Appendix A: Laplace Transforms

•  Important analytical method for solving linear ordinary differential equations.

- Application to nonlinear ODEs? Must linearize first.

•  Laplace transforms play a key role in important process control concepts and techniques.

- Examples:

•  Transfer functions

•  Frequency response

•  Control system design

•  Stability analysis

[ ] ( )0

( ) ( ) (3-1)stF s f t f t e dt∞ −= = ∫L

Definition The Laplace transform of a function, f(t), is defined as:

where F(s) is the symbol for the Laplace transform, L is the Laplace transform operator, and f(t) is some function of time, t.

Note: The L operator transforms a time domain function f(t) into an s domain function, F(s). s is a complex variable: s = a + bj, 1j ≡ −

Inverse Laplace Transform

By definition, the inverse Laplace transform operator, L-1, converts an s-domain function back to the corresponding time domain function:

( ) ( )1f t F s− " #= $ %L

Important Properties:

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

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

ax t by t a x t b y t

aX s bY s

! " ! " ! "+ = +# $ # $ # $

= +

L L L

where:

- x(t) and y(t) are arbitrary functions

- a and b are constants

- X s( ) L x t( )!"

#$andY s( ) L y t( )!

"#$

Similarly,

( ) ( ) ( ) ( )1 aX s bY s ax t by t− " #+ = +$ %L

Laplace Transforms of Common Functions

1.  Constant Function

Let f(t) = a (a constant). Then from the definition of the Laplace transform in (3-1),

( )0

0

0 (3-4)st sta a aa ae dt es s s

∞∞ − − # $= = − = − − =% &

' (∫L

2.  Step Function

The unit step function is widely used in the analysis of process control problems. It is defined as:

S t( ) 0 for t < 01 for t ≥ 0

"#$

%$(3-5)

Because the step function is a special case of a “constant”, it follows from (3-4) that

( ) 1 (3-6)S ts

! " =# $L

3.  Derivatives

This is a very important transform because derivatives appear in the ODEs we wish to solve. In the text (p.53), it is shown that

( ) ( )0 (3-9)df sF s fdt! " = −$ %& '

L

initial condition at t = 0

'0

0

00

00

( )[ ( )] ( ) ( )

( ) ( )( )

( ) ( )

0 (0) [ ( )]

stst

st st

st st

d eL f t f t e f t dtdt

f t e f t s e dt

f t e s f t e dt

f sL f t

∞ −− ∞

∞− ∞ −

∞− ∞ −

= −

= − −

= +

= − +

∫

∫

∫

Proof:

2nd order derivatives

Ld fdt

Lddt

( ) ( )2

2 =Φ

Φ = f t' ( )Where

use the result for the 1st-order derivative

Lddt

s s( ) ( ) ( )Φ

Φ Φ= − 0

Φ( ) ( ) ( )s sF s f= − 0

Ld fdt

s sF s f( ) [ ( ) ( )] ( )2

2 0 0= − −Φ

Ld fdt

s F s sf f( ) ( ) ( ) ( )'2

22 0 0= − −

f k( ) 0( ) dk fdtk t=0

( )n

nn

d f s F sdt

! "=# $

# $% &L

Similarly, for higher order derivatives:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

11 2

2 1

0 0

... 0 0 (3-14)

nn n n

n

n n

d f s F s s f s fdt

sf f

− −

− −

" #= − − −$ %

$ %& '

− − −

L

where: - n is an arbitrary positive integer

-

Special Case: When all initial Conditions are Zero, then:

In process control problems, we often assume zero initial conditions. Reason: This corresponds to the nominal steady state when “deviation variables” are used, as shown in Ch. 4.

