LAPLACE TRANSFORMS

Chapter 3

Learning Outcomes

Upon the completion of this chapter, students are able to:Convert Ordinary Differential Equation, ODE equations to algebraic equations using Laplace Transform

1. Standard notation in dynamics and control2. Converts ordinary differential equation to algebraic operations3. 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)

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.

Definitions and Properties of LT Laplace Transform

Inverse Laplace Transform

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

Similarly,

0

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

1f 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 b y 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 dfL(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)fsf(0)sF(s)sdt

fdL

1)(n2)(n

(1)2n1nnn

n

3.4

3.16

3.9

Rectangular Pulse Function

h

f t

wt

Time, t

Step Function

1(3-6)S t

s L

0 1

0 0)(

tfor

tfortS

0 for 0

for 0 (3-20)

0 forw

w

t

f t h t t

t t

1 (3-22)wt shF s e

s

Table 3.1 Laplace Transforms for Various Time-Domain Functionsa

9

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

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 s

s

Rearrange, 5 2

(3-34)5 4

sY s

s s

Take L-1, 1 5 2

5 4

sy t

s s

L

From Table 3.1,

0.80.5 0.5 (3-37)ty t e

Example 3.1

Example 3.2Solve the ordinary differential equation

With initial condition y(0)=y’(0)=y”(0)=0

SolutionTake LTs,term by term,using Table 3.1

Rearranging and factoring out Y(s) ,we obtain

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).

Consider a general PFE (purposely for no complex and repeated factors appear)

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 ss b

1

1

(3-46b)n

in

iii

i

N sY s

s bs b

Method 1 : by multiply both side by

Three methods to solve PFECalculate the value of for the below equation

Method 2:Because the above eq must be valid for all value if s,we can specify two values of s and solve for the two constant

Method 3:The fastest and most popular method is call HEAVISIDE EXPANSION.In this method multiply both side of the equation by one of the denominator term (s+bi) and then set s=-bi, which causes all terms except one to be multiplied by zero.

Method 1 : by multiply both side by

Method 2:Because the above eq must be valid for all value if s,we can specify two values of s and solve for the two constant

Method 3:The fastest and most popular method is call HEAVISIDE EXPANSION.In this method multiply both side of the equation by one of the denominator term (s+bi) and then set s=-bi, which causes all terms except one to be multiplied by zero.

Example 3.2-continued

Solve for PFE

20

Example 3.2 (continued)

Recall that the ODE, d3y/dt3 + 6d2y/dt2 + 11dy/dt + 6y = 1

, with zero initial conditions resulted in the expression

3 2

1(3-40)

6 11 6Y s

s 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

21

Solve for coefficients to get

1 2 3 41 1 1 1

, , ,6 2 2 6

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

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

1 2 3Y s

s s s s

Take L-1 of both sides:

1 1 1 1 11/ 6 1/ 2 1/ 2 1/ 6

1 2 3Y s

s 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 6t t ty t e e e

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

3 2

3 26 11 6 4

0 0 0 0

d y d y dyy

dt dt dty( )= y ( )= y ( )=

Example 3.3

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 3

4 2

1 2 3 3

ss

αα αα s

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

α

For 2, multiply by (s+1), set s=-1 (same procedure for 3, 4)

2 3 4

22 2

3α , α , α

2 32 22 2

3 32

0 (0) 0.3

t t ty(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

Exercise Using partial fraction expansion where required, find x(t) for

=1

= -6

=6

SOLUTION:using PFE

By doing Inverse Laplace for both side.

Final equation function of t is:

Repeated Factor

If a term s+b occurs r times in the denominators terms must be included in the expansion that incorporate the s+b factor

Example for repeated factor problem

Eq A

To find value of ,multiplying eq A by s and letting s=0 yields

To find value of ,substituting the values=-1 in eq A yields

To find value of ,multiplying eq A by and letting s=-2 yields

Exercise: Using partial fraction expansion where required,find x(t) for Eq 1

Solution

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

sy y t

s

θ θ time delayy t

θθ sy 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: