contsys1 l3 laplace trans
TRANSCRIPT
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Mechanical Engineering Science 8
Dr. Daniil Yurchenko
Laplace Transform
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Block Diagram
+
Y
Y
X
X
Closed-Loop
Open Loop
?
?
?
Here we consider systems described by ODEs
ya xadt
xd a
dt xd
a nn
nn
n
n
011
1
1 .......
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Definition of the Laplace
Transform For the given real f(t)
0)()( dt et f s F st
F(s) Laplace transform of f(t)
Capital letters Laplace transform f(t)
)]([)( t f L s F
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Examples of Laplace Transform
Unit Step Function
f(t)
t0 t,00 t,1)()( t ut f
s se
sdt e s F st st 1)1(011)(
00
st u L /1)]([
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Examples of Laplace Transform
Exponential Function 0,)( t t t f
20
200
00
10
1
1)(
s s
edt e
s s
te
tde s
dt te s F
st st
st
st st
1
!][ n
n
s
nt L
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Examples of Laplace Transform
Exponential Function 0,)( t et f at
a sa se
a sdt ee s F t sa st at
1)
1(0
1)(
0
)(
0
a se L at
1][
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Properties of L - Transform
Theorem 1if k is a constant, then
)()]([ skF t kf L
)()()]()([ sG s F t g t f L
Theorem 2if F(s) and G(s) are L-transform of f(t) and g(t), then
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Properties of L - Transform
Theorem 3 (Initial Value of f(t))
)(lim)0()(lim0
s sF f t f st
Theorem 4 (Final Value of f(t))
)(lim)()(lim 0 s sF f t f st
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Properties of L - Transform
Theorem 5 (L-transform of differentiation)
)0()()(
f s sF dt
t df L
For higher order derivatives
)0(')0()()(
)0()()( 2
0
2
2
2
f sf s F sdt
t df sf s F s
dt
t f d L
t
)0(.. .)0()0()()( )1()1(21 nnnn
n
n
f f s f s s F sdt
t f d L
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Properties of L - Transform
Theorem 6 (L-transform of integration)
s s F
d f L
t )()(
0
)()( a s F t f e L at
Theorem 7
Table of Laplace transform is attached
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ExampleVibration of a Single-Degree-of-Freedom System
b xa x
t y Kx xC x M
)0(,)0(
),(
Y(s) X(s)?
)]([][ t y L Kx xC x M L
)()()]0()([)]0()0()([][][][ 2
sY s KX x s sX C x sx s X s M x KL xCL x ML
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Example
)()()}0()({)}0()0()({ 2 sY s KX x s sX C x sx s X s M
)()(})({})({ 2 sY s KX a s sX C b sa s X s M
)(}{)(}{ 2 sY b sa M Ca s X K Cs Ms
K Cs Ms sY s X
sY s X K Cs Ms
when
2
2
1)()(
)()(}{
0, ba
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Example
K Cs Ms sY s X
2
1)()(
Y(s) X(s)
By definition Transfer Function is the ratioof the L-transform of the output variable tothe L-transform of the input variable (with
zero ICs)
K Cs Ms 21
How to find x(t) ? Use inverse L-transform
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Inverse Laplace Transform
const cdse s F i
t f ic
ic
st -,)(2
1)(
)]([)( t f L s F )]([)( 1 s F Lt f
Instead of using this formula, wewill use Partial-Fraction expansion
or the table of L-transform
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Example: Response to a unit step
K Cs Ms sY
s X 2)(
)(
Y(s) X(s)
K Cs Ms 21
s sY t U t y
1)( ),()(
23
,2,3,122 s s K Cs Ms
K C M Let
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Example: Response to a unit step
0232 s s
To find the roots of this quadratic equation we equateit to zero
Roots of a quadratic equation
02 cbsasare found as:
a
acbb s
a
acbb s
2
4 ;
2
4 2
2
2
1
In our case a=1,b=3,c=2, thus
22
833 ;1
2
833 22
2
1 s s
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Example: Response to a unit step
s s s s X
1)1)(2(
1)(
Look in the table of L-transform
t t ee s s s
Lt x s X L 2121
)1)(2(1
)()( 211
The response of the system to a unit step excitation
0)0(,0)0(
),(23
x x
t u x x x t t eet x 212
1)( 2
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Partial-Fraction Expansion
)1)(2(1
)( s s s
s X
What to do if we do not have L-transform tables?
mmmm
nnnn
m
n
b sb sb sba sa sa sa
s B s A
s F 1
110
11
10
...
...)()(
)(
In general, if:
We consider the case n
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Partial-Fraction ExpansionIf F(s) involves distinct real poles
m
m
p sq
p sq
p sq
s F ...)(2
2
1
1
Then
There are two ways of calculating unknown coefficients
mnn
nn
mmmm
nnnn
p s p s p sa sa sa sa
b sb sb sba sa sa sa
......
...
...
21
11
10
11
10
11
10
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Partial-Fraction Expansion1) Get F(s) to the common denominator
)...(...))...(())...((
)(21
1221
m
mm
p s p s p s p s p sq p s p sq
s F
And compare the numerator to the original one, sothat
Equating the coefficients with the same power we obtain aset of linear equations with respect to q 1,q2, q3,, qm
nnnn
mm
a sa sa sa
p s p sq p s p sq
11
10
1221
...
...))...(())...((
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Partial-Fraction Expansion
)1)(2(1
)( s s s
s X If
)1)(2()2()1()1)(2(
12)1)(2(1
321
321
s s s s sq s sq s sq
sq
sq
sq
s s sThen
12:
023:
0:
10
3211
3212
q s
qqq s
qqq s
1
2/1,2/1
3
2
1
q
qq
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Partial-Fraction Expansion
11
22/1
21
)1)(2(1
)( s s s s s s
s X Thus
Then
t t t t eeee s s s
L s X Lt x
2121
21
21
11
22/1
21
)()(
22
11
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Partial-Fraction Expansion2) To multiply the original F(s) ))((lim k p sk p s s F q k
In our case of
1)2(
1lim)1(
)1)(2(
1lim
,21
)1(1
lim)2()1)(2(
1lim
,21
)1)(2(1
lim)1)(2(
1lim
113
222
001
s s s
s s sq
s s s
s s sq
s s s
s s sq
s s
s s
s s
12)1)(2(
1)( 321
s
q
s
q
s
q
s s s s X
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Partial-Fraction Expansion If the second order polynomial has two complexroots, we DO NOT EXPAND it. LEAVE it as is.
1)1(1)( 2
3212 s s
q sq sq
s s s s X
The polynomial in the numerator has to be of the
first order. CLASS! GET UNKNOWN COEFFICIENTS If the denominators roots are multiple, the process offinding the coefficients is slightly more complicated.
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Partial-Fraction Expansion formultiple poles
na s s X
)(1
)(If
nn
n a sq
a sq
a sq
a sq
a s ...
)(1
33
221
Then
For instance
3
2
)1(32
)( s
s s s X
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Partial-Fraction Expansion formultiple poles
Then
3:
22:
1:
3210
211
12
qqq s
qq s
q sThus
33
221
3
2
111)1(32
)( s
q s
q s
q s
s s s X
3
322
13
2
1)1(1
)1(32
sq sq sq
s s s
201
3
2
1
qqq
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Partial-Fraction Expansion formultiple poles
n
nnn
bas s
p sq
bas s
p sqbas s
p sqbas s 222
222
112 ...)(
1
Then
nbas s s X
)(1
)( 2If
and0
2
bas shas complex-conjugate roots
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Partial-Fraction Expansion Guide
Y X
b sb sb sba sa sa sa
s B s A
s F
mm
mmnn
nn
m
n
1
1
10
11
10
...
...)()(
)(
1. Equate the denominator to zero and find roots ofthe equation.
2. You will write your denominator as multiplicationof first and second order terms.
3. If the second order polynomial has two complexroots, we DO NOT EXPAND it! LEAVE it as is!
0... 11
10 mmmm b sb sb sb
)...()( 222
21 k c sc sb sac s
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Partial-Fraction Expansion Guide
has
4. Use the Laplace transform table or partialfraction expansion to get inverse Laplace transform
To recover the system response )(t x
042
acb
0222
2 c sb sa
This happens when