4.  Exponential Functions

Consider where b > 0. Then, ( ) btf t e−=

( )

( )

0 0

0

1 1 (3-16)

b s tbt bt st

b s t

e e e dt e dt

eb s s b

∞ ∞ − +− − −

∞− +

# $ = =% &

# $= − =% &+ +

∫ ∫L

cos( ) ( )wt e ejwt jwt= + −12

0 0

1[cos( )] [

2jwt st jwt stL wt e e dt e e dt

∞ ∞− −= +∫ ∫

2 21 1 1 1 2[ ]2 2

ss jw s jw s w

= + =− + +

L wts

s w[cos( )] =

+2 2

5. Trigonometric Functions

Find L[sin(wt)] by yourself

h

( )f t

wt Time, t

The Laplace transform of the rectangular pulse is given by

( ) ( )1 (3-22)wt shF s es

−= −

6.  Rectangular Pulse Function

It is defined by:

( )0 for 0for 0

0 forw

w

tf t h t t

t t

<!"

= ≤ <$" ≥&

00 0

1[ ( )] 0 ( )w w w

w

t t tst st st st

t

L f t h e dt e dt h e dt h es

∞− − − −= ⋅ + ⋅ = ⋅ = −∫ ∫ ∫

L f ths

e t sw[ ( )] [ ]= − −1

7.  Impulse Function (or Dirac Delta Function) The impulse function is obtained by taking the limit of the rectangular pulse as its width, tw, goes to zero but holding the area under the pulse constant at one. (i.e., let )

Let,

Then,

1

wh

t=

δ t( ) impulse function

( ) 1tδ" # =$ %L

Solution of ODEs by Laplace Transforms Procedure:

1.  Take the L of both sides of the ODE.

2.  Rearrange the resulting algebraic equation in the s domain to solve for the L of the output variable, e.g., Y(s).

3.  Perform a partial fraction expansion.

4.  Use the L-1 to find y(t) from the expression for Y(s).

Table&A.1.&Laplace&Transforms&

See&page&A53&of&the&textbook.&

Example 3.1 Solve the ODE,

( )5 4 2 0 1 (3-26)dy y ydt+ = =

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

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

− + =

Rearrange,

( )( )5 2 (3-34)5 4sY s

s s+

=+

Take L-1, ( )

( )1 5 2

5 4sy t

s s− " #+

= $ %+& '

L

From Table 3.1,

( ) 0.80.5 0.5 (3-37)ty t e−= +

Partial Fraction Expansions

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).

Example:

( )( )( )

5 (3-41)1 4sY s

s s+

=+ +

Perform a partial fraction expansion (PFE)

( )( )1 25 (3-42)

1 4 1 4s

s s s sα α+

= ++ + + +

where coefficients and have to be determined. 1α 2α

2α

To find : Multiply both sides by s + 1 and let s = -1 1α

11

5 44 3s

ss

α=−

+∴ = =

+

To find : Multiply both sides by s + 4 and let s = -4

24

5 11 3s

ss

α=−

+∴ = = −

+

A General PFE Consider a general expression,

( ) ( )( )

( )

( )1

(3-46a)ni

i

N s N sY s

D s s bπ=

= =+

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:

( )is b= −

( ) ( )

( ) 11

(3-46b)n

in

iii

i

N sY s

s bs b

α

π ==

= =+

+∑

Note: D(s) is called the “characteristic polynomial”.

Heaviside expansion method

αi i s bs bN sD s i

= + =−( )( )( )

Where:

Example 3.2 (continued)

Recall that the ODE, , with zero initial conditions resulted in the expression

( )( )3 2

1 (3-40)6 11 6

Y ss s s s

=+ + +

The denominator can be factored as

( ) ( )( )( )3 26 11 6 1 2 3 (3-50)s s s s s s s s+ + + = + + +

Note: Normally, numerical techniques are required in order to calculate the roots.

The PFE for (3-40) is

( )( )( )( )

31 2 41 (3-51)1 2 3 1 2 3

Y ss s s s s s s s

αα α α= = + + +

+ + + + + +

y ++6y +11 y +6y =1

Solve for coefficients to get

1 2 3 41 1 1 1, , ,6 2 2 6

α α α α= = − = = −

(For example, find , by multiplying both sides by s and then setting s = 0.)

1/6 1/ 2 1/ 2 1/ 6( )1 2 3

Y ss s s s

= − + ++ + +

Take L-1 of both sides:

( )1 1 1 1 11/ 6 1/ 2 1/ 2 1/ 61 2 3

Y ss s s s

− − − − −" # " # " # " #" # = − + +$ % & ' & ' & ' & '+ + +$ % $ % $ % $ %L L L L L

Substitute numerical values into (3-51):

From Table 3.1,

( ) 2 31 1 1 1 (3-52)6 2 2 6

t t ty t e e e− − −= − + −

α

General Procedure for Solving Differential Equations

Special Situations: Two other types of situations commonly occur when D(s) has:

i)  Complex roots: e.g., ii) Repeated roots (e.g., )

( )3 4 1ib j j= ± −@1 2 3b b= = −

Repeated Factors (different from the text)

When the denominator D(S) has a term (s+b) occurring r times, r number of terms must be included in the expansion.

Y sN s

s b s bi

k

ir

( )( )

[ ( )]( )=

+ +=1Π

Where: n=k+r

Y ss b s b s b s b

i

ii

kr

r( )( ) ( ) ( )

=+

++

++

+ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅++=

∑α β β β1 2

21

αi �can be obtained by Heaviside method as before. Heaviside rule has to be modified to determine β j

( )s b r+Let both sides multiplied by

1

1 21 2

( ) ( ) ( )( ) ( )

( ) ( )

kr r i

i ir r

r

N s s b s bD s s b

s b s b

α

β β β=

− −

+ = + ⋅+

+ + + + + ⋅ ⋅ ⋅+

∑

Q sN sD s

s b r( )( )( )( )= +Let

rrrr

k

i i

ir

bsbsbsbs

bssQ

ββββ

α

+++⋅⋅⋅+++++

+⋅+=

−−−

=∑

)()()(

])(

[)()(

12

21

1

1

s b= −Let

βr s brs bQ s

N sD s

s b= = +=− =−( )( )( )( )

Take derivative with respect to s to both sides of the above eqn.

0))(2()()1(

])(

[)()(

13

22

1

1

++⋅⋅⋅++−++−+

#$%

&'(

+⋅+

=

−−−

=∑

rrr

k

i i

ir

bsrbsrds

bsbsd

dssdQ

βββ

α

s b= −Let

βr s b

dQ sds− =−=1( )

Similarly: ( )

( )1!

( )l

r l ls b

d Ql

sds

β −

=−

=

Therefore βj can be determined by:

( )

( )1

( )!( )r j

j r js b

d Q sdsr j

β−

−=−

−=

,

j r= ⋅ ⋅ ⋅ ⋅1

Repeated factor - example

Y ss

s s( )

( )( )

=++12 2

Y ss s s

( )( )

= ++

++

α β β1 1 222 2

α1 2 0

12

14

= ⋅+

+==s

ss s s

( )( )

21)2(

)2()1(

22

22 =+⋅++

= −=sssss

β

22

1 2 2 22

( 1) 1( 2)( 2) 1 1( )

4s s s

s sd s ds s sds ds s

β =− =− =−

# $+ +# $⋅ +& ' & '+( ) ( )= = = − = −

2)2(21

241

41

)(+

++

−+=

ssssY

21

)!12(41

41)(

2122 ×

−+⋅−=

−−−

tt etety

tt teety 22

21

41

41)( −− +⋅−=

Complex Factors Another possibility is that the factor of the characteristic polynomial involves complex numbers.

[ ]Y s

N sD s

N s

s b s b s d s di

k

ir

( )( )( )

( )

( ) ( ) ( )= =

+ + + +=Π1

21 0

dd1

2

04<Where

( ) ( )( )s d s d s a jw s a jw21 0+ + = + + + −

,

ad

= 1

2 w dd

= −012

4where and

Y(s) can be expanded into:

Y sN sD s

s b

s b s b s bc jcs a jw

c jcs a jw

i

ki

i

rr

r i r i

( )( )( )

( ) ( )

=

=+

++

++

+ ⋅ ⋅ ⋅++

++

+ ++

−+ −

=∑1

12

21

α

ββ β

cr and ci �can be determined by the modified Heaviside rule as well.

Multiply (s+a-jw) for the both sides

N sD s

s a jw s a jws b s b

i

i

jj

j

r

i

k( )( )

( ) ( )[( )

]⋅ + − = + −+

++==

∑∑α β

11

irir jccjwasjwasjcc

−+−+++

++ )(

)(

s a jw= − +Let

c jcN sD s

s a jwr i s a jw− = + − =− +

( )( )( )

cN sD s

s a jwr s a jw= + −"#$

%&'

=− +Re [( )( )( )]

cN sD s

s a jwi s a jw= − + −"#$

%&'

=− +Im [( )( )( )]

Therefore

Alternative: (complex factor)

[ ][ ]

( ) ( )

Y sN sD s

N s

s b s b s a w

s b s b s b s bc s cs a w

i

k

ir

i

ki

i

rr

( )( )( )

( )

( ) ( ) ( )

...

( )

=

=+!

"#$%&

+ + +

=+

'()

*+,+

++

++ +

+

'()

*+,

++

+ +

=

=∑

Π1

2 2

1

1 22

1 22 2

α β β β

Multiply both sides by [ ][ ]Πi

k

irs b s b s a w

=+"

#$%&'

+ + +()*

+,-1

2 2( ) ( ) ( )

collect all the terms with the same order, and solve for all the coefficients

1 2 1 2 12 2 2 2 2 2

( ) ( )( ) ... ...( ) ( ) ( ) ( )N s c s c c s a c acY sD s s a w s a w s a w

! "+ + −= = + = + +$ %

+ + + + + +& '

Example: find y(t)

Y ss s

( )( )[( ) ]

=+ + +

12 1 12

Y ss

c ics i

c ics i

r i r i( ) =+

++

+ ++

−

+ −

α12 1 1

α1 2 2 222

2 1 112

= ⋅ +#

$%

&

'( =

++ + +

==− =−

N sD s

ss

s ss s( )( )

( )( )[( ) ]

1 1

2

( ) 1( 1 ) [ ]( ) ( 2)( 1 )

1 1 1 1 1 1 1(1 )( 1 1 ) (1 )(2 ) 2 2 2 1 2 1 ( )1 1 14 4 4

r i s i s i

r i

N sc ic s iD s s s i

ii i i i i i i ii i

c ic

=− + =− +− = + − =+ + +

+= = = = − = −

+ − + + + + − − −

− −= = − −

= −

1 1 1

2 (1 ) (1 ) 2

2 2

1 1 1(1 ) (1 )2 4 4( ) ( )2 1 1

1 1 1 1 1 1(1 ) (1 ) (1 ) (1 )

2 4 4 2 4 41 1 1 1(1 ) (1 ) (2 4 2 4

t i t i t t t it t it

t t it it t t i

i iy t L L L

s s i s i

e i e i e e i e e i e e

e e i e i e e e e

− − −

− − + − − − − − −

− − − − − −

" # " #− − − +$ % $ %= + +$ % $ %+ + + + −$ % $ %

& ' & '

− − − + = − − − +

( )− − + + = −* +

���

���

[ ]2

) ( )

1 1 cos( ) sin( )2 2

t it it it

t t

e i e e

e e t t

−

− −

( )+ − −* +

− −���

,

cr = −14 ci =

14

Therefore: and

1 1 1(1 ) (1 )2 4 4( )2 1 1

i iY s

s s i s i

− − − += + +

+ + + + −

Alternative method

1 1 22 2

21 1 2

2

1( )( 2)[( 1) 1] 2 ( 1) 1[ 2 2] ( 2)( )

( 2)[( 1) 1]

c s cY ss s s ss s s c s c

s s

α

α

+= = +

+ + + + + +

+ + + + +=

+ + +

s c21 1 0: α + = α1

12

=

s c c: 2 2 01 1 2α + + =➜� c112

= −

s c01 22 2 1: α + = c2 0=

Y ss

ss s

( )( ) ( )

=+

−+

+ ++

+ +

1212

12

11 1

12

11 12 2

y t e e t e tt t t( ) cos sin= − +− − −12

12

12

2

Same result as before

Important Properties of Laplace Transforms

1. Final Value Theorem

It can be used to find the steady-state value of a closed loop system (providing that a steady-state value exists.

Statement of FVT:

( ) ( )0

limlimt s

sY sy t→∞ →

# $= % &

providing that the limit exists (is finite) for all where Re (s) denotes the real part of complex

variable, s. ( )Re 0,s ≥

Proof:

Laplace transform for the 1st-order derivative

)0()(][0

yssYdtedtdy

dtdyL st −== ∫

∞−

\______/ \_______/ (1) (2)

Take limit for terms (1) and (2), as s! 0

lim lim[ ( ) ( )]s

st

s

dydte dt sY s y

→

−∞

→∫ = −00

00

[ ])0()()( lim0

0ySsYty

s−=

>−

∞

y y sY s ys

( ) ( ) lim ( ) ( )∞ − = −→

0 00

y sY ss

( ) lim ( )∞ =→0

Example:

Suppose,

( )( )5 2 (3-34)5 4sY s

s s+

=+

Then,

( ) ( )0

5 2lim 0.5lim 5 4t s

sy y ts→∞ →

+# $∞ = = =% &+' (

You may use FVT to final steady value without doing the inverse Laplace transform; also, it may be used to check if an inversion is correctly performed.

2. Initial Value Theorem (IVT) Similar to the final value theorem, initial value theorem (IVT) can be stated as:

y sy ss

( ) lim[ ( )]0 =→∞

Prof.: dydte dt sy s yst−

∞

∫ = −0

0( ) ( )

take the limit for s→ ∞

lim lim[ ( ) ( )]s

st

s

dydte dt sy s y

→∞

−∞

→∞∫ = −0

0

y sY ss

( ) lim[ ( )]0 =→∞

Example: y ss

s s( )

( )=

+

+

5 25 4

Find y(0)?

y ss

s ssss s

( ) lim[( )

] lim05 25 4

5 25 4

1= ⋅+

+=

+

+=

→∞ →∞

3. Transform of an integral

It is also some times necessary to find Laplace transform of an integral.

L f t dt f t dt e dtt

stt

[ ( ) ] { ( ) }* * * *

0 00∫ ∫∫= −

∞

use integration-by-part udv uv vdu∫ ∫= −

= − ⋅−∞

∫∫1

00sf t dt

d edt

stt

{ ( ) }( )* *

* *

* * 00

0 0

( )1 1[( ( ) ) ]

t

tst st

d f t dtf t dt e e

s s dt

∞∞− −

# $% %# $ % %

= − +& ' & '( ) % %

% %( )

∫∫ ∫

0 0

1 10 0 ( ) ( )st ste f t dt f t e dts s

∞ ∞− −# $

= − − − =% &' (

∫ ∫

1 ( )F ss

=

4. Time Delay

Time delays occur due to fluid flow, time required to do an analysis (e.g., gas chromatograph). The delayed signal can be represented as

( )θ θ time delayy t − =Also,

( ) ( )θθ sy t e Y s−" #− =$ %L

Motor

Heater

V

Ti=0 w

TC 1

Q

TC 2 At time t = 0, heater turns on delay may also be represented by td.

( )( )

0d d

dd

f t t t tf t

t t− ≥#

= $<%

( ) ( ) ( )d d df t f t t s t t= − ⋅ −

or

0 0

[( ( )] ( ) 0 ( )d

d

tst st st

d d dt

L f t f t e dt e dt f t t e dt∞ ∞

− − −= = ⋅ + −∫ ∫ ∫

= − −∞

∫ f t t e dtdst

td

( )

Let t t td* = −

* ** * * *

0 0

[( ( )] ( ) ( ) ( )d d dst st stst stdL f t f t e e dt e f t e dt e F s

∞ ∞− − −− −= = =∫ ∫

( ) [ ( ) ( )] ( )dt sd d dF s L f t t s t t e F s−= − − =

21( )(4 1)(3 1)

seY ss s

−+=

+ +

How to find fd(t) from ( ) ( )dt sdF s e F s−=

Two steps: (1)  Find f(t)

(2)  Replace f(t) by ( ) ( ) ( )d d df t f t t s t t= − −

Example exercise

Find y(t)? (do it by yourself)