PC. Lecture 4

8/7/2019 PC. Lecture 4

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LAPLACETRANSFORMS

Chapter 3



Learning Outcomes

Upon the completion of this chapter, students are able

to:

Convert Ordinary Differential Equation (ODE) to

algebraic equations using Laplace Transform



1. Standard notation in dynamics and control (shorthand

notation)

2. Converts ordinary differential equation to algebraic operations

3. Laplace transforms play a key role in important process

control concepts and techniques. Examples:

• Transfer functions•

Frequency response• Control system design• Stability analysis

Laplace Transforms (LT)



Definitions and Properties of

LT

Laplace Transform

Inverse Laplace Transform

Both L and L-1 are linear operators. Thus,

Similarly,

∫ ∞

=0

L dt f (t)e(f (t))= F(s) -st

( ) ( )1 f t F s− = L

( ) ( ) ( ) ( )( ) ( ) (3-3)

ax t by t a x t b y t

aX s bY s + = + = +

L L L

( ) ( ) ( ) ( )1

aX s bY s ax t by t

−

+ = + L



-st st

00

-bt -bt -st -(b+s)t ( s)t

00 0

-st

0

a a aL(a)= ae dt e 0

s s s

1 1L(e )= e e dt e dt -e b+s s+b

df df L(f ) L e dt

dt dt

b

∞ ∞−

∞ ∞ ∞− +

∞

= − = − − =

= = =

′ = = =

∫

∫ ∫

∫ Usually define f(0) = 0 (e.g., the error)

f(0)-sF(s)f(0)sL(f) =−

Exponential Function

Constant

Function

Derivative

Function

LT of Common Functions

Higher order

Derivative

Function

(0)f (0)sf

(0)f sf(0)sF(s)sdt

f dL

1)(n2)(n

(1)2n1nn

n

n

−−

−−

−−

−−−=



Rectangular Pulse

Function

h

( ) f t

wt

Time, t

Step Function

( )1

(3-6)S t s

= L

≥<

= 0 1

0 0)( t for

t for t S

( )

0 for 0

for 0 (3-20)0 for

w

w

t

f t h t t t t

<

= ≤ < ≥

( ) ( )1 (3-22)wt sh

F s e s

−= −



Table 3.1 Laplace Transforms for Various Time-Domain Functionsa



The general procedure

for solving ODE using LT



Solve the ODE,

( )5 4 2 0 1 (3-26)dy

y ydt

+ = =

First, take L of both sides of (3-26),

( )( ) ( ) 25 1 4 sY s Y s s

− + =

Rearrange, ( )( )

5 2(3-34)

5 4

sY s

s s

+=

+

Take L-1, ( )( )

1 5 25 4 s y t

s s− += +

L

From Table 3.1,

( )

0.8

0.5 0.5 (3-37)

t

y t e

−

= +

Example 3.1



Partial Fraction Expansion

(PFE)

Basic idea: Expand a complex expression for Y (s) into simpler

terms, each of which appears in the Laplace Transform table. Then

you can take the L-1 of both sides of the equation to obtain y (t ).

1. Unrepeated real factors

Here D(s) is an n-th order polynomial with the roots all being real

numbers which are distinct so there are no repeated roots.

The PFE is:

( )( )

( )

( )

( )1

(3-46a)n

ii

N s N sY s

D s s bπ

=

= =+

( )( )

( ) 11

(3-46b)n

in

iiii

N sY s

s b s b

α

π ==

= =+

+

∑



Step 1 Take L.T. (note zero initial conditions)

3 2

3 26 11 6 4

0 0 0 0

d y d y dy y

dt dt dt

y( )= y ( )= y ( )=

+ + + =

′ ′′

Example 3.2

3 2 46 11 6 ( ) s Y(s)+ s Y(s)+ sY(s) Y s =

s+

3 2

4

( 6 11 6)Y(s)= s s s s+ + +

Rearranging

Step 2a. Factor denominator of Y(s)

) )(s+ )(s+ )=s(s+ s++ s+ s(s 3216116 23



Step 2b. Use partial fraction decomposition

Multiply by s, set s = 0

31 2 44

1 2 3 1 2 3

αα α α

s(s+ )(s+ )(s+ ) s s s s= + + +

+ + +

32 41

00

1

4

1 2 3 1 2 34 2

1 2 3 3

s s

αα αα s

(s+ )(s+ )(s+ ) s s s

α

==

= + + + + + +

= =⋅ ⋅

For α 2

, multiply by (s+1), set s=-1 (same procedure for α 3, α 4

)

2 3 4

22 2

3α , α , α=− = =−



2 32 22 2

3 3

2 0 (0) 0.3

t t t y(t)= e e e

t y(t) t y

− − −− + −

→ ∞ → = =

Step 3. Take inverse of L.T.

2 2 2 2 / 3( + )3 1 2 3Y(s) = s s s s

− −+ + +

2. Repeated factors (see page 61&62)3. Unreal factors (see page 63-65)



A. Final value theorem

Example:Example:

B. Time-shift theorem

sY(s) )= y( s 0lim

→∞

Important Properties of LT

( )( )

5 2(3-34)

5 4

sY s

s s

+=+

( ) ( )0

5 2lim 0.5lim

5 4t

s

s y y t

s→∞

→

+ ∞ = = = +

( )θ θ time delay y t − =

( ) ( )θ

θs

y t e Y s−

− = L



C. Initial value theorem

by initial value theorem

by final value theorem

sY(s) y(t)= st ∞→→limlim

0

) )(s+ )(s+ s(s+

s+= For Y(s)

321

24

00 )= y(

3

1 )= y( ∞

Example:Example:



17

Transfer Functions

( )

( )

( )

( )system

x t y t

X s Y s→ →

Transfer Functions•The transfer function (TF) of a system is defined as the

Laplace transform of the output variables ,Y(t), divided by Laplace transform of the input variable ,X(t), with all

initial conditions equal to zero.

•Transfer function can be derived only for linear differential equation model .

)(

)()(

s X

sY sGnctionTransferFu ==



Development of Transfer

FunctionsCha

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19

Figure 2.3 Stirred-tank heating process with constant

holdup, V .

Development of Transfer Functions

Example: Stirred Tank Heating System



20

Recall the previous dynamic model, assuming constant

liquid holdup and flow rates:

( ) (1)i

dT

V C wC T T Qdt ρ = − +

( ) ( ) ( ) ( )0 , 0 , 0 2i iT T T T Q Q= = =

Suppose the process is initially at steady state:

where steady-state value of T, etc. For steady-state

conditions:T @

( )0 (3)iwC T T Q= − +

Subtract (3) from (1):

( ) ( ) ( ) (4)i i

dT V C wC T T T T Q Q

dt ρ = − − − + −



21

But,

( ) because is a constant (5)

d T T dT T

dt dt

−=

Thus we can substitute into (4-2) to get,

( ) (6)i

dT V C wC T T Q

dt ρ

′′ ′ ′= − +

where we have introduced the following “deviation

variables”, also called “perturbation variables”:

, , (7)i i iT T T T T T Q Q Q′ ′ ′− − −@ @ @

( ) ( ) ( ) ( ) ( )0 (8)iV C sT s T t wC T s T s Q s ρ ′ ′ ′ ′ ′ − = = − −

Take L of (6):



22

Evaluate ( )0 .T t ′ =

By definition, Thus at time, t = 0,.T T T ′ −@

( ) ( )0 0 (9)T T T ′ = −

But since our assumed initial condition was that the

process was initially at steady state, i.e., it

follows from (9) that

Note: The advantage of using deviation variables is that

the initial condition term becomes zero. This simplifies

the later analysis.

( )0T T =( )0 0.T ′ =

Rearrange (8) to solve for

( ) ( ) ( )1

(10)1 1

i

K T s Q s T s

s sτ τ

′ ′ ′= + + +

( ) :T s′



23

where two new symbols are defined:

( )1

and 11V

K wC w

ρ τ @ @

Transfer Function Between andQ′ T ′

Suppose is constant at the steady-state value. Then,

Then we can substitute into(10) and rearrange to get the desired TF:

iT

( ) ( ) ( )0 0.i i i iT t T T t T s′ ′= ⇒ = ⇒ =

( )

( ) (12)1

T s K

Q s sτ

′

=′ +



24

Thus, rearranging

( )( )

1(13)

1i

T s

T s sτ

′ =′ +

Transfer Function Between andT ′ :iT ′

Suppose that Q is constant at its steady-state value:

( ) ( ) ( )0 0Q t Q Q t Q s′ ′= ⇒ = ⇒ =



25

Properties of Transfer Function

Models

1. Steady-State Gain

The steady-state of a TF can be used to calculate the

steady-state change in an output due to a steady-state

change in the input. For example, suppose we know two

steady states for an input, u , and an output, y . Then wecan calculate the steady-state gain, K , from:

2 1

2 1

(4-38) y y

K

u u

−=

−

For a linear system, K is a constant. But for a nonlinear

system, K will depend on the operating condition ( ), .u y



26

Calculation of K from the TF Model:

If a TF model has a steady-state gain, then:

( )0

lim (14) s

K G s→

=

• This important result is a consequence of the Final

Value Theorem

• Note: Some TF models do not have a steady-state gain

(e.g., integrating process in Ch. 5)



27

2. Order of a TF Model

Consider a general n-th order, linear ODE:

1

1 1 01

1

1 1 01(4-39)

n n m

n n mn n m

m

m m

d y dy dy d ua a a a y bdt dt dt dt

d u dub b b u

dt dt

−

− −

−

− −

+ + + = +

+ + +

K

K

Take L, assuming the initial conditions are all zero.

Rearranging gives the TF:

( )( )

( )0

0

(4-40)

mi

iin

ii

i

b sY sG s

U sa s

=

=

= = ∑∑



3. Two IMPORTANT properties (L.T.)

A. Multiplicative Rule

B. Additive Rule

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Example(4.4): Surge tank in series



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Linearization of Nonlinear Models

•So far, we have emphasized linear models which can betransformed into TF models.

• But most physical processes and physical models are nonlinear.

- But over a small range of operating conditions, the behavior

may be approximately linear.

- Conclude: Linear approximations can be useful, especially

for purpose of analysis.

• Approximate linear models can be obtained analytically by amethod called “linearization”. It is based on a Taylor Series

Expansion of a nonlinear function about a specified operating

point.

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• Consider a nonlinear dynamic model relating two process variables u and y:



• Consider a nonlinear, dynamic model relating two process variables, u and y:

( ), (4-60)dy

f y udt

=

Perform a Taylor Series Expansion about and and truncate after the

first order terms,u u= y y=

( ) ( ), , (4-61) y y

f f f u y f u y u yu y

∂ ∂′ ′= + +∂ ∂

where and . .Note that the partial derivative terms are

actually constants because they have been evaluated at the nominal operating point,

Substitute (4-61) into (4-60) gives:

u u u′ = − y y y′ = −

( ), .u y

( ), y y

dy f f f u y u y

dt u y

∂ ∂′ ′= + +∂ ∂

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The steady state version of (4 60) is:



The steady-state version of (4-60) is:

( )0 , f u y=

Substitute into (7) and recall that ,dy dy

dt dt

′=

(4-62) y y

dy f f u y

dt u y

′ ∂ ∂′ ′= +∂ ∂

Linearized model

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Summary:

In order to linearize a nonlinear, dynamic model:

1. Perform a Taylor Series Expansion of each nonlinear term and truncate after thefirst-order terms.

2. Subtract the steady-state version of the equation.

3. Introduce deviation variables.

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Example: Liquid Storage System (nonlinear)

Mass balance:

Valve relation:

A = area, C v = constant

(1)i

dh A q q

dt = −

(2)vq C h=

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More realistically, if q0 is manipulated by a flow

control valve,

hC q v=0

nonlinear element

C bi (1) d (2)



Combine (1) and (2),

(3)i v

dh A q C h

dt = −

Linearize term,

( )

1

(4)2h h h hh≈ − −Or

1

(5)h h h R ′≈ −

2 R h

h h h′ −

@

@

where:

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S b tit t li i d i (5) i t (3)



Substitute linearized expression (5) into (3):

1(6)i v

dh A q C h h

dt R

′= − −

The steady-state version of (3) is:

0 (7)i vq C h= −

Subtract (7) from (6) and let , noting that gives the

linearized model:

1(8)i

dh A q h

dt R′ ′ ′= −

i i iq q q′ −@ dh dh

dt dt

′=

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PC. Lecture 4

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