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Feedback Control

Laplace Transforms Use

EGR 386

8/31/2011

Common Laplace Transforms

Process for using Laplace Transforms

to solve differential equations

• Determine governing differential equation as function

of time ‘t’

• Use Laplace transform table to convert each term to

algebraic equation as function of ‘s’

• Apply initial conditions

• Express as “ Ouput(s)=…” or “Y(s)=“…

• Use algebra to have each term of Y(s) able to be

found on Laplace transform table

• Convert each term of Y(s) usingLaplace transform

table , obtain y(t)

Laplace Transform yields complete solution

Laplace Transform yields complete

solution (continued)

x(t)x(t)x(t)x(t)

Transfer Function

)(

)(

)(

)()(

sU

sY

sInput

sOutputsG ==≡

Initial Value Theorem (IVT)

{ })()0( ssFs

Limtf

∞→==

Final Value Theorem (FVT)

{ })(0

)( ssFs

Limtf

→=∞=

FVT is typically used to find the steady state value of f(t)FVT is typically used to find the steady state value of f(t)

FVT is more often used than the IVT

Final value may not exist*

*Note: There is no final value for oscillatory functions,

such as)sin()( tCtf ω=

Partial Fraction Expansion

PFE continued

Substituting for R & P we have easily converted Laplace Transforms

So taking the inverse Laplace Transforms we get h(t)

PFE w/ TF with ‘s’ in the numerator

Transfer Function

Terminology : Poles and Zeros

• Poles [at a ‘pole’ G(s) →∞ ]

• values ‘s’ of D(s)=0 ,

))()((

))((

)(

)()(

esdscs

bsas

sD

sNsG

−−−

−−==

• values ‘s’ of D(s)=0 ,

• roots of the denominator: s= c, d, e

• Zeros [at a ‘zero’ G(s) →0 ]

• values ‘s’ of N(s)=0 ,

• roots of the numerator : s= a, b

Poles and Zeros

• Later we will see that poles and zeros of a

transfer function will characterize

performance of system, and will show what

effects feedback may haveeffects feedback may have

Examples