me330 lecture4
TRANSCRIPT
ME 330 Control Systems
FA 2010
Lecture 4
ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10,zscan=0):ScanYZTest(dsc,bl=0.1,rr=1.0,yscan=10,zscan=0):
Recall from last lecture Differential equations and forcing functions combine
algebraically in the Laplace domain. Time domain representation requires taking the
inverse Laplace transform.
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
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
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
Case 3: Complex Roots If F(s) denominator has complex roots, can be
solved with complex Ki’s then recognizing that
cos2
jj ee
)()())((
)()(
232
1
12
1 bass
KsK
ps
K
bassps
sNsF
Or, F(s) denominator has complex roots and
sin2
j
ee jj
Solve K1 then multiply lowest common denominator
))(()()( 1322
1 psKsKbassKsN
2nd order 2nd order Combine like terms
Solve for Ki’s by balancing the 2 unknown coefficients
Case 3: Complex Roots Once K2, K3 are known,
complete the square on the quadratic
)1(
)1(
)1(
cHWproblemB
bHWproblemA
aHWproblem
g
hk
g
hxgkhxgx
42)(
22
2
221
1
232
1
1
)(
)(
)(
)()()(
as
BasA
ps
K
bass
KsK
ps
KsF
Example Assign m = 1, c = 2, k = 5, and
f(t)
0,1
0,0)(
t
ttf
),(52
1)(
2sF
sssX
1)( sF
2121522 jsjsss note:
22 2)1(
2
2
1)(
ssX
22
22
)()cos(
)()sin(
as
asteL
asteL
at
atrecognize:
)2sin(2
1)( tetx t
Matlab Rational function of polynomials in s
nn
nnn
asas
bsbsb
sA
sB
1
110
)(
)(
num = [b0 b1 … bn] den = [1 a1 … an]
Matlab representation
Partial fraction expansion
)()(
)(
)2(
)2(
)1(
)1(
)(
)(sk
nps
nr
ps
r
ps
r
sA
sB
Matlab command[r,p,k] = residue(num,den)
r,p,k are vectors whose coefficients define the partial fraction expansion
Example Case 3 If F(s) denominator has complex roots
)22(
1)(
2
ssssF
)22( 2321
ss
KsK
s
K
)1)(1( jsjs Multiply both sides by the greatest common denominator
)22( 2 sss
sKsKssK )()22(1 322
1
1312
21 2)2()(1 KsKKsKK
210 KK 3120 KK 121 K
1,2
1,2
1321 KKK
Example Case 3 After polynomial division and partial fraction expansion
)22(2
2
2
1)(
2
ss
s
ssF
2222 1)1(
1
2
1
1)1(
1
2
1
2
1)(
s
s
sssF
222 1)1()22( sss
Completing the square
1)1()2( ss
and factoring the numerator
)(2
1)( tutf
stu
1)]([ L
22)()]sin([
aste atL
)sin(2
1te t
22)()]cos([
as
aste atL
)cos(2
1te t
Next Lecture
Derivation of dynamical system models