4. Chapter-2,6,9-AKW

4. Chapter-2,6,9-AKW - ELEC211: Signals and Systems Lecture...

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1 AKW Fall 2010 Chapter 9 ELEC211: Signals and Systems Lecture 22 Differential Equations as LTI Systems ODE as LTI System Spring-Damper System
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2 AKW Fall 2010 Chapter 9 Differential Equation Revisited Why difference and differential equations? - Many systems can be modeled by differential equations. - Ex_1: An RC circuit - Ex_2: Spring-Damper System ) ( ) ( ) ( t V dt t dV CR t V c c s + = dt t dy D t Ky dt t y d M t x dt t dy D t Ky t x dt t y d M ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 2 2 2 + + = = R + _ C V c i V s () x t yt
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3 AKW Fall 2010 Chapter 9 Recall what are ODE and PDE. We will focus on constant coefficient ODE because they can be analyzed using the techniques that we have learned in this course. An ODE is in the general form: ODE describes an implicit relationship between input and output. In general we cannot describe the output explicitly in terms of the input When describing a physical system with time t as the independent variable, usually a differential equation should represent a causal system . In this lecture we will assume that is the case. In the upcoming lectures, we will see how Laplace transform allows us to discuss right-sided as well as left-sided solution. Also, we will focus on non-boundary value ODE which is LTI. 00 () nm NM dy t d x t ab dt == = ∑∑ Ordinary Differential Equations
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4 AKW Fall 2010 Chapter 9 1. Non-Boundary Value formulation . Find y ( t ) given specification of x ( t ) for all t . (Example, , etc.) 2. Boundary-Value formulation . Sometimes we are given the specification of x ( t ) only for t >0, with t =0 being the time at which the system becomes of interest to us. In formulation 2, a set of initial conditions must be given for the output to be determined. The exact x ( t ) for t <0 may not be known or may not be relevant if we have a sufficient set of initial conditions that fully specifies the state of the system at t =0. Sometimes the system is specified as “ initial rest ”, which means that x ( t ) = 0 for all t < 0. This is equivalent to specification of x ( t ) for all t . A constant coefficient differential equation is by nature LTI, but with initial conditions that is not initial rest, the LTI property is destroyed as a matter of problem model . This is explained in the next two slides. () cos ( ) , () cos ( ) () x tt x u t ω = = Two Scenarios in Analyzing an ODE
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5 AKW Fall 2010 Chapter 9 1. Linearity If is the solution to input , and the solution to , then is the solution to input 1 () yt 12 y t α β =+ 1 x t 2 2 x t xt x t = + Every derivative of x ( t ) can be decomposed: 00 0 0 mm m MM M m m m nn n NN N n n n dxt dx t bb b dt dy t t t aa a αβ == = = =+= ∑∑ 2. Time Invariant It is obvious because the equality will not be affected by shifting x ( t ) and y ( t ) by the same amount in time Proof of LTI property of Constant Coefficient ODE With specification of x ( t ) for all t (including initial rest), ODE is clearly LTI Proof ( ) m m t x t + hence y 1 ( t ) solution to x 1 ( t ) and y 2 ( t ) solution to x 2 ( t )
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6 AKW Fall 2010 Chapter 9 Initial conditions is something that is inherited from the past (t < 0). It destroys linearity.
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This note was uploaded on 01/28/2011 for the course ELEC 211 taught by Professor Albertk.wong during the Fall '09 term at HKUST.

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4. Chapter-2,6,9-AKW - ELEC211: Signals and Systems Lecture...

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