(New page: ==Problem 5== 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 ...) |
|||
(3 intermediate revisions by one other user not shown) | |||
Line 1: | Line 1: | ||
− | + | [[Category: ECE]] | |
+ | [[Category: ECE 301]] | ||
+ | [[Category: Summer]] | ||
+ | [[Category: 2008]] | ||
+ | [[Category: asan]] | ||
+ | [[Category: Exams]] | ||
+ | =Problem= | ||
+ | The unit impulse response of an LTI system is the CT signal | ||
+ | |||
+ | <math> h(t)=e^{-t}u(t). \ </math> | ||
+ | |||
+ | What is the system's response to the input | ||
+ | |||
+ | <math>x(t)= u(t-1) ? \ </math> | ||
+ | =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. | ||
Line 29: | Line 43: | ||
==Alternative Solutions== | ==Alternative Solutions== | ||
− | [[Problem 5 - Alternate Solution]] | + | [[Problem 5 - Alternate Solution_(ECE301Summer2008asan)|Problem 5 - Alternate Solution]] |
− | [[Problem 5 - Graphical Solution]] | + | [[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} $