Control Systems - Lehigh Universityinconsy/lab/css/ME389/lectures/lecture02_Laplace.pdfClassical...
Transcript of Control Systems - Lehigh Universityinconsy/lab/css/ME389/lectures/lecture02_Laplace.pdfClassical...
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Classical Control – Prof. Eugenio Schuster – Lehigh University 1
Control Systems
Lecture 2 Laplace Transform
Classical Control – Prof. Eugenio Schuster – Lehigh University 2
Laplace Transform
Function f(t) of time Piecewise continuous and exponential order
0- limit is used to capture transients and discontinuities at t=0 s is a complex variable (σ+jω)
There is a need to worry about regions of convergence of the integral
Units of s are sec-1=Hz A frequency
If f(t) is volts (amps) then F(s) is volt-seconds (amp-seconds)
btKetf <)(
∫ ∞
� = 0-
) ( ) ( dt e t f s F st ( ) [ ] ∫ ∞ +
∞ � � = =
j
j st ds e s F
j t f s F
α
α π ) ( 2 1 ) ( 1 L
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Classical Control – Prof. Eugenio Schuster – Lehigh University 3
Laplace Transform Examples Step function – unit Heavyside Function
After Oliver Heavyside (1850-1925)
Exponential function After Oliver Exponential (1176 BC- 1066 BC)
Delta (impulse) function δ(t)
0if1)()(0
)(
000>=
+−=−===
∞+−∞−∞
−
−∞
−
−∫∫ σ
ωσ
ωσ
sje
sedtedtetusF
tjststst
% & '
≥ <
= 0 for , 1 0 for , 0
) ( t t
t u
∫ ∫∞ ∞ ∞+−
+−−− >+
=+
−===0 0 0
)()( if1)( ασ
αα
ααα
ssedtedteesF
tstsstt
sdtetsF st allfor1)()(0
== ∫∞
−
−δ
Classical Control – Prof. Eugenio Schuster – Lehigh University 4
Laplace Transform Table Signal Waveform Transform impulse step
ramp
exponential
damped ramp
sine
cosine
damped sine
damped cosine
)(tδ
22)( βα
α
++
+
s
s
22)( βα
β
++s
22 β
β
+s
22 β+s
s
1
s1
21
s
α+s1
2)(
1
α+s
)(tu
)(ttu
)(tue tα−
)(tutte α−
( ) )(sin tutβ
( ) )(cos tutβ
( ) )(sin tutte βα−
( ) )(cos tutte βα−
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Classical Control – Prof. Eugenio Schuster – Lehigh University 5
Laplace Transform Properties
{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLL
Linearity: (absolutely critical property)
( )ssFdf
t=
!"#
$%&∫0
)( ττLIntegration property:
)0()()(−−=
"#$
%&' fssFdttdf
LDifferentiation property:
)0()0()(22)(2
−"−−−=#$
#%&
#'
#()
fsfsFsdt
tfdL
)0()0()0()()( )(21 −−−−"−−−=#$
#%&
#'
#() −− mmm
m
mffsfssF
dttfd
msL
Classical Control – Prof. Eugenio Schuster – Lehigh University 6
Laplace Transform Properties
)()}({ αα +=− sFtfe tL
{ } 0)()()( >=−− − asFeatuatf as forL
Translation properties:
s-domain translation:
t-domain translation:
Initial Value Property: )(lim)(lim0
ssFtfst ∞→+→
=
Final Value Property: )(lim)(lim0
ssFtfst →∞→
=
If all poles of F(s) are in the LHP
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Classical Control – Prof. Eugenio Schuster – Lehigh University 7
Laplace Transform Properties )(1)}({asF
aatf =LTime Scaling:
Multiplication by time:
Convolution:
dssdFttf )()}({ −=L
)()(})()({0
sGsFdtgft
=−∫ τττL
Time product: λλπ
ωσ
ωσdsGsF
jtgtf
j
j∫+
−−= )()(
21)}()({L
Classical Control – Prof. Eugenio Schuster – Lehigh University 8
Laplace Transform
Exercise: Find the Laplace transform of the following waveform
[ ] )()2cos(2)2sin(22)( tutttf −+=( )( )4
24)( 2 ++
=ssssF
Exercise: Find the Laplace transform of the following waveform
( )
[ ] ( )tudttedtuetf
dxxtuetft
t
tt
4040
0
4
5)(5)(
4sin5)()(−
−
−
+=
+= ∫ ( )( )
( )2
2
3
4020010)(
1648036)(
+
+=
++++
=
sssF
ssssssF
Exercise: Find the Laplace transform of the following waveform
)2()(2)()( TtAuTtAutAutf −+−−=( )seAsFTs 21)(−−
=
5
Classical Control – Prof. Eugenio Schuster – Lehigh University 9
Laplace Transform
The diagram commutes Same answer whichever way you go
Linear system
Differential equation
Classical techniques
Response signal
Laplace transform L
Inverse Laplace transform L-1
Algebraic equation
Algebraic techniques
Response transform
Tim
e do
mai
n (t
dom
ain)
Complex frequency domain (s domain)
Classical Control – Prof. Eugenio Schuster – Lehigh University 10
Solving LTI ODE�s via Laplace Transform
011
1
1
0
)1(
0
1
0
)1(1
0
011
1
011
1
)0()0()()(
asasas
subsyasU
asasasbsbsbsbsY n
nn
ji
j
jim
ii
ji
j
jin
ii
nn
n
mm
mm
++++
−
+++++++++
= −−
−
=
−−
=
−
=
−−−
=−
−
−−
∑∑∑∑
( ) ( ) ( ) ( ) ubububyayay mm
mm
nn
n0
110
11 +++=+++ −
−−
−
Initial Conditions:
!"
#$%
&−=!
"
#$%
&−+− ∑∑∑∑∑
−
=
−−
=
−
=
−−−
=
−
=
−− ji
j
jiim
ii
ji
j
jiin
ii
n
j
jjnn susUsbsysYsasysYs1
0
)1(
0
1
0
)1(1
0
1
0
)1( )0()()0()()0()(
( )( ) ( ) ( )( ) ( )0,,0,0,,0 11 uuyy mn …… −−
Recall jk
j
jkkk
k
sfsFdttfd
∑−
=
−−−=#$%
&'( 1
0
)1( )0()()(sL
For a given rational U(s) we get Y(s)=Q(s)/P(s)
6
Classical Control – Prof. Eugenio Schuster – Lehigh University 11
Laplace Transform
Exercise: Find the Laplace transform V(s)
3)0(
)(4)(6)(
−=−
=+
v
tutvdttdv
( ) 63
64)(
+−
+=
ssssV
Exercise: Find the Laplace transform V(s)
( )( )( ) 12
3215)(
+−
+++=
sssssV
2)0(',2)0(
5)(3)(4)( 22
2
=−−=−
=++ −
vv
etvdttdv
dttvd t
What about v(t)?
Classical Control – Prof. Eugenio Schuster – Lehigh University 12
Computing Transfer Functions via Laplace Transform
)())(()())((
)()()(
)()()()()(
21
21
011
1
011
1
011
1
011
1
n
m
nn
n
mm
nn
n
mm
pspspszszszsK
asasasbsbsb
sUsYsH
sUsAsBsU
asasasbsbsbsY
−−−−−−
=
+++++++
==
=++++
+++=
−−
−−
−−
−−
( ) ( ) ( ) ububyayay mm
nn
n0
110
11 ++=+++ −
−−
−
Assume all Initial Conditions Zero:
( ) ( ) )()( 011
1011
1 sUbsbsbsYasasas mm
nn
n +++=++++ −−
−−
Input Output
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Classical Control – Prof. Eugenio Schuster – Lehigh University 13
Rational Functions
We shall mostly be dealing with TFs which are rational functions – ratios of polynomials in s pi are the poles and zi are the zeros of the function K is the scale factor or (sometimes) gain
A proper rational function has n≥m A strictly proper rational function has n>m An improper rational function has n<m
)())(()())((
)(
21
21
011
1
011
1
n
m
nn
nn
mm
mm
pspspszszszsK
asasasabsbsbsbsF
−−−−−−
=
++++++++
= −−
−−
Classical Control – Prof. Eugenio Schuster – Lehigh University 14
Partial Fraction Expansion - Residues at Simple Poles
)()(lim sFpsk ipsii
−=→
)()()()())(()())(()(
2
2
1
1
21
21
n
n
n
m
psk
psk
psk
pspspszszszsKsF
−++
−+
−=
−−−−−−
=
( ))()(
)()(
)()()(
2
2
1
1
n
ini
iii ps
pskkpspsk
pspsksFps
−−
++++−−
+−−
=−
Functions of a complex variable with isolated, finite order poles have residues at the poles
Residue at a simple pole:
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Classical Control – Prof. Eugenio Schuster – Lehigh University 15
Partial Fraction Expansion - Residues at multiple poles
rjsFrpsjrds
jrdpsjr
k ii
j 1,)()(lim)!(
1=!"
#$%& −
−
−
→−=
( )31522
+
+
s
ssExample:
( ) ( ) ( )( ) )(32
13
11
12
152 2
321
3
21 tutte
sssL
sssL t −+=""
#
$%%&
'
+−
++
+=""#
$%%&
'
+
+ −−−
( ) ( )31211321
++
++
+=
s
k
s
ksk
rr
rm
psk
psk
psk
pszszszsKsF
)()()()()())(()(
12
1
2
1
1
1
21
−++
−+
−=
−−−−
=
Residue at a multiple pole:
2 ) 1 (
) 5 2 ( ) 1 ( lim ! 2 1
3 2 3
2 2
1 = ) ) * +
, , - .
+ + +
� → s s s s
ds d
s 1k
1 ) 1 (
) 5 2 ( ) 1 ( lim 3 2 3
1 = ) ) * +
, , - .
+ + +
� → s s s s
ds d
s 2k
3 ) 1 (
) 5 2 ( ) 1 ( lim 3 2 3
3 � =
+ + +
� → s s s s
s 3k
Classical Control – Prof. Eugenio Schuster – Lehigh University 16
Partial Fraction Expansion - Residues at Complex Poles
Compute residues at the poles Bundle complex conjugate pole pairs into second-order terms if you want … but you will need to be careful!
Inverse Laplace Transform is a sum of complex exponentials. But the answer will be real.
( )[ ]222 2))(( βααβαβα ++−=+−−− ssjsjs
)()(lim sFasas
−→
9
Classical Control – Prof. Eugenio Schuster – Lehigh University 17
Inverting Laplace Transforms in Practice
We have a table of inverse LTs Write F(s) as a partial fraction expansion Now appeal to linearity to invert via the table
Surprise! Nastiness: computing the partial fraction expansion is best done by calculating the residues
€
F(s) =bms
m + bm−1sm−1 ++ b1s+ b0
ansn + an−1s
n−1 ++ a1s+ a0= K (s− z1)(s− z2)(s− zm)
(s− p1)(s− p2)(s− pn )
=α1s− p1( )
+α2s− p2( )
+α31(s− p3)
+α32s− p3( )2
+α33s− p3( )3
+ ...+αqs− pq( )
Classical Control – Prof. Eugenio Schuster – Lehigh University 18
Inverse Laplace Transform
) 5 2 )( 1 ( ) 3 ( 20 ) ( 2 + + +
+ =
s s s s s F
21211)(
*221js
kjs
ksksF
+++
−++
+=
π45
255521)21)(1(
)3(20)()21(21
lim2
101522
)3(20)()1(1
lim1
jej
jsjssssFjs
jsk
sss
ssFss
k
=−−=+−=+++
+=−+
+−→=
=−=++
+=+
−→=
)()452cos(21010
)(252510)( 45)21(
45)21(
tutee
tueeetf
tt
jtjjtjt
!"#
$%& ++=
!!
"
#
$$
%
&++=
−−
−−−++−−
π
ππ
Exercise: Find the Inverse Laplace transform of
10
Classical Control – Prof. Eugenio Schuster – Lehigh University 19
Inverse Laplace Transform
Exercise: Find v(t)
3)0(
)(4)(6)(
−=−
=+
v
tutvdttdv
( ) 63
64)(
+−
+=
ssssV
Exercise: Find v(t)
( )( )( ) 12
3215)(
+−
+++=
sssssV
2)0(',2)0(
5)(3)(4)( 22
2
=−−=−
=++ −
vv
etvdttdv
dttvd t
What about v(t)?
)(311)(
32)( 6 tuetutv t−−=
)(255
21)( 32 tueeetv ttt !
"
#$%
& +−= −−−
Classical Control – Prof. Eugenio Schuster – Lehigh University 20
Not Strictly Proper Laplace Transforms
Find the inverse LT of Convert to polynomial plus strictly proper rational function
Use polynomial division Invert as normal
3 4 8 12 6 ) ( 2
2 3
+ + + + +
= s s
s s s s F
35.0
15.02
3422)( 2
++
+++=
++
+++=
sss
sssssF
)(5.05.0)(2)()( 3 tueetdttdtf tt
!"#
$%& +++= −−δδ