Laplace Transform BIOE 4200. Why use Laplace Transforms? Find solution to differential equation using algebra Relationship to Fourier Transform allows.

Laplace Transform

BIOE 4200

Why use Laplace Transforms?

Find solution to differential equation using algebra

Relationship to Fourier Transform allows easy way to characterize systems

No need for convolution of input and differential equation solution

Useful with multiple processes in system

How to use Laplace

Find differential equations that describe system

Obtain Laplace transformPerform algebra to solve for output or

variable of interestApply inverse transform to find solution

What are Laplace transforms?

j

j

st1

0

st

dse)s(Fj2

1)}s(F{L)t(f

dte)t(f)}t(f{L)s(F

t is real, s is complex! Inverse requires complex analysis to solve Note “transform”: f(t) F(s), where t is integrated and

s is variable Conversely F(s) f(t), t is variable and s is

integrated Assumes f(t) = 0 for all t < 0

Evaluating F(s) = L{f(t)}

Hard Way – do the integral

0

st

0 0

t)as(stat

at

0

st

dt)tsin(e)s(F

tsin)t(f

as

1dtedtee)s(F

e)t(f

s

1)10(

s

1dte)s(F

1)t(flet

let

let

Integrate by parts

Evaluating F(s)=L{f(t)}- Hard Way

remember vduuvudv

)tcos(v,dt)tsin(dv

dtsedu,eu stst

0

stst

0 0

st

0

stst

dt)tcos(es)1(e

dt)tcos(es)tcos(e[dt)tsin(e ]

)tsin(v,dt)tcos(dv

dtsedu,eu stst

0

stst

0

st

0

st

0

st

dt)tsin(es)0(edt)tsin(es)tsin(e[

dt)tcos(e

]

20

st

0

st2

0 0

st2st

s1

1dt)tsin(e

1dt)tsin(e)s1(

dt)tsin(es1dt)tsin(se

let

let

Substituting, we get:

It only gets worse…

Evaluating F(s) = L{f(t)}

This is the easy way ...Recognize a few different transformsSee table 2.3 on page 42 in textbookOr see handout .... Learn a few different properties Do a little math

Table of selected Laplace Transforms

1s

1)s(F)t(u)tsin()t(f

1s

s)s(F)t(u)tcos()t(f

as

1)s(F)t(ue)t(f

s

1)s(F)t(u)t(f

2

2

at

More transforms

1nn

s

!n)s(F)t(ut)t(f

665

2

1

s

120

s

!5)s(F)t(ut)t(f,5n

s

!1)s(F)t(tu)t(f,1n

s

1

s

!0)s(F)t(u)t(f,0n

1)s(F)t()t(f

Note on step functions in Laplace

0

stdte)t(f)}t(f{L

0t,0)t(u

0t,1)t(u

Unit step function definition:

Used in conjunction with f(t) f(t)u(t) because of Laplace integral limits:

Properties of Laplace Transforms

LinearityScaling in timeTime shift“frequency” or s-plane shiftMultiplication by tn

IntegrationDifferentiation

Properties: Linearity

)s(Fc)s(Fc)}t(fc)t(fc{L 22112211 Example :

1s

1)

1s

)1s()1s((

2

1

)1s

1

1s

1(

2

1

}e{L2

1}e{L

2

1

}e2

1e

2

1{y

)}t{sinh(L

22

tt

tt

Proof :

)s(Fc)s(Fc

dte)t(fcdte)t(fc

dte)]t(fc)t(fc[

)}t(fc)t(fc{L

2211

0

st22

0

st11

st22

0

11

2211

)a

s(F

a

1)}at(f{L

Example :

22

22

2

2

s

)s

(1

)1)s(

1(

1

)}t{sin(L

Proof :

)a

s(F

a

1

due)u(fa

1

dua

1dt,

a

ut,atu

dte)at(f

)}at(f{L

a

0

u)a

s(

0

st

let

Properties: Scaling in Time

Properties: Time Shift

)s(Fe)}tt(u)tt(f{L 0st00

Example :

as

e

)}10t(ue{Ls10

)10t(a

Proof :

)s(Fedue)u(fe

due)u(f

tut,ttu

dte)tt(f

dte)tt(u)tt(f

)}tt(u)tt(f{L

00

0

0

0

st

0

sust

t

0

)tu(s

00

t

st0

0

st00

00

let

Properties: S-plane (frequency) shift

)as(F)}t(fe{L at

Example :

22

at

)as(

)}tsin(e{L

Proof :

)as(F

dte)t(f

dte)t(fe

)}t(fe{L

0

t)as(

0

stat

at

Properties: Multiplication by tn

)s(Fds

d)1()}t(ft{L

n

nnn

Example :

1n

n

nn

n

s

!n

)s

1(

ds

d)1(

)}t(ut{L

Proof :

)s(Fs

)1(dte)t(fs

)1(

dtes

)t(f)1(

dtet)t(f

dte)t(ft)}t(ft{L

n

nn

0

stn

nn

0

stn

nn

0

stn

0

stnn

The “D” Operator

1. Differentiation shorthand

2. Integration shorthand)t(f

dt

d)t(fD

dt

)t(df)t(Df

2

22

)t(f)t(Dg

dt)t(f)t(gt

)t(fD)t(g

dt)t(f)t(g

1a

t

a

if

then then

if

Properties: Integrals

s

)s(F)}t(fD{L 1

0

Example :

)}t{sin(L1s

1)

1s

s)(

s

1(

)}tcos(D{L

22

10

Proof :

let

stst

0

st

10

es

1v,dtedv

dt)t(fdu),t(gu

dte)t(g)}t{sin(L

)t(fD)t(g

t

0

st0

st

dt)t(f)t(g

s

)s(Fdte)t(f

s

1]e)t(g

s

1[

0

)()( dtetft st

If t=0, g(t)=0

for so

slower than

0

)()( tgdttf 0 ste

Properties: Derivatives(this is the big one)

)0(f)s(sF)}t(Df{L Example :

)}tsin({L1s

11s

)1s(s

11s

s

)0(f1s

s

)}tcos(D{L

2

2

22

2

2

2

2

Proof :

)s(sF)0(f

dte)t(fs)]t(fe[

)t(fv,dt)t(fdt

ddv

sedu,eu

dte)t(fdt

d)}t(Df{L

0

st0

st

stst

0

st

let

Difference in

The values are only different if f(t) is not continuous @ t=0

Example of discontinuous function: u(t)

)0(f&)0(f),0(f

1)0(u)0(f

1)t(ulim)0(f

0)t(ulim)0(f

0t

0t

?)}t(fD{L 2

)0('f)0(sF)s(Fs)0('f))0(f)s(sF(s

)0('f)}t(Df{sL)0(g)s(sG)}t(gD{L

)0('f)0(Df)0(g),t(Df)t(g

2

2

let

)0(f)0(sf)0('fs)0(fs)s(Fs)}t(fD{L )'1n()'2n()2n()1n(nn

NOTE: to takeyou need the value @ t=0 for

called initial conditions!We will use this to solve differential equations!

)t(f),t(Df),...t(fD),t(fD 2n1n

)}t(fD{L n

Properties: Nth order derivatives

Properties: Nth order derivatives

)0(f)s(sF)}t(Df{L )}t(fD{L 2

)0(f)s(sF)}t(Df{L)}t(g{L)s(G

)0('f)0(g

)t(Df)t(g

)0(g)s(sG)}t(Dg{L

)t(fD)t(Dgand)t(Df)t(g 2

)0('f)0(sf)s(Fs)0('f)]0(f)s(sF[s)0(g)s(sG)}t(Dg{L 2

.etc),t(fD),t(fD 43

Start with

Now apply again

letthen

remember

Can repeat for

)0(f)0(sf)0('fs)0(fs)s(Fs)}t(fD{L )'1n()'2n()2n()1n(nn

Relevant Book Sections

Modeling - 2.2Linear Systems - 2.3, page 38 onlyLaplace - 2.4Transfer functions – 2.5 thru ex 2.4