Chapter 3 Laplace Transform
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Transcript of Chapter 3 Laplace Transform
![Page 1: Chapter 3 Laplace Transform](https://reader038.fdocuments.in/reader038/viewer/2022102503/551d37be4979594b198b49a4/html5/thumbnails/1.jpg)
LAPLACE TRANSFORMS
Chapter 3
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Learning Outcomes
Upon the completion of this chapter, students are able to:Convert Ordinary Differential Equation, ODE equations to algebraic equations using Laplace Transform
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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)
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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.
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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
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-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
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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
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Table 3.1 Laplace Transforms for Various Time-Domain Functionsa
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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).
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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
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Example 3.2Solve the ordinary differential equation
With initial condition y(0)=y’(0)=y”(0)=0
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SolutionTake LTs,term by term,using Table 3.1
Rearranging and factoring out Y(s) ,we obtain
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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
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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.
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Method 1 : by multiply both side by
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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
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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.
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Example 3.2-continued
Solve for PFE
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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
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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
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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
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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
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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
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Exercise Using partial fraction expansion where required, find x(t) for
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=1
= -6
=6
SOLUTION:using PFE
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By doing Inverse Laplace for both side.
Final equation function of t is:
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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
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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
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Exercise: Using partial fraction expansion where required,find x(t) for Eq 1
Solution
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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
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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: