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Lecture 4
Laplace Transform: Motivation
Differential equations model dynamic systems
Control system design requires simple methods for solving these equations!
Laplace Transforms allow us to
– systematically solve linear time invariant (LTI) differential equations for arbitrary inputs.
– easily combine coupled differential equations into one equation.
– use with block diagrams to find representations for systems that are made up of smaller subsystems.
uxbxm =+⇔ &&&
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
1 2
Laplace Transforms:
Laplace transform of a time domain function f(t) is
(Doubled-sided Laplace transform)
where s = σ + jω is the complex variable.
In most engineering problems, f(t) is causal, i.e.
Thus, single-sided Laplace transform is used, i.e.
{ } ∫∞
∞−
−== ττ τdefsFtf
s)()()(L
0for 0)( <= ttf
{ } ∫∞
−==
0
)()()( ττ τdefsFtf
sL
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
Laplace Transform Example 1
Example:
Show that
Notation for “unit step”
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
3
Laplace Transform Example 1 contd..
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
4
Laplace Transform of a Unit Step
Find the Laplace Transform for the following function
∞≤≤
=otherwise0
01)(
ttu s
[ ]
s
s
es
dtesFstst
1
101
11)(
00
=
−−
=
−==
∞
−
∞
−
−−
∫
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
5
Example 2
Find the Laplace Transform for the following function
≤≤
=otherwise0
103)(
ttf
[ ]
[ ]s
s
stst
es
es
es
dtesF
−
−
−−
−=
−−
=
−==
−−
∫
13
13
33)(
1
0
1
0
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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The function with the simplest Laplace Transform (1)
A special input (class) has a very simple Laplace Transform
The impulse function:
– Has unit “energy”
– Is zero except at t=0
Think of pulse in the limit
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
( ) ( )( ) 1t
sFtf
δ
7
LT Properties: Scaling and Linearity
Proof: Both properties inherited from linearity of integration and
the Laplace Transform definition
( ) ( )( ) ( )
( ) ( ) ( ) ( )sFsFtftf
saFtaf
sFtf
2121 ++
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example 3
Find the following Laplace Transforms
Use Euler’s Formula
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
( )tjtjee
jt
ωωω −−=2
1sin
[ ]
22
11
2
1sin
ω
ω
ωωω
+=
+−
−=
s
jsjsjtL [ ]
22
11
2
1cos
ω
ωωω
+=
++
−=
s
s
jsjstL
( )tjtjeet
ωωω −+=2
1cos
( ) ( )
( )
( )22
22
cos
sin
ωω
ω
ωω
+
+
s
st
st
sFtf
9
LT Properties: Time and Frequency Shift
Proof of frequency shift: Combine exponentials
( ) ( )( ) ( ) ( )
( ) ( )σ
ττσ
τ
+
−−−
−
sFtfe
sFetutf
sFtf
t
s
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example 4
[ ]
22
22
)(
cos
ω
ωω
++
+=
+=
+=
−
as
as
s
ste
ass
atL
[ ]
22
22
)(
sin
ω
ω
ω
ωω
++=
+=
+=
−
as
ste
ass
atL
( ) ( )
( )( )
( )( ) 22
22
cos
sin
ωω
ω
ωω
++
+++
−
−
as
aste
aste
sFtf
at
at
Find the following Laplace Transforms
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
11
LT Properties: Integration & Differentiation
Proof of Differentiation Theorem: Integration by parts
∫∫ −= vduuvudv
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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LT Properties: Integration & Differentiation
( ) ( )
( ) ( ) ( )
( )( )
( )∫∫ ∞−∞−
−
+
−
01
0
ττττ dfss
sFdf
fssFtfdt
d
sFtf
t
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Some Examples:
[ ] [ ]2
111)(
ssstdtut === ∫LL
1)0(1)(
=−=
−u
ss
dt
tduL Impulse!
10
1
1)sin(22 +
=−+
=
s
s
ss
dt
tdL Cosine!
[ ]( )22
11
asste
ass
at
+==
+=
−L
( ) ( )
( )
( )
( )1
sin
1
1
1
2
2
2
+
+
−
s
st
dt
d
tudt
d
aste
st
sFtf
at
Derivative of a step?
Derivative of sine?
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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15
Properties of the Laplace Transforms
Let F(s) = LLLL {f(t)}.
(1) Time Differentiation
(2) Time Integration
(3) Complex Translation (Shifting in the s-domain)
in
inn
i
in
n
n
dt
fdssFstf
dt
d
−−
−−−
=∑−=
1
11
0
)0()()(L
n
t t t
ns
sFddf
n )()(
1 2
0 0 0
11 =
∫ ∫ ∫ τττ LLL
{ } )()( asFtfe at ±=mL
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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(4) Real Translation (Shifting in the time domain)
where u(t – T) is a shifted unit-step function.
(5) Real Convolution (Complex Multiplication)
(6) Initial Value Theorem
{ } )()()( sFeTtuTtfsT−=−−L
∫∫ −=−=∗
tt
dtffdtfftftf0
12
0
2121 )()( )()()()( ττττττ
{ } )()()()( 2121 sFsFtftf =∗L
)(lim)0()(lim0
ssFftfst ∞→→
==
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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(7) Final Value Theorem
provided sF(s) is analytic in the closed right-half of the s–plane
(i.e. the denominator of sF(s) has no roots on the jω−axis or in the
right-half of the s–plane).
Example: Consider
i.e.
and hence So, yss = 1.
Alternatively, by final value theorem, we have
)(lim)(lim0
ssFtfst →∞→
=
ssR
sssT
sR
sY 1)( ;
)2)(1(
2)(
)(
)(=
++==
2( ) 1 2 .
t ty t e e
− −= + −
1)(lim)(lim0
===→∞→
ssYtyyst
ss
2 1 2 1( )
( 1)( 2) 1 2=Y s
s s s s s s= − +
+ + + +
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example:
In this case, denominator of F(s) has 2 roots on the jω−axis, so the
f.v.t. cannot be used. (If we applied the f.v.t., it gives a final value of
zero which is wrong!)
ttfs
sF ωω
ωsin)( i.e )(
22=
+=
Example:
Clearly, sF(s) is analytic in the closed right-half of the s–plane, so
the f.v.t. can be applied to give
)2(
5)(
2 ++=
ssssF
2
5)(lim)(lim
0===
→∞→ssFtff
stss
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
Basic Laplace Transform Pairs
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Inverse Laplace Transform
1σ• is such that the integral is taken over a line in the region
of convergence
• Very difficult to apply directly
• Instead, we convert X(s) to a form such that we can easily
find the inverse
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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The differentiation theorem
Higher order derivatives
Laplace Differentiation Theorem
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Differentiation Theorem
Differentiation Theorem when initial conditions are zero
( ) ( ) ( ) ( ) ( )−
−
−−−−− −−−−↔ 000
1
121 f
dt
df
dt
dsfssFstf
dt
dn
nnnn
n
n
K
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Solving Differential Equations: an Example
Consider0,1 ≥= t
dt
dx
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Solving Differential Equations: an Example
Solution Summary
– Use differentiation theorem to take Laplace Transform of the
differential equation
– Solve for the unknown Laplace Transform Function
– Find the inverse Laplace Transform
( ) ( ) 0,0 ≥+= − txttx
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example 5
Find the Laplace Transform for the solution to
( ) ( ) ( ) ( )( ) ( )( )
( ) ( ) ( ) ( ) ( )[ ] ( ) 120300
00...0
2
121
=+−+−−⇒
−−−−=
−−−
sXxssXxsxsXs
fsffssFstfdt
dL
nnnn
n
n
&
Notation:
( ) ( ) 30,100,123 −==≥=++ xxtxxx &&&&
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Partial Fraction Expansions
In general, ODEs can be transformed into a function that is expressed
as a ratio of polynomials
In a partial fraction expansion we try to break it into its parts, so we can
use a table to go back to the time domain:
Three ways of finding coefficients
– Put partial fraction expansion over common denominator and
equate coefficients of s (Example 1)
– Residue formula
– Equate both sides for several values of s (not covered)
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Partial Fraction Expansions
Have to consider that in general we can encounter:
– Real, distinct roots
– Real repeated roots
– Complex conjugate pair roots (2nd order terms)
– Repeated complex conjugate roots
222
2
222
11 ))(()()()(
)()(
bas
GFsEs
bas
DCs
ps
B
ps
AK
sD
sNsX
++
+++
++
++
++
++==
Dr. Kalyana Veluvolu
Example 6
Given X(s), find x(t).
Step 1:
Factorize the denominator, then use partial fraction expansion:
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Step 2: Finding A, B, and C
To solve, re-combine RHS and equate numerator coefficients (“Equate
coefficients” method)
Example 6 contd..Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
30
Finally,
Since
By inspection,
α
α
+↔≥−
s
KtKe
t 0,
( ) 0,2
52
2
1 2 ≥+−= −−teetx
tt
Example 6 contd..
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Residue Formula (1)
The residue formula allows us to find one coefficient at a time by
multiplying both sides of the equation by the appropriate factor.
Returning to Example 1:
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Residue Formula (2)
For Laplace Transform with non-repeating roots,
The general residue formula is:
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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)2(
1
)1(
2)(
+
−+
+=
sssX 0,2)( 2 ≥−= −− teetx tt
( ) ( ) ( )( )
( ) 0)(21)(3)(
0)(20)(300)(
2
2
=+−+−
=+−+−−
sXssXssXs
sXxssXxsxsXs &
3)()23( 2 +=++ ssXss
)2()1()2)(1(
3)(
++
+=
++
+=
s
B
s
A
ss
ssX
( ) ( ) 221
311
1=
+−
+−=+=
−=ssXsA ( ) ( ) 1
12
322
2−=
+−
+−=+=
−=ssXsB
Example 7: Find the solution to the following differential equation:
0)0( 1(0) 023 ===++ xxxxx &&&&
Solution:
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Inverse Laplace Transform with Repeated Roots
Now we will consider partial fraction expansion rules for functions
with repeated (real) roots:
– # of constants = order of repeated roots
Example:
23223
4
)3()3(3)3(
)1(
s
E
s
D
s
C
s
B
s
A
ss
s++
++
++
+=
+
+
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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1)3()3(3)1()3(
)1(23223
4
−+++
++
++
+=
−+
+
s
F
s
E
s
D
s
C
s
B
s
A
sss
s
The easiest way to take an inverse Laplace transform is to use a table
of Laplace transform pairs.
Repeated real roots in Laplace transform table
( )
( )222
222
1
2
)(
)(2)sin(
2)sin(
)(
1
!
)(
1
)()(
ω
ωω
ω
ωω
++
++
+
+
−
+
−
−
as
astte
s
stt
asn
et
aste
sFtf
at
n
atn
at
Repeated Real Roots
Repeated Imaginary Roots
(also use cosine term)
Repeated Complex Roots
(also use cosine term)
}
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example with repeated roots
Find x(t)
Take Laplace Transform of both sides:
tef
2−=
21
1x
( ) ( ) 10,00
2 2
==
=++ −
xx
exxx t
&
&&&
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example with repeated roots
Terms with repeated roots:
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example with repeated roots
C = 1B = 2
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example
Find the solution to the following differential equation
0)0(1)0(044 ===++ xxxxx &&&&
( ) ( ) ( ) ( ) ( )[ ] ( )
( ) ( ) ( )
( )[ ]
( )( ) ( )
( ) ( ) ( )
( )( )( )
( ) tt
ss
teetx
Ass
sA
s
s
ssXsB
s
B
s
A
s
ssX
ssssX
sXssXssXs
sXxssXxsxsXs
22
22
22
2
22
2
2
2
2
1:2
22
2
4
242
222
4
444
04440
040400
−−
−=−=
+=
=⇒+
++=
+
+
=+=+=
++
+=
+
+=
+=++
=+−+−−
=+−+−− &
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
40
Inverse Laplace Transform with Complex Roots
To simplify your algebra, don’t use first-order denominators such as
Instead, rename variables
So that
21 KKB += ( ) ( )( )21 11 KjKjC −++=
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
41
More Laplace transform pairs (complex roots):
Also, see the table in your textbook and most other control systems
textbooks.
Laplace Transform Pairs for Complex Roots
22
22
)()sin(
)()cos(
)()(
ωσ
ωω
ωσ
σω
σ
σ
++
++
+
−
−
ste
s
ste
sFtf
t
t
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
42
Example with complex roots
Example: find x(t)
Laplace Transform
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
43
Example with complex roots (2)
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example with complex roots (3)
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example with complex roots (4)
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
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Example
Find solution to the following differential equation
0)0(1)0(084 ===++ xxxxx &&&&
( ) ( ) ( ) ( ) ( )[ ] ( )
( ) ( ) ( )
( )
( )
( ) ( )
( ) ( ) ( )tetetx
ss
s
s
s
ss
ssX
sXssXssXs
sXxssXxsxsXs
tt 2sin2cos
22
2
22
2
42
4
84
4
0844
080400
22
2222
2
2
2
2
−− +=
+++
++
+=
++
+=
++
+=
=+−−−
=+−+−− &
Lecture 4
Laplace and Inverse Laplace
Dr. Kalyana Veluvolu
47
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