me330 lecture3
TRANSCRIPT
ME 330 Control Systems
SP2011
Lecture 3
ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10,zscan=0):ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10,zscan=0):
Example Process Solving differential equations to get the time-domain
response of the dynamical system.
)(1
)(2
sFkcsms
sX
)()()()( tkxtxctxmtf f(t)
Step 2) Transform f(t) to F(s)
Step 1) Represent the system differential equation in Laplace domain “transfer function”
Step 3) Multiply “transfer function” with Laplace domain input F(s)
Step 4) Solve for time-domain response with inverse Laplace transform
Partial Fraction Expansion Typically, won’t directly find the Laplace transform for
solution to differential equations Example:
52
122)(
2
ss
ssX Nothing in tables for general
quadratic equation
)21)(21(522 jsjsss Sometime even less straightforward when complex roots are involved
Simplify the functions F(s) in order to easily find inverse Laplace transforms from the table. Convert original function F(s) into sum of simpler terms
)(...)()()( 21 sFsFsFsF n
)(...)()(
)(...)()()(
21
21
tftftf
sFsFsFsF
n
n
-1-1-1-1 LLLL
Requirement Original function F(s) must be strictly proper
)(
)()(
sD
sNsF order of N(s) < order of D(s)
Divide N(s) by D(s) until remainder is strictly proper
5
21
5
2)5)(1(
5
762)(
22
2
2
23
sss
ss
sss
ss
ssssF
Then perform partial fraction expansion on remaining function terms
Methods for Partial Fraction Expansion Three special cases for simplification Depends on the roots of the denominator polynomial.
Propose equivalent simplified form for Laplace domain representation and solve for coefficients.
For each of the special cases: simplification coefficients solved using its own
technique. general method for solving coefficients based on
powers of s.
Case 1: Real Roots If F(s) denominator has real and distinct roots
)()()()(
)()())((
)()(
2
2
1
1
21
n
n
m
m
nm
ps
K
ps
K
ps
K
ps
K
pspspsps
sNsF
)(
)(
)(
)(
)(
)(
)()())((
)()()()(
2
2
1
1
21
n
mnm
mm
nm
mm
ps
psKK
ps
psK
ps
psK
pspspsps
sNpssFps
Multiply F(s) by (s+pm) to find residue Km
Set s = -pm to evaluate residue Km 0 0 0
)())((
)(
21 mnmm
mm pppppp
pNK
Example Case 1 If F(s) denominator has
real and distinct roots )()()()(
)()())((
)()(
2
2
1
1
21
n
n
m
m
nm
ps
K
ps
K
ps
K
ps
K
pspspsps
sNsF
)23(
795)(
2
23
ss
ssssF
)23(
32
2
ss
ss polynomial long division
(shown on board)
21)2)(1(
3)( 21
s
K
s
K
ss
ssG
22
3
)2)(1(
3)1(
11
1
sss
s
ss
ssK
11
3
)2)(1(
3)2(
22
2
sss
s
ss
ssK
)2)(1( ss
Example Case 1 After polynomial division and partial fraction expansion
2
1
1
22)(
ss
ssF
2
1
1
2][]2[)( 1111
ssstf LLLL
)(2 t
1)]([ tL
dt
td )(
sdt
td
)(
L
as
e at
1
L
tt ee 22
Case 2: Real Repeated Roots If F(s) denominator has real and repeated roots
)()()()()(
)()()(
)()(
2
1
11
1
2
1
1
21
n
nrrrr
nr
ps
K
ps
K
ps
K
ps
K
ps
K
pspsps
sNsF
)(
)(
)(
)()()(
)()()(
)()()()(
1
2
1111211
21
11
n
rn
rr
rr
nr
rr
ps
psK
ps
psKKpsKpsK
pspsps
sNpssFps
Multiply F(s) by (s+p1)r to find residue K1
Differentiate (s+p1)rF(s) to find residues K2 through Kr
rids
sFpsd
iK
ps
i
ri
i ,,2,1)]()[(
)!1(
1
1
11
1
Set s=-p1 to find residues K1
1ps 0 000
Example Case 2 If F(s) denominator has
real and repeated roots
)()()()()(
)()()(
)()(
2
1
11
1
2
1
1
21
n
nrrrr
nr
ps
K
ps
K
ps
K
ps
K
ps
K
pspsps
sNsF
rids
sFpsd
iK
ps
i
ri
i ,,2,1)]()[(
)!1(
1
1
11
1
3
2
)1(
32)(
s
sssF 3
32
21
)1()1()1(
s
K
s
K
s
K
repeated root
1
3
23
3 )1(
32)1(
ss
sssK
322
13
23 )1()1(
)1(
32)1( KsKsK
s
sss
1
3
23
)1(
32)1(
!0
1
ss
sss
Example Case 2 Find K2 and K3 with by taking derivatives
213
23 )1(2
)1(
32)1( KsK
s
sss
ds
d
1
3
23
2 )1(
32)1(
ss
sss
ds
dK
13
23
2
2
2)1(
32)1( K
s
sss
ds
d
1
3
23
2
2
1 )1(
32)1(
2
1
ss
sss
ds
dK
Example Case 2 After polynomial division and partial fraction expansion
32 )1(
2
)1(
0
1
1)(
ssssF
31
211
)1(
2
)1(
0
1
1)(
ssstf LLL
te
ase at
1][L
natn
aset
n )(
1
)!1(
1 1
L
tet 2
Next Lecture
Derivation of dynamical system models