(New page: ==Problem 5 - Alternate 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 ...)
 
 
(2 intermediate revisions by one other user not shown)
Line 1: Line 1:
==Problem 5 - Alternate Solution==
+
[[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.
  

Latest revision as of 09:57, 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

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva