laplace transforms1

7/29/2019 Laplace Transforms1

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LAPLACE TRANSFORMS

INTRODUCTION


2/110

Definition Transforms -- a mathematical conversion

from one way of thinking to another tomake a problem easier to solve

transformsolution

in transformway ofthinking

inversetransform

solutionin original

way ofthinking

problemin original

way ofthinking

2. Transforms


3/110

Laplacetransform

solutionin

s domain

inverseLaplace

transform

solutionin timedomain

problemin timedomain

Other transforms Fourier z-transform wavelets

2. Transforms


4/110

Laplace transformation

linear

differentialequation

time

domainsolution

Laplace

transformedequation

Laplace

solution

time domain

Laplace domain orcomplex frequency domain

algebra

Laplace transform

inverse Laplacetransform

4. Laplace transforms


5/110

Basic Tool For Continuous Time:

Laplace Transform

Convert time-domain functions and operations into

frequency-domain

f(t) F(s) (tR, sC

Linear differential equations (LDE) algebraic expressionin Complex plane

Graphical solution for key LDE characteristics

Discrete systems use the analogous z-transform

0

)()()]([ dtetfsFtfst

L


6/110

The Laplace Transform

The Laplace Transform of a function, f(t), is defined as;

0

)()()]([ dtetfsFtfL st

The Inverse Laplace Transform is defined by

j

j

tsdsesFj

tfsFL

)(

2

1)()]([

1

*notes

Eq A

Eq B


7/110


We generally do not use Eq B to take the inverse Laplace. However,

this is the formal way that one would take the inverse. To use

Eq B requires a background in the use of complex variables and

the theory of residues. Fortunately, we can accomplish the same

goal (that of taking the inverse Laplace) by using partial fraction

expansionand recognizing transform pairs.

*notes


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Laplace Transform of the unit step.

*notes

|00

11)]([

stst es

dtetuL

s

tuL1

)]([

The Laplace Transform of a unit step is:

s

1


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The Laplace transform of a unit impulse:

Pictorially, the unit impulse appears as follows:

0 t0

f(t) (t t0)

Mathematically:

(t t0) = 0 t 0

*note

01)(

0

0

0

dttt

t

t


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An important property of the unit impulse is a sifting

or sampling property. The following is an important.

2

12010

2010

0,0

)()()(

t

tttttttttfdttttf


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In particular, if we let f(t) = (t) and take the Laplace

1)()]([0

0

sst

edtettL


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An important point to remember:

)()( sFtf

The above is a statement that f(t) and F(s) are

transform pairs. What this means is that for

each f(t) there is a unique F(s) and for each F(s)

there is a unique f(t). If we can remember the

Pair relationships between approximately 10 of the

Laplace transform pairs we can go a long way.


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Building transform pairs:

eL(

e

tasstatat dtedteetueL0

)(

0

)]([

asas

etueL

stat

1

)(

)]([ |0

as

tue at

1)(A transform

pair


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0

)]([ dttettuL st

0 00

| vduuvudvu = t

dv = e-

stdt

2

1)(

s

ttu A transform

pair


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22

0

11

2

1

2

)()][cos(

ws

s

jwsjws

dteee

wtL stjwtjwt

22)()cos(

ws

stuwt

A transformpair


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Time Shift

0 0

)()()(

,.,0,

,

)()]()([

dxexfedxexf

SoxtasandxatAsaxtanddtdxthenatxLet

eatfatuatfL

sxasaxs

a

st

)()]()([ sFeatuatfL as


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Frequency Shift

0

)(

0

)()(

)]([)]([

asFdtetf

dtetfetfeL

tas

statat

)()]([ asFtfeL at


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Example: Using Frequency Shift

Find the L[e-atcos(wt)]

In this case, f(t) = cos(wt)

so,

22

22

)(

)(

)(

)(

was

as

asFand

ws

ssF

22)()(

)()]cos([

was

aswteL at


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Time Integration:

The property is:

stst

t

st

t

e

s

vdtedv

and

dttfdudxxfuLet

partsbyIntegrate

dtedxxfdttfL

1,

)(,)(

:

)()(

0

0 00


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Time Integration:

Making these substitutions and carrying out

The integration shows that

)(1

)(1

)(

00

sFs

dtetfs

dttfL st


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Time Differentiation:

If the L[f(t)] = F(s), we want to show:

)0()(]

)(

[ fssFdt

tdf

L

Integrate by parts:

)(),()(

,

tfvsotdfdt

dt

tdfdv

anddtsedueu stst

*note


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Making the previous substitutions gives,

0

00

)()0(0

)()( |

dtetfsf

dtsetfetfdt

df

L

st

stst

So we have shown:

)0()()(

fssF

dt

tdfL


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We can extend the previous to show;

)0(...

)0(')0()()(

)0('')0(')0()()(

)0(')0()()(

)1(

21

23

3

3

2

2

2

n

nnn

n

n

f

fsfssFsdt

tdfL

casegeneral

fsffssFsdt

tdfL

fsfsFs

dt

tdfL


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Transform Pairs:

____________________________________)()( sFtf

f(t)

F(s)

1

2

!

1

1

1

)(

1)(

n

n

st

s

nt

s

t

as

e

stu

t


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Transform Pairs:

f(t)

F(s)

22

22

1

2

)cos(

)sin(

)(

!

1

ws

swt

ws

wwt

as

net

aste

n

atn

at


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Transform Pairs:

f(t)

F(s)

22

22

22

22

sincos)cos(

cossin

)sin(

)()cos(

)()sin(

ws

wswt

ws

ws

wt

was

aswte

was

wwte

at

at

Yes !


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Common Transform Properties:

f(t) F(s)

)(1

)(

)()(

)0(...)0(')0()()(

)()(

)([0),()(

)(0),()(

0

1021

00

000

sFs

df

ds

sdFttf

ffsfsfssFsdt

tfd

asFtfe

ttfLetttutf

sFetttuttf

t

nnnn

n

n

at

sot

sot


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Using Matlab with Laplace transform:

Example Use Matlab to find the transform of tte 4

The following is written in italic to indicate Matlab code

syms t,s

laplace(t*exp(-4*t),t,s)

ans =

1/(s+4)^2


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Using Matlab with Laplace transform:

Example Use Matlab to find the inverse transform of

19.12.)186)(3(

)6()(

2prob

sss

sssF

syms s t

ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18)))

ans =

-exp(-3*t)+2*exp(-3*t)*cos(3*t)


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The Laplace TransformTheorem: Initial Value

If the function f(t) and its first derivative are Laplace transformable and f(t)

Has the Laplace transform F(s), and the exists, then)(lim ssF

0

)0()(lim)(lim

tsftfssF

The utility of this theorem lies in not having to take the inverse of F(s)in order to find out the initial condition in the time domain. This is

particularly useful in circuits and systems.

Theorem:

s

Initial ValueTheorem

Th L l T f


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Initial Value Theorem:Example:

Given;

225)1(

)2()(

s

ssF

Find f(0)

1)26(2

2lim

2512

2lim

5)1(

)2(lim)(lim)0(

2222

222

2

2

22

sssss

ssss

ss

ss

s

ssssFf ss s

s

Th L l T f


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Theorem: Final Value Theorem:

If the function f(t) and its first derivative are Laplace transformable and f(t)

has the Laplace transform F(s), and the exists, then)(lim ssFs

)()(lim)(lim ftfssF0s t

Again, the utility of this theorem lies in not having to take the inverseof F(s) in order to find out the final value of f(t) in the time domain.

This is particularly useful in circuits and systems.

Final Value

Theorem

Th L l T f


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Final Value Theorem:Example:

Given:

ttesFnotesssF t 3cos)(3)2( 3)2()( 2122 22 Find )(f .

03)2( 3)2(lim)(lim)( 22 22 sssssFf 0s0s


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The Complex Plane (review)

Imaginary axis (j)

Real axis

jyxu

x

y

r

r

jyxu

(complex) conjugate

y

22

1

||||

tan

yxuru

x

yu


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Laplace Transforms of Common

Functions

Name f(t) F(s)

Impulse

Step

Ramp

Exponential

Sine

1

s

1

2

1

s

as

1

22

1

s

1)( tf

ttf )(

atetf )(

)sin()( ttf

00

01)(

t

ttf


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Laplace Transform Properties

)(lim)(lim

)(lim)0(

)()()

)(1)(

)(

)0()()(

)()()]()([

0

0

2121

0

2121

ssFtf-

ssFf-

sFsFd()f(tf

dttfss

sFdttfL

fssFtf

dt

dL

sbFsaFtbftafL

st

s

t

t

theoremvalueFinal

theoremvalueInitial

nConvolutio

nIntegratio

ationDifferenti

calingAddition/S


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LAPLACE TRANSFORMS

SIMPLE TRANSFORMATIONS

Tr f r (1 f 11)


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Transforms (1 of 11)

Impulse -- (to)

F(s) =

0

e-st (to) dt

= e-sto

f(t)

t

(to)


Tr nsf rms (2 f 11)


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Transforms (2 of 11)

Step -- u (to)

F(s) =

0

e-st u (to) dt

= e-sto/sf(t)

t

u (to)1



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Transforms (3 of 11) e-at

F(s) =

0

e-st e-at dt

= 1/(s+a)



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Transforms (4 of 11)f1(t) f2(t)

a f(t)

eat f(t)

f(t - T)

f(t/a)

F1(s) F2(s)

a F(s)

F(s-a)

eTs F(as)

a F(as)

Linearity

Constant multiplication

Complex shift

Real shift

Scaling



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Transforms (5 of 11) Most mathematical handbooks have tables

of Laplace transforms



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LAPLACE TRANSFORMS

PARTIAL FRACTION EXPANSION


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Definition

Definition -- Partial fractions are several

fractions whose sum equals a given fraction

Purpose -- Working with transforms requires

breaking complex fractions into simpler

fractions to allow use of tables of transforms


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Partial Fraction Expansions

32)3()2(

1

s

B

s

A

ss

s Expand into a term for

each factor in the

denominator.

Recombine RHS

Equate terms in s and

constant terms. Solve.

Each term is in a form sothat inverse Laplace

transforms can be

applied.

)3()2(

2)3(

)3()2(

1

ss

sBsA

ss

s

3

2

2

1

)3()2(

1

ssss

s

1BA 123 BA


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Example of Solution of an ODE

0)0(')0(2862

2

yyydt

dy

dt

yd ODE w/initial conditions

Apply Laplace transform

to each term Solve for Y(s)

Apply partial fraction

expansion

Apply inverse Laplace

transform to each term

ssYsYssYs /2)(8)(6)(2

)4()2(

2)(

ssssY

)4(4

1

)2(2

1

4

1

)(

ssssY

424

1)(

42 tt eety


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Different terms of 1st degree

To separate a fraction into partial fractions

when its denominator can be divided intodifferent terms of first degree, assume an

unknown numerator for each fraction

Example -- (11x-1)/(X2 - 1) = A/(x+1) + B/(x-1)

= [A(x-1) +B(x+1)]/[(x+1)(x-1))]

A+B=11 -A+B=-1

A=6, B=5

Repeated terms of 1st degree (1 of 2)


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When the factors of the denominator are of

the first degree but some are repeated,assume unknown numerators for each

factor

If a term is present twice, make the fractions

the corresponding term and its second power

If a term is present three times, make the

fractions the term and its second and third

powers

3. Partial fractions



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Example --

(x2+3x+4)/(x+1)3=A/(x+1) + B/(x+1)2 +C/(x+1)3

x2+3x+4 = A(x+1)2 + B(x+1) + C

= Ax2 + (2A+B)x + (A+B+C)

A=1

2A+B = 3

A+B+C = 4

A=1, B=1, C=2



51/110

Different quadratic terms When there is a quadratic term, assume a

numerator of the form Ax + B Example --

1/[(x+1) (x2 + x + 2)] = A/(x+1) + (Bx +C)/ (x2 +

x + 2)

1 = A (x2 + x + 2) + Bx(x+1) + C(x+1)

1 = (A+B) x2 + (A+B+C)x +(2A+C)

A+B=0

A+B+C=0 2A+C=1

A=0.5, B=-0.5, C=0



52/110

Repeated quadratic terms Example --

1/[(x+1) (x2

+ x + 2)2

] = A/(x+1) + (Bx +C)/ (x2

+ x + 2) + (Dx +E)/ (x2 + x + 2)2

1 = A(x2 + x + 2)2 + Bx(x+1) (x2 + x + 2) +

C(x+1) (x2 + x + 2) + Dx(x+1) + E(x+1)

A+B=0 2A+2B+C=0

5A+3B+2C+D=0

4A+2B+3C+D+E=0

4A+2C+E=1

A=0.25, B=-0.25, C=0, D=-0.5, E=0


A l I i i l d Fi l V l


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Apply Initial- and Final-ValueTheorems to this Example

Laplace

transform of the

function.

Apply final-value

theorem

Apply initial-

value theorem

)4()2(

2)(

ssssY

4

1

)40()20()0(

)0(2)(lim

tft

0)4()2()(

)(2)(lim 0

tft


54/110

LAPLACE TRANSFORMS

SOLUTION PROCESS


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Solution process (1 of 8)

Any nonhomogeneous linear differentialequation with constant coefficients can be

solved with the following procedure, which

reduces the solution to algebra



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Step 1: Put differential equation into

standard form

D2 y + 2D y + 2y = cos t

y(0) = 1

D y(0) = 0

l f


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Step 2: Take the Laplace transform of both

sides

L{D2 y} + L{2D y} + L{2y} = L{cos t}

S l i (4 f 8)


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Step 3: Use table of transforms to expressequation in s-domain

L{D2 y} + L{2D y} + L{2y} = L{cos t}

L{D2

y} = s2

Y(s) - sy(0) - D y(0) L{2D y} = 2[ s Y(s) - y(0)]

L{2y} = 2 Y(s)

L{cos t} = s/(s2 + 1)

s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s /(s2 + 1)

S l i (5 f 8)


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Step 4: Solve for Y(s) s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s/(s2 + 1)

(s2 + 2s + 2) Y(s) = s/(s2 + 1) + s + 2

Y(s) = [s/(s2 + 1) + s + 2]/ (s2 + 2s + 2)

= (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)]

S l i (6 f 8)


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Step 5: Expand equation into format covered by

table Y(s) = (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)]

= (As + B)/ (s2 + 1) + (Cs + E)/ (s2 + 2s + 2)

(A+C)s3

+ (2A + B + E) s2

+ (2A + 2B + C)s + (2B+E)

1 = A + C

2 = 2A + B + E

2 = 2A + 2B + C

2 = 2B + E

A = 0.2, B = 0.4, C = 0.8, E = 1.2

S l i (7 f 8)


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(0.2s + 0.4)/ (s2 + 1)

= 0.2 s/ (s2 + 1) + 0.4 / (s2 + 1)

(0.8s + 1.2)/ (s2 + 2s + 2)

= 0.8 (s+1)/[(s+1)2

+ 1] + 0.4/ [(s+1)2

+ 1]

S l i (8 f 8)


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Step 6: Use table to convert s-domain to

time domain

0.2 s/ (s2 + 1) becomes 0.2 cos t

0.4 / (s2 + 1) becomes 0.4 sin t

0.8 (s+1)/[(s+1)2 + 1] becomes 0.8 e-t cos t

0.4/ [(s+1)2 + 1] becomes 0.4 e-t sin t

y(t) = 0.2 cos t + 0.4 sin t + 0.8 e-t cos t + 0.4

e

-t

sin t


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LAPLACE TRANSFORMS

TRANSFER FUNCTIONS

I d i


64/110

Introduction Definition -- a transfer function is an

expression that relates the output to theinput in the s-domain

differential

equation

r(t) y(t)

transferfunction

r(s) y(s)

5. Transfer functions

T f F i


65/110

Transfer Function

Definition H(s) = Y(s) / X(s)

Relates the output of a linear system (or

component) to its input Describes how a linear system responds to

an impulse

All linear operations allowed Scaling, addition, multiplication

H(s)X(s) Y(s)

Bl k Di


66/110

Block Diagrams

Pictorially expresses flows and relationshipsbetween elements in system

Blocks may recursively be systems

Rules Cascaded (non-loading) elements: convolution

Summation and difference elements

Can simplify

T i l bl k di


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Typical block diagram

controlGc(s)

plantGp(s)

feedbackH(s)

pre-filterG1(s)

post-filterG2(s)

reference input, R(s)error, E(s)

plant inputs, U(s)output, Y(s)

feedback, H(s)Y(s)


E l


68/110

Example

v(t)

R

C

L

v(t) = R I(t) + 1/C I(t) dt + L di(t)/dt

V(s) = [R I(s) + 1/(C s) I(s) + s L I(s)]

Note: Ignore initial conditions5. Transfer functions

Block diagram and transfer function


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g

V(s)

= (R + 1/(C s) + s L ) I(s) = (C L s2 + C R s + 1 )/(C s) I(s)

I(s)/V(s) = C s / (C L s2 + C R s + 1 )

C s / (C L s2 + C R s + 1 )

V(s) I(s)


Bl k di d ti l


70/110

Block diagram reduction rules

G1

G2

G1

G2

U Y U Y

G1

G2

U Y+

+ G1 + G2U Y

G1

G2

U Y+

- G1 /(1+G1 G2)U Y

Series

Parallel

Feedback


R ti l L pl Tr f r


71/110

Rational Laplace Transforms

m

sFsAs

sFsBs

bsbsbsB

asasasA

sBsAsF

mm

n

n

poles#systemofOrder

complexarezeroesandPoles

(So,:Zeroes

(So,:Poles

)0*)(0*)(*

)*)(0*)(*

...)(

...)(

)()()(

01

01

First Ord r S st m


72/110

First Order System

Reference

)(sY

)(sR

S)(sE

1)(sB

)(sU

sT1

1K

sT

K

sTK

K

sR

sY

11)(

)(

First Order S stem


73/110

First Order System

Impulseresponse

Exponential

Step response Step,exponential

Ramp response Ramp, step,

exponential

1 sT

K

/1

2 Ts

KT-

s

KT-

s

K

/1

Ts

K-

s

K

No oscillations (as seen by poles)

Second Order System


74/110

Second Order System

:frequencynaturalUndamped

where:ratioDamping

(ie,partimaginaryzero-nonhavepolesifOscillates

:responseImpulse

JK

JKBB

B

JKB

ssKBsJs

K

sR

sY

N

c

c

NN

N

2

)04

2)(

)(

2

22

2

2

Second Order System: Parameters


75/110

Second Order System: Parameters

noscillatiotheoffrequencythegives

frequencynaturalundampedoftionInterpreta

0)Im0,(ReOverdamped1

Im)(RedUnderdampe

0)Im0,(RenoscillatioUndamped

ratiodampingoftionInterpreta

N

:

0:10

:0

Transient Response Characteristics


76/110

Transient Response Characteristics

statesteadyof%specifiedwithinstaystimeSettling:

reachedisvaluepeakwhichatTime:

valuestatesteadyreachfirstuntildelaytimeRise:

valuestatesteadyof50%reachuntilDelay:

s

p

r

d

t

t

t

t

0.5 1 1.5 2 2.5 3

0.25

0.5

0.75

1

1.25

1.5

1.75

2

rt

overshoot maximum p M

pt stdt

Transient Response


77/110

Transient Response

Estimates the shape of the curve based onthe foregoing points on the x and y axis

Typically applied to the following inputs

Impulse Step

Ramp

Quadratic (Parabola)

Effect of pole locations


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Effect of pole locations

Faster Decay Faster Blowup

Oscillations(higher-freq)

Im(s)

Re(s)(e-at) (eat)

Basic Control Actions: u(t)


79/110

Basic Control Actions: u(t)

:controlalDifferenti

:controlIntegral

:controlalProportion

sKsE

sU

tedt

d

Ktu

s

K

sE

sUdtteKtu

KsE

sUteKtu

dd

i

t

i

pp

)(

)(

)()(

)(

)()()(

)(

)()()(

0

Effect of Control Actions


80/110

Effect of Control Actions

Proportional Action Adjustable gain (amplifier)

Integral Action Eliminates bias (steady-state error)

Can cause oscillations

Derivative Action (rate control) Effective in transient periods

Provides faster response (higher sensitivity)

Never used alone

Basic Controllers


81/110

Basic Controllers

Proportional control is often used by itself

Integral and differential control are typically

used in combination with at least proportional

control eg, Proportional Integral (PI) controller:

sTK

s

KK

sE

sUsG

i

pI

p

11

)(

)()(


82/110

Root locus Analysis


83/110

Root-locus Analysis

Based on characteristic eqn of closed-loop transferfunction

Plot location ofroots of this eqn Same as poles of closed-loop transfer function

Parameter (gain) varied from 0 to Multiple parameters are ok

Vary one-by-one

Plot a root contour (usually for 2-3 params)

Quickly get approximate results Range of parameters that gives desired response


84/110

LAPLACE TRANSFORMS

LAPLACE APPLICATIONS

Initial value


85/110

Initial value In the initial value of f(t) as t approaches 0

is given by

f(0 ) = Lim s F(s)

s

f(t) = e -t

F(s) = 1/(s+1)

f(0 ) = Lim s /(s+1) = 1

s

Example

6. Laplace applications

Final value


86/110

Final value In the final value of f(t) as t approaches

is given by

f(0 ) = Lim s F(s)

s 0

f(t) = e -t

F(s) = 1/(s+1)

f(0 ) = Lim s /(s+1) = 0

s 0

Example


Apply Initial- and Final-Value


87/110

Apply Initial and Final ValueTheorems to this Example

Laplace

transform of the

function.

Apply final-value

theorem

Apply initial-

value theorem

)4()2(

2)(

ssssY

4

1

)40()20()0(

)0(2)(lim

tft

0)4()2()(

)(2)(lim 0

tft

Poles


88/110

Poles The poles of a Laplace function are the

values of s that make the Laplace functionevaluate to infinity. They are therefore the

roots of the denominator polynomial

10 (s + 2)/[(s + 1)(s + 3)] has a pole at s =

-1 and a pole at s = -3

Complex poles always appear in complex-

conjugate pairs

The transient response of system is

determined by the location of poles


Zeros


89/110

Zeros The zeros of a Laplace function are the

values of s that make the Laplace functionevaluate to zero. They are therefore the

zeros of the numerator polynomial

10 (s + 2)/[(s + 1)(s + 3)] has a zero at s =

-2

Complex zeros always appear in complex-

conjugate pairs


Stability


90/110

Stability A system is stable if bounded inputs

produce bounded outputs The complex s-plane is divided into two

regions: the stable region, which is the left

half of the plane, and the unstable region,

which is the right half of the s-plane

s-plane

stable unstable

x

x

xx x

x

x

j


91/110

LAPLACE TRANSFORMS

FREQUENCY RESPONSE

Introduction


92/110

Introduction Many problems can be thought of in the

time domain, and solutions can bedeveloped accordingly.

Other problems are more easily thought of

in the frequency domain.

A technique for thinking in the frequency

domain is to express the system in terms

of a frequency response

7. Frequency response

Definition


93/110

Definition The response of the system to a sinusoidal

signal. The output of the system at eachfrequency is the result of driving the system

with a sinusoid of unit amplitude at that

frequency.

The frequency response has both amplitude

and phase


Process


94/110

Process The frequency response is computed by

replacing s with j in the transfer function

f(t) = e -t

F(s) = 1/(s+1)

Example

F(j ) = 1/(j +1)

Magnitude = 1/SQRT(1 + 2)

Magnitude in dB = 20 log10 (magnitude)

Phase = argument = ATAN2(- , 1)

magnitude in dB


Graphical methods


95/110

Graphical methods Frequency response is a graphical method

Polar plot -- difficult to construct Corner plot -- easy to construct


Constant K


96/110

Constant K

+180o+90o

0o

-270o-180o

-90o

60 dB40 dB

20 dB0 dB

-20 dB-40 dB-60 dB

magnitude

phase

0.1 1 10 100, radians/sec

20 log10 K

arg K


Simple pole or zero at origin, 1/ (j)n


97/110

+180o+90o

0o

-270o-180o

-90o

60 dB40 dB

20 dB0 dB

-20 dB-40 dB

-60 dB

magnitude

phase

0.1 1 10 100, radians/sec

1/1/2

1/3

1/1/21/3

G(s) = n2/(s2 + 2ns + n2)

Simple pole or zero, 1/(1+j)


98/110

S p e po e o e o, /( j)

+180o+90o

0o

-270o

-180o-90o

60 dB40 dB

20 dB0 dB-20 dB-40 dB-60 dB

magnitude

phase

0.1 1 10 100T


Error in asymptotic approximation


99/110

y p pp

T0.01

0.1

0.5

0.76

1.0

1.31

1.73

2.0

5.0

10.0

dB

0

0.043

1

2

3

4.3

6.0

7.0

14.2

20.3

arg (deg)

0.5

5.7

26.6

37.4

45.0

52.7

60.0

63.4

78.7

84.3


Quadratic pole or zero


100/110

Q p

+180o+90o

0o

-270o-180o

-90o

60 dB40 dB

20 dB0 dB

-20 dB-40 dB-60 dB

magnitude

phase

0.1 1 10 100T


Transfer Functions


101/110

Defined as G(s) = Y(s)/U(s)

Represents a normalized model of a process,

i.e., can be used with any input.

Y(s) and U(s) are both written in deviationvariable form.

The form of the transfer function indicates the

dynamic behavior of the process.

Derivation of a Transfer Function


102/110

TFFTFTFdtdTM )( 212211 Dynamic model of

CST thermal

mixer

Apply deviation

variables

Equation in termsof deviation

variables.

0220110 TTTTTTTTT

TFFTFTFdt

TdM )( 212211

Derivation of a Transfer Function


103/110

21

1

1)(

)()(

FFsM

F

sT

sTsG

Apply Laplace transform

to each term considering

that only inlet and outlet

temperatures change. Determine the transfer

function for the effect of

inlet temperature

changes on the outlettemperature.

Note that the response

is first order.

212211 )()()(FFsM

sTFsTFsT

Poles of the Transfer Function


104/110

Indicate the Dynamic Response

For a, b, c, and d positive constants, transfer

function indicates exponential decay, oscillatory

response, and exponential growth, respectively.

)()()()(

2 ds

C

cbss

B

as

A

sY

dtptat eCteBeAty )sin()(

)()()(

1)(

2 dscbssassG

Poles on a Complex Plane


105/110

p

Re

Im

Exponential Decay


106/110

p y

Re

Im

Time

y

Damped Sinusoidal


107/110

p

Re

Im

Time

y

Exponentially Growing Sinusoidal


108/110

p y g

Behavior (Unstable)

Re

Im

Time

y


109/110

Unstable Behavior


110/110

If the output of a process grows withoutbound for a bounded input, the process is

referred to a unstable.

If the real portion of any pole of a transferfunction is positive, the process

corresponding to the transfer function is

unstable.

If any pole is located in the right half plane,

the process is unstable.

laplace transforms1

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