simply use convolution on x(t) and h(t) to find the system's output.

$ y(t) = h(t) * x(t) = \int_{-\infty}^\infty h(t-\tau)x(t)d\tau $  (COMMUTATIVE PROPERTY)


Plugging in the given x(t) and h(t) values results in:

$ \begin{align} y(t) & = \int_{-\infty}^\infty e^{-(t-\tau)}u(t-\tau)u(\tau-1)d\tau \\ & = \int_1^\infty e^{-(t-\tau)}u(t-\tau)d\tau \\ & = \int_1^{t} e^{-(t-\tau)}d\tau \\ & = e^{-t}\int_1^{t} e^{\tau}d\tau \\ & = e^{-t}(e^{t} - e) \\ & = 1-e^{-(t-1)}\, \mbox{ for } t > 1 \end{align} $


Since x(t) = 0 when t < 1:

$ y(t) = 0\, \mbox{ for } t < 1 $


$ \therefore y(t) = \begin{cases} 1-e^{-(t-1)}, & \mbox{if }t\mbox{ is} > 1 \\ 0, & \mbox{if }t\mbox{ is} < 1 \end{cases} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood