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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.
  
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==Alternative Solutions==
 
==Alternative Solutions==
[[Problem 5 - Alternate Solution]]
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[[Problem 5 - Alternate Solution_(ECE301Summer2008asan)|Problem 5 - Alternate Solution]]
  
[[Problem 5 - Graphical Solution]]
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[[Problem 5 - Graphical Solution_(ECE301Summer2008asan)|Problem 5 - Graphical Solution]]

Latest revision as of 09:55, 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(\tau)x(t-\tau)d\tau $


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

$ \begin{align} y(t) & = \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t-\tau-1)d\tau \\ & = \int_0^\infty e^{-\tau}u(t-\tau-1)d\tau \\ & = \int_0^{t-1} e^{-\tau}d\tau \\ & = 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} $

Alternative Solutions

Problem 5 - Alternate Solution

Problem 5 - Graphical Solution

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn