Embed Size (px)
Citation preview
EL2620 Nonlinear Control
Lecture notesKarl Henrik Johansson, Bo Wahlberg and Elling W. Jacobsen
This revision December 2011
Automatic ControlKTH, Stockholm, Sweden
Preface
Many people have contributed to these lecture notes in nonlinear control.Originally it was developed by Bo Bernhardsson and Karl Henrik Johansson,and later revised by Bo Wahlberg and myself. Contributions and commentsby Mikael Johansson, Ola Markusson, Ragnar Wallin, Henning Schmidt,Krister Jacobsson, Bjorn Johansson and Torbjorn Nordling are gratefullyacknowledged.
Elling W. JacobsenStockholm, December 2011
EL2
620
2011
EL2
620
Non
linea
rC
ontr
olA
utom
atic
Con
trol
Lab,
KTH
•D
ispo
sitio
n7.
5cr
edits
,lp
2
28h
lect
ures
,28h
exer
cise
s,3
hom
e-w
orks
•In
stru
ctor
sE
lling
W.J
acob
sen,
lect
ures
and
cour
sere
spon
sibl
PerH
agg,
Farh
adFa
rokh
i,te
achi
ngas
sist
ants
Han
naH
olm
qvis
t,co
urse
adm
inis
tratio
STE
X(e
ntra
nce
floor
,Osq
ulda
sv.
10),
cour
sem
ater
ial
Lect
ure
11
EL2
620
2011
Cou
rse
Goa
l
Topr
ovid
epa
rtic
ipan
tsw
itha
solid
theo
retic
alfo
unda
tion
ofno
nlin
ear
cont
rols
yste
ms
com
bine
dw
itha
good
engi
neer
ing
unde
rsta
ndin
g
You
shou
ldaf
tert
heco
urse
beab
leto
•un
ders
tand
com
mon
nonl
inea
rcon
trolp
heno
men
a
•ap
ply
the
mos
tpow
erfu
lnon
linea
rana
lysi
sm
etho
ds
•us
eso
me
prac
tical
nonl
inea
rcon
trold
esig
nm
etho
ds
Lect
ure
12
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
1
•Pra
ctic
alin
form
atio
n
•Cou
rse
outli
ne
•Lin
earv
sN
onlin
earS
yste
ms
•Non
linea
rdiff
eren
tiale
quat
ions
Lect
ure
13
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•D
escr
ibe
dist
inct
ive
phen
omen
ain
nonl
inea
rdyn
amic
syst
ems
•M
athe
mat
ical
lyde
scrib
eco
mm
onno
nlin
earit
ies
inco
ntro
lsy
stem
s
•Tr
ansf
orm
diffe
rent
iale
quat
ions
tofir
st-o
rder
form
•D
eriv
eeq
uilib
rium
poin
ts
Lect
ure
14
EL2
620
2011
Cou
rse
Info
rmat
ion
•A
llin
foan
dha
ndou
tsar
eav
aila
ble
atth
eco
urse
hom
epag
eat
KTH
Soc
ial
•C
ompu
lsor
yco
urse
item
s
–3
hom
ewor
ks,h
ave
tobe
hand
edin
ontim
e(w
ear
est
ricto
nth
is!)
–5h
writ
ten
exam
onD
ecem
ber1
120
12
Lect
ure
15
EL2
620
2011
Mat
eria
l
•Te
xtbo
ok:
Kha
lil,N
onlin
earS
yste
ms,
Pre
ntic
eH
all,
3rd
ed.,
2002
.O
ptio
nalb
uthi
ghly
reco
mm
ende
d.
•Le
ctur
eno
tes:
Cop
ies
oftra
nspa
renc
ies
(from
prev
ious
year
)
•E
xerc
ises
:C
lass
room
and
hom
eex
erci
ses
•H
omew
orks
:3
com
pute
rexe
rcis
esto
hand
in(a
ndre
view
)
•S
oftw
are:
Mat
lab
Alte
rnat
ive
text
book
s(d
ecre
asin
gm
athe
mat
ical
rigou
r):
Sas
try,
Non
linea
rSys
tem
s:A
naly
sis,
Sta
bilit
yan
dC
ontro
l;V
idya
saga
r,N
onlin
earS
yste
ms
Ana
lysi
s;S
lotin
e&
Li,A
pplie
dN
onlin
earC
ontro
l;G
lad
&Lj
ung,
Reg
lert
eori,
flerv
aria
bla
och
olin
jara
met
oder
.O
nly
refe
renc
esto
Kha
lilw
illbe
give
n.
Two
cour
seco
mpe
ndia
sold
byS
TEX
.
Lect
ure
16
EL2
620
2011
Cou
rse
Out
line
•In
trod
uctio
n:no
nlin
earm
odel
san
dph
enom
ena,
com
pute
rsi
mul
atio
n(L
1-L2
)
•Fe
edba
ckan
alys
is:
linea
rizat
ion,
stab
ility
theo
ry,d
escr
ibin
gfu
nctio
ns(L
3-L6
)
•C
ontr
olde
sign
:co
mpe
nsat
ion,
high
-gai
nde
sign
,Lya
puno
vm
etho
ds(L
7-L1
0)
•A
ltern
ativ
es:
gain
sche
dulin
g,op
timal
cont
rol,
neur
alne
twor
ks,
fuzz
yco
ntro
l(L1
1-L1
3)
•S
umm
ary
(L14
)
Lect
ure
17
EL2
620
2011
Line
arS
yste
ms
Defi
nitio
n:Le
tMbe
asi
gnal
spac
e.Th
esy
stem
S:M
!M
islin
eari
ffor
allu
,v"
Man
d!
"R
S(!
u)
=!S
(u)
scal
ing
S(u
+v)
=S
(u)+
S(v
)su
perp
ositi
on
Exa
mpl
e:Li
near
time-
inva
riant
syst
ems
x(t
)=
Ax(t
)+
Bu(t
),y(t
)=
Cx(t
),x(0
)=
0
y(t
)=
g(t
)"
u(t
)=
!t
0
g(#
)u(t
##)d
#
Y(s
)=
G(s
)U(s
)
Not
ice
the
impo
rtan
ceto
have
zero
initi
alco
nditi
ons
Lect
ure
18
EL2
620
2011
Line
arS
yste
ms
Hav
eN
ice
Pro
pert
ies
Loca
lsta
bilit
y=gl
obal
stab
ility
Sta
bilit
yif
alle
igen
valu
esof
A(o
rpo
les
ofG
(s))
are
inth
ele
ftha
lf-pl
ane
Sup
erpo
sitio
nE
noug
hto
know
ast
ep(o
rim
puls
e)re
spon
se
Freq
uenc
yan
alys
ispo
ssib
leS
inus
oida
linp
uts
give
sinu
soid
alou
tput
s:Y
(i$)
=G
(i$)U
(i$)
Lect
ure
19
EL2
620
2011
Line
arM
odel
sm
aybe
too
Cru
deA
ppro
xim
atio
ns
Exa
mpl
e:Po
sitio
ning
ofa
ball
ona
beam
%x
Non
linea
rmod
el:m
x(t
)=
mg
sin
%(t
),Li
near
mod
el:x(t
)=
g%(t
)
Lect
ure
110
EL2
620
2011
Can
the
ball
mov
e0.
1m
eter
in0.
1se
cond
sfro
mst
eady
stat
e?
Line
arm
odel
(ste
pre
spon
sew
ith%
=%
0)g
ives
x(t
)$
10t2 2
%0
$0.
05%
0
soth
at
%0
$0.
1
0.05
=2
rad
=11
4!
Unr
ealis
tican
swer
.Cle
arly
outs
ide
linea
rre
gion
!
Line
arm
odel
valid
only
ifsi
n%
$%
Mus
tcon
side
rnon
linea
rmod
el.
Poss
ibly
also
incl
ude
othe
rno
nlin
earit
ies
such
asce
ntrip
etal
forc
e,sa
tura
tion,
fric
tion
etc.
Lect
ure
111
EL2
620
2011
Line
arM
odel
sar
eno
tRic
hE
noug
h
Line
arm
odel
sca
nno
tdes
crib
em
any
phen
omen
ase
enin
nonl
inea
rsy
stem
s
Lect
ure
112
EL2
620
2011
Sta
bilit
yC
anD
epen
don
Ref
eren
ceS
igna
lE
xam
ple:
Con
trols
yste
mw
ithva
lve
char
acte
ristic
f(u
)=
u2
PS
fr
!1 s
1(s+
1)2
Mot
orVa
lve
Pro
cess
!1
ry
Sim
ulin
kbl
ock
diag
ram
:
1(s
+1)(s
+1)
ys1
-1u2
Lect
ure
113
EL2
620
2011
ST
EP
RE
SP
ON
SE
S
05
1015
2025
300
0.2
0.4
Tim
e t
Output y
05
1015
2025
30024
Tim
e t
Output y
05
1015
2025
300510
Tim
e t
Output y
r=
0.2
r=
1.6
8
r=
1.7
2
Sta
bilit
yde
pend
son
ampl
itude
ofth
ere
fere
nce
sign
al!
(The
linea
rized
gain
ofth
eva
lve
incr
ease
sw
ithin
crea
sing
ampl
itude
)
Lect
ure
114
EL2
620
2011
Mul
tiple
Equ
ilibr
ia
Exa
mpl
e:ch
emic
alre
acto
r
x1
=!
x1ex
p
!!
1 x2
"+
f(1
!x1)
x2
=x1ex
p
!!
1 x2
"!
!f(x
2!
xc)
f=
0.7
,!
=0.4
050
100
150
200
101520
Coolant temp xc
Inpu
t
050
100
150
200
050100
150
200
time
[s]
Temperature x2
Out
put
Exi
sten
ceof
mul
tiple
stab
leeq
uilib
riafo
rthe
sam
ein
putg
ives
hyst
eres
isef
fect
Lect
ure
115
EL2
620
2011
Sta
ble
Per
iodi
cS
olut
ions
Exa
mpl
e:Po
sitio
nco
ntro
lofm
otor
with
back
-lash
y
Sum
5
P-co
ntro
ller
15s
+s
2
Mot
or
0Co
nsta
ntBa
ckla
sh
-1
Mot
or:G
(s)
=1
s(1+
5s)
Con
trolle
r:K
=5
Lect
ure
116
EL2
620
2011
Bac
k-la
shin
duce
san
osci
llatio
n
Perio
dan
dam
plitu
dein
depe
nden
tofi
nitia
lcon
ditio
ns:
010
2030
4050
-0.50
0.5
Tim
e t
Output y
010
2030
4050
-0.50
0.5
Tim
e t
Output y
010
2030
4050
-0.50
0.5
Tim
e t
Output y
How
pred
icta
ndav
oid
osci
llatio
ns?
Lect
ure
117
EL2
620
2011
Aut
omat
icTu
ning
ofP
IDC
ontr
olle
rsR
elay
indu
ces
ade
sire
dos
cilla
tion
who
sefre
quen
cyan
dam
plitu
dear
eus
edto
choo
seP
IDpa
ram
eter
s
ΣP
roce
ss
PID
Rela
y
A T
uy
−1
05
10
−101
Tim
e
uy
Lect
ure
118
EL2
620
2011
Har
mon
icD
isto
rtio
nE
xam
ple:
Sin
usoi
dalr
espo
nse
ofsa
tura
tion
asi
nt
y(t
)=
"" k=
1A
ksi
n(k
t)S
atur
atio
n
10
0
10
-2
10
0
Fre
qu
en
cy (
Hz)
Amplitude y
01
23
45
-2-1
012
Tim
e t
10
0
10
-2
10
0
Fre
qu
en
cy (
Hz)
Amplitude y
01
23
45
-2-1
012
Tim
e t
a=
1
a=
2a
=2
a=
2a
=2
Lect
ure
119
EL2
620
2011
Exa
mpl
e:E
lect
rical
pow
erdi
strib
utio
n
Non
linea
ritie
ssu
chas
rect
ifier
s,sw
itche
del
ectro
nics
,and
trans
form
ers
give
rise
toha
rmon
icdi
stor
tion
Tota
lHar
mon
icD
isto
rtio
n=
"" k=
2E
nerg
yin
tone
k
Ene
rgy
into
ne1
Exa
mpl
e:E
lect
rical
ampl
ifier
s
Effe
ctiv
eam
plifi
ers
wor
kin
nonl
inea
rreg
ion
Intro
duce
ssp
ectr
umle
akag
e,w
hich
isa
prob
lem
ince
llula
rsys
tem
s
Trad
e-of
fbet
wee
nef
fect
ivity
and
linea
rity
Lect
ure
120
EL2
620
2011
Sub
harm
onic
s
Exa
mpl
e:D
uffin
g’s
equa
tion
y+
y+
y#
y3
=a
sin($
t)
05
1015
2025
30-0
.50
0.5 0
510
1520
2530
-1
-0.50
0.51
Tim
et
Tim
et
y asin$t
Lect
ure
121
EL2
620
2011
Non
linea
rD
iffer
entia
lEqu
atio
ns
Defi
nitio
n:A
solu
tion
to
x(t
)=
f(x
(t))
,x(0
)=
x0
(1)
over
anin
terv
al[0
,T]i
saC
1fu
nctio
nx
:[0
,T]!
Rn
such
that
(1)
isfu
lfille
d.
•W
hen
does
ther
eex
ists
aso
lutio
n?
•W
hen
isth
eso
lutio
nun
ique
?
Exa
mpl
e:x
=A
x,x
(0)
=x
0,g
ives
x(t
)=
exp(A
t)x
0
Lect
ure
122
EL2
620
2011
Exi
sten
ceP
robl
ems
Exa
mpl
e:Th
edi
ffere
ntia
lequ
atio
nx
=x
2,x
(0)
=x
0
has
solu
tion
x(t
)=
x0
1#
x0t,
0%
t<
1 x0
Sol
utio
nno
tdefi
ned
for
t f=
1 x0
Sol
utio
nin
terv
alde
pend
son
initi
alco
nditi
on!
Rec
allt
hetr
ick:
x=
x2
&dx x2
=dt
Inte
grat
e&
#1
x(t
)#
#1
x(0
)=
t&
x(t
)=
x0
1#
x0t
Lect
ure
123
EL2
620
2011
Fini
teE
scap
eTi
me
Sim
ulat
ion
forv
ario
usin
itial
cond
ition
sx
0
01
23
45
0
0.51
1.52
2.53
3.54
4.55
Tim
e t
x(t)
Fini
te e
scap
e tim
e of
dx/
dt =
x2
Lect
ure
124
EL2
620
2011
Uni
quen
ess
Pro
blem
s
Exa
mpl
e:x
='
x,x
(0)
=0,
has
man
yso
lutio
ns:
x(t
)=
#(t
#C
)2/4
t>
C
0t%
C
01
23
45
-1
-0.50
0.51
1.52
Tim
e t
x(t)
Lect
ure
125
EL2
620
2011
Phy
sica
lInt
erpr
etat
ion
Con
side
rthe
reve
rse
exam
ple,
i.e.,
the
wat
erta
nkla
bpr
oces
sw
ith
x=
#'
x,
x(T
)=
0
whe
rex
isth
ew
ater
leve
l.It
isth
enim
poss
ible
tokn
owat
wha
ttim
et<
Tth
ele
velw
asx(t
)=
x0
>0.
Hin
t:R
ever
setim
es
=T
#t&
ds
=#
dt
and
thus
dx ds
=#
dx dt
Lect
ure
126
EL2
620
2011
Lips
chitz
Con
tinui
ty
Defi
nitio
n:f
:R
n!
Rn
isLi
psch
itzco
ntin
uous
ifth
ere
exis
tsL
,r>
0su
chth
atfo
rall
x,y
"B
r(x
0)
={z
"R
n:
(z#
x0(
<r}
,
(f(x
)#
f(y
)(%
L(x
#y(
Euc
lidea
nno
rmis
give
nby
(x(2
=x
2 1+
···+
x2 n
Slo
peL
Lect
ure
127
EL2
620
2011
Loca
lExi
sten
cean
dU
niqu
enes
s
Theo
rem
:If
fis
Lips
chitz
cont
inuo
us,t
hen
ther
eex
ists
&>
0su
chth
at
x(t
)=
f(x
(t))
,x(0
)=
x0
has
aun
ique
solu
tion
inB
r(x
0)
over
[0,&
].P
roof
:S
eeK
halil
,
App
endi
xC
.1.
Bas
edon
the
cont
ract
ion
map
ping
theo
rem
Rem
arks
•&
=&(
r,L
)
•f
bein
gC
0is
nots
uffic
ient
(cf.,
tank
exam
ple)
•f
bein
gC
1im
plie
sLi
psch
itzco
ntin
uity
(L=
max
x#B
r(x
0)f
$ (x))
Lect
ure
128
EL2
620
2011
Sta
te-S
pace
Mod
els
Sta
tex
,inp
utu
,out
puty
Gen
eral
:f(x
,u,y
,x,u
,y,.
..)
=0
Exp
licit:
x=
f(x
,u),
y=
h(x
)
Affi
nein
u:
x=
f(x
)+
g(x
)u,
y=
h(x
)
Line
ar:
x=
Ax
+B
u,
y=
Cx
Lect
ure
129
EL2
620
2011
Tran
sfor
mat
ion
toA
uton
omou
sS
yste
m
Ano
naut
onom
ous
syst
em
x=
f(x
,t)
isal
way
spo
ssib
leto
trans
form
toan
auto
nom
ous
syst
emby
intro
duci
ngx
n+
1=
t:
x=
f(x
,xn+
1)
xn+
1=
1
Lect
ure
130
EL2
620
2011
Tran
sfor
mat
ion
toFi
rst-
Ord
erS
yste
m
Giv
ena
diffe
rent
iale
quat
ion
iny
with
high
estd
eriv
ativ
ed
ny
dtn
,
expr
ess
the
equa
tion
inx
=$ y
dy dt
...
dn
!1y
dtn
!1
% TE
xam
ple:
Pend
ulum
MR
2'
+k'
+M
gR
sin
'=
0
x=
& ''' T
give
s x1
=x
2
x2
=#
k
MR
2x
2#
g Rsi
nx
1
Lect
ure
131
EL2
620
2011
Equ
ilibr
ia
Defi
nitio
n:A
poin
t(x
% ,u
% ,y
% )is
aneq
uilib
rium
,ifa
solu
tion
star
ting
in(x
% ,u
% ,y
% )st
ays
ther
efo
reve
r.
Cor
resp
onds
topu
tting
alld
eriv
ativ
esto
zero
:
Gen
eral
:f(x
% ,u
% ,y
% ,0,
0,..
.)=
0
Exp
licit:
0=
f(x
% ,u
% ),
y%
=h(x
% )
Affi
nein
u:
0=
f(x
% )+
g(x
% )u
% ,y
%=
h(x
% )
Line
ar:
0=
Ax
%+
Bu
% ,y
%=
Cx
%
Ofte
nth
eeq
uilib
rium
isde
fined
only
thro
ugh
the
stat
ex
%
Lect
ure
132
EL2
620
2011
Mul
tiple
Equ
ilibr
ia
Exa
mpl
e:Pe
ndul
um MR
2'
+k'
+M
gR
sin
'=
0
'=
'=
0gi
vessi
n'
=0
and
thus
'%=
k(
Alte
rnat
ivel
yin
first
-ord
erfo
rm:
x1
=x
2
x2
=#
k
MR
2x
2#
g Rsi
nx
1
x1
=x
2=
0gi
vesx
% 2=
0an
dsi
n(x
% 1)
=0
Lect
ure
133
EL2
620
2011
Som
eC
omm
onN
onlin
eari
ties
inC
ontr
olS
yste
ms
Sign
Satu
ratio
n
Rela
y
eu
Mat
hFu
nctio
n
Look
-Up
Tabl
eDe
ad Z
one
Coul
omb
&Vi
scou
s Fr
ictio
nBa
ckla
sh
|u|
Abs
Lect
ure
134
EL2
620
2011
Whe
ndo
we
need
Non
linea
rA
naly
sis
&D
esig
n?
•W
hen
the
syst
emis
stro
ngly
nonl
inea
r
•W
hen
the
rang
eof
oper
atio
nis
larg
e
•W
hen
dist
inct
ive
nonl
inea
rphe
nom
ena
are
rele
vant
•W
hen
we
wan
tto
push
perfo
rman
ceto
the
limit
Lect
ure
135
EL2
620
2011
Nex
tLec
ture
•S
imul
atio
nin
Mat
lab
•Li
near
izat
ion
•P
hase
plan
ean
alys
is
Lect
ure
136
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
2
•Wra
p-up
ofLe
ctur
e1:
Non
linea
rsys
tem
san
dph
enom
ena
•Mod
elin
gan
dsi
mul
atio
nin
Sim
ulin
k
•Pha
se-p
lane
anal
ysis
Lect
ure
21
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•M
odel
and
sim
ulat
ein
Sim
ulin
k
•Li
near
ize
usin
gS
imul
ink
•D
oph
ase-
plan
ean
alys
isus
ing
ppla
ne(o
roth
erto
ol)
Lect
ure
22
EL2
620
2011
Ana
lysi
sTh
roug
hS
imul
atio
n
Sim
ulat
ion
tool
s:
OD
E’s
x=
f(t
,x,u
)
•A
CS
L,S
imno
n,S
imul
ink
DA
E’s
F(t
,x,x
,u)
=0
•O
msi
m,D
ymol
a,M
odel
ica
http://www.modelica.org
Spe
cial
purp
ose
sim
ulat
ion
tool
s
•S
pice
,EM
TP,A
DA
MS
,gP
RO
MS
Lect
ure
23
EL2
620
2011
Sim
ulin
k
>matlab
>>simulink
Lect
ure
24
EL2
620
2011
An
Exa
mpl
ein
Sim
ulin
k
File->New->Model
Double
clickonContinous
Transfer
Fcn
Step(inSources)
Scope(inSinks)
Connect
(mouse-left)
Simulation
->Parameters
1 s+1
Tran
sfer
Fcn
Step
Scop
e
Lect
ure
25
EL2
620
2011
Cho
ose
Sim
ulat
ion
Par
amet
ers
Don
’tfo
rget
“App
ly”
Lect
ure
26
EL2
620
2011
Sav
eR
esul
tsto
Wor
kspa
cest
epm
odel
.mdl
1 s+1
Tran
sfer
Fcn
u
To W
orks
pace
2
y
To W
orks
pace
1
t
To W
orks
pace
Step
Cloc
k
Che
ck“S
ave
form
at”o
fout
putb
lock
s(“
Arr
ay”i
nste
adof
“Str
uctu
re”)
>>plot(t,y)
Lect
ure
27
EL2
620
2011
How
ToG
etB
ette
rA
ccur
acy
Mod
ifyR
efine
,Abs
olut
ean
dR
elat
ive
Tole
ranc
es,I
nteg
ratio
nm
etho
d
Refi
nead
dsin
terp
olat
ion
poin
ts:
01
23
45
67
89
10-1
-0.8
-0.6
-0.4
-0.20
0.2
0.4
0.6
0.81
Refin
e =
1
01
23
45
67
89
10-1
-0.8
-0.6
-0.4
-0.20
0.2
0.4
0.6
0.81
Refin
e =
10
Lect
ure
28
EL2
620
2011
Use
Scr
ipts
toD
ocum
entS
imul
atio
ns
Ifth
ebl
ock-
diag
ram
issa
ved
tostepmodel.mdl
,th
efo
llow
ing
Scr
ipt-fi
lesimstepmodel.m
sim
ulat
esth
esy
stem
:
open_system(’stepmodel’)
set_param(’stepmodel’,’RelTol’,’1e-3’)
set_param(’stepmodel’,’AbsTol’,’1e-6’)
set_param(’stepmodel’,’Refine’,’1’)
tic
sim(’stepmodel’,6)
toc
subplot(2,1,1),plot(t,y),title(’y’)
subplot(2,1,2),plot(t,u),title(’u’)
Lect
ure
29
EL2
620
2011
Non
linea
rC
ontr
olS
yste
mE
xam
ple:
Con
trols
yste
mw
ithva
lve
char
acte
ristic
f(u
)=
u2
PS
fr
!1 s
1(s+
1)2
Mot
orVa
lve
Pro
cess
!1
ry
Sim
ulin
kbl
ock
diag
ram
:
1(s
+1)(s
+1)
ys1
-1u2
Lect
ure
210
EL2
620
2011
Exa
mpl
e:Tw
o-Ta
nkS
yste
mTh
esy
stem
cons
ists
oftw
oid
entic
alta
nkm
odel
s:
h=
(u!
q)/A
q=
a!
2g"
h
2 h1 qSu
ms1
Inte
grat
or
1/A Gai
n
f(u)
Fcn
1 In
1 Out
Inq h
Subs
yste
m2
Inq h
Subs
yste
m
1 In
Lect
ure
211
EL2
620
2011
Line
ariz
atio
nin
Sim
ulin
k
Line
ariz
eab
oute
quili
briu
m(x
0,u
0,y
0):
>>
A=2.7e-3;a=7e-6,g=9.8;
>>
[x0,u0,y0]=trim(’twotank’,[0.10.1]’,[],0.1)
x0
=0.1000
0.1000
u0
= 9.7995e-006
y0
=0.1000
>>
[aa,bb,cc,dd]=linmod(’twotank’,x0,u0);
>>
sys=ss(aa,bb,cc,dd);
>>
bode(sys)
Lect
ure
212
EL2
620
2011
Diff
eren
tialE
quat
ion
Edi
tor
dee
isa
Sim
ulin
k-ba
sed
diffe
rent
iale
quat
ion
edito
r
>>dee
deed
emo4
deed
emo3
deed
emo2
deed
emo1
Diffe
rent
ial E
quat
ion
Edi
tor
DEE
Run
the
dem
onst
ratio
ns
Lect
ure
213
EL2
620
2011
Pha
se-P
lane
Ana
lysi
s
•D
ownl
oad
ICTo
ols
from
http://www.control.lth.se/˜ictools
•D
own
load
DFI
ELD
and
PP
LAN
Efro
m
http://math.rice.edu/˜dfield
This
was
the
pref
erre
dto
olla
stye
ar!
Lect
ure
214
EL2
620
2011
Hom
ewor
k1
•U
seyo
urfa
vorit
eph
ase-
plan
ean
alys
isto
ol
•Fo
llow
inst
ruct
ions
inE
xerc
ise
Com
pend
ium
onho
wto
writ
eth
ere
port
.
•S
eeth
eco
urse
hom
epag
efo
rare
port
exam
ple
•Th
ere
port
shou
ldbe
shor
tand
incl
ude
only
nece
ssar
ypl
ots.
Writ
ein
Eng
lish.
Lect
ure
215
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
3
•Sta
bilit
yde
finiti
ons
•Lin
eariz
atio
n
•Pha
se-p
lane
anal
ysis
•Per
iodi
cso
lutio
ns
Lect
ure
31
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•E
xpla
inlo
cala
ndgl
obal
stab
ility
•Li
near
ize
arou
ndeq
uilib
riaan
dtra
ject
orie
s
•S
ketc
hph
ase
port
raits
fort
wo-
dim
ensi
onal
syst
ems
•C
lass
ifyeq
uilib
riain
tono
des,
focu
ses,
sadd
lepo
ints
,and
cent
erpo
ints
•A
naly
zest
abili
tyof
perio
dic
solu
tions
thro
ugh
Poin
care
map
s
Lect
ure
32
EL2
620
2011
Loca
lSta
bilit
yC
onsi
derx
=f(x
)w
ithf(0
)=
0
Defi
nitio
n:Th
eeq
uilib
rium
x!
=0
isst
able
iffo
rall!
>0
ther
eex
ists
"=
"(!)
>0
such
that
!x(0
)!<
""
!x(t
)!<
!,#t
$0
x(t
)
"
!
Ifx
!=
0is
nots
tabl
eit
isca
lled
unst
able
.
Lect
ure
33
EL2
620
2011
Asy
mpt
otic
Sta
bilit
y
Defi
nitio
n:Th
eeq
uilib
rium
x=
0is
asym
ptot
ical
lyst
able
ifit
isst
able
and
"ca
nbe
chos
ensu
chth
at
!x(0
)!<
""
lim
t"#
x(t
)=
0
The
equi
libriu
mis
glob
ally
asym
ptot
ical
lyst
able
ifit
isst
able
and
lim
t"#
x(t
)=
0fo
rallx(0
).
Lect
ure
34
EL2
620
2011
Line
ariz
atio
nA
roun
da
Traj
ecto
ry
Let(
x0(t
),u
0(t
))de
note
aso
lutio
nto
x=
f(x
,u)
and
cons
ider
anot
hers
olut
ion
(x(t
),u(t
))=
(x0(t
)+
x(t
),u
0(t
)+
u(t
)):
x(t
)=
f(x
0(t
)+
x(t
),u
0(t
)+
u(t
))
=f(x
0(t
),u
0(t
))+
#f
#x
(x0(t
),u
0(t
))x(t
)
+#f
#u
(x0(t
),u
0(t
))u(t
)+
O(!
x,u
!2)
(x0(t
),u
0(t
))
(x0(t
)+
x(t
),u
0(t
)+
u(t
))
Lect
ure
35
EL2
620
2011
Hen
ce,f
orsm
all(
x,u
),ap
prox
imat
ely
˙ x(t
)=
A(x
0(t
),u
0(t
))x(t
)+
B(x
0(t
),u
0(t
))u(t
)
whe
re
A(x
0(t
),u
0(t
))=
#f
#x
(x0(t
),u
0(t
))
B(x
0(t
),u
0(t
))=
#f
#u
(x0(t
),u
0(t
))
Not
eth
atA
and
Bar
etim
ede
pend
ent.
How
ever
,if
(x0(t
),u
0(t
))%
(x0,u
0)
then
Aan
dB
are
cons
tant
.
Lect
ure
36
EL2
620
2011
Exa
mpl
e
h(t
)
m(t
)
h(t
)=
v(t
)
v(t
)=
&g
+v e
u(t
)/m
(t)
m(t
)=
&u(t
)
Letx
0(t
)=
(h0(t
),v 0
(t),
m0(t
))T
,u0(t
)%
u0
>0,
bea
solu
tion.
Then
, ˙ x(t
)=
! " #0
10
00
&v
eu0
(m0$
u0t)
2
00
0
$ % &x(t
)+
! #0 ve
m0$
u0t
1
$ &u(t
)
Lect
ure
37
EL2
620
2011
Poin
twis
eLe
ftH
alf-
Pla
neE
igen
valu
esof
A(t
)D
oN
otIm
pose
Sta
bilit
y
A(t
)=
'&
1+
$co
s2t
1&
$si
ntco
st
&1
&$
sin
tco
st
&1
+$
sin
2t
(,
$>
0
Poin
twis
eei
genv
alue
sar
egi
ven
by
%(t
)=
%=
$&
2±
'$
2&
4
2
whi
char
est
able
for0
<$
<2.
How
ever
,
x(t
)=
'e(
!$
1)t
cost
e$tsi
nt
&e(
!$
1)t
sin
te$
tco
st
(x(0
),
isun
boun
ded
solu
tion
for$
>1.
Lect
ure
38
EL2
620
2011
Lyap
unov
’sLi
near
izat
ion
Met
hod
Theo
rem
:Le
tx0
bean
equi
libriu
mof
x=
f(x
)w
ithf
(C
1.
Den
ote
A=
"f
"x(x
0)
and
$(A
)=
max
Re(
%(A
)).
•If
$(A
)<
0,th
enx
0is
asym
ptot
ical
lyst
able
•If
$(A
)>
0,th
enx
0is
unst
able
The
fund
amen
talr
esul
tfor
linea
rsy
stem
sth
eory
!
The
case
$(A
)=
0ne
eds
furt
heri
nves
tigat
ion.
The
theo
rem
isal
soca
lled
Lyap
unov
’sIn
dire
ctM
etho
d.
Apr
oofi
sgi
ven
next
lect
ure.
Lect
ure
39
EL2
620
2011
Exa
mpl
e
The
linea
rizat
ion
of
x1
=&
x2 1+
x1+
sin
x2
x2
=co
sx
2&
x3 1&
5x2
atth
eeq
uilib
rium
x0
=(1
,0)T
isgi
ven
by
A=
' &1
1
&3
&5(
,%(A
)=
{&2,
&4}
x0
isth
usan
asym
ptot
ical
lyst
able
equi
libriu
mfo
rthe
nonl
inea
rsy
stem
.
Lect
ure
310
EL2
620
2011
Line
arS
yste
ms
Rev
ival
d dt
) x1
x2
*=
A
) x1
x2
*
Ana
lytic
solu
tion:
x(t
)=
eAt x
(0),
t$
0.
IfA
isdi
agon
aliz
able
,the
n
eAt=
Ve!
t V$
1=
+ v 1v 2
,) e#1t
0
0e#
2t* + v 1
v 2, $
1
whe
rev 1
,v2
are
the
eige
nvec
tors
ofA
(Av 1
=%
1v 1
etc)
.
This
impl
ies
that
x(t
)=
c 1e#
1t v
1+
c 2e#
2t v
2,
whe
reth
eco
nsta
ntsc 1
and
c 2ar
egi
ven
byth
ein
itial
cond
ition
s
Lect
ure
311
EL2
620
2011
Exa
mpl
e:Tw
ore
alne
gativ
eei
genv
alue
s
Giv
enth
eei
genv
alue
s%
1<
%2
<0,
with
corr
espo
ndin
gei
genv
ecto
rsv 1
and
v 2,r
espe
ctiv
ely.
Sol
utio
n:x(t
)=
c 1e#
1t v
1+
c 2e#
2t v
2
Slo
wei
genv
alue
/vec
tor:
x(t
))
c 2e#
2t v
2fo
rlar
get.
Mov
esal
ong
the
slow
eige
nvec
tort
owar
dsx
=0
forl
arge
t
Fast
eige
nval
ue/v
ecto
r:x(t
))
c 1e#
1t v
1+
c 2v 2
fors
mal
lt.
Mov
esal
ong
the
fast
eige
nvec
torf
orsm
allt
Lect
ure
312
EL2
620
2011
Pha
se-P
lane
Ana
lysi
sfo
rLi
near
Sys
tem
s
The
loca
tion
ofth
eei
genv
alue
s%(A
)de
term
ines
the
char
acte
ristic
sof
the
traje
ctor
ies.
Six
case
s:
stab
leno
deun
stab
leno
desa
ddle
poin
t
Im%
i=
0:
%1,%
2<
0%
1,%
2>
0%
1<
0<
%2
Im%
i*=
0:
Re%
i<
0R
e%
i>
0R
e%
i=
0
stab
lefo
cus
unst
able
focu
sce
nter
poin
t
Lect
ure
313
EL2
620
2011
Equ
ilibr
ium
Poin
tsfo
rLi
near
Sys
tem
s
Re
!
Im!
x1
x2
Lect
ure
314
EL2
620
2011
Exa
mpl
e—U
nsta
ble
Focu
s
x=
) &&
'
'&
* x,
&,'
>0,
%1,2
=&
±i'
x(t
)=
eAt x
(0)
=
)1
1
&i
i*)e$
t ei%
t0
0e$
t e$
i%t*)
11
&i
i* $1
x(0
)
Inpo
larc
oord
inat
esr
=-
x2 1+
x2 2,(
=ar
ctan
x2/x
1
(x1
=rco
s(,
x2
=rsi
n()
:
r=
&r
(=
'
Lect
ure
315
EL2
620
2011
%1,2
=1
±i
%1,2
=0.
3±
i
Lect
ure
316
EL2
620
2011
Exa
mpl
e—S
tabl
eN
ode
x=
) &1
1
0&
2* x
(%1,%
2)
=(&
1,&
2)an
d+ v 1
v 2, =
) 1&
1
01
*
v 1is
the
slow
dire
ctio
nan
dv 2
isth
efa
st.
Lect
ure
317
EL2
620
2011
Fast
:x
2=
&x
1+
c 3S
low
:x
2=
0
Lect
ure
318
EL2
620
2011
5m
inut
eex
erci
se:
Wha
tis
the
phas
epo
rtra
itif%
1=
%2?
Hin
t:Tw
oca
ses;
only
one
linea
rind
epen
dent
eige
nvec
toro
rall
vect
ors
are
eige
nvec
tors
Lect
ure
319
EL2
620
2011
Pha
se-P
lane
Ana
lysi
sfo
rN
onlin
ear
Sys
tem
s
Clo
seto
equi
libriu
mpo
ints
“non
linea
rsys
tem
”)“li
near
syst
em”
Theo
rem
:A
ssum
e
x=
f(x
)=
Ax
+g(x
),
with
lim
%x%"
0!g
(x)!
/!x!
=0.
Ifz
=A
zha
sa
focu
s,no
de,o
rsa
ddle
poin
t,th
enx
=f(x
)ha
sth
esa
me
type
ofeq
uilib
rium
atth
eor
igin
.
Rem
ark:
Ifth
elin
eariz
edsy
stem
has
ace
nter
,the
nth
eno
nlin
ear
syst
emha
sei
ther
ace
nter
ora
focu
s.
Lect
ure
320
EL2
620
2011
How
toD
raw
Pha
sePo
rtra
its
By
hand
:
1.Fi
ndeq
uilib
ria
2.S
ketc
hlo
calb
ehav
iora
roun
deq
uilib
ria
3.S
ketc
h(x
1,x
2)
fors
ome
othe
rpoi
nts.
Not
ice
that
dx
2
dx
1
=x
1
x2
4.Tr
yto
find
poss
ible
perio
dic
orbi
ts
5.G
uess
solu
tions
By
com
pute
r:
1.M
atla
b:dee
orpplane
Lect
ure
321
EL2
620
2011
Pha
se-L
ocke
dLo
op
AP
LLtra
cks
phas
e( i
n(t
)of
asi
gnal
s in(t
)=
Asi
n['
t+
( in(t
)].
Pha
seD
etec
tor
Filte
rV
CO
s in
“(ou
t”
sin(·)
&eK
1+
sT
1 s
( in
( out
( out
Lect
ure
322
EL2
620
2011
Pha
se-P
lane
Ana
lysi
sof
PLL
Let(
x1,x
2)
=((
out,( o
ut),
K,T
>0,
and
( in(t
)%
( in.
x1(t
)=
x2(t
)
x2(t
)=
&T
$1x
2(t
)+
KT
$1si
n((
in&
x1(t
))
Equ
ilibr
iaar
e((
in+
n),0
)si
nce
x1
=0
"x
2=
0
x2
=0
"si
n((
in&
x1)
=0
"x
1=
( in+
n)
Lect
ure
323
EL2
620
2011
Cla
ssifi
catio
nof
Equ
ilibr
ia
Line
ariz
atio
ngi
ves
the
follo
win
gch
arac
teris
ticeq
uatio
ns:
nev
en:
%2+
T$
1%
+K
T$
1=
0
K>
(4T
)$1
give
sst
able
focu
s0
<K
<(4
T)$
1gi
ves
stab
leno
de
nod
d:%
2+
T$
1%
&K
T$
1=
0
Sad
dle
poin
tsfo
rallK
,T>
0
Lect
ure
324
EL2
620
2011
Pha
se-P
lane
for
PLL
(K,T
)=
(1/2,1
):fo
cuse
s. 2
k!,0
/ ,sad
dle
poin
ts. (2
k+
1)!,0
/
Lect
ure
325
EL2
620
2011
Per
iodi
cS
olut
ions
Exa
mpl
eof
anas
ympt
otic
ally
stab
lepe
riodi
cso
lutio
n:
x1
=x
1&
x2&
x1(x
2 1+
x2 2)
x2
=x
1+
x2&
x2(x
2 1+
x2 2)
(1)
Lect
ure
326
EL2
620
2011
Per
iodi
cso
lutio
n:Po
lar
coor
dina
tes.
x1
=rco
s(
"x
1=
cos(r
&rsi
n((
x2
=rsi
n(
"x
2=
sin
(r+
rco
s((
impl
ies
'r (
(=
1 r
'rco
s(
rsi
n(
&si
n(
cos(
('
x1
x2
(
Now
,fro
m(1
)x
1=
r(1
&r2
)co
s(
&rsi
n(
x2
=r(
1&
r2)si
n(
+rco
s(
give
sr
=r(
1&
r2)
(=
1
Onl
yr
=1
isa
stab
leeq
uilib
rium
!
Lect
ure
327
EL2
620
2011
Asy
stem
has
ape
riod
icso
lutio
nif
fors
ome
T>
0
x(t
+T
)=
x(t
),#t
$0
Ape
riod
icor
biti
sth
eim
age
ofx
inth
eph
ase
port
rait.
•W
hen
does
ther
eex
ista
perio
dic
solu
tion?
•W
hen
isit
stab
le?
Not
eth
atx(t
)%
cons
tis
byco
nven
tion
notr
egar
ded
perio
dic
Lect
ure
328
EL2
620
2011
Flow
The
solu
tion
ofx
=f(x
)is
som
etim
esde
note
d
*t(
x0)
toem
phas
ize
the
depe
nden
ceon
the
initi
alpo
intx
0(
Rn
*t(
·)is
calle
dth
eflo
w.
Lect
ure
329
EL2
620
2011
Poin
care
Map
Ass
ume
*t(
x0)
isa
perio
dic
solu
tion
with
perio
dT
.
Let!
+R
nbe
ann
&1-
dim
hype
rpla
netra
nsve
rse
tof
atx
0.
Defi
nitio
n:Th
ePo
inca
rem
apP
:!
,!
is
P(x
)=
*&(x
)(x)
whe
re+(x
)is
the
time
offir
stre
turn
.
x*
t(x)
! P(x
)
Lect
ure
330
EL2
620
2011
Exi
sten
ceof
Per
iodi
cO
rbits
Apo
intx
!su
chth
atP
(x! )
=x
!co
rres
pond
sto
ape
riodi
cor
bit.
P(x
! )=
x!
x!
isca
lled
afix
edpo
into
fP.
Lect
ure
331
EL2
620
2011
1m
inut
eex
erci
se:
Wha
tdoe
sa
fixed
poin
tofP
kco
rres
pond
sto
?
Lect
ure
332
EL2
620
2011
Sta
ble
Per
iodi
cO
rbit
The
linea
rizat
ion
ofP
arou
ndx
!gi
ves
am
atrix
Wsu
chth
at
P(x
))
Wx
ifx
iscl
ose
tox
! .
•%
j(W
)=
1fo
rsom
ej
•If
|%i(W
)|<
1fo
ralli*=
j,th
enth
eco
rres
pond
ing
perio
dic
orbi
tis
asym
ptot
ical
lyst
able
•If
|%i(W
)|>
1fo
rsom
ei,
then
the
perio
dic
orbi
tis
unst
able
.
Lect
ure
333
EL2
620
2011
Exa
mpl
e—S
tabl
eU
nitC
ircl
e
Rew
rite
(1)i
npo
larc
oord
inat
es:
r=
r(1
&r2
)
(=
1
Cho
ose
!=
{(r,
():
r>
0,(
=2)
k}.
The
solu
tion
is
*t(
r 0,(
0)
=
' [1+
(r$
20
&1)
e$2t ]
$1/2,t
+( 0
(
Firs
tret
urn
time
from
any
poin
t(r 0
,(0)(
!is
+(r
0,(
0)
=2)
.
Lect
ure
334
EL2
620
2011
The
Poin
care
map
is
P(r
0,(
0)
=
' [1+
(r$
20
&1)
e$2·2
']$
1/2
( 0+
2)
(
(r0,(
0)
=(1
,2)k)
isa
fixed
poin
t.
The
perio
dic
solu
tion
that
corr
espo
nds
to(r
(t),
((t)
)=
(1,t
)is
asym
ptot
ical
lyst
able
beca
use
W=
dP
d(r
0,(
0)(1
,2)k)
=
' e$4'
0
01(
"S
tabl
epe
riodi
cor
bit(
asw
eal
read
ykn
ewfo
rthi
sex
ampl
e)!
Lect
ure
335
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
5
•Inp
ut–o
utpu
tsta
bilit
y
uy
S
Lect
ure
51
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•de
rive
the
gain
ofa
syst
em
•an
alyz
est
abili
tyus
ing
–S
mal
lGai
nTh
eore
m
–C
ircle
Crit
erio
n
–P
assi
vity
Lect
ure
52
EL2
620
2011
His
tory
!r
yG
(s)
f(·)
y
f(y
)
Forw
hatG
(s)
and
f(·)
isth
ecl
osed
-loop
syst
emst
able
?
•Lu
rean
dPo
stni
kov’
spr
oble
m(1
944)
•A
izer
man
’sco
njec
ture
(194
9)(F
alse
!)
•K
alm
an’s
conj
ectu
re(1
957)
(Fal
se!)
•S
olut
ion
byPo
pov
(196
0)(L
edto
the
Circ
leC
riter
ion)
Lect
ure
53
EL2
620
2011
Gai
n
Idea
:G
ener
aliz
eth
eco
ncep
tofg
ain
tono
nlin
eard
ynam
ical
syst
ems
uy
S
The
gain
!of
Sis
the
larg
esta
mpl
ifica
tion
from
uto
y
Her
eS
can
bea
cons
tant
,am
atrix
,alin
eart
ime-
inva
riant
syst
em,e
tc
Que
stio
n:H
owsh
ould
we
mea
sure
the
size
ofu
and
y?
Lect
ure
54
EL2
620
2011
Nor
ms
Ano
rm"·
"mea
sure
ssi
ze
Defi
nitio
n:A
norm
isa
func
tion
"·"
:!
#R
+,s
uch
that
fora
llx,y
$!
•"x
"%
0an
d"x
"=
0&
x=
0
•"x
+y"
'"x
"+"y
"•
""x"
=|"
|·"x
",fo
rall"
$R
Exa
mpl
es:
Euc
lidea
nno
rm:"x
"=
!x
2 1+
···+
x2 n
Max
norm
:"x
"=
max
{|x
1|,.
..,|x
n|}
Lect
ure
55
EL2
620
2011
Gai
nof
aM
atri
x
Eve
rym
atrix
M$
Cn!
nha
sa
sing
ular
valu
ede
com
posi
tion
M=
U"
V"
whe
re
"=
diag
{#1,.
..,#
n};
U" U
=I
;V
" V=
I
#i
-the
sing
ular
valu
es
The
“gai
n”of
Mis
the
larg
ests
ingu
larv
alue
ofM
:
#m
ax(M
)=
#1
=su
px#R
n
"Mx"
"x"
whe
re"·
"is
the
Euc
lidea
nno
rm.
Lect
ure
56
EL2
620
2011
Eig
enva
lues
are
notg
ains
The
spec
tralr
adiu
sof
am
atrix
M
$(M
)=
max i
|%i(M
)|
isno
taga
in(n
ora
norm
).
Why
?W
hata
mpl
ifica
tion
isde
scrib
edby
the
eige
nval
ues?
Lect
ure
57
EL2
620
2011
Sig
nalN
orm
s
Asi
gnal
xis
afu
nctio
nx
:R
+#
R.
Asi
gnal
norm
"·" k
isa
norm
onth
esp
ace
ofsi
gnal
sx
.
Exa
mpl
es:
2-no
rm(e
nerg
yno
rm):
"x" 2
="
# $ 0|x
(t)|2
dt
sup-
norm
:"x
" $=
sup
t#R
+|x
(t)|
Lect
ure
58
EL2
620
2011
Par
seva
l’sTh
eore
mL 2
deno
tes
the
spac
eof
sign
als
with
boun
ded
ener
gy:"x
" 2<
(Th
eore
m:
Ifx,y
$L 2
have
the
Four
iert
rans
form
s
X(i
&)
=
$$
0
e%i!
t x(t
)dt,
Y(i
&)
=
$$
0
e%i!
t y(t
)dt,
then
$$
0
y(t
)x(t
)dt=
1 2'
$$ %$
Y" (
i&)X
(i&)d
&.
Inpa
rtic
ular
,
"x"2 2
=
$$
0
|x(t
)|2dt=
1 2'
$$ %$
|X(i
&)|2
d&.
The
pow
erca
lcul
ated
inth
etim
edo
mai
neq
uals
the
pow
erca
lcul
ated
inth
efre
quen
cydo
mai
n
Lect
ure
59
EL2
620
2011
Sys
tem
Gai
n
Asy
stem
Sis
am
apfro
mL 2
toL 2
:y
=S
(u).
uy
S
The
gain
ofS
isde
fined
as!(S
)=
sup
u#L
2
"y" 2
"u" 2
=su
pu#L
2
"S(u
)"2
"u" 2
Exa
mpl
e:Th
ega
inof
asc
alar
stat
icsy
stem
y(t
)=
"u(t
)is
!("
)=
sup
u#L
2
""u" 2
"u" 2
=su
pu#L
2
|"|"
u" 2
"u" 2
=|"
|
Lect
ure
510
EL2
620
2011
2m
inut
eex
erci
se:
Sho
wth
at!(S
1S
2)'
!(S
1)!
(S2).
uy
S2
S1
Lect
ure
511
EL2
620
2011
Gai
nof
aS
tatic
Non
linea
rity
Lem
ma:
Ast
atic
nonl
inea
rityf
such
that
|f(x
)|'
K|x
|and
f(x
" )=
Kx
"ha
sga
in!(f
)=
K.
u(t
)y(t
)f(·)
xx
"Kx
f(x
)
Pro
of:"y
"2 2=
# $ 0f
2% u
(t)& d
t'
# $ 0K
2u
2(t
)dt=
K2"u
"2 2,
whe
reu(t
)=
x" ,t
$(0
,1),
give
seq
ualit
y,so
!(f
)=
sup
u#L
2
"y" 2
' "u" 2
=K
Lect
ure
512
EL2
620
2011
Gai
nof
aS
tabl
eLi
near
Sys
tem
Lem
ma:
!% G
& =su
pu#L
2
"Gu" 2
"u" 2
=su
p!
#(0,$
)
|G(i
&)|
10-1
100
101
10-2
10-1
100
101
!(G
)|G
(i&)|
Pro
of:
Ass
ume
|G(i
&)|
'K
for&
$(0
,()
and
|G(i
&" )
|=K
fors
ome
&" .
Par
seva
l’sth
eore
mgi
ves
"y"2 2
=1 2'
$$ %$
|Y(i
&)|2
d&
=1 2'
$$ %$
|G(i
&)|2
|U(i
&)|2
d&
'K
2"u
"2 2
Arb
itrar
ycl
ose
toeq
ualit
yby
choo
sing
u(t
)cl
ose
tosi
n&
" t.
Lect
ure
513
EL2
620
2011
BIB
OS
tabi
lity
Defi
nitio
n:S
isbo
unde
d-in
putb
ound
ed-o
utpu
t(B
IBO
)sta
ble
if!(S
)<
(.
uy
S!% S
& =su
pu#L
2
"S(u
)"2
"u" 2
Exa
mpl
e:If
x=
Ax
isas
ympt
otic
ally
stab
leth
enG
(s)
=C
(sI
!A
)%1B
+D
isB
IBO
stab
le.
Lect
ure
514
EL2
620
2011
The
Sm
allG
ain
Theo
rem
r 1
r 2
e 1
e 2
S1
S2
Theo
rem
:A
ssum
eS
1an
dS
2ar
eB
IBO
stab
le.
If!(S
1)!
(S2)<
1,th
enth
ecl
osed
-loop
syst
emis
BIB
Ost
able
from
(r1,r
2)
to(e
1,e
2)
Lect
ure
515
EL2
620
2011
Exa
mpl
e—S
tatic
Non
linea
rFe
edba
ck
!r
yG
(s)
f(·)
y
Ky
f(y
)
G(s
)=
2
(s+
1)2,
0'
f(y
)
y'
K,
)y*=
0,f(0
)=
0
!(G
)=
2an
d!(f
)'
K.
Sm
allG
ain
Theo
rem
give
sB
IBO
stab
ility
forK
$(0
,1/2
).
Lect
ure
516
EL2
620
2011
“Pro
of”
ofth
eS
mal
lGai
nTh
eore
m
"e1" 2
'"r
1" 2
+!(S
2)["r
2" 2
+!(S
1)"
e 1" 2
]
give
s
"e1" 2
'"r
1" 2
+!(S
2)"
r 2" 2
1!
!(S
2)!
(S1)
!(S
2)!
(S1)<
1,"r
1" 2
<(
,"r 2
" 2<
(gi
ve"e
1" 2
<(
.S
imila
rlyw
ege
t
"e2" 2
'"r
2" 2
+!(S
1)"
r 1" 2
1!
!(S
1)!
(S2)
soal
soe 2
isbo
unde
d.
Not
e:Fo
rmal
proo
freq
uire
s"·
" 2e,s
eeK
halil
Lect
ure
517
EL2
620
2011
The
Nyq
uist
Theo
rem
!G
(s)
-1-0
.50
0.5
1-1
-0.50
0.51
From
: U(1
)
To: Y(1)
Theo
rem
:If
Gha
sno
pole
sin
the
right
half
plan
ean
dth
eN
yqui
stcu
rve
G(i
&),
&$
[0,(
),do
esno
tenc
ircle
!1,
then
the
clos
ed-lo
opsy
stem
isst
able
.
Lect
ure
518
EL2
620
2011
Sm
allG
ain
Theo
rem
can
beC
onse
rvat
ive
Letf
(y)
=K
yin
the
prev
ious
exam
ple.
Then
the
Nyq
uist
Theo
rem
prov
esst
abili
tyfo
rallK
$[0
,(),
whi
leth
eS
mal
lGai
nTh
eore
mon
lypr
oves
stab
ility
forK
$(0
,1/2
).
!r
yG
(s)
f(·)
-1-0
.50
0.5
11.
52
-1
-0.50
0.51
G(i
&)
Lect
ure
519
EL2
620
2011
The
Cir
cle
Cri
teri
on
!r
yG
(s)
f(·)
yk1y
k2y
f(y
)
!1 k1
!1 k2
G(i
&)
Theo
rem
:A
ssum
eth
atG
(s)
has
nopo
les
inth
erig
htha
lfpl
ane,
and
0'
k1
'f(y
)
y'
k2,
)y*=
0,f(0
)=
0
Ifth
eN
yqui
stcu
rve
ofG
(s)
does
note
ncirc
leor
inte
rsec
tthe
circ
lede
fined
byth
epo
ints
!1/
k1
and
!1/
k2,t
hen
the
clos
ed-lo
opsy
stem
isB
IBO
stab
le.
Lect
ure
520
EL2
620
2011
Exa
mpl
e—S
tatic
Non
linea
rFe
edba
ck(c
ont’d
)
y
Ky
f(y
)
-1-0
.50
0.5
11.
52
-1
-0.50
0.51
!1 K
G(i
&)
The
“circ
le”i
sde
fined
by!
1/k
1=
!(
and
!1/
k2
=!
1/K
.S
ince
min
ReG
(i&)
=!
1/4
the
Circ
leC
riter
ion
give
sth
atth
esy
stem
isB
IBO
stab
leifK
$(0
,4).
Lect
ure
521
EL2
620
2011
Pro
ofof
the
Cir
cle
Cri
teri
on
Letk
=(k
1+
k2)/
2,( f(
y)
=f(y
)!
ky
,and
(r 1=
r 1!
kr 2
:
) ) ) )( f(y)
y
) ) ) )'k
2!
k1
2=
:R
,)y
*=0,
( f(0)
=0
y 1
y 2
r 1
r 2
e 1
e 2
G(s
)
!f(·)
(r 1
( G
G!k
y 1
r 2!
( f(·)
Lect
ure
522
EL2
620
2011
(r 1
r 2
( G(s
)
!( f(
·)
!k
R
1
G(i
&)
Sm
allG
ain
Theo
rem
give
sst
abili
tyif|( G
(i&)|R
<1,
whe
re
( G=
G
1+
kG
isst
able
(Thi
sha
sto
bech
ecke
dla
ter)
.H
ence
,
1
|( G(i
&)|
=
) ) ) )1
G(i
&)
+k
) ) ) )>R
Lect
ure
523
EL2
620
2011
The
curv
eG
%1(i
&)
and
the
circ
le{z
$C
:|z
+k|>
R}m
appe
dth
roug
hz
+#1/
zgi
ves
the
resu
lt:
!k2
!k1
R
1
G(i
&)
!1 k1
!1 k2
G(i
&)
Not
eth
atG
1+
kG
isst
able
sinc
e!
1/k
isin
side
the
circ
le.
Not
eth
atG
(s)
may
have
pole
son
the
imag
inar
yax
is,e
.g.,
inte
grat
ors
are
allo
wed
Lect
ure
524
EL2
620
2011
Pas
sivi
tyan
dB
IBO
Sta
bilit
y
The
mai
nre
sult:
Feed
back
inte
rcon
nect
ions
ofpa
ssiv
esy
stem
sar
epa
ssiv
e,an
dB
IBO
stab
le(u
nder
som
ead
ditio
nalm
ildcr
iteria
)
Lect
ure
525
EL2
620
2011
Sca
lar
Pro
duct
Sca
lar
prod
uctf
orsi
gnal
sy
and
u
,y,u
- T=
$T
0
yT(t
)u(t
)dt
uy
S
Ifu
and
yar
ein
terp
rete
das
vect
ors
then
,y,u
- T=
|y| T
|u| T
cos(
|y| T
=!
,y,y
- T-l
engt
hof
y,(
-ang
lebe
twee
nu
and
y
Cau
chy-
Sch
war
zIn
equa
lity:
,y,u
- T'
|y| T
|u| T
Exa
mpl
e:u
=si
nt
and
y=
cost
are
orth
ogon
alifT
=k'
,be
caus
e
cos(
=,y
,u- T
|y| T
|u| T
=0
Lect
ure
526
EL2
620
2011
Pas
sive
Sys
tem
uy
S
Defi
nitio
n:C
onsi
ders
igna
lsu,y
:[0
,T]#
Rm
.Th
esy
stem
Sis
pass
ive
if,y
,u- T
%0,
fora
llT
>0
and
allu
and
stri
ctly
pass
ive
ifth
ere
exis
ts)
>0
such
that
,y,u
- T%
)(|y
|2 T+
|u|2 T
),fo
rallT
>0
and
allu
War
ning
:Th
ere
exis
tman
yot
herd
efini
tions
fors
tric
tlypa
ssiv
e
Lect
ure
527
EL2
620
2011
2m
inut
eex
erci
se:
Isth
epu
rede
lay
syst
emy(t
)=
u(t
!*)
pass
ive?
Con
side
rfor
inst
ance
the
inpu
tu(t
)=
sin
(('/*
)t).
Lect
ure
528
EL2
620
2011
Exa
mpl
e—P
assi
veE
lect
rica
lCom
pone
nts
u(t
)=
Ri(
t):,u
,i- T
=
$T
0
Ri2
(t)d
t%
R,i,
i- T%
0
i=
Cdu dt
:,u
,i- T
=
$T
0
u(t
)Cdu dtdt=
Cu
2(T
)
2%
0
u=
Ldi
dt
:,u
,i- T
=
$T
0
Ldi
dti(
t)dt=
Li2
(T)
2%
0
Lect
ure
529
EL2
620
2011
Feed
back
ofP
assi
veS
yste
ms
isP
assi
ve
!r 1
r 2
y 1
y 2
e 1
e 2
S1
S2
Lem
ma:
IfS
1an
dS
2ar
epa
ssiv
eth
enth
ecl
osed
-loop
syst
emfro
m(r
1,r
2)
to(y
1,y
2)
isal
sopa
ssiv
e.
Pro
of: ,y,e
- T=
,y1,e
1- T
+,y
2,e
2- T
=,y
1,r
1!
y 2- T
+,y
2,r
2+
y 1- T
=,y
1,r
1- T
+,y
2,r
2- T
=,y
,r- T
Hen
ce,,
y,r
- T%
0if,y
1,e
1- T
%0
and
,y2,e
2- T
%0.
Lect
ure
530
EL2
620
2011
Pas
sivi
tyof
Line
arS
yste
ms
Theo
rem
:A
nas
ympt
otic
ally
stab
lelin
ears
yste
mG
(s)
ispa
ssiv
eif
and
only
ifR
eG
(i&)%
0,)&
>0
Itis
stri
ctly
pass
ive
ifan
don
lyif
ther
eex
ists
)>
0su
chth
at
ReG
(i&
!))
%0,
)&>
0
Exa
mpl
e:
G(s
)=
1
s+
1is
stric
tlypa
ssiv
e,
G(s
)=
1 sis
pass
ive
but
not
stric
tly
pass
ive.
00.
20.
40.
60.
81
-0.4
-0.20
0.2
0.4
0.6
G(i
&)
Lect
ure
531
EL2
620
2011
AS
tric
tlyP
assi
veS
yste
mH
asFi
nite
Gai
n
uy
S
Lem
ma:
IfS
isst
rictly
pass
ive
then
!(S
)=
sup
u#L
2
&y& 2
&u& 2
<(
.
Pro
of: )(|y
|2 $+
|u|2 $
)'
,y,u
- $'
|y| $
·|u| $
="y
" 2·"
u" 2
Hen
ce,)
"y"2 2
'"y
" 2·"
u" 2
,so
"y" 2
'1 )"u
" 2
Lect
ure
532
EL2
620
2011
The
Pas
sivi
tyTh
eore
m
!r 1
r 2
y 1
y 2
e 1
e 2
S1
S2
Theo
rem
:If
S1
isst
rictly
pass
ive
and
S2
ispa
ssiv
e,th
enth
ecl
osed
-loop
syst
emis
BIB
Ost
able
from
rto
y.
Lect
ure
533
EL2
620
2011
Pro
ofS
1st
rictly
pass
ive
and
S2
pass
ive
give
)% |y 1|2 T
+|e 1
|2 T
& ',y
1,e
1- T
+,y
2,e
2- T
=,y
,r- T
Ther
efor
e
|y 1|2 T
+,r
1!
y 2,r
1!
y 2- T
'1 ),y
,r- T
or
|y 1|2 T
+|y 2
|2 T!
2,y 2
,r2- T
+|r 1
|2 T'
1 ),y
,r- T
Hen
ce
|y|2 T
'2,
y 2,r
2- T
+1 ),y
,r- T
'* 2
+1 )
+ |y| T
|r|T
LetT
#(
and
the
resu
ltfo
llow
s.
Lect
ure
534
EL2
620
2011
The
Pas
sivi
tyTh
eore
mis
a“S
mal
lPha
seTh
eore
m”
!r 1
r 2
y 1
y 2e 1
e 2
S1
S2
(2
(1
S2
pass
ive
.co
s(
2%
0.
|(2|'
'/2
S1
stric
tlypa
ssiv
e.
cos(
1>
0.
|(1|<
'/2
Lect
ure
535
EL2
620
2011
!G
(s)
2m
inut
eex
erci
se:
App
lyth
eP
assi
vity
Theo
rem
and
com
pare
itw
ithth
eN
yqui
stTh
eore
m.
Wha
tabo
utco
nser
vativ
enes
s?[C
ompa
reth
edi
scus
sion
onth
eS
mal
lGai
nTh
eore
m.]
Lect
ure
536
EL2
620
2011
Exa
mpl
e—G
ain
Ada
ptat
ion
App
licat
ions
inte
leco
mm
unic
atio
nch
anne
lest
imat
ion
and
inno
ise
canc
ella
tion
etc.
Mod
el
Pro
cess
u*"
*(t)
G(s
)
G(s
)
y y m
Ada
ptat
ion
law
: d* dt
=!
cu(t
)[y m
(t)!
y(t
)],
c>
0.
Lect
ure
537
EL2
620
2011
Gai
nA
dapt
atio
n—C
lose
d-Lo
opS
yste
m
repl
acem
ents
u
!!
c s
*"
*(t)
G(s
)
G(s
)
y y m
*
Lect
ure
538
EL2
620
2011
Gai
nA
dapt
atio
nis
BIB
OS
tabl
e
uS
*" *
(*!
*" )u
y m!
y
!!
c s
G(s
)
Sis
pass
ive
(see
exer
cise
s).
IfG
(s)
isst
rict
lypa
ssiv
e,th
ecl
osed
-loop
syst
emis
BIB
Ost
able
Lect
ure
539
EL2
620
2011
Sim
ulat
ion
ofG
ain
Ada
ptat
ion
LetG
(s)
=1
s+
1,c
=1,
u=
sin
t,an
d*(
0)=
0.
05
1015
20-202 0
510
1520
0
0.51
1.5
y,y
m *
Lect
ure
540
EL2
620
2011
Sto
rage
Func
tion
Con
side
rthe
nonl
inea
rcon
trols
yste
m
x=
f(x
,u),
y=
h(x
)
Ast
orag
efu
nctio
nis
aC
1fu
nctio
nV
:R
n#
Rsu
chth
at
•V
(0)
=0
and
V(x
)%
0,)x
*=0
•V
(x)'
uTy
,)x
,u
Rem
ark:
•V
(T)
repr
esen
tsth
est
ored
ener
gyin
the
syst
em
•V
(x(T
)),
-./
stor
eden
ergy
att=
T
'$
T
0
y(t
)u(t
)dt
,-.
/ab
sorb
eden
ergy
+V
(x(0
)),
-./
stor
eden
ergy
att=
0
,)T
>0
Lect
ure
541
EL2
620
2011
Sto
rage
Func
tion
and
Pas
sivi
ty
Lem
ma:
Ifth
ere
exis
tsa
stor
age
func
tion
Vfo
rasy
stem
x=
f(x
,u),
y=
h(x
)
with
x(0
)=
0,th
enth
esy
stem
ispa
ssiv
e.
Pro
of:
Fora
llT
>0,
,y,u
- T=
$T
0
y(t
)u(t
)dt%
V(x
(T))
!V
(x(0
))=
V(x
(T))
%0
Lect
ure
542
EL2
620
2011
Lyap
unov
vs.P
assi
vity
Sto
rage
func
tion
isa
gene
raliz
atio
nof
Lyap
unov
func
tion
Lyap
unov
idea
:“E
nerg
yis
decr
easi
ng”
V'
0
Pas
sivi
tyid
ea:
“Incr
ease
inst
ored
ener
gy'
Add
eden
ergy
”
V'
uTy
Lect
ure
543
EL2
620
2011
Exa
mpl
e—K
YP
Lem
ma
Con
side
ran
asym
ptot
ical
lyst
able
linea
rsys
tem
x=
Ax
+B
u,
y=
Cx
Ass
ume
ther
eex
ists
posi
tive
defin
item
atric
esP,Q
such
that
ATP
+P
A=
!Q
,B
TP
=C
Con
side
rV=
0.5x
TP
x.
Then
V=
0.5(
xTP
x+
xTP
x)
=0.
5xT(A
TP
+P
A)x
+uB
TP
x
=!
0.5x
TQ
x+
uy
<uy,
x*=
0
and
henc
eth
esy
stem
isst
rictly
pass
ive.
This
fact
ispa
rtof
the
Kal
man
-Yak
ubov
ich-
Popo
vle
mm
a.
Lect
ure
544
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•de
rive
the
gain
ofa
syst
em
•an
alyz
est
abili
tyus
ing
–S
mal
lGai
nTh
eore
m
–C
ircle
Crit
erio
n
–P
assi
vity
Lect
ure
545
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
6
•Des
crib
ing
func
tion
anal
ysis
Lect
ure
61
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•D
eriv
ede
scrib
ing
func
tions
fors
tatic
nonl
inea
ritie
s
•A
naly
zeex
iste
nce
and
stab
ility
ofpe
riodi
cso
lutio
nsby
desc
ribin
gfu
nctio
nan
alys
is
Lect
ure
62
EL2
620
2011
Mot
ivat
ing
Exa
mpl
e
re
uy
!G
(s)
05
1015
20
-2-1012y
u
G(s
)=
4
s(s
+1)
2an
du
=sa
tegi
vea
stab
leos
cilla
tion.
•H
owca
nth
eos
cilla
tion
bepr
edic
ted?
Lect
ure
63
EL2
620
2011
AFr
eque
ncy
Res
pons
eA
ppro
ach
Nyq
uist
/Bod
e:
A(li
near
)fee
dbac
ksy
stem
will
have
sust
aine
dos
cilla
tions
(cen
ter)
ifth
elo
op-g
ain
is1
atth
efre
quen
cyw
here
the
phas
ela
gis
!18
0o
But
,can
we
talk
abou
tthe
frequ
ency
resp
onse
,in
term
sof
gain
and
phas
ela
g,of
ast
atic
nonl
inea
rity?
Lect
ure
64
EL2
620
2011
Four
ier
Ser
ies
Ape
riodi
cfu
nctio
nu(t
)=
u(t
+T
)ha
sa
Four
iers
erie
sex
pans
ion
u(t
)=
a0 2
+! ! n=
1
(anco
sn!t+
b nsi
nn!t)
=a
0 2+
! ! n=
1
"a
2 n+
b2 nsi
n[n
!t+
arct
an(a
n/b
n)]
whe
re!
=2"
/Tan
d
an(!
)=
2 T
#T
0
u(t
)co
sn!tdt,
b n(!
)=
2 T
#T
0
u(t
)si
nn!tdt
Not
e:S
omet
imes
we
mak
eth
ech
ange
ofva
riabl
et"
#/!
Lect
ure
65
EL2
620
2011
The
Four
ier
Coe
ffici
ents
are
Opt
imal
The
finite
expa
nsio
n
$u k(t
)=
a0 2
+k ! n=
1
(anco
sn!t+
b nsi
nn!t)
solv
es
min u
2 T
#T
0
% u(t
)!
$u k(t
)& 2dt
Lect
ure
66
EL2
620
2011
Key
Idea
re
uy
!N
.L.
G(s
)
e(t)
=A
sin
!t
give
s
u(t
)=
! ! n=
1
"a
2 n+
b2 nsi
n[n
!t+
arct
an(a
n/b
n)]
If|G
(in!)|
#|G
(i!)|
forn
$2,
then
n=
1su
ffice
s,so
that
y(t
)%
|G(i
!)|'
a2 1+
b2 1si
n[!
t+
arct
an(a
1/b
1)+
arg
G(i
!)]
That
is,w
eas
sum
eal
lhig
her
harm
onic
sar
efil
tere
dou
tby
G
Lect
ure
67
EL2
620
2011
Defi
nitio
nof
Des
crib
ing
Func
tion
The
desc
ribi
ngfu
nctio
nis
N(A
,!)
=b 1
(!)+
ia1(!
)
A
e(t)
u(t
)N
.L.
e(t)
$u 1(t
)N
(A,!
)
IfG
islo
wpa
ssan
da
0=
0,th
en
$u 1(t
)=
|N(A
,!)|A
sin[!
t+
arg
N(A
,!)]
can
beus
edin
stea
dof
u(t
)to
anal
yze
the
syst
em.
Am
plitu
dede
pend
entg
ain
and
phas
esh
ift!
Lect
ure
68
EL2
620
2011
Des
crib
ing
Func
tion
for
aR
elay
H !H
u
e
01
23
45
6-1
-0.50
0.51
eu
#=
2"t/
T
a1
=1 !
#2!
0
u("
)co
s"
d"
=0
b 1=
1 !
#2!
0
u("
)si
n"
d"
=2 !
#!
0
Hsi
n"
d"
=4H !
The
desc
ribin
gfu
nctio
nfo
rare
lay
isth
us
N(A
)=
b 1(!
)+
ia1(!
)
A=
4H "A
Lect
ure
69
EL2
620
2011
Odd
Sta
ticN
onlin
eari
ties
Ass
ume
f(·)
and
g(·)
are
odd
(i.e.
f(!
e)=
!f(e
))st
atic
nonl
inea
ritie
sw
ithde
scrib
ing
func
tions
Nf
and
Ng.
Then
,
•Im
Nf(A
,!)
=0
•Nf(A
,!)
=N
f(A
)
•N!
f(A
)=
$N
f(A
)
•Nf+
g(A
)=
Nf(A
)+
Ng(A
)
Lect
ure
610
EL2
620
2011
Exi
sten
ceof
Per
iodi
cS
olut
ions
repl
acem
ents
0e
uy
!f(·)
G(s
)
!1/N
(A) AG
(i#)
Pro
posa
l:su
stai
ned
osci
llatio
nsif
loop
-gai
n1
and
phas
e-la
g!
180o
G(i
!)N
(A)
=!
1&
G(i
!)
=!
1/N
(A)
The
inte
rsec
tions
ofth
ecu
rves
G(i
!)
and
!1/
N(A
)gi
ve!
and
Afo
rapo
ssib
lepe
riodi
cso
lutio
n.
Lect
ure
611
EL2
620
2011
Per
iodi
cS
olut
ions
inR
elay
Sys
tem
re
uy
!G
(s)
-1-0.8
-0.6
-0.4
-0.2
0
-0.5
-0.4
-0.3
-0.2
-0.10
0.1
G(i
#)
!1/N
(A)
G(s
)=
3
(s+
1)3
with
feed
back
u=
!sg
ny
No
phas
ela
gin
f(·)
,arg
G(i
!)
=!
"fo
r!=
'3
=1.
7
G(i
'3)
=!
3/8
=!
1/N
(A)
=!
"A
/4(
A=
12/8
"%
0.48
Lect
ure
612
EL2
620
2011
The
pred
ictio
nvi
ath
ede
scrib
ing
func
tion
agre
esve
ryw
ellw
ithth
etr
ueos
cilla
tions
:
02
46
810
-1
-0.50
0.51
uy
Not
eth
atG
filte
rsou
talm
osta
llhi
gher
-ord
erha
rmon
ics.
Lect
ure
613
EL2
620
2011
Des
crib
ing
Func
tion
for
aS
atur
atio
n
H !H
D!
D
u
e
01
23
45
6-1
-0.50
0.51
eu
#
Lete
(t)
=A
sin
!t=
Asi
n#
.Fi
rsts
etH
=D
.Th
enfo
r#
)(0
,")
u(#
)=
(A
sin
#,
#)
(0,#
0)*
("!
#0,"
)
D,
#)
(#0,"
!#
0)
whe
re#
0=
arcs
inD
/A.
Lect
ure
614
EL2
620
2011
a1
=1 "
#2"
0
u(#
)co
s#
d#
=0
b 1=
1 "
#2"
0
u(#
)si
n#
d#
=4 "
#"/2
0
u(#
)si
n#
d#
=4A "
##
0
0
sin
2#
d#
+4D "
#"/2
#0
sin
#d#
=A "
) 2#0+
sin
2#0
*
Lect
ure
615
EL2
620
2011
Hen
ce,i
fH=
D,t
hen
N(A
)=
1 "
) 2#0+
sin
2#0
* .
IfH
+=D
,the
nth
eru
leN
!f(A
)=
$N
f(A
)gi
ves
N(A
)=
H D"
) 2#0+
sin
2#0
*
02
46
810
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.91
1.1
N(A
)fo
rH=
D=
1
Lect
ure
616
EL2
620
2011
5m
inut
eex
erci
se:
Wha
tosc
illat
ion
ampl
itude
and
frequ
ency
doth
ede
scrib
ing
func
tion
anal
ysis
pred
ictf
orth
e“M
otiv
atin
gE
xam
ple”
?
Lect
ure
617
EL2
620
2011
The
Nyq
uist
Theo
rem
!K
G(s
)!
1/KG
(i#)
Ass
ume
that
Gis
stab
le,a
ndK
isa
posi
tive
gain
.
•If
G(i
!)
goes
thro
ugh
the
poin
t!1/
Kth
ecl
osed
-loop
syst
emdi
spla
yssu
stai
ned
osci
llatio
ns
•If
G(i
!)
enci
rcle
sth
epo
int!
1/K
,the
nth
ecl
osed
-loop
syst
emis
unst
able
(gro
win
gam
plitu
deos
cilla
tions
).
•If
G(i
!)
does
note
ncirc
leth
epo
int!
1/K
,the
nth
ecl
osed
-loop
syst
emis
stab
le(d
ampe
dos
cilla
tions
)
Lect
ure
618
EL2
620
2011
Sta
bilit
yof
Per
iodi
cS
olut
ions
!
!1/
N(A
)
G(!
)
Ass
ume
that
G(s
)is
stab
le.
•If
G(!
)en
circ
les
the
poin
t!1/
N(A
),th
enth
eos
cilla
tion
ampl
itude
isin
crea
sing
.
•If
G(!
)do
esno
tenc
ircle
the
poin
t!1/
N(A
),th
enth
eos
cilla
tion
ampl
itude
isde
crea
sing
.
Lect
ure
619
EL2
620
2011
An
Uns
tabl
eP
erio
dic
Sol
utio
n
!1/N
(A)
G(!
)
An
inte
rsec
tion
with
ampl
itude
A0
isun
stab
leifA
<A
0le
ads
tode
crea
sing
ampl
itude
and
A>
A0
lead
sto
incr
easi
ng.
Lect
ure
620
EL2
620
2011
Sta
ble
Per
iodi
cS
olut
ion
inR
elay
Sys
tem
re
uy
!G
(s)
-5-4
-3-2
-10
-0.2
-0.15
-0.1
-0.050
0.050.1
0.150.2
G(i
!) !1/
N(A
)
G(s
)=
(s+
10)2
(s+
1)3
with
feed
back
u=
!sg
ny
give
son
est
able
and
one
unst
able
limit
cycl
e.Th
ele
ftm
ost
inte
rsec
tion
corr
espo
nds
toth
est
able
one.
Lect
ure
621
EL2
620
2011
Aut
omat
icTu
ning
ofP
IDC
ontr
olle
rPe
riod
and
ampl
itude
ofre
lay
feed
back
limit
cycl
eca
nbe
used
for
auto
tuni
ng:
ΣP
roce
ss
PID
Rela
y
A T
uy
−1
05
10
−101
Tim
e
uy
Lect
ure
622
EL2
620
2011
Des
crib
ing
Func
tion
for
aQ
uant
izer
1
D
u
e
01
23
45
6-1
-0.8
-0.6
-0.4
-0.20
0.2
0.4
0.6
0.81
e
u
Lete
(t)
=A
sin
!t=
Asi
n#
.Th
enfo
r#)
(0,"
)
u(#
)=
(0,
#)
(0,#
0)
1,#
)(#
0,"
!#
0)
whe
re#
0=
arcs
inD
/A.
Lect
ure
623
EL2
620
2011
a1
=1 "
#2"
0
u(#
)co
s#
d#
=0
b 1=
1 "
#2"
0
u(#
)si
n#d#
=4 "
#"/2
#0
sin
#d#
=4 "
cos#
0=
4 "
"1
!D
2/A
2
N(A
)=
+0,
A<
D4 "A
"1
!D
2/A
2,
A$
D
Lect
ure
624
EL2
620
2011
Plo
tofD
escr
ibin
gFu
nctio
nQ
uant
izer
02
46
810
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
N(A
)fo
rD=
1
Not
ice
that
N(A
)%
1.3/
Afo
rlar
geam
plitu
des
Lect
ure
625
EL2
620
2011
Des
crib
ing
Func
tion
Pitf
alls
Des
crib
ing
func
tion
anal
ysis
can
give
erro
neou
sre
sults
.
•A
DF
may
pred
icta
limit
cycl
eev
enif
one
does
note
xist
.
•A
limit
cycl
em
ayex
iste
ven
ifth
eD
Fdo
esno
tpre
dict
it.
•Th
epr
edic
ted
ampl
itude
and
frequ
ency
are
only
appr
oxim
atio
nsan
dca
nbe
farf
rom
the
true
valu
es.
Lect
ure
626
EL2
620
2011
Acc
urac
yof
Des
crib
ing
Func
tion
Ana
lysi
s
Con
troll
oop
with
fric
tion
F=
sgny
: _
_G
C
Fric
tion
y ref
F
y
Cor
resp
onds
to
G
1+
GC
=s(
s!
z)
s3+
2s2+
2s+
1w
ithfe
edba
cku
=!
sgny
The
osci
llatio
nde
pend
son
the
zero
ats
=z.
Lect
ure
627
EL2
620
2011
05
1015
2025
30-101 0
510
1520
2530
-0.4
-0.20
0.2
0.4
y y
z=
1/3
z=
4/3
-1-0.5
00.5
-0.4
-0.20
0.2
0.4
0.6
0.81
1.2
z=
1/3
z=
4/3
DF
give
spe
riod
times
and
ampl
itude
s(T
,A)
=(1
1.4,
1.00
)an
d(1
7.3,
0.23
),re
spec
tivel
y.
Acc
urat
ere
sults
only
ify
iscl
ose
tosi
nuso
idal
!
Lect
ure
628
EL2
620
2011
2m
inut
eex
erci
se:
Wha
tisN
(A)
forf
(x)
=x
2?
Lect
ure
629
EL2
620
2011
Har
mon
icB
alan
ce
e(t)
=A
sin
!t
u(t
)f(·)
Afe
wm
ore
Four
ierc
oeffi
cien
tsin
the
trun
catio
n
$u k(t
)=
a0 2
+k ! n=
1
(anco
sn!t+
b nsi
nn!t)
may
give
muc
hbe
tterr
esul
t.D
escr
ibin
gfu
nctio
nco
rres
pond
sto
k=
1an
da
0=
0.
Exa
mpl
e:f(x
)=
x2
give
su(t
)=
(1!
cos2!
t)/2
.H
ence
byco
nsid
erin
ga
0=
1an
da
2=
1/2
we
gett
heex
actr
esul
t.
Lect
ure
630
EL2
620
2011
Ana
lysi
sof
Osc
illat
ions
—A
Sum
mar
y
Tim
e-do
mai
n:
•Po
inca
rem
aps
and
Lyap
unov
func
tions
•R
igor
ous
resu
ltsbu
tonl
yfo
rsim
ple
exam
ples
•H
ard
tous
efo
rlar
gepr
oble
ms
Freq
uenc
y-do
mai
n:
•D
escr
ibin
gfu
nctio
nan
alys
is
•A
ppro
xim
ate
resu
lts
•Po
wer
fulg
raph
ical
met
hods
Lect
ure
631
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•D
eriv
ede
scrib
ing
func
tions
fors
tatic
nonl
inea
ritie
s
•A
naly
zeex
iste
nce
and
stab
ility
ofpe
riodi
cso
lutio
nsby
desc
ribin
gfu
nctio
nan
alys
is
Lect
ure
632
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
7
•Com
pens
atio
nfo
rsat
urat
ion
(ant
i-win
dup)
•Fric
tion
mod
els
•Com
pens
atio
nfo
rfric
tion
Lect
ure
71
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
toan
alyz
ean
dde
sign
•A
nti-w
indu
pfo
rPID
and
stat
e-sp
ace
cont
rolle
rs
•C
ompe
nsat
ion
forf
rictio
n
Lect
ure
72
EL2
620
2011
The
Pro
blem
with
Sat
urat
ing
Act
uato
r
�
PS
fr
ye
rv
GP
IDu
•Th
efe
edba
ckpa
this
brok
enw
hen
usa
tura
tes!
Ope
nlo
opbe
havi
or!
•Le
ads
topr
oble
mw
hen
syst
eman
d/or
the
cont
rolle
rare
unst
able
–E
xam
ple:
I-par
tin
PID
Rec
all:
CP
ID(s
)=
K
! 1+
1 Tis
+T
ds"
Lect
ure
73
EL2
620
2011
Exa
mpl
e—W
indu
pin
PID
Con
trol
ler
020
40
60
80
012
020
40
60
80
!0.10
0.1
Tim
e
y u
PID
cont
rolle
rwith
out(
dash
ed)a
ndw
ith(s
olid
)ant
i-win
dup
Lect
ure
74
EL2
620
2011
Ant
i-Win
dup
for
PID
Con
trol
ler
Ant
i-win
dup
(a)w
ithac
tuat
orou
tput
avai
labl
ean
d(b
)with
out
∑
∑1 s 1 T
t
e s
−y
vu
e
−+
∑∑
−y
eu
v
−+
e s
1 Tt1 s
Act
uato
r m
od
el
Act
uato
r
Act
uato
r
∑
K
KT
dsK
K Ti
KT
ds
K Ti
∑
(a)
(b)
Lect
ure
75
EL2
620
2011
Ant
i-Win
dup
isB
ased
onTr
acki
ng
Whe
nth
eco
ntro
lsig
nals
atur
ates
,the
inte
grat
ion
stat
ein
the
cont
rolle
rtra
cks
the
prop
erst
ate
The
track
ing
time
Tt
isth
ede
sign
para
met
erof
the
anti-
win
dup
Com
mon
choi
ces
ofT
t:
•T
t=
Ti
•T
t=
"T
iTd
Rem
ark:
If0
<T
t#
Ti,
then
the
inte
grat
orst
ate
beco
mes
sens
itive
toth
ein
stan
ces
whe
ne s
$=0:
I(t
)=
#t
0
$ Ke(
!)
Ti
+e s
(!)
Tt
% d!
%1 Tt
#t
0
e s(!
)d!
Lect
ure
76
EL2
620
2011
Ant
i-Win
dup
for
Obs
erve
r-B
ased
Sta
teFe
edba
ckC
ontr
olle
r
Act
uato
ry
Σxm
−ˆ x
Lsa
t v
Obse
rver
Sta
te f
eed
back
˙ x=
Ax
+B
satv
+K
(y&
Cx)
v=
L(x
m&
x)
xis
estim
ate
ofpr
oces
sst
ate,
xm
desi
red
(mod
el)s
tate
Nee
dac
tuat
orm
odel
ifsa
tv
isno
tmea
sura
ble
Lect
ure
77
EL2
620
2011
Ant
i-Win
dup
for
Gen
eral
Sta
te-S
pace
Con
trol
ler
Sta
te-s
pace
cont
rolle
r:
xc=
Fx
c+
Gy
u=
Cx
c+
Dy
Win
dup
poss
ible
ifF
unst
able
and
usa
tura
tes
Idea
:R
ewrit
ere
pres
enta
tion
ofco
ntro
llaw
from
(a)t
o(b
)with
the
sam
ein
put–
outp
utre
latio
n,bu
twhe
reth
eun
stab
leS
Ais
repl
aced
bya
stab
leS
B.
Ifu
satu
rate
s,th
en(b
)beh
aves
bette
rtha
n(a
).
yS
A
yu
uS
B
(a)
(b)
Lect
ure
78
EL2
620
2011
Mim
icth
eob
serv
er-b
ased
cont
rolle
r:
xc=
Fx
c+
Gy
+K
(u&
Cx
c&
Dy)
=(F
&K
C)x
c+
(G&
KD
)y+
Ku
Cho
ose
Ksu
chth
atF
0=
F&
KC
has
desi
red
(sta
ble)
eige
nval
ues.
Then
use
cont
rolle
r
xc=
F0x
c+
G0y
+K
u
u=
sat(
Cx
c+
Dy)
whe
reG
0=
G&
KD
.
Lect
ure
79
EL2
620
2011
Sta
te-s
pace
cont
rolle
rwith
outa
ndw
ithan
ti-w
indu
p:
yu
∑∑
∑∑
sat
u G
−K
Dy
vs−1 K
F−
KC
CD
G
F s−1
CD
xc xc
Lect
ure
710
EL2
620
2011
Con
trol
lers
with
”Sta
ble”
Zero
s
Mos
tcon
trolle
rsar
em
inim
umph
ase,
i.e.h
ave
zero
sst
rictly
inLH
P
xc=
Fx
c+
Gy
u=
Cx
c+
Dy
!u=
0x
c=
zero
dynamics
&'(
)(F
&G
C/D
)x
c
y=
&C
/Dx
c
Thus
,cho
ose
”obs
erve
r”ga
in
K=
G/D
!F
&K
C=
F&
GC
/D
and
the
eige
nval
ues
ofth
e”o
bser
ver”
base
dco
ntro
llerb
ecom
eseq
ual
toth
eze
ros
ofF
(s)
whe
nu
satu
rate
s
Not
eth
atth
isim
plie
sG
&K
D=
0in
the
figur
eon
the
prev
ious
slid
e,an
dw
eth
usob
tain
P-fe
edba
ckw
ithga
inD
unde
rsat
urat
ion.
Lect
ure
711
EL2
620
2011
Con
trol
lerF
(s)
with
”Sta
ble”
Zero
s
LetD
=lim
s!"
F(s
)an
dco
nsid
erth
efe
edba
ckim
plem
enta
tion
y
!
sat
u
D
1F
(s)
!1 D
!+
Itis
easy
tosh
owth
attra
nsfe
rfun
ctio
nfro
my
tou
with
nosa
tura
tion
equa
lsF
(s)!
Ifth
etra
nsfe
rfun
ctio
n(1
/F(s
)&
1/D
)in
the
feed
back
loop
isst
able
(sta
ble
zero
s)!
No
stab
ility
prob
lem
sin
case
ofsa
tura
tion
Lect
ure
712
EL2
620
2011
Inte
rnal
Mod
elC
ontr
ol(IM
C)
IMC
:app
lyfe
edba
ckon
lyw
hen
syst
emG
and
mod
elG
diffe
r!
&
&
ru
y
*y
Q(s
)
C(s
)
G(s
)
* G(s
)
Ass
ume
Gst
able
.N
ote:
feed
back
from
the
mod
eler
rory
&*y.
Des
ign:
assu
me
* G%
Gan
dch
oose
Qst
able
with
Q%
G#
1.
Lect
ure
713
EL2
620
2011
Exa
mpl
e
* G(s
)=
1
T1s
+1
Cho
ose
Q=
T1s
+1
!s
+1
,!
<T
1
Giv
esth
eco
ntro
ller F=
Q
1&
Q* G
!
F=
T1s
+1
!s
=T
1 !
! 1+
1 T1s
"
PI-c
ontro
ller!
Lect
ure
714
EL2
620
2011
IMC
with
Sta
ticN
onlin
eari
ty
Incl
ude
nonl
inea
rity
inm
odel
&
&r
u
v
yQ
G * G
Cho
ose
Q%
G#
1.
Lect
ure
715
EL2
620
2011
Exa
mpl
e(c
ont’d
)
Ass
ume
r=
0an
dab
use
ofLa
plac
etra
nsfo
rmno
tatio
n
u=
&Q
(y&
* Gv)
=&
T1s
+1
!s
+1
y+
1
!s
+1v
if|u
|<u
max
(v=
u):
PIc
ontro
lleru
=&
(T1s
+1)
!s
y
If|u
|>u
max
(v=
±u
max
):
u=
&T
1s
+1
!s
+1
y±
um
ax
!s
+1
No
inte
grat
ion.
An
alte
rnat
ive
way
toim
plem
enta
nti-w
indu
p!
Lect
ure
716
EL2
620
2011
Oth
erA
nti-W
indu
pS
olut
ions
Sol
utio
nsab
ove
are
allb
ased
ontra
ckin
g.
Oth
erso
lutio
nsin
clud
e:
•Tu
neco
ntro
llert
oav
oid
satu
ratio
n
•D
on’t
upda
teco
ntro
llers
tate
sat
satu
ratio
n
•C
ondi
tiona
llyre
seti
nteg
ratio
nst
ate
toze
ro
•A
pply
optim
alco
ntro
lthe
ory
(Lec
ture
12)
Lect
ure
717
EL2
620
2011
Bum
ples
sTr
ansf
er
Ano
ther
appl
icat
ion
ofth
etra
ckin
gid
eais
inth
esw
itchi
ngbe
twee
nau
tom
atic
and
man
ualc
ontro
lmod
es.
PID
with
anti-
win
dup
and
bum
ples
stra
nsfe
r: −+
M
A
∑
u∑
y uc
PD 1 s
∑∑
∑
+−
1
Tm
1 Tr
uc
e
1 Tr
1 Tr
1 s
Not
eth
ein
crem
enta
lfor
mof
the
man
ualc
ontro
lmod
e(u
%u
c/T
m)
Lect
ure
718
EL2
620
2011
Fric
tion
Fric
tion
ispr
esen
talm
oste
very
whe
re
•O
ften
bad:
–Fr
ictio
nin
valv
esan
dot
hera
ctua
tors
•S
omet
imes
good
:
–Fr
ictio
nin
brak
es
•S
omet
imes
too
smal
l:
–E
arth
quak
es
Pro
blem
s:
•H
owto
mod
elfr
ictio
n?
•H
owto
com
pens
ate
forf
rictio
n?
•H
owto
dete
ctfr
ictio
nin
cont
roll
oops
?
Lect
ure
719
EL2
620
2011
Stic
k-S
lipM
otio
n
Ff
Fp
xy
010
20012 0
1020
0
0.2
0.4
010
2001
Tim
e
Tim
e
Tim
ex y
y
Fp
Ff
Lect
ure
720
EL2
620
2011
Posi
tion
Con
trol
ofS
ervo
with
Fric
tion
Fri
ctio
n
ux
r
v
x
F
1s
ΣP
ID1
ms
−
020
4060
80100
0
0.51 0
2040
6080
100
−0.20.2
020
4060
80100
−101
Tim
e
Tim
e
Tim
e
Lect
ure
721
EL2
620
2011
5m
inut
eex
erci
se:
Whi
char
eth
esi
gnal
sin
the
prev
ious
plot
s?
Lect
ure
722
EL2
620
2011
Fric
tion
Mod
elin
g
Lect
ure
723
EL2
620
2011
Str
ibec
kE
ffec
tFr
ictio
nin
crea
ses
with
decr
easi
ngve
loci
ty(fo
rlow
velo
citie
s)
Str
ibec
k(1
902)
-0.0
4-0
.03
-0.0
2-0
.01
00.
010.
020.
030.
04
-200
-150
-100-50050100
150
200
Stea
dy s
tate
fric
tion:
Joi
nt 1
Velo
city
[rad
/sec
]Friction [Nm]
Lect
ure
724
EL2
620
2011
Cla
ssic
alFr
ictio
nM
odel
s
Adv
ance
dm
odel
sca
ptur
eva
rious
fric
tion
phen
omen
abe
tter
Lect
ure
725
EL2
620
2011
Fric
tion
Com
pens
atio
n
•Lu
bric
atio
n
•In
tegr
alac
tion
•D
ither
sign
al
•M
odel
-bas
edfr
ictio
nco
mpe
nsat
ion
•A
dapt
ive
fric
tion
com
pens
atio
n
•Th
eK
nock
er
Lect
ure
726
EL2
620
2011
Inte
gral
Act
ion
•In
tegr
alac
tion
com
pens
ates
fora
nyex
tern
aldi
stur
banc
e
•W
orks
iffr
ictio
nfo
rce
chan
ges
slow
ly(v
(t)%
cons
t)
•If
fric
tion
forc
ech
ange
squ
ickl
y,th
enla
rge
inte
gral
actio
n(s
mal
lTi)
nece
ssar
y.M
ayle
adto
stab
ility
prob
lem
��
xr
uv
x
FFr
ictio
n
PID
1 ms
1 s
Lect
ure
727
EL2
620
2011
Mod
ified
Inte
gral
Act
ion
Mod
ifyth
ein
tegr
alpa
rtto
I=
K Ti
+ te(
!)d
!w
here
e(t)
e(t)
"
Adv
anta
ge:
Avo
idth
atsm
alls
tatic
erro
rint
rodu
ces
osci
llatio
n
Dis
adva
ntag
e:E
rror
won
’tgo
toze
ro
Lect
ure
728
EL2
620
2011
Dith
erS
igna
lA
void
stic
king
atv
=0
(whe
reth
ere
ishi
ghfr
ictio
n)by
addi
nghi
gh-fr
eque
ncy
mec
hani
calv
ibra
tion
(dith
er)
��
repl
acem
ents
xr
uv
x
FFr
ictio
n
PID
1 ms
1 s
Cf.,
mec
hani
calm
aze
puzz
le(la
byrin
tspe
l)
Lect
ure
729
EL2
620
2011
Mod
el-B
ased
Fric
tion
Com
pens
atio
n
Forp
roce
ssw
ithfr
ictio
nF
: mx
=u
&F
use
cont
rols
igna
l
u=
uP
ID+
F
whe
reu
PID
isth
ere
gula
rcon
trols
igna
land
Fan
estim
ate
ofF
.
Poss
ible
if:
•A
nes
timat
eF
%F
isav
aila
ble
•u
and
Fdo
esap
ply
atth
esa
me
poin
t
Lect
ure
730
EL2
620
2011
Ada
ptiv
eFr
ictio
nC
ompe
nsat
ion
uP
ID1 ms
Fric
tion
estim
ator
Fric
tion
v+ +&
F F
x1 s
Cou
lom
bfr
ictio
nm
odel
:F
=a
sgn
v
Fric
tion
estim
ator
:
z=
ku
PID
sgn
v
a=
z&
km
|v|
F=
asg
nv
Lect
ure
731
EL2
620
2011
Ada
ptat
ion
conv
erge
s:e
=a
&a
'0
ast'
(P
roof
:
de dt
=&
da dt
=&
dz dt
+km
d dt|v
|=
&ku
PID
sgn
v+
km
vsg
nv
=&
ksg
nv(u
PID
&m
v)
=&
ksg
nv(F
&F
)
=&
k(a
&a)
=&
ke
Rem
ark:
Car
eful
with
d dt|v
|atv
=0.
Lect
ure
732
EL2
620
2011
01
23
45
67
8-1
-0.50
0.51 0
12
34
56
78
-10-50510
v ref
v
u
P-c
ontro
ller
01
23
45
67
8-1
-0.50
0.51 0
12
34
56
78
-10-50510
PI-c
ontro
ller
01
23
45
67
8-1
-0.50
0.51 0
12
34
56
78
-10-50510
P-c
ontro
llerw
ithad
aptiv
efr
ictio
nco
mpe
nsat
ion
Lect
ure
733
EL2
620
2011
The
Kno
cker
Cou
lom
bfr
ictio
nco
mpe
nsat
ion
and
squa
rew
ave
dith
erTy
pica
lcon
trols
igna
lu
01
23
45
67
89
10-0
.50
0.51
1.52
2.5
Hag
glun
d:P
aten
tand
Inno
vatio
nC
upw
inne
r
Lect
ure
734
EL2
620
2011
Det
ectio
nof
Fric
tion
inC
ontr
olLo
ops
•Fr
ictio
nis
due
tow
eara
ndin
crea
ses
with
time
•Q
:Whe
nsh
ould
valv
esbe
mai
ntai
ned?
•Id
ea:
Mon
itorl
oops
auto
mat
ical
lyan
des
timat
efr
ictio
n
5010
015
020
025
030
035
0
9899100
101
102
5010
015
020
025
030
035
0
10.511
11.512y u
time
Hor
ch:
PhD
thes
is(2
000)
and
pate
nt
Lect
ure
735
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
toan
alyz
ean
dde
sign
•A
nti-w
indu
pfo
rPID
,sta
te-s
pace
,and
poly
nom
ialc
ontro
llers
•C
ompe
nsat
ion
forf
rictio
n
Lect
ure
736
EL2
620
2011
Nex
tLec
ture
•B
ackl
ash
•Q
uant
izat
ion
Lect
ure
737
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
8
•Bac
klas
h
•Qua
ntiz
atio
n
Lect
ure
81
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
toan
alyz
ean
dde
sign
for
•B
ackl
ash
•Q
uant
izat
ion
Lect
ure
82
EL2
620
2011
Line
aran
dA
ngul
arB
ackl
ash
!F
in
!F
out
xin
xout
"!
D"
!D
Tin
Tout
! in
! out
2D
Lect
ure
83
EL2
620
2011
Bac
klas
h
Bac
klas
h(g
lapp
)is
•pr
esen
tin
mos
tmec
hani
cala
ndhy
drau
licsy
stem
s
•in
crea
sing
with
wea
r
•ne
cess
ary
fora
gear
box
tow
ork
inhi
ghte
mpe
ratu
re
•ba
dfo
rcon
trolp
erfo
rman
ce
•so
met
imes
indu
cing
osci
llatio
ns
Lect
ure
84
EL2
620
2011
Bac
klas
hM
odel
xout=
! xin
,in
cont
act
0,ot
herw
ise
“inco
ntac
t”if|x
out!
xin
|=D
and
xin
(xin
!x
out)
>0.
xin
! in
D
D
xout
! out
•M
ultiv
alue
dou
tput
;cur
rent
outp
utde
pend
son
hist
ory.
Thus
,ba
ckla
shis
ady
nam
icph
enom
ena.
Lect
ure
85
EL2
620
2011
Alte
rnat
ive
Mod
el
xin
!x
out
(!in
!! o
ut)
D
Forc
e
(Tor
que)
Not
equi
vale
ntto
“Bac
klas
hM
odel
”
Lect
ure
86
EL2
620
2011
Eff
ects
ofB
ackl
ash
P-c
ontro
lofm
otor
angl
ew
ithge
arbo
xha
ving
back
lash
with
D=
0.2
! ref
!
eu
! out
! in
! in
1
1+
sT
1 sK
Lect
ure
87
EL2
620
2011
05
1015
2025
3035
400
0.51
1.5
No
back
lash
K=
0.25
,1,4
05
1015
2025
3035
400
0.2
0.4
0.6
0.81
1.2
1.4
1.6
Bac
klas
hK
=0.
25,1
,4
•O
scill
atio
nsfo
rK=
4bu
tnot
forK
=0.
25or
K=
1.W
hy?
•N
ote
that
the
ampl
itude
will
decr
ease
with
decr
easi
ngD
,bu
tnev
erva
nish
Lect
ure
88
EL2
620
2011
Des
crib
ing
Func
tion
for
Bac
klas
h
IfA
<D
then
N(A
)=
0el
se
ReN
(A)
=1 "
" " 2+
arcs
in(1
!2D
/A)
+2(
1!
2D
/A)#
D A
$ 1!
D A
%&
ImN
(A)
=!
4D
"A
$ 1!
D A
%
Lect
ure
89
EL2
620
2011
!1/
N(A
)fo
rD=
0.2:
-4-3
.5-3
-2.5
-2-1
.5-1
-0.5
00.
51
-16
-14
-12
-10-8-6-4-20
Re
-1/N
(A)
Im -1/N(A)
Not
eth
at!
1/N
(A)"
!1
asA
"#
(phy
sica
lint
erpr
etat
ion?
)
Lect
ure
810
EL2
620
2011
Des
crib
ing
Func
tion
Ana
lysi
s
Rea
l Axi
s
Imaginary Axis
Nyq
uist
Dia
gram
s
-4-2
02
4-4-3-2-101234
K=
0.25
K=
1K
=4!1/
N(A
)
05
1015
2025
3035
400
0.2
0.4
0.6
0.81
1.2
1.4
1.6
1.8
Inpu
tand
outp
utof
back
lash
K=
4,D
=0.
2:
DF
anal
ysis
:In
ters
ectio
nat
A=
0.33
,"=
1.24
Sim
ulat
ion:
A=
0.33
,"=
2#/5
.0=
1.26
Des
crib
ing
func
tion
pred
icts
osci
llatio
nw
ell
Lect
ure
811
EL2
620
2011
Sta
bilit
yP
roof
for
Bac
klas
hS
yste
m
The
desc
ribin
gfu
nctio
nm
etho
dis
only
appr
oxim
ate.
Do
ther
eex
istc
ondi
tions
that
guar
ante
est
abili
ty(o
fthe
stea
dy-s
tate
)?
repl
acem
ents ! r
efu
!
e! o
ut! i
n! i
n1
1+
sT
1 sK
Not
eth
at! i
nan
d! o
ut
will
notc
onve
rge
toze
ro
Q:W
hata
bout
! in
and
! out?
Lect
ure
812
EL2
620
2011
Rew
rite
the
syst
em:
G(s
)
! in
! out
G(s
)
! in
! out
BL
The
bloc
k“B
L”sa
tisfie
s ! out=
!! i
nin
cont
act
0ot
herw
ise
Lect
ure
813
EL2
620
2011
Hom
ewor
k2
Ana
lyze
this
back
lash
syst
emw
ithin
put–
outp
utst
abili
tyre
sults
:
G(s
)
! in
! out
BL
Pas
sivi
tyTh
eore
mB
Lis
pass
ive
Sm
allG
ain
Theo
rem
BL
has
gain
$(B
L)
=1
Cir
cle
Cri
teri
onB
Lco
ntai
ned
inse
ctor
[0,1
]
Lect
ure
814
EL2
620
2011
Bac
klas
hC
ompe
nsat
ion
•M
echa
nica
lsol
utio
ns
•D
eadz
one
•Li
near
cont
rolle
rdes
ign
•B
ackl
ash
inve
rse
Lect
ure
815
EL2
620
2011
Line
arC
ontr
olle
rD
esig
n
Intro
duce
phas
ele
adco
mpe
nsat
ion:
! ref
!
eu
! out
! in
! in
1
1+
sT
1 sK
1+
sT
2
1+
sT
1
Lect
ure
816
EL2
620
2011
F(s
)=
K1+
sT2
1+
sT1
with
T1
=0.
5,T
2=
2.0:
Rea
l Axi
s
Imaginary Axis
Nyq
uist
Dia
gram
s
-10
-50
510
-10-8-6-4-20246810
with
filte
r
with
out f
ilter
05
1015
200
0.51
1.5
y, w
ith/w
ithou
t filt
er
05
1015
20-2-1012
u, w
ith/w
ithou
t filt
er
Osc
illat
ion
rem
oved
!
Lect
ure
817
EL2
620
2011
Bac
klas
hIn
vers
e
Idea
:Le
txin
jum
p±
2Dw
hen
xout
shou
ldch
ange
sign
###
#$ !
##
##$ !
!!
!u
xin
xout
xin
D
D
xout
Lect
ure
818
EL2
620
2011
xin
(t)
=
' ( ) ( *
u+
+ Difu(t
)>
u(t
!)
u!
+ Difu(t
)<
u(t
!)
xin
(t!
)ot
herw
ise
•If
+ D=
Dth
enpe
rfect
com
pens
atio
n(x
out=
u)
•If
+ D<
Dth
enun
der-
com
pens
atio
n(d
ecre
ased
back
lash
)
•If
+ D>
Dth
enov
er-c
ompe
nsat
ion
(may
give
osci
llatio
n)
Lect
ure
819
EL2
620
2011
Exa
mpl
e—P
erfe
ctC
ompe
nsat
ion
Mot
orw
ithba
ckla
shon
inpu
tin
feed
back
with
PD
-con
trolle
r
02
46
810
0
0.4
0.8
1.2 0
24
68
10-20246
Lect
ure
820
EL2
620
2011
Exa
mpl
e—U
nder
-Com
pens
atio
n
020
4060
800
0.4
0.8
1.2 0
2040
6080
024
Lect
ure
821
EL2
620
2011
Exa
mpl
e—O
ver-
Com
pens
atio
n
01
23
45
0
0.4
0.8
1.2 0
12
34
5-20246
Lect
ure
822
EL2
620
2011
Qua
ntiz
atio
n
•W
hatp
reci
sion
isne
eded
inA
/Dan
dD
/Aco
nver
ters
?(8
–14
bits
?)
•W
hatp
reci
sion
isne
eded
inco
mpu
tatio
ns?
(8–6
4bi
ts?)
yu
GD
/AA
/D!
/2
!
•Q
uant
izat
ion
inA
/Dan
dD
/Aco
nver
ters
•Q
uant
izat
ion
ofpa
ram
eter
s
•R
ound
off,
over
flow
,und
erflo
win
com
puta
tions
Lect
ure
823
EL2
620
2011
Line
arM
odel
ofQ
uant
izat
ion
Mod
elqu
antiz
atio
ner
rora
sa
unifo
rmly
dist
ribut
edst
ocha
stic
sign
ale
inde
pend
ento
fuw
ith
Var
(e)
=
,! "!
e2f e
de
=
,!
/2
"!
/2
e2 !de
=!
2
12
yy
uu
e
Q+
•M
aybe
reas
onab
leif!
issm
allc
ompa
red
toth
eva
riatio
nsin
u
•B
ut,a
dded
nois
eca
nne
vera
ffect
stab
ility
whi
lequ
antiz
atio
nca
n!
Lect
ure
824
EL2
620
2011
Des
crib
ing
Func
tion
for
Dea
dzon
eR
elay
1
D
u
e
01
23
45
6-1
-0.8
-0.6
-0.4
-0.20
0.2
0.4
0.6
0.81
e
u
Lect
ure
6$
N(A
)=
-0,
A<
D4 #A
.1
!D
2/A
2,
A>
D
Lect
ure
825
EL2
620
2011
Des
crib
ing
Func
tion
for
Qua
ntiz
er
yu
Qu
y
!/2
!
N(A
)=
' ( ) ( *
0,A
<! 2
4! #A
n / i=1
#1
!$
2i!
1
2A!
% 2,
2n"
12
!<
A<
2n+
12
!
Lect
ure
826
EL2
620
2011
02
401
A/
Del
ta
N(A)
•Th
em
axim
umva
lue
is4/
#%
1.27
atta
ined
atA
%0.
71!
.
•P
redi
cts
osci
llatio
nif
Nyq
uist
curv
ein
ters
ects
nega
tive
real
axis
toth
ele
ftof
!#/4
%!
0.79
•C
ontro
llerw
ithga
inm
argi
n>
1/0.
79=
1.27
avoi
dsos
cilla
tion
•R
educ
ing
!re
duce
son
lyth
eos
cilla
tion
ampl
itude
Lect
ure
827
EL2
620
2011
Exa
mpl
e—M
otor
with
P-c
ontr
olle
r.
Qua
ntiz
atio
nof
proc
ess
outp
utw
ith!
=0.
2
Nyq
uist
oflin
earp
art(
K&
ZOH
&G
(s))
inte
rsec
tsat
!0.
5K:
Sta
bilit
yfo
rK<
2w
ithou
tQ.
Sta
ble
osci
llatio
npr
edic
ted
forK
>2/
1.27
=1.
57w
ithQ
.
r
!u
yK
1"
e!s
sG
(s)
Q
Lect
ure
828
EL2
620
2011
05
001
Output
(a)
05
001
Output(b
)
05
001
Tim
e
Output
(c)
K=
0.8
K=
1.2
K=
1.6
Lect
ure
829
EL2
620
2011
Exa
mpl
e—1/
s2&
2nd-
Ord
erC
ontr
olle
r
!6
!4
!2
0
!202
Imaginary axis
Rea
l a
xis
A/D
D/A
FG
Lect
ure
830
EL2
620
2011
Qua
ntiz
atio
n!
=0.
02in
A/D
conv
erte
r:
050
100
150
01
Output
050
100
150
0.9
8
1.0
2
Output
050
100
150
!0.0
50
0.0
5
Tim
e
Input
Des
crib
ing
func
tion:
Ay
=0.
01an
dT
=39
Sim
ulat
ion:
Ay
=0.
01an
dT
=28
Lect
ure
831
EL2
620
2011
Qua
ntiz
atio
n!
=0.
01in
D/A
conv
erte
r:
050
100
150
01
Output
050
100
150
!0.0
50
0.0
5
Unquantized
050
100
150
!0.0
50
0.0
5
Tim
e
Input
Des
crib
ing
func
tion:
Au
=0.
005
and
T=
39
Sim
ulat
ion:
Au
=0.
005
and
T=
39
Lect
ure
832
EL2
620
2011
Qua
ntiz
atio
nC
ompe
nsat
ion
•Im
prov
eac
cura
cy(la
rger
wor
dle
ngth
)
•A
void
unst
able
cont
rolle
rand
gain
mar
gins
<1.
3
•U
seth
etra
ckin
gid
eafro
man
ti-w
indu
pto
impr
ove
D/A
conv
erte
r
cont
rolle
rD
/AD
igita
lA
nalo
g
+!
•U
sean
alog
dith
er,
over
sam
-pl
ing,
and
digi
tal
low
pass
fil-
tert
oim
prov
eac
cura
cyof
A/D
conv
erte
rA
/Dfil
ter
deci
m.
+
Lect
ure
833
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldno
wbe
able
toan
alyz
ean
dde
sign
for
•B
ackl
ash
•Q
uant
izat
ion
Lect
ure
834
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
9
•Non
linea
rcon
trold
esig
nba
sed
onhi
gh-g
ain
cont
rol
Lect
ure
91
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
toan
alyz
ean
dde
sign
•H
igh-
gain
cont
rols
yste
ms
•S
lidin
gm
ode
cont
rolle
rs
Lect
ure
92
EL2
620
2011
His
tory
ofth
eFe
edba
ckA
mpl
ifier
New
York
–San
Fran
cisc
oco
mm
unic
atio
nlin
k19
14.
Hig
hsi
gnal
ampl
ifica
tion
with
low
dist
ortio
nw
asne
eded
.
f(·)
f(·)
!r
yf(·) k
Feed
back
ampl
ifier
sw
ere
the
solu
tion!
Bla
ck,B
ode,
and
Nyq
uist
atB
ellL
abs
1920
–195
0.
Lect
ure
93
EL2
620
2011
Line
ariz
atio
nTh
roug
hH
igh
Gai
nFe
edba
ck
!r
yf(·) K
e!1e
!2e
f(e
)
!1
"f(e
)
e"
!2
#!
1
1+
!1K
r"
y"
!2
1+
!2K
r
choo
seK
$1/
!1,y
ield
s
y%
1 Kr
Lect
ure
94
EL2
620
2011
AW
ord
ofC
autio
n
Nyq
uist
:hi
ghlo
op-g
ain
may
indu
ceos
cilla
tions
(due
tody
nam
ics)
!
Lect
ure
95
EL2
620
2011
Inve
rtin
gN
onlin
eari
ties
Com
pens
atio
nof
stat
icno
nlin
earit
yth
roug
hin
vers
ion:
F(s
)f
!1(·)
f(·)
G(s
)!
Con
trolle
r
Sho
uld
beco
mbi
ned
with
feed
back
asin
the
figur
e!
Lect
ure
96
EL2
620
2011
Rem
ark:
How
toO
btai
nf
!1
from
fus
ing
Feed
back
!
vu
k s
f(·)
u
f(u
)
u=
k! v
!f(u
)"
Ifk
>0
larg
ean
ddf
/du
>0,
then
u&
0an
d
0=
k! v
!f(u
)"'
f(u
)=
v'
u=
f!
1(v
)
Lect
ure
97
EL2
620
2011
Exa
mpl
e—Li
near
izat
ion
ofS
tatic
Non
linea
rity
re
uy
!K
f(·)
00.2
0.4
0.6
0.8
10
0.2
0.4
0.6
0.81
y(r
)
f(u
)
Line
ariz
atio
nof
f(u
)=
u2
thro
ugh
feed
back
.
The
case
K=
100
issh
own
inth
epl
ot:y(r
)%
r.
Lect
ure
98
EL2
620
2011
The
Sen
sitiv
ityFu
nctio
nS
=(1
+G
F)!
1
The
clos
ed-lo
opsy
stem
is
Gcl
=G
1+
GF
!r
yG F
Sm
allp
ertu
rbat
ions
dG
inG
give
s
dG
cl
dG
=1
(1+
GF
)2#
dG
cl
Gcl
=1
1+
GF
dG G
=S
dG G
Sis
the
clos
ed-lo
opse
nsiti
vity
toop
en-lo
oppe
rtur
batio
ns.
Lect
ure
99
EL2
620
2011
Dis
tort
ion
Red
uctio
nvi
aFe
edba
ck
The
feed
back
redu
ces
dist
ortio
nin
each
link.
Sev
eral
links
give
dist
ortio
n-fre
ehi
ghga
in.
!!
f(·)
f(·)
KK
Lect
ure
910
EL2
620
2011
Exa
mpl
e—D
isto
rtio
nR
educ
tion
LetG
=10
00,
dist
ortio
ndG
/G=
0.1
!r
yG K
Cho
ose
K=
0.1
#S
=(1
+G
K)!
1%
0.01
.Th
en
dG
cl
Gcl
=S
dG G
%0.
001
100
feed
back
ampl
ifier
sin
serie
sgi
veto
tala
mpl
ifica
tion
Gto
t=
(Gcl)1
00
%10
100
and
tota
ldis
tort
ion dG
tot
Gto
t=
(1+
10!
3)1
00!
1%
0.1
Lect
ure
911
EL2
620
2011
Tran
scon
tinen
talC
omm
unic
atio
nR
evol
utio
n
The
feed
back
ampl
ifier
was
pate
nted
byB
lack
1937
.
Year
Cha
nnel
sLo
ss(d
B)
No
amp’
s
1914
160
3–6
1923
1–4
150–
400
6–20
1938
1610
0040
1941
480
3000
060
0
Lect
ure
912
EL2
620
2011
Sen
sitiv
ityan
dth
eC
ircl
eC
rite
rion
r!1
GF
(i")
!r
yG
F
f(·)
Con
side
raci
rcle
C:=
{z(
C:
|z+
1|=
r},r
((0
,1).
GF
(i")
stay
sou
tsid
eC
if
|1+
GF
(i")|
>r
'|S
(i")|
"r!
1
Then
,the
Circ
leC
riter
ion
give
sst
abili
tyif
1
1+
r"
f(y
)
y"
1
1!
r
Lect
ure
913
EL2
620
2011
Sm
allS
ensi
tivity
Allo
ws
Larg
eU
ncer
tain
ty
If|S
(i")|
issm
all,
we
can
choo
ser
larg
e(c
lose
toon
e).
This
corr
espo
nds
toa
larg
ese
ctor
forf
(·).
Hen
ce,|S
(i")|
smal
lim
plie
slo
wse
nsiti
vity
tono
nlin
earit
ies.
k1
=1
1+
r
k2
=1
1!
r
yk1y
k2y
f(y
)
Lect
ure
914
EL2
620
2011
On–
Off
Con
trol
On–
offc
ontro
lis
the
sim
ples
tcon
trols
trate
gy.
Com
mon
inte
mpe
ratu
reco
ntro
l,le
velc
ontro
letc
.
re
uy
!G
(s)
The
rela
yco
rres
pond
sto
infin
itely
high
gain
atth
esw
itchi
ngpo
int.
Lect
ure
915
EL2
620
2011
AC
ontr
olD
esig
nId
eaA
ssum
eV
(x)
=x
TP
x,P
=P
T>
0,re
pres
ents
the
ener
gyof
x=
Ax
+B
u,
u(
[!1,
1]
Cho
ose
usu
chth
atV
deca
ysas
fast
aspo
ssib
le:
V=
xT(A
TP
+P
A)x
+2B
TP
xu
ism
inim
ized
ifu
=!
sgn(B
TP
x)
(Not
ice
that
V=
a+
bu,i
.e.j
ust
ase
gmen
tofl
ine
inu
,!1
<u
<1.
Hen
ceth
elo
wes
tval
ueis
atan
endp
oint
,dep
endi
ngon
the
sign
ofth
esl
ope
b.)
BTQ
Px
=0
x=
Ax
!B x
=A
x+
B
Lect
ure
916
EL2
620
2011
Slid
ing
Mod
es
x=
# f+(x
),#(x
)>
0
f!(x
),#(x
)<
0
#(x
)>
0
#(x
)<
0f
+
f!
The
slid
ing
mod
eis
x=
!f
++
(1!
!)f
!,w
here
!sa
tisfie
s!f
+ n+
(1!
!)f
! n=
0fo
rthe
norm
alpr
ojec
tions
off
+,f
!
!f
++
(1!
!)f
!
f+
f!
The
slid
ing
surf
ace
isS
={x
:#(x
)=
0}.
Lect
ure
917
EL2
620
2011
Exa
mpl
e
x=
$ 0!
1
1!
1%x
+
$ 1 1%u
=A
x+
Bu
u=
!sg
n#(x
)=
!sg
nx
2=
!sg
n(C
x)
iseq
uiva
lent
to
x=
# Ax
!B
,x
2>
0
Ax
+B
,x
2<
0
Lect
ure
918
EL2
620
2011
Fors
mal
lx2
we
have
& ' ' ' ( ' ' ' )
x2(t
)%
x1!
1,dx
2
dx
1
%1
!x
1x
2>
0
x2(t
)%
x1+
1,dx
2
dx
1
%1
+x
1x
2<
0
This
impl
ies
the
follo
win
gbe
havi
or
x2
>0
x2
<0
x2
=0
x1
=!
1x1
=1
Lect
ure
919
EL2
620
2011
Slid
ing
Mod
eD
ynam
ics
The
dyna
mic
sal
ong
the
slid
ing
surfa
ceS
isob
tain
edby
setti
ngu
=u
eq(
[!1,
1]su
chth
atx(t
)st
ays
onS
.
ueq
isca
lled
the
equi
vale
ntco
ntro
l.
Lect
ure
920
EL2
620
2011
Exa
mpl
e(c
ont’d
)
Find
ing
u=
ueq
such
that
#(x
)=
x2
=0
on#(x
)=
x2
=0
give
s
0=
x2
=x
1!
x2
*+,-
=0
+u
eq=
x1+
ueq
#u
eq=
!x
1
Inse
rtth
isin
the
equa
tion
forx
1:
x1
=!
x2
*+,-
=0
+u
eq=
!x
1
give
sth
edy
nam
ics
onth
esl
idin
gsu
rface
S=
{x:
x2
=0}
.
Lect
ure
921
EL2
620
2011
Der
ivin
gth
eE
quiv
alen
tCon
trol
Ass
ume
x=
f(x
)+
g(x
)u
u=
!sg
n#(x
)
has
ast
able
slid
ing
surfa
ceS
={x
:#(x
)=
0}.
Then
,for
x(
S,
0=
#(x
)=
d# dx
·dx dt
=d# dx
$ f(x
)+
g(x
)u
%
The
equi
vale
ntco
ntro
lis
thus
give
nby
ueq
=!
$d# dx
g(x
)% !1d# dx
f(x
)
ifth
ein
vers
eex
ists
.
Lect
ure
922
EL2
620
2011
Equ
ival
entC
ontr
olfo
rLi
near
Sys
tem
x=
Ax
+B
u
u=
!sg
n#(x
)=
!sg
n(C
x)
Ass
ume
CB
>0.
The
slid
ing
surfa
ceS
={x
:C
x=
0}so
0=
#(x
)=
d# dx
$ f(x
)+
g(x
)u
%=
C! A
x+
Bu
eq
"
give
su
eq=
!C
Ax/C
B.
Exa
mpl
e(c
ont’d
):Fo
rthe
exam
ple:
ueq
=!
CA
x/C
B=
!! 1
!1" x
=!
x1,
beca
use
#(x
)=
x2
=0.
(Sam
ere
sult
asbe
fore
.)
Lect
ure
923
EL2
620
2011
Slid
ing
Dyn
amic
s
The
dyna
mic
son
S=
{x:
Cx
=0}
isgi
ven
by
x=
Ax
+B
ueq
=
$ I!
1
CB
BC
% Ax,
unde
rthe
cons
train
tCx
=0,
whe
reth
eei
genv
alue
sof
(I!
BC
/CB
)Aar
eeq
ualt
oth
eze
ros
ofsG
(s)
=sC
(sI
!A
)!1B
.
Rem
ark:
The
cond
ition
that
Cx
=0
corr
espo
nds
toth
eze
roat
s=
0,an
dth
usth
isdy
nam
icdi
sapp
ears
onS
={x
:C
x=
0}.
Lect
ure
924
EL2
620
2011
Pro
of
x=
Ax
+B
u
y=
Cx
#y
=C
Ax
+C
Bu
#u
=1
CB
CA
x!
1
CB
y#
x=
$ I!
1
CB
BC
% Ax
!1
CB
By
Hen
ce,t
hetra
nsfe
rfun
ctio
nfro
my
tou
equa
ls
!1
CB
+1
CB
CA
(sI
!((
I!
1
CB
BC
)A))
!1!
1
CB
B
butt
his
trans
ferf
unct
ion
isal
so1/
(sG
(s))
Hen
ce,t
heei
genv
alue
sof
(I!
BC
/CB
)Aar
eeq
ualt
oth
eze
ros
ofsG
(s).
Lect
ure
925
EL2
620
2011
Des
ign
ofS
lidin
gM
ode
Con
trol
ler
Idea
:D
esig
na
cont
roll
awth
atfo
rces
the
stat
eto
#(x
)=
0.C
hoos
e#(x
)su
chth
atth
esl
idin
gm
ode
tend
sto
the
orig
in.
Ass
ume
d dt
. / / / 0
x1
x2 . . . xn
1 2 2 2 3=
. / / / 0
f 1(x
)+
g 1(x
)u
x1 . . .
xn!
1
1 2 2 2 3=
f(x
)+
g(x
)u
Cho
ose
cont
roll
aw u=
!pT
f(x
)
pTg(x
)!
µ
pTg(x
)sg
n#(x
),
whe
reµ
>0
isa
desi
gnpa
ram
eter
,#(x
)=
pTx
,and
pT=
! p 1..
.p n
" are
the
coef
ficie
nts
ofa
stab
lepo
lyno
mia
l.
Lect
ure
926
EL2
620
2011
Clo
sed-
Loop
Sta
bilit
yC
onsi
derV
(x)
=#
2(x
)/2
with
#(x
)=
pTx
.Th
en,
V=
#T(x
)#(x
)=
xTp! pT
f(x
)+
pTg(x
)u"
With
the
chos
enco
ntro
llaw
,we
get
V=
!µ#(x
)sg
n#(x
)<
0
sox
tend
to#(x
)=
0.
0=
#(x
)=
p 1x
1+
···+
p n!
1x
n!
1+
p nx
n
=p 1
x(n
!1)
n+
···+
p n!
1x
(1)
n+
p nx
(0)
n
whe
rex
(k)
deno
tetim
ede
rivat
ive.
Now
pco
rres
pond
sto
ast
able
diffe
rent
iale
quat
ion,
and
xn
&0
expo
nent
ially
ast&
).
The
stat
ere
latio
nsx
k!
1=
xk
now
give
x&
0ex
pone
ntia
llyas
t&
).
Lect
ure
927
EL2
620
2011
Tim
eto
Sw
itch
Con
side
ran
initi
alpo
intx
0su
chth
at#
0=
#(x
0)>
0.S
ince
#(x
)#(x
)=
!µ#(x
)sg
n#(x
)
itfo
llow
sth
atas
long
as#(x
)>
0:
#(x
)=
!µ
Hen
ce,t
hetim
eto
the
first
switc
h(#
(x)
=0)
is
t s=
#0 µ
<)
Not
eth
att s
&0
asµ
&)
.
Lect
ure
928
EL2
620
2011
Exa
mpl
e—S
lidin
gM
ode
Con
trol
ler
Des
ign
stat
e-fe
edba
ckco
ntro
llerf
or
x=
$ 10
10%
x+
$ 1 0%u
y=
! 01" x
Cho
ose
p 1s
+p 2
=s
+1
soth
at#(x
)=
x1+
x2.
The
cont
rolle
ris
give
nby
u=
!pT
Ax
pTB
!µ
pTB
sgn
#(x
)
=2x
1!
µsg
n(x
1+
x2)
Lect
ure
929
EL2
620
2011
Pha
sePo
rtra
it
Sim
ulat
ion
with
µ=
0.5.
Not
eth
esl
idin
gsu
rface
#(x
)=
x1+
x2.
−2−1
01
2
−101
x1
x2
Lect
ure
930
EL2
620
2011
Tim
eP
lots
Initi
alco
nditi
onx(0
)=
! 1.5
0" T.
Sim
ulat
ion
agre
esw
ellw
ithtim
eto
switc
h
t s=
#0 µ
=3
and
slid
ing
dyna
mic
s
y=
!y
x1
x2
u
Lect
ure
931
EL2
620
2011
The
Slid
ing
Mod
eC
ontr
olle
ris
Rob
ust
Ass
ume
that
only
am
odel
x=
4 f(x)+
4g(x)u
ofth
etr
uesy
stem
x=
f(x
)+
g(x
)uis
know
n.S
till,
how
ever
,
V=
#(x
)5 pT(f
4gT!
4 fgT)p
pT4g
!µ
pTg
pT4gsg
n#(x
)6<
0
ifsg
n(p
Tg)
=sg
n(p
T4g)
and
µ>
0is
suffi
cien
tlyla
rge.
The
clos
ed-lo
opsy
stem
isth
usro
bust
agai
nstm
odel
erro
rs!
(Hig
hga
inco
ntro
lwith
stab
leop
enlo
opze
ros)
Lect
ure
932
EL2
620
2011
Com
men
tson
Slid
ing
Mod
eC
ontr
ol
•E
ffici
enth
andl
ing
ofm
odel
unce
rtai
ntie
s
•O
ften
impo
ssib
leto
impl
emen
tinfi
nite
fast
switc
hing
•S
moo
thve
rsio
nth
roug
hlo
wpa
ssfil
tero
rbou
ndar
yla
yer
•A
pplic
atio
nsin
robo
tics
and
vehi
cle
cont
rol
•C
ompa
repu
ls-w
idth
mod
ulat
edco
ntro
lsig
nals
Lect
ure
933
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
toan
alyz
ean
dde
sign
•H
igh-
gain
cont
rols
yste
ms
•S
lidin
gm
ode
cont
rolle
rs
Lect
ure
934
EL2
620
2011
Nex
tLec
ture
•Ly
apun
ovde
sign
met
hods
•E
xact
feed
back
linea
rizat
ion
Lect
ure
935
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
10
•E
xact
feed
back
linea
rizat
ion
•In
put-o
utpu
tlin
eariz
atio
n
•Ly
apun
ov-b
ased
cont
rold
esig
nm
etho
ds
Lect
ure
101
EL2
620
2011
Out
putF
eedb
ack
and
Sta
teFe
edba
ck
x=
f(x
,u)
y=
h(x
)
•O
utpu
tfee
dbac
k:Fi
ndu
=k(y
)su
chth
atth
ecl
osed
-loop
syst
emx
=f! x
,k! h
(x)""
has
nice
prop
ertie
s.
•S
tate
feed
back
:Fi
ndu
=!(
x)
such
that
x=
f(x
,!(x
))
has
nice
prop
ertie
s.
kan
d!
may
incl
ude
dyna
mic
s.
Lect
ure
102
EL2
620
2011
Non
linea
rC
ontr
olle
rs
•N
onlin
eard
ynam
ical
cont
rolle
r:z
=a(z
,y),
u=
c(z)
•Li
near
dyna
mic
s,st
atic
nonl
inea
rity:
z=
Az
+B
y,u
=c(
z)
•Li
near
cont
rolle
r:z
=A
z+
By
,u=
Cz
Lect
ure
103
EL2
620
2011
Non
linea
rO
bser
vers
Wha
tifx
isno
tmea
sura
ble?
x=
f(x
,u),
y=
h(x
)
Sim
ples
tobs
erve
r#x
=f(#x
,u)
Feed
back
corr
ectio
n,as
inlin
earc
ase,
#x=
f(#x
,u)+
K(y
!h(#x
))
Cho
ices
ofK
•Li
near
ize
fat
x0,fi
ndK
fort
helin
eariz
atio
n
•Li
near
ize
fat
#x(t)
,find
K=
K(#x
)fo
rthe
linea
rizat
ion
Sec
ond
case
isca
lled
Ext
ende
dK
alm
anFi
lter
Lect
ure
104
EL2
620
2011
Som
est
ate
feed
back
cont
rola
ppro
ache
s
•E
xact
feed
back
linea
rizat
ion
•In
put-o
utpu
tlin
eariz
atio
n
•Ly
apun
ov-b
ased
desi
gn-b
acks
tepp
ing
cont
rol
Lect
ure
105
EL2
620
2011
Exa
ctFe
edba
ckLi
near
izat
ion
Con
side
rthe
nonl
inea
rcon
trol-a
ffine
syst
em
x=
f(x
)+
g(x
)u
idea
:us
ea
stat
e-fe
edba
ckco
ntro
lleru
(x)
tom
ake
the
syst
emlin
ear
Exa
mpl
e1:
x=
cosx
!x
3+
u
The
stat
e-fe
edba
ckco
ntro
ller
u(x
)=
!co
sx
+x
3!
kx
+v
yiel
dsth
elin
ears
yste
m
x=
!kx
+v
Lect
ure
106
EL2
620
2011
Ano
ther
Exa
mpl
e
x1
=a
sin
x2
x2
=!
x2 1+
u
How
dow
eca
ncel
the
term
sin
x2?
Perfo
rmtra
nsfo
rmat
ion
ofst
ates
into
linea
rizab
lefo
rm:
z 1=
x1,
z 2=
x1
=a
sin
x2
yiel
dsz 1
=z 2
,z 2
=a
cosx
2(!
x2 1+
u)
and
the
linea
rizin
gco
ntro
lbec
omes
u(x
)=
x2 1+
v
aco
sx
2
,x
2"
[!"/2
,"/2
]
Lect
ure
107
EL2
620
2011
Diff
eom
orph
ism
sA
nonl
inea
rsta
tetra
nsfo
rmat
ion
z=
T(x
)w
ith
•T
inve
rtib
lefo
rxin
the
dom
ain
ofin
tere
st
•T
and
T!
1co
ntin
uous
lydi
ffere
ntia
ble
isca
lled
adi
ffeom
orph
ism
Defi
nitio
n:A
nonl
inea
rsys
tem
x=
f(x
)+
g(x
)u
isfe
edba
cklin
eari
zabl
eif
ther
eex
ista
diffe
omor
phis
mT
who
sedo
mai
nco
ntai
nsth
eor
igin
and
trans
form
sth
esy
stem
into
the
form
x=
Ax
+B
#(x
)(u
!$(x
))
with
(A,B
)co
ntro
llabl
ean
d#(x
)no
nsin
gula
rfor
allx
inth
edo
mai
nof
inte
rest
.
Lect
ure
108
EL2
620
2011
Exa
ctvs
.Inp
ut-O
utpu
tLin
eari
zatio
nTh
eex
ampl
eag
ain,
butn
oww
ithan
outp
ut
x1
=a
sin
x2,
x2
=!
x2 1+
u,
y=
x2
•Th
eco
ntro
llaw
u=
x2 1+
v/a
cosx
2yi
elds
z 1=
z 2;
z 2=
v;
y=
sin
!1(z
2/a
)
whi
chis
nonl
inea
rin
the
outp
ut.
•If
we
wan
talin
eari
nput
-out
putr
elat
ions
hip
we
coul
din
stea
dus
e
u=
x2 1+
v
toob
tain
x1
=a
sin
x2,
x2
=v,
y=
x2
whi
chis
linea
rfro
mv
toy
Cau
tion:
the
cont
rolh
asre
nder
edth
est
ate
x1
unob
serv
able
Lect
ure
109
EL2
620
2011
Inpu
t-O
utpu
tLin
eari
zatio
n
Use
stat
efe
edba
cku(x
)to
mak
eth
eco
ntro
l-affi
nesy
stem
x=
f(x
)+
g(x
)u
y=
h(x
)
linea
rfro
mth
ein
putv
toth
eou
tput
y
The
gene
rali
dea:
diffe
rent
iate
the
outp
ut,y
=h(x
),p
times
until
lthe
cont
rolu
appe
ars
expl
icitl
yin
y(p
) ,and
then
dete
rmin
eu
soth
at
y(p
)=
v
i.e.,
G(s
)=
1/sp
Lect
ure
1010
EL2
620
2011
Exa
mpl
e:co
ntro
lled
van
derP
oleq
uatio
n
x1
=x
2
x2
=!
x1+
%(1
!x
2 1)x
2+
u
y=
x1
Diff
eren
tiate
the
outp
ut
y=
x1
=x
2
y=
x2
=!
x1+
%(1
!x
2 1)x
2+
u
The
stat
efe
edba
ckco
ntro
ller
u=
x1!
%(1
!x
2 1)x
2+
v#
y=
v
Lect
ure
1011
EL2
620
2011
Lie
Der
ivat
ives
Con
side
rthe
nonl
inea
rSIS
Osy
stem
x=
f(x
)+
g(x
)u;
x"
Rn,u
"R
1
y=
h(x
),y
"R
The
deriv
ativ
eof
the
outp
ut
y=
dh
dxx
=dh
dx
(f(x
)+
g(x
)u)!
Lfh(x
)+
Lgh(x
)u
whe
reL
fh(x
)an
dL
gh(x
)ar
eLi
ede
rivat
ives
(Lfh
isth
ede
rivat
ive
ofh
alon
gth
eve
ctor
field
ofx
=f(x
))
Rep
eate
dde
rivat
ives
Lk fh(x
)=
d(L
k!
1f
h)
dx
f(x
),L
gL
fh(x
)=
d(L
fh)
dx
g(x
)
Lect
ure
1012
EL2
620
2011
Lie
deri
vativ
esan
dre
lativ
ede
gree
•Th
ere
lativ
ede
gree
pof
asy
stem
isde
fined
asth
enu
mbe
rof
inte
grat
ors
betw
een
the
inpu
tand
the
outp
ut(th
enu
mbe
roft
imes
ym
ustb
edi
ffere
ntia
ted
fort
hein
putu
toap
pear
)
•A
linea
rsys
tem
Y(s
)
U(s
)=
b 0sm
+..
.+b m
sn+
a1sn
!1+
...+
an
has
rela
tive
degr
eep
=n
!m
•A
nonl
inea
rsys
tem
has
rela
tive
degr
eep
if
LgL
i!1
fh(x
)=
0,i=
1,..
.,p!
1;
LgL
p!
1f
h(x
)$=
0%x
"D
Lect
ure
1013
EL2
620
2011
Exa
mpl
e
The
cont
rolle
dva
nde
rPol
equa
tion
x1
=x
2
x2
=!
x1+
%(1
!x
2 1)x
2+
u
y=
x1
Diff
eren
tiatin
gth
eou
tput y
=x
1=
x2
y=
x2
=!
x1+
%(1
!x
2 1)x
2+
u
Thus
,the
syst
emha
sre
lativ
ede
gree
p=
2
Lect
ure
1014
EL2
620
2011
The
inpu
t-ou
tput
linea
rizi
ngco
ntro
l
Con
side
rant
hor
derS
ISO
syst
emw
ithre
lativ
ede
gree
p
x=
f(x
)+
g(x
)u
y=
h(x
)
Diff
eren
tiatin
gth
eou
tput
repe
ated
ly
y=
dh
dxx
=L
fh(x
)+
=0
$%&
'L
gh(x
)u
. . .
y(p
)=
Lp fh(x
)+
LgL
p!
1f
h(x
)u
Lect
ure
1015
EL2
620
2011
and
henc
eth
est
ate-
feed
back
cont
rolle
r
u=
1
LgL
p!
1f
h(x
)
! !L
p fh(x
)+
v"
resu
ltsin
the
linea
rinp
ut-o
utpu
tsys
tem
y(p
)=
v
Lect
ure
1016
EL2
620
2011
Zero
Dyn
amic
s
•N
ote
that
the
orde
roft
helin
eariz
edsy
stem
isp,
corr
espo
ndin
gto
the
rela
tive
degr
eeof
the
syst
em
•Th
us,i
fp<
nth
enn
!p
stat
esar
eun
obse
rvab
lein
y.
•Th
edy
nam
ics
ofth
en
!p
stat
esno
tobs
erva
ble
inth
elin
eariz
eddy
nam
ics
ofy
are
calle
dth
eze
rody
nam
ics.
Cor
resp
onds
toth
edy
nam
ics
ofth
esy
stem
whe
ny
isfo
rced
tobe
zero
fora
lltim
es.
•A
syst
emw
ithun
stab
leze
rody
nam
ics
isca
lled
non-
min
imum
phas
e(a
ndsh
ould
notb
ein
put-o
utpu
tlin
eariz
ed!)
Lect
ure
1017
EL2
620
2011
van
der
Pola
gain
x1
=x
2
x2
=!
x1+
%(1
!x
2 1)x
2+
u
•W
ithy
=x
1th
ere
lativ
ede
gree
p=
n=
2an
dth
ere
are
noze
rody
nam
ics,
thus
we
can
trans
form
the
syst
emin
toy
=v
.Tr
yit
your
self!
•W
ithy
=x
2th
ere
lativ
ede
gree
p=
1<
nan
dth
eze
rody
nam
ics
are
give
nby
x1
=0,
whi
chis
nota
sym
ptot
ical
lyst
able
(but
boun
ded)
Lect
ure
1018
EL2
620
2011
Lyap
unov
-Bas
edC
ontr
olD
esig
nM
etho
ds
x=
f(x
,u)
•Fi
ndst
abili
zing
stat
efe
edba
cku
=u(x
)
•Ve
rify
stab
ility
thro
ugh
Con
trolL
yapu
nov
func
tion
•M
etho
dsde
pend
onst
ruct
ure
off
Her
ew
elim
itdi
scus
sion
toB
ack-
step
ping
cont
rold
esig
n,w
hich
requ
irece
rtai
nf
disc
usse
dla
ter.
Lect
ure
1019
EL2
620
2011
Asi
mpl
ein
trod
ucto
ryex
ampl
e
Con
side
rx
=co
sx
!x
3+
u
App
lyth
elin
eariz
ing
cont
rol
u=
!co
sx
+x
3!
kx
Cho
ose
the
Lyap
unov
cand
idat
eV
(x)
=x
2/2
V(x
)>
0,
V=
!kx
2<
0
Thus
,the
syst
emis
glob
ally
asym
ptot
ical
lyst
able
But
,the
term
x3
inth
eco
ntro
llaw
may
requ
irela
rge
cont
rolm
oves
!
Lect
ure
1020
EL2
620
2011
The
sam
eex
ampl
ex
=co
sx
!x
3+
u
Now
try
the
cont
roll
aw
u=
!co
sx
!kx
Cho
ose
the
sam
eLy
apun
ovca
ndid
ate
V(x
)=
x2/2
V(x
)>
0,
V=
!x
4!
kx
2<
0
Thus
,als
ogl
obal
lyas
ympt
otic
ally
stab
le(a
ndm
ore
nega
tive
V)
Lect
ure
1021
EL2
620
2011
Sim
ulat
ing
the
two
cont
rolle
rsS
imul
atio
nw
ithx(0
)=
10
Sta
tetra
ject
ory
Con
troli
nput
00.5
11.5
22.5
33.5
44.5
5012345678910
time [s]
State x
linearizing
non-linearizing
00.5
11.5
22.5
33.5
44.5
5-200
0
200
400
600
800
1000
time [s]
Input u
linearizing
non-linearizing
The
linea
rizin
gco
ntro
lis
slow
eran
dus
esex
cess
ive
inpu
t.Th
us,
linea
rizat
ion
can
have
asi
gnifi
cant
cost
!
Lect
ure
1022
EL2
620
2011
Bac
k-S
tepp
ing
Con
trol
Des
ign
We
wan
tto
desi
gna
stat
efe
edba
cku
=u(x
)th
atst
abili
zes
x1
=f(x
1)+
g(x
1)x
2
x2
=u
(1)
atx
=0
with
f(0
)=
0.
Idea
:S
eeth
esy
stem
asa
casc
ade
conn
ectio
n.D
esig
nco
ntro
llerfi
rst
fort
hein
nerl
oop
and
then
fort
heou
ter.
ux
2x
1(
g(x
1)
f()
(
Lect
ure
1023
EL2
620
2011
Sup
pose
the
part
ials
yste
m
x1
=f(x
1)+
g(x
1)v
can
best
abili
zed
byv
=&(x
1)
and
ther
eex
ists
Lyap
unov
fcn
V1
=V
1(x
1)
such
that
V1(x
1)
=dV
1
dx
1
) f(x
1)+
g(x
1)&
(x1)*
&!
W(x
1)
fors
ome
posi
tive
defin
itefu
nctio
nW
.
This
isa
criti
cala
ssum
ptio
nin
back
step
ping
cont
rol!
Lect
ure
1024
EL2
620
2011
The
Tric
k
Equ
atio
n(1
)can
bere
writ
ten
as
x1
=f(x
1)+
g(x
1)&
(x1)+
g(x
1)[x
2!
&(x
1)]
x2
=u
!!(x
1)
ux
2x
1(
g(x
1)
f+
g!
(
Lect
ure
1025
EL2
620
2011
Intro
duce
new
stat
e'
=x
2!
&(x
1)
and
cont
rolv
=u
!&
:
x1
=f(x
1)+
g(x
1)&
(x1)+
g(x
1)'
'=
v
whe
re
&(x
1)
=d&
dx
1
x1
=d&
dx
1
) f(x
1)+
g(x
1)x
2
*
!!(x
1)
u"
x1
(g(x
1)
f+
g!
(
Lect
ure
1026
EL2
620
2011
Con
side
rV2(x
1,x
2)
=V
1(x
1)+
'2/2
.Th
en,
V2(x
1,x
2)
=dV
1
dx
1
) f(x
1)+
g(x
1)&
(x1)*
+dV
1
dx
1
g(x
1)'
+'v
&!
W(x
1)+
dV
1
dx
1
g(x
1)'
+'v
Cho
osin
g
v=
!dV
1
dx
1
g(x
1)!
k',
k>
0
give
sV
2(x
1,x
2)&
!W
(x1)!
k'
2
Hen
ce,x
=0
isas
ympt
otic
ally
stab
lefo
r(1)
with
cont
roll
awu(x
)=
&(x
)+
v(x
).
IfV
1ra
dial
lyun
boun
ded,
then
glob
alst
abili
ty.
Lect
ure
1027
EL2
620
2011
Bac
k-S
tepp
ing
Lem
ma
Lem
ma:
Letz
=(x
1,.
..,x
k!
1)T
and
z=
f(z
)+
g(z
)xk
xk
=u
Ass
ume
&(0
)=
0,f(0
)=
0,
z=
f(z
)+
g(z
)&(z
)
stab
le,a
ndV
(z)
aLy
apun
ovfc
n(w
ithV
&!
W).
Then
,
u=
d& dz
) f(z
)+
g(z
)xk
*!
dV dz
g(z
)!
(xk!
&(z
))
stab
ilize
sx
=0
with
V(z
)+
(xk!
&(z
))2/2
bein
ga
Lyap
unov
fcn.
Lect
ure
1028
EL2
620
2011
Str
ictF
eedb
ack
Sys
tem
s
Bac
k-st
eppi
ngLe
mm
aca
nbe
appl
ied
tost
abili
zesy
stem
son
stric
tfe
edba
ckfo
rm: x
1=
f 1(x
1)+
g 1(x
1)x
2
x2
=f 2
(x1,x
2)+
g 2(x
1,x
2)x
3
x3
=f 3
(x1,x
2,x
3)+
g 3(x
1,x
2,x
3)x
4
. . .
xn
=f n
(x1,.
..,x
n)+
g n(x
1,.
..,x
n)u
whe
reg k
$=0
Not
e:x
1,.
..,x
kdo
notd
epen
don
xk+
2,.
..,x
n.
Lect
ure
1029
EL2
620
2011
2m
inut
eex
erci
se:
Giv
ean
exam
ple
ofa
linea
rsys
tem
x=
Ax
+B
uon
stric
tfee
dbac
kfo
rm.
Lect
ure
1030
EL2
620
2011
Bac
k-S
tepp
ing
Bac
k-S
tepp
ing
Lem
ma
can
beap
plie
dre
curs
ivel
yto
asy
stem
x=
f(x
)+
g(x
)u
onst
rictf
eedb
ack
form
.
Bac
k-st
eppi
ngge
nera
tes
stab
ilizi
ngfe
edba
cks&
k(x
1,.
..,x
k)
(equ
alto
uin
Bac
k-S
tepp
ing
Lem
ma)
and
Lyap
unov
func
tions
Vk(x
1,.
..,x
k)
=V
k!
1(x
1,.
..,x
k!
1)+
[xk!
&k!
1]2
/2
by“s
tepp
ing
back
”fro
mx
1to
u(s
eeK
halil
pp.5
93–5
94fo
rdet
ails
).
Bac
k-st
eppi
ngre
sults
inth
efin
alst
ate
feed
back
u=
&n(x
1,.
..,x
n)
Lect
ure
1031
EL2
620
2011
Exa
mpl
e
Des
ign
back
-ste
ppin
gco
ntro
llerf
or
x1
=x
2 1+
x2,
x2
=x
3,
x3
=u
Ste
p0
Verif
yst
rictf
eedb
ack
form
Ste
p1
Con
side
rfirs
tsub
syst
em
x1
=x
2 1+
&1(x
1),
x2
=u
1
whe
re&
1(x
1)
=!
x2 1!
x1
stab
ilize
sth
efir
steq
uatio
n.W
ithV
1(x
1)
=x
2 1/2
,Bac
k-S
tepp
ing
Lem
ma
give
s
u1
=(!
2x1!
1)(x
2 1+
x2)!
x1!
(x2+
x2 1+
x1)
=&
2(x
1,x
2)
V2
=x
2 1/2
+(x
2+
x2 1+
x1)2
/2
Lect
ure
1032
EL2
620
2011
Ste
p2
App
lyin
gB
ack-
Ste
ppin
gLe
mm
aon
x1
=x
2 1+
x2
x2
=x
3
x3
=u
give
s
u=
u2
=d&
2
dz
) f(z
)+
g(z
)xn
*!
dV
2
dz
g(z
)!
(xn
!&
2(z
))
=(&
2
(x
1
(x2 1+
x2)+
(&
2
(x
2
x3!
(V
2
(x
2
!(x
3!
&2(x
1,x
2))
whi
chgl
obal
lyst
abili
zes
the
syst
em.
Lect
ure
1033
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
11
•N
onlin
earc
ontro
llabi
lity
•G
ain
sche
dulin
g
Lect
ure
111
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•D
eter
min
eif
ano
nlin
ears
yste
mis
cont
rolla
ble
•A
pply
gain
sche
dulin
gto
sim
ple
exam
ples
Lect
ure
112
EL2
620
2011
Con
trol
labi
lity
Defi
nitio
n:x
=f(x
,u)
isco
ntro
llabl
eif
fora
nyx
0,x
1th
ere
exis
tsT
>0
and
u:[0
,T]!
Rsu
chth
atx(0
)=
x0
and
x(T
)=
x1.
Lect
ure
113
EL2
620
2011
Line
arS
yste
ms
Lem
ma:
x=
Ax
+B
u
isco
ntro
llabl
eif
and
only
if
Wn
=! B
AB
...
An!
1B
"
has
full
rank
.
Isth
ere
aco
rres
pond
ing
resu
ltfo
rno
nlin
ear
syst
ems?
Lect
ure
114
EL2
620
2011
Con
trol
labl
eLi
near
izat
ion
Lem
ma:
Let
z=
Az
+B
u
beth
elin
eariz
atio
nof
x=
f(x
)+
g(x
)u
atx
=0
with
f(0
)=
0.If
the
linea
rsys
tem
isco
ntro
llabl
eth
enth
eno
nlin
ears
yste
mis
cont
rolla
ble
ina
neig
hbor
hood
ofth
eor
igin
.
Rem
ark:
•H
ence
,ifr
ank
Wn
=n
then
ther
eis
an!
>0
such
that
fore
very
x1
"B
!(0)
ther
eex
ists
u:[0
,T]!
Rso
that
x(T
)=
x1
•A
nonl
inea
rsys
tem
can
beco
ntro
llabl
e,ev
enif
the
linea
rized
syst
emis
notc
ontro
llabl
e
Lect
ure
115
EL2
620
2011
Car
Exa
mpl
e
"#
x
y
(x,y
)
Inpu
t:u
1st
eerin
gw
heel
velo
city
,u2
forw
ard
velo
city
d dt
# $ $ $ %
x y " #
& ' ' ' (=
# $ $ $ %
0 0 0 1& ' ' ' (u
1+
# $ $ $ %
cos(
"+
#)
sin("
+#)
sin(#
)
0
& ' ' ' (u
2=
g 1(z
)u1+
g 2(z
)u2
Lect
ure
116
EL2
620
2011
Line
ariz
atio
nfo
ru1
=u
2=
0gi
ves
z=
Az
+B
1u
1+
B2u
2
with
A=
0an
d
B1
=
# $ $ $ %
0 0 0 1& ' ' ' (,
B2
=
# $ $ $ %
cos(
"0+
# 0)
sin("
0+
# 0)
sin(#
0)
0
& ' ' ' (
rank
Wn
=ra
nk
! BA
B..
.A
n!
1B
" =2
<4,
soth
elin
eariz
atio
nis
notc
ontro
llabl
e.S
tillt
heca
ris
cont
rolla
ble!
Line
ariz
atio
ndo
esno
tcap
ture
the
cont
rolla
bilit
ygo
oden
ough
Lect
ure
117
EL2
620
2011
Lie
Bra
cket
s
Lie
brac
ketb
etw
een
vect
orfie
ldsf,g
:R
n!
Rn
isa
vect
orfie
ldde
fined
by
[f,g
]=
$g
$xf
#$f
$x
g
Exa
mpl
e:
f=
) cosx
2
x1
*,
g=
) x1 1
*
[f,g
]=
) 10
00*
) cosx
2
x1
*#
) 0#
sin
x2
10
*) x
1 1
*
=
) cosx
2+
sin
x2
#x
1
*
Lect
ure
118
EL2
620
2011
Lie
Bra
cket
Dir
ectio
n
Fort
hesy
stem
x=
g 1(x
)u1+
g 2(x
)u2
the
cont
rol
(u1,u
2)
=
+ , , , - , , , .
(1,0
),t"
[0,!
)
(0,1
),t"
[!,2
!)
(#1,
0),
t"
[2!,
3!)
(0,#
1),
t"
[3!,
4!)
give
sm
otio
n
x(4
!)=
x(0
)+
!2[g
1,g
2]+
O(!
3)
The
syst
emca
nm
ove
inth
e[g
1,g
2]d
irec
tion!
Lect
ure
119
EL2
620
2011
Pro
of
1.Fo
rt"
[0,!
],as
sum
ing
!sm
alla
ndx(0
)=
x0,T
aylo
rser
ies
yiel
ds
x(!
)=
x0+
g 1(x
0)!
+1 2
dg 1 dx
g 1(x
0)!
2+
O(!
3)
(1)
2.S
imila
rily,
fort
"[!
,2!]
x(2
!)=
x(!
)+
g 2(x
(!))
!+
1 2
dg 2 dx
g 2(x
(!))
!2
and
with
x(!
)fro
m(1
),an
dg 2
(x(!
))=
g 2(x
0)+
dg2
dx
!g1(x
0)
x(2
!)=
x0+
!(g 1
(x0)+
g 2(x
0))
+
!2
)1 2
dg 1 dx
(x0)g
1(x
0)+
dg 2 dx
(x0)g
1(x
0)+
1 2
dg 2 dx
(x0)g
2(x
0)*
Lect
ure
1110
EL2
620
2011
Pro
of,c
ontin
ued
3.S
imila
rily,
fort
"[2
!,3!]
x(3
!)=
x0+
!g2+
!2
)dg 2 dx
g 1#
dg 1 dx
g 2+
1 2
dg 2 dx
g 2
*
4.Fi
nally
,for
t"
[3!,
4!]
x(4
!)=
x0+
!2
)dg 2 dx
g 1#
dg 1 dx
g 2
*
Lect
ure
1111
EL2
620
2011
Car
Exa
mpl
e(C
ont’d
)
g 3:=
[g1,g
2]=
$g 2 $x
g 1#
$g 1 $x
g 2
=
# $ $ $ %
00
#si
n("
+#)
#si
n("
+#)
00
cos(
"+
#)co
s("
+#)
00
0co
s(#)
00
00
& ' ' ' (
# $ $ $ %
0 0 0 1& ' ' ' (#
0
=
# $ $ $ %
#si
n("
+#)
cos(
"+
#)
cos(
#)
0
& ' ' ' (
Lect
ure
1112
EL2
620
2011
We
can
henc
em
ove
the
cari
nth
eg 3
dire
ctio
n(“
wrig
gle”
)by
appl
ying
the
cont
rols
eque
nce
(u1,u
2)
={(
1,0)
,(0,
1),(
#1,
0),(
0,#
1)}
"
#
x
y
(x,y
)
Lect
ure
1113
EL2
620
2011
The
carc
anal
som
ove
inth
edi
rect
ion
g 4:=
[g3,g
2]=
"g 2 "x
g 3#
"g 3 "x
g 2=
...=
# $ $ $ %
#si
n(#
+2$)
cos(
#+
2$)
0 0
& ' ' ' (
g 4di
rect
ion
corr
espo
nds
tosi
dew
ays
mov
emen
t
(#si
n("
),co
s("))
Lect
ure
1114
EL2
620
2011
Par
king
Theo
rem
You
can
geto
utof
any
park
ing
lott
hati
s!
>0
bigg
erth
anyo
urca
rby
appl
ying
cont
rolc
orre
spon
ding
tog 4
,tha
tis,
byap
plyi
ngth
eco
ntro
lseq
uenc
e
Wrig
gle,
Driv
e,#
Wrig
gle,
#D
rive
Lect
ure
1115
EL2
620
2011
2m
inut
eex
erci
se:
Wha
tdoe
sth
edi
rect
ion
[g1,g
2]c
orre
spon
dto
for
alin
ears
yste
mx
=g 1
(x)u
1+
g 2(x
)u2
=B
1u
1+
B2u
2?
Lect
ure
1116
EL2
620
2011
The
Lie
Bra
cket
Tree
[g1,g
2]
[g1,[
g 1,g
2]]
[g2,[
g 1,g
2]]
[g1,[
g 1,[
g 1,g
2]]]
[g2,[
g 1,[
g 1,g
2]]]
[g1,[
g 2,[
g 1,g
2]]]
[g2,[
g 2,[
g 1,g
2]]]
Lect
ure
1117
EL2
620
2011
Con
trol
labi
lity
Theo
rem
Theo
rem
:The
syst
em
x=
g 1(x
)u1+
g 2(x
)u2
isco
ntro
llabl
eif
the
Lie
brac
kett
ree
(toge
ther
with
g 1an
dg 2
)spa
nsR
nfo
rallx
Rem
ark:
•Th
esy
stem
can
best
eere
din
any
dire
ctio
nof
the
Lie
brac
kett
ree
Lect
ure
1118
EL2
620
2011
Exa
mpl
e—U
nicy
cle
d dt
# %x
1
x2 #
& (=
# %co
s#
sin
#
0
& (u
1+
# %0 0 1& (
u2
(x1,x
2)#
g 1=
# %co
s#
sin
#
0
& (,
g 2=
# %0 0 1& (
,[g
1,g
2]=
# %si
n#
#co
s#
0
& (
Con
trolla
ble
beca
use
{g1,g
2,[
g 1,g
2]}
span
sR
3
Lect
ure
1119
EL2
620
2011
2m
inut
eex
erci
se:
•S
how
that
{g1,g
2,[
g 1,g
2]}
span
sR
3fo
rthe
unic
ycle
•Is
the
linea
rizat
ion
ofth
eun
icyc
leco
ntro
llabl
e?
Lect
ure
1120
EL2
620
2011
Exa
mpl
e—R
ollin
gP
enny
(x1,x
2)
#
%
d dt
# $ $ $ %
x1
x2 # %
& ' ' ' (=
# $ $ $ %
cos#
sin
#
0 1
& ' ' ' (u
1+
# $ $ $ %
0 0 1 0& ' ' ' (u
2
Con
trolla
ble
beca
use
{g1,g
2,[
g 1,g
2],
[g2,[
g 1,g
2]]}s
pans
R4
Lect
ure
1121
EL2
620
2011
Whe
nis
Feed
back
Line
ariz
atio
nPo
ssib
le?
Q:W
hen
can
we
trans
form
x=
f(x
)+
g(x
)uin
toz
=A
z+
bvby
mea
nsof
feed
back
u=
&(x
)+
'(x
)van
dch
ange
ofva
riabl
esz
=T
(x)
(see
prev
ious
lect
ure)
?
A:T
hean
swer
requ
ires
Lie
brac
kets
and
furt
herc
once
pts
from
diffe
rent
ialg
eom
etry
(see
Kha
lilan
dP
hDco
urse
)
vx
zu
x=
f%
&
T
Lect
ure
1122
EL2
620
2011
Gai
nS
ched
ulin
g
Con
trolp
aram
eter
sde
pend
onop
erat
ing
cond
ition
s:
Pro
cess
s ch
ed
ule
Ga
in
Ou
tpu
t
Con
trol
sign
al
Con
troll
er
para
mete
rs
Op
era
tin
gco
nd
itio
n
Com
man
dsi
gn
al
Con
troll
er
Exa
mpl
e:P
IDco
ntro
llerw
ithK
=K
(&),
whe
re&
isth
esc
hedu
ling
varia
ble.
Exa
mpl
esof
sche
dulin
gva
riabl
ear
epr
oduc
tion
rate
,mac
hine
spee
d,M
ach
num
ber,
flow
rate
Lect
ure
1123
EL2
620
2011
Valv
eC
hara
cter
istic
s
Flo
�
Po
sit
ion
Qu
ick
�op
en
ing
Lin
ea
r
E�u
al�p
erc
en
tag
e
Lect
ure
1124
EL2
620
2011
Non
linea
rVa
lve
ΣP
Ic
uf
vy
Pro
cess
−1 f −1
uc
G0
(s)
Valv
ech
arac
teris
tics
00.
51
1.5
2010
f(u
)f(u
)
Lect
ure
1125
EL2
620
2011
With
outg
ain
sche
dulin
g:
010
2030
400.
2
0.3
010
2030
401.
0
1.1
010
2030
40
5.0
5.2
Tim
e
Tim
e
Tim
e
uc
y
uc
y
y
uc
Lect
ure
1126
EL2
620
2011
With
gain
sche
dulin
g:
020
4060
8010
00.
2
0.3
020
4060
8010
01.
0
1.1
020
4060
8010
05.
0
5.1
Tim
e
Tim
e
Tim
e
uc
y
uc y
uc
y
Lect
ure
1127
EL2
620
2011
Flig
htC
ontr
ol
Pitc
hdy
nam
ics
α
Vθ
q=
˙ θ
Nz
δ e
Lect
ure
1128
EL2
620
2011
Flig
htC
ontr
ol
Ope
ratin
gco
nditi
ons:
00.
40.
81.
21.
62.
02.
4
80 60 40 20 01
234
Mac
h nu
mbe
r
Altitude (x1000 ft)
Lect
ure
1129
EL2
620
2011
The
Pitc
hC
ontr
olC
hann
el
Fil
ter
Fil
ter
Fil
ter
A/D
A/D
A/D
D/A
D/A
Fil
ter
−
H
HM
H
M
Pit
ch s
tick
Posi
tion
Acc
ele
rati
on
Pit
ch r
ate
Σ
Σ
Σ
ΣΣ
Gea
r
To s
erv
os
Σ
VIA
S
VIA
S
VIA
SM
H
KD
SE
KS
G
T1s
1+
T1s
11
+T
3s
KQ
1 K
NZ
M H
KQ
D
T2s
1+
T2s
Lect
ure
1130
EL2
620
2011
Toda
y’s
Goa
l
You
shou
ldbe
able
to
•D
eter
min
eif
ano
nlin
ears
yste
mis
cont
rolla
ble
•A
pply
gain
sche
dulin
gto
sim
ple
exam
ples
Lect
ure
1131
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
12
•O
ptim
alco
ntro
l
Lect
ure
121
EL2
620
2011
Toda
y’s
Goa
lYo
ush
ould
beab
leto
•D
esig
nco
ntro
llers
base
don
optim
alco
ntro
lthe
ory
Lect
ure
122
EL2
620
2011
Opt
imal
Con
trol
Pro
blem
sId
ea:
form
ulat
eth
eco
ntro
ldes
ign
prob
lem
asan
optim
izat
ion
prob
lem
min
u(t
)J(x
,u,t
),x
=f(t
,x,u
)
+pr
ovid
esa
syst
emat
icde
sign
fram
ewor
k
+ap
plic
able
tono
nlin
earp
robl
ems
+ca
nde
alw
ithco
nstra
ints
-di
fficu
ltto
form
ulat
eco
ntro
lobj
ectiv
esas
asi
ngle
obje
ctiv
efu
nctio
n
-de
term
inin
gth
eop
timal
cont
rolle
rcan
beha
rd
Lect
ure
123
EL2
620
2011
Exa
mpl
e—B
oati
nS
trea
mS
aila
sfa
ras
poss
ible
inx
1di
rect
ion
Spe
edof
wat
erv(x
2)
with
dv/d
x2
=1
Rud
dera
ngle
cont
rol:
u(t
)!
U=
{(u
1,u
2)
:u
2 1+
u2 2
=1}
x1
x2
v(x
2)
max
u:[0,t
f]!
Ux
1(t
f)
x1(t
)=
v(x
2)+
u1(t
)
x2(t
)=
u2(t
)
x1(0
)=
x2(0
)=
0
Lect
ure
124
EL2
620
2011
Exa
mpl
e—R
esou
rce
Allo
catio
nM
axim
izat
ion
ofst
ored
profi
t
x(t
)!
[0,"
)pr
oduc
tion
rate
u(t
)!
[0,1
]po
rtio
nof
xre
inve
sted
1#
u(t
)po
rtio
nof
xst
ored
!u(t
)x(t
)ch
ange
ofpr
oduc
tion
rate
(!>
0)
[1#
u(t
)]x(t
)am
ount
ofst
ored
profi
t
max
u:[0,t
f]!
[0,1
]
!t f
0
[1#
u(t
)]x(t
)dt
x(t
)=!u(t
)x(t
)
x(0
)=
x0
>0
Lect
ure
125
EL2
620
2011
Exa
mpl
e—M
inim
alC
urve
Leng
thFi
ndth
ecu
rve
with
min
imal
leng
thbe
twee
na
give
npo
inta
nda
line
Cur
ve:(t
,x(t
))w
ithx(0
)=
a
Line
:Ve
rtic
alth
roug
h(t
f,0
)
t
t f
x(t
)
a
min
u:[0,t
f]!
R
!t f
0
"1
+u
2(t
)dt
x(t
)=
u(t
)
x(0
)=
a
Lect
ure
126
EL2
620
2011
Opt
imal
Con
trol
Pro
blem
Sta
ndar
dfo
rm: m
inu:[0,t
f]!
U
!t f
0
L(x
(t),
u(t
))dt+"(x
(tf))
x(t
)=
f(x
(t),
u(t
)),
x(0
)=
x0
Rem
arks
:
•U
$R
mse
tofa
dmis
sibl
eco
ntro
l
•In
finite
dim
ensi
onal
optim
izat
ion
prob
lem
:O
ptim
izat
ion
over
func
tions
u:[0
,tf]%
U
•C
onst
rain
tson
xfro
mth
edy
nam
ics
•Fi
nalt
ime
t ffix
ed(fr
eela
ter)
Lect
ure
127
EL2
620
2011
Pont
ryag
in’s
Max
imum
Pri
ncip
leTh
eore
m:
Intro
duce
the
Ham
ilton
ian
func
tion
H(x
,u,#
)=
L(x
,u)+#
Tf(x
,u)
Sup
pose
the
optim
alco
ntro
lpro
blem
abov
eha
sth
eso
lutio
nu
":[0
,tf]%
Uan
dx
":[0
,tf]%
Rn
.Th
en,
min
u#U
H(x
" (t)
,u,#
(t))
=H
(x" (
t),u
" (t)
,#(t
)),
&t!
[0,t
f]
whe
re#(t
)so
lves
the
adjo
inte
quat
ion
#(t
)=
#$H
T
$x
(x" (
t),u
" (t)
,#(t
)),#(t
f)
=$"
T
$x
(x" (
t f))
Mor
eove
r,th
eop
timal
cont
roli
sgi
ven
by
u" (
t)=
arg
min
u#U
H(x
" (t)
,u,#
(t))
Lect
ure
128
EL2
620
2011
Rem
arks
•S
eete
xtbo
ok,e
.g.,
Gla
dan
dLj
ung,
forp
roof
.The
outli
neis
sim
ply
tono
teth
atev
ery
chan
geof
u(t
)fro
mth
eop
timal
u" (
t)m
ust
incr
ease
the
crite
rium
.Th
enpe
rform
acl
ever
Tayl
orex
pans
ion.
•Po
ntry
agin
’sM
axim
umP
rinci
ple
prov
ides
nece
ssar
yco
nditi
on:
ther
em
ayex
istm
any
orno
neso
lutio
ns(c
f.,m
inu:[0,1
]!R
x(1
),x
=u
,x(0
)=
0)
•Th
eM
axim
umP
rinci
ple
prov
ides
allp
ossi
ble
cand
idat
es.
•S
olut
ion
invo
lves
2nO
DE
’sw
ithbo
unda
ryco
nditi
onsx(0
)=
x0
and#(t
f)
=$"
T/$
x(x
" (t f
)).
Ofte
nha
rdto
solv
eex
plic
itly.
•“m
axim
um”i
sdu
eto
Pont
ryag
in’s
orig
inal
form
ulat
ion
Lect
ure
129
EL2
620
2011
Exa
mpl
e—B
oati
nS
trea
m(c
ont’d
)H
amilt
onia
nsa
tisfie
s
H=#
Tf
=# #
1#
2
$% v(x
2)+
u1
u2
&
$H $x
=# 0
#1
$ ,"(x
)=
#x
1
Adj
oint
equa
tions
#1(t
)=
0,#
1(t
f)
=#
1
#2(t
)=
##
1(t
),#
2(t
f)
=0
have
solu
tion
#1(t
)=
#1,
#2(t
)=
t#
t f
Lect
ure
1210
EL2
620
2011
Opt
imal
cont
rol
u" (
t)=
arg
min
u2 1+
u2 2=
1#
1(t
)(v(x
" 2(t
))+
u1)+#
2(t
)u2
=ar
gm
inu2 1+
u2 2=
1#
1(t
)u1+#
2(t
)u2
Hen
ce,
u1(t
)=
##
1(t
)"#
2 1(t
)+#
2 2(t
),u
2(t
)=
##
2(t
)"#
2 1(t
)+#
2 2(t
)
or
u1(t
)=
1"
1+
(t#
t f)2
,u
2(t
)=
t f#
t"
1+
(t#
t f)2
Lect
ure
1211
EL2
620
2011
Exa
mpl
e—R
esou
rce
Allo
catio
n(c
ont’d
)
min
u:[0,t
f]!
[0,1
]
!t f
0
[u(t
)#
1]x(t
)dt
x(t
)=!u(t
)x(t
),x(0
)=
x0
Ham
ilton
ian
satis
fies
H=
L+#
Tf
=(u
#1)
x+#!ux
Adj
oint
equa
tion
#(t
)=
1#
u" (
t)##(t
)!u
" (t)
,#(t
f)
=0
Lect
ure
1212
EL2
620
2011
Opt
imal
cont
rol
u" (
t)=
arg
min
u#[
0,1
](u#
1)x
" (t)
+#(t
)!ux
" (t)
=ar
gm
inu#[
0,1
]u(1
+#(t
)!),
(x" (
t)>
0)
=
' 0,#(t
)'
#1/!
1,#(t
)<
#1/!
Fort
(t f
,we
have
u" (
t)=
0(w
hy?)
and
thus#(t
)=
1.
Fort
<t f
#1/!
,we
have
u" (
t)=
1an
dth
us#(t
)=
#!#(t
).
Lect
ure
1213
EL2
620
2011
00.2
0.4
0.6
0.8
11.2
1.4
1.6
1.8
2
-1
-0.8
-0.6
-0.4
-0.20 0
0.2
0.4
0.6
0.8
11.2
1.4
1.6
1.8
2
0
0.2
0.4
0.6
0.81
#(t
)
u" (
t)
•u
" (t)
=
' 1,t!
[0,t
f#
1/!]
0,t!
(tf
#1/!,t
f]
•It’
sop
timal
tore
inve
stin
the
begi
nnin
g
Lect
ure
1214
EL2
620
2011
5m
inut
eex
erci
se:
Find
the
curv
ew
ithm
inim
alle
ngth
byso
lvin
g
min
u:[0,t
f]!
R
!t f
0
"1
+u
2(t
)dt
x(t
)=
u(t
),x(0
)=
a
Lect
ure
1215
EL2
620
2011
5m
inut
eex
erci
seII:
Sol
veth
eop
timal
cont
rolp
robl
em
min
!1
0
u4dt+
x(1
)
x=
#x
+u
x(0
)=
0
Lect
ure
1216
EL2
620
2011
His
tory
—C
alcu
lus
ofVa
riat
ions
•B
rach
isto
chro
ne(s
hort
estt
ime)
prob
lem
(169
6):
Find
the
(fric
tionl
ess)
curv
eth
atta
kes
apa
rtic
lefro
mA
toB
insh
orte
sttim
e
dt=
ds v
=
"dx
2+
dy
2
v=
"1
+y
$ (x)2
"2g
y(x
)dx
Min
imiz
e
J(y
)=
!B
A
"1
+y
$ (x)2
"2g
y(x
)dx
Sol
ved
byJo
hnan
dJa
mes
Ber
noul
li,N
ewto
n,l’H
ospi
tal
•Fi
ndth
ecu
rve
encl
osin
gla
rges
tare
a(E
uler
)
Lect
ure
1217
EL2
620
2011
His
tory
—O
ptim
alC
ontr
ol•
The
spac
era
ce(S
putn
ik,1
957)
•Po
ntry
agin
’sM
axim
umP
rinci
ple
(195
6)
•B
ellm
an’s
Dyn
amic
Pro
gram
min
g(1
957)
•H
uge
influ
ence
onen
gine
erin
gan
dot
hers
cien
ces:
–R
obot
ics—
traje
ctor
yge
nera
tion
–A
eron
autic
s—sa
telli
teor
bits
–P
hysi
cs—
Sne
ll’sla
w,c
onse
rvat
ion
law
s
–Fi
nanc
e—po
rtfo
lioth
eory
Lect
ure
1218
EL2
620
2011
God
dard
’sR
ocke
tPro
blem
(191
0)H
owto
send
aro
cket
ashi
ghup
inth
eai
ras
poss
ible
?
d dt
( ) *v h m
+ , -=
( ) ) *
u#
D
m#
g
v #!u
+ , , -
h
m
(v(0
),h(0
),m
(0))
=(0
,0,m
0),
g,!
>0
um
otor
forc
e,D
=D
(v,h
)ai
rres
ista
nce
Con
stra
ints
:0
)u
)u
max
and
m(t
f)
=m
1(e
mpt
y)
Opt
imiz
atio
ncr
iterio
n:m
axuh(t
f)
Lect
ure
1219
EL2
620
2011
Gen
eral
ized
form
:
min
u:[0,t
f]!
U
!t f
0
L(x
(t),
u(t
))dt+"(x
(tf))
x(t
)=
f(x
(t),
u(t
)),
x(0
)=
x0
%(x
(tf))
=0
Not
eth
edi
ffenc
esco
mpa
red
tost
anda
rdfo
rm:
•E
ndtim
et f
isfre
e
•Fi
nals
tate
isco
nstra
ined
:%
(x(t
f))
=x
3(t
f)#
m1
=0
Lect
ure
1220
EL2
620
2011
Sol
utio
nto
God
dard
’sP
robl
emG
odda
rd’s
prob
lem
ison
gene
raliz
edfo
rmw
ith
x=
(v,h
,m)T
,L
*0,
"(x
)=
#x
2,%
(x)
=x
3#
m1
D(v
,h)*
0:
•E
asy:
letu
(t)
=u
max
until
m(t
)=
m1
•B
urn
fuel
asfa
stas
poss
ible
,bec
ause
itco
sts
ener
gyto
lifti
t
D(v
,h)+*
0:
•H
ard:
e.g.
,itc
anbe
optim
alto
have
low
spee
dw
hen
air
resi
stan
ceis
high
,in
orde
rto
burn
fuel
athi
gher
leve
l
•To
ok50
year
sbe
fore
aco
mpl
ete
solu
tion
was
pres
ente
d
Lect
ure
1221
EL2
620
2011
Gen
eral
Pont
ryag
in’s
Max
imum
Pri
ncip
leTh
eore
m:
Sup
pose
u"
:[0
,tf]%
Uan
dx
":[0
,tf]%
Rn
are
solu
tions
to
min
u:[0,t
f]!
U
!t f
0
L(x
(t),
u(t
))dt+"(t
f,x
(tf))
x(t
)=
f(x
(t),
u(t
)),
x(0
)=
x0
%(t
f,x
(tf))
=0
Then
,the
reex
ists
n0
'0,
µ!
Rn
such
that
(n0,µ
T)+=
0an
d
min
u#U
H(x
" (t)
,u,#
(t),
n0)
=H
(x" (
t),u
" (t)
,#(t
),n
0),
t!
[0,t
f]
whe
reH
(x,u
,#,n
0)
=n
0L
(x,u
)+#
Tf(x
,u)
Lect
ure
1222
EL2
620
2011
#(t
)=
#$H
T
$x
(x" (
t),u
" (t)
,#(t
),n
0)
#T(t
f)
=n
0$"
$x
(tf,x
" (t f
))+
µT$%
$x
(tf,x
" (t f
))
H(x
" (t f
),u
" (t f
),#(t
f),
n0)
=#
n0$" $t(t
f,x
" (t f
))#
µT$% $t(t
f,x
" (t f
))
Rem
arks
:
•t f
may
bea
free
varia
ble
•W
ithfix
edt f
:H
(x" (
t f),
u" (
t f),#(t
f),
n0)
=0
•%
defin
esen
dpo
intc
onst
rain
ts
Lect
ure
1223
EL2
620
2011
Exa
mpl
e—M
inim
umTi
me
Con
trol
Bri
ngth
est
ates
ofth
edo
uble
inte
grat
orto
the
orig
inas
fast
aspo
ssib
le
min
u:[0,t
f]!
[%1,1
]
!t f
0
1dt=
min
u:[0,t
f]!
[%1,1
]t f
x1(t
)=
x2(t
),x
2(t
)=
u(t
)
%(x
(tf))
=(x
1(t
f),
x2(t
f))
T=
(0,0
)T
Opt
imal
cont
roli
sth
eba
ng-b
ang
cont
rol
u" (
t)=
arg
min
u#[
%1,1
]1
+#
1(t
)x" 2(t
)+#
2(t
)u
=
' 1,#
2(t
)<
0
#1,
#2(t
)'
0
Lect
ure
1224
EL2
620
2011
Adj
oint
equa
tions#
1(t
)=
0,#
2(t
)=
##
1(t
)gi
ves
#1(t
)=
c 1,#
2(t
)=
c 2#
c 1t
With
u(t
)=&
=±
1,w
eha
ve
x1(t
)=
x1(0
)+
x2(0
)t+&t2
/2
x2(t
)=
x2(0
)+&t
Elim
inat
ing
tgi
ves
curv
es
x1(t
)±
x2(t
)2/2
=co
nst
Thes
ede
fine
the
switc
hcu
rve,
whe
reth
eop
timal
cont
rols
witc
h
Lect
ure
1225
EL2
620
2011
Ref
eren
ceG
ener
atio
nus
ing
Opt
imal
Con
trol
•O
ptim
alco
ntro
lpro
blem
mak
esno
dist
inct
ion
betw
een
open
-loop
cont
rolu
" (t)
and
clos
ed-lo
opco
ntro
lu" (
t,x).
•W
em
ayus
eth
eop
timal
open
-loop
solu
tion
u" (
t)as
the
refe
renc
eva
lue
toa
linea
rreg
ulat
or,w
hich
keep
sth
esy
stem
clos
eto
the
wan
ted
traje
ctor
y
•E
ffici
entd
esig
nm
etho
dfo
rnon
linea
rpro
blem
s
Lect
ure
1226
EL2
620
2011
Line
arQ
uadr
atic
Con
trol
min
u:[0,&
)!R
m
!&
0
(xTQ
x+
uTR
u)dt
with
x=
Ax
+B
u
has
optim
also
lutio
nu
=#
Lx
whe
reL
=R
%1B
TS
and
S>
0is
the
solu
tion
to
SA
+A
TS
+Q
#SB
R%
1B
TS
=0
Lect
ure
1227
EL2
620
2011
Pro
pert
ies
ofLQ
Con
trol
•S
tabi
lizin
g
•C
lose
d-lo
opsy
stem
stab
lew
ithu
=#'(t
)Lx
for
'(t
)!
[1/2
,")
(infin
itega
inm
argi
n)
•P
hase
mar
gin
60de
gree
s
Ifx
isno
tmea
sura
ble,
then
one
may
use
aK
alm
anfil
ter;
lead
sto
linea
rqua
drat
icG
auss
ian
(LQ
G)c
ontro
l.
•B
ut,t
hen
syst
emm
ayha
vear
bitra
rily
poor
robu
stne
ss!
(Doy
le,
1978
)
Lect
ure
1228
EL2
620
2011
Tetr
aP
akM
ilkR
ace
Mov
em
ilkin
min
imum
time
with
outs
pilli
ng
[Gru
ndel
ius
&B
ernh
ards
son,
1999
]
Lect
ure
1229
EL2
620
2011
Giv
endy
nam
ics
ofsy
stem
and
max
imum
slos
h"
=0.
63,s
olve
min
u:[0,t
f]!
[%10,1
0]
. tf
01dt,
whe
reu
isth
eac
cele
ratio
n.
00.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
-15
-10-5051015
Acceleration
00.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
-1
-0.50
0.51
Slosh
Opt
imal
time
=37
5m
s,Te
traP
ak=
540m
s
Lect
ure
1230
EL2
620
2011
Pro
s&
Con
sfo
rO
ptim
alC
ontr
ol+
Sys
tem
atic
desi
gnpr
oced
ure
+A
pplic
able
tono
nlin
earc
ontro
lpro
blem
s
+C
aptu
res
limita
tions
(as
optim
izat
ion
cons
train
ts)
#H
ard
tofin
dsu
itabl
ecr
iteria
#H
ard
toso
lve
the
equa
tions
that
give
optim
alco
ntro
ller
Lect
ure
1231
EL2
620
2011
SF2
852
Opt
imal
Con
trol
Theo
ry•
Per
iod
3,7.
5cr
edits
•O
ptim
izat
ion
and
Sys
tem
sTh
eory
http://www.math.kth.se/optsyst/
Dyn
amic
Pro
gram
min
g:D
iscr
ete
&co
ntin
uous
;Prin
cipl
eof
optim
ality
;Ham
ilton
-Jac
obi-B
ellm
aneq
uatio
n
Pont
ryag
in’s
Max
imum
prin
cipl
e:M
ain
resu
lts;S
peci
alca
ses
such
astim
eop
timal
cont
rola
ndLQ
cont
rol
Num
eric
alM
etho
ds:
Num
eric
also
lutio
nof
optim
alco
ntro
lpro
blem
s
App
licat
ions
:A
eron
autic
s,R
obot
ics,
Pro
cess
Con
trol,
Bio
engi
neer
ing,
Eco
nom
ics,
Logi
stic
s
Lect
ure
1232
EL2
620
2011
Toda
y’s
Goa
lYo
ush
ould
beab
leto
•D
esig
nco
ntro
llers
base
don
optim
alco
ntro
lthe
ory
for
–S
tand
ard
form
–G
ener
aliz
edfo
rm
•U
nder
stan
dpo
ssib
ilitie
san
dlim
itatio
nsof
optim
alco
ntro
l
Lect
ure
1233
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
13
•Fuz
zylo
gic
and
fuzz
yco
ntro
l
•Art
ifici
alne
ural
netw
orks
Som
esl
ides
copi
edfro
mK
.-E.A
rzen
and
M.J
ohan
sson
Lect
ure
131
EL2
620
2011
Toda
y’s
Goa
lYo
ush
ould
•un
ders
tand
the
basi
csof
fuzz
ylo
gic
and
fuzz
yco
ntro
llers
•un
ders
tand
sim
ple
neur
alne
twor
ks
Lect
ure
132
EL2
620
2011
Fuzz
yC
ontr
ol•
Man
ypl
ants
are
man
ually
cont
rolle
dby
expe
rienc
edop
erat
ors
•Tr
ansf
erpr
oces
skn
owle
dge
toco
ntro
lalg
orith
mis
diffi
cult
Idea
:
•M
odel
oper
ator
’sco
ntro
lact
ions
(inst
ead
ofth
epl
ant)
•Im
plem
enta
sru
les
(inst
ead
ofas
diffe
rent
iale
quat
ions
)
Exa
mpl
eof
aru
le:
IFSpeedisHighANDTraffic
isHeavy
THENReduce
GasABit
Lect
ure
133
EL2
620
2011
Mod
elC
ontr
olle
rIn
stea
dof
Pla
ntC
onve
ntio
nalc
ontr
olde
sign
Mod
elpl
antP
!A
naly
zefe
edba
ck!
Syn
thes
ize
cont
rolle
rC!
Impl
emen
tcon
trola
lgor
ithm
Fuzz
yco
ntro
ldes
ign
Mod
elm
anua
lcon
trol!
Impl
emen
tcon
trolr
ules
re
uy
"C
P
Lect
ure
134
EL2
620
2011
Fuzz
yS
etTh
eory
Spe
cify
how
wel
lan
obje
ctsa
tisfie
sa
(vag
ue)d
escr
iptio
n
Con
vent
iona
lset
theo
ry:x
#A
orx
$#A
Fuzz
yse
tthe
ory:
x#
Ato
ace
rtai
nde
gree
µA(x
)
Mem
bers
hip
func
tion:
µA
:!
![0
,1]e
xpre
sses
the
degr
eex
belo
ngs
toA
025
10
01
x
µC
µW
Col
dW
arm
Afu
zzy
seti
sde
fined
as(A
,µA)
Lect
ure
135
EL2
620
2011
Exa
mpl
eQ
1:Is
the
tem
pera
ture
x=
15co
ld?
A1:
Itis
quite
cold
sinc
eµ
C(1
5)=
2/3.
Q2:
Isx
=15
war
m?
A2:
Itis
notr
eally
war
msi
nce
µW
(15)
=1/
3.
15
1 0
10
25
0x
µC
µW
Col
dW
arm
Lect
ure
136
EL2
620
2011
Fuzz
yLo
gic
How
toca
lcul
ate
with
fuzz
yse
ts(A
,µA)?
Con
vent
iona
llog
ic:
AND
:A
%B
OR
:A
&B
NOT
:A
!
Fuzz
ylo
gic:
AND
:µ
A"B
(x)
=m
in(µ
A(x
),µ
B(x
))OR
:µ
A#B
(x)
=m
ax(µ
A(x
),µ
B(x
))NOT
:µ
A! (x)
=1
"µ
A(x
)
Defi
nes
logi
cca
lcul
atio
nsas
XAND
YOR
Z
Mim
ichu
man
lingu
istic
(app
roxi
mat
e)re
ason
ing
[Zad
eh,1
965]
Lect
ure
137
EL2
620
2011
Exa
mpl
eQ
1:Is
itco
ldAND
war
m?
Q2:
Isit
coldOR
war
m?
1 0
10
25
0x
µC
"W
Col
dW
arm
025
10
01
x
µC
#W
Col
dW
arm
Lect
ure
138
EL2
620
2011
Fuzz
yC
ontr
olS
yste
m
Plant
Controller
Fuzzy
ry
u
r,y,u
:[0
,')(!
Rar
eco
nven
tiona
lsig
nals
Fuzz
yco
ntro
lleri
sa
nonl
inea
rmap
ping
from
y(a
ndr)
tou
Lect
ure
139
EL2
620
2011
Fuzz
yC
ontr
olle
r
yu
Fuzz
ifier
Def
uzzi
fier
Fuzz
yC
ontro
ller
Fuzz
yIn
fere
nce
Fuzz
ifier
:Fu
zzy
sete
valu
atio
nof
y(a
ndr)
Fuzz
yIn
fere
nce:
Fuzz
yse
tcal
cula
tions
Def
uzzi
fier:
Map
fuzz
yse
tto
u
Fuzz
ifier
and
defu
zzifi
erac
tas
inte
rface
sto
the
cris
psi
gnal
s
Lect
ure
1310
EL2
620
2011
Fuzz
ifier
Fuzz
yse
teva
luat
ion
ofin
puty
Exa
mpl
ey
=15
:µ
C(1
5)=
2/3
and
µW
(15)
=1/
3
15
1 0
10
25
0y
µC
µW
Col
dW
arm
Lect
ure
1311
EL2
620
2011
Fuzz
yIn
fere
nce
Fuzz
yIn
fere
nce:
1.C
alcu
late
degr
eeof
fulfi
llmen
tfor
each
rule
2.C
alcu
late
fuzz
you
tput
ofea
chru
le
3.A
ggre
gate
rule
outp
uts
Exa
mpl
esof
fuzz
yru
les:
Rul
e1:
IF
yis
Col
d!
"#$
1.
THEN
uis
Hig
h!
"#$
2.
Rul
e2:
IF
yis
War
m!
"#$
1.
THEN
uis
Low
!"#
$2.
Lect
ure
1312
EL2
620
2011
1.C
alcu
late
degr
eeof
fulfi
llmen
tfor
rule
s
Lect
ure
1313
EL2
620
2011
2.C
alcu
late
fuzz
you
tput
ofea
chru
le
Not
eth
at”m
u”is
stan
dard
fuzz
y-lo
gic
nom
encl
atur
efo
r”tr
uth
valu
e”:
Lect
ure
1314
EL2
620
2011
3.A
ggre
gate
rule
outp
uts
Lect
ure
1315
EL2
620
2011
Def
uzzi
fier
Lect
ure
1316
EL2
620
2011
Fuzz
yC
ontr
olle
r—S
umm
ary
yu
Fuzz
ifier
Def
uzzi
fier
Fuzz
yC
ontro
ller
Fuzz
yIn
fere
nce
Fuzz
ifier
:Fu
zzy
sete
valu
atio
nof
y(a
ndr)
Fuzz
yIn
fere
nce:
Fuzz
yse
tcal
cula
tions
1.C
alcu
late
degr
eeof
fulfi
llmen
tfor
each
rule
2.C
alcu
late
fuzz
you
tput
ofea
chru
le
3.A
ggre
gate
rule
outp
uts
Def
uzzi
fier:
Map
fuzz
yse
tto
u
Lect
ure
1317
EL2
620
2011
Exa
mpl
e—Fu
zzy
Con
trol
ofS
team
Eng
ine
http://isc.faqs.org/docs/air/ttfuzzy.html
Lect
ure
1318
EL2
620
2011
Lect
ure
1319
EL2
620
2011
Lect
ure
1320
EL2
620
2011
Lect
ure
1321
EL2
620
2011
Lect
ure
1322
EL2
620
2011
Rul
e-B
ased
Vie
wof
Fuzz
yC
ontr
ol
Lect
ure
1323
EL2
620
2011
Non
linea
rV
iew
ofFu
zzy
Con
trol
Lect
ure
1324
EL2
620
2011
Pro
san
dC
ons
ofFu
zzy
Con
trol
Adv
anta
ges
•U
ser-
frie
ndly
way
tode
sign
nonl
inea
rcon
trolle
rs
•E
xplic
itre
pres
enta
tion
ofop
erat
or(p
roce
ss)k
now
ledg
e
•In
tuiti
vefo
rnon
-exp
erts
inco
nven
tiona
lcon
trol
Dis
adva
ntag
es
•Li
mite
dan
alys
isan
dsy
nthe
sis
•S
omet
imes
hard
toco
mbi
new
ithcl
assi
calc
ontro
l
•N
otob
viou
sho
wto
incl
ude
dyna
mic
sin
cont
rolle
r
Fuzz
yco
ntro
lis
aw
ayto
obta
ina
clas
sof
nonl
inea
rco
ntro
llers
Lect
ure
1325
EL2
620
2011
Neu
ralN
etw
orks
•H
owdo
esth
ebr
ain
wor
k?
•A
netw
ork
ofco
mpu
ting
com
pone
nts
(neu
rons
)
Lect
ure
1326
EL2
620
2011
Neu
rons
Bra
inne
uron
Art
ifici
alne
uron
y
b
x1
x2
xn
w1
w2
wn
!!(·)
Lect
ure
1327
EL2
620
2011
Mod
elof
aN
euro
nIn
puts
:x
1,x
2,.
..,x
n
Wei
ghts
:w
1,w
2,.
..,w
n
Bia
s:b
y=
!% b
+&
n i=1w
ixi'
Non
linea
rity:
!(·)
Out
put:
y
y
b
x1
x2
xn
w1
w2
wn
!!(·)
Lect
ure
1328
EL2
620
2011
AS
impl
eN
eura
lNet
wor
kN
eura
lnet
wor
kco
nsis
ting
ofsi
xne
uron
s:
y
u1
u2
u3
Inpu
tLay
erH
idde
nLa
yer
Out
putL
ayer
Rep
rese
nts
ano
nlin
ear
map
ping
from
inpu
tsto
outp
uts
Lect
ure
1329
EL2
620
2011
Neu
ralN
etw
ork
Des
ign
1.H
owm
any
hidd
enla
yers
?
2.H
owm
any
neur
ons
inea
chla
yer?
3.H
owto
choo
seth
ew
eigh
ts?
The
choi
ceof
wei
ghts
are
ofte
ndo
nead
aptiv
ely
thro
ugh
lear
ning
Lect
ure
1330
EL2
620
2011
Suc
cess
Sto
ries
Fuzz
yco
ntro
ls:
•Za
deh
(196
5)
•C
ompl
expr
oble
ms
butw
ithpo
ssib
lelin
guis
ticco
ntro
ls
•A
pplic
atio
nsto
okof
fin
mid
70’s
–C
emen
tkiln
s,w
ashi
ngm
achi
nes,
vacu
umcl
eane
rs
Art
ifici
alne
ural
netw
orks
:
•M
cCul
loch
&P
itts
(194
3),M
insk
y(1
951)
•C
ompl
expr
oble
ms
with
unkn
own
and
high
lyno
nlin
ears
truc
ture
•A
pplic
atio
nsto
okof
fin
mid
80’s
–P
atte
rnre
cogn
ition
(e.g
.,sp
eech
,vis
ion)
,dat
acl
assi
ficat
ion
Lect
ure
1331
EL2
620
2011
Toda
y’s
Goa
lYo
ush
ould
•un
ders
tand
the
basi
csof
fuzz
ylo
gic
and
fuzz
yco
ntro
llers
•un
ders
tand
sim
ple
neur
alne
twor
ks
Lect
ure
1332
EL2
620
2011
Nex
tLec
ture
•E
L262
0N
onlin
earC
ontro
lrev
isite
d
•S
prin
gco
urse
sin
cont
rol
•M
aste
rthe
sis
proj
ects
•P
hDth
esis
proj
ects
Lect
ure
1333
EL2
620
2011
EL2
620
Non
linea
rC
ontr
ol
Lect
ure
14
•S
umm
ary
and
repe
titio
n
•S
prin
gco
urse
sin
cont
rol
•M
aste
rthe
sis
proj
ects
Lect
ure
141
EL2
620
2011
Exa
m•
Reg
ular
writ
ten
exam
(inE
nglis
h)w
ithfiv
epr
oble
ms
•S
ign
upon
cour
seho
mep
age
•Yo
um
aybr
ing
lect
ure
note
s,G
lad
&Lj
ung
“Reg
lert
ekni
k”,a
ndTE
FYM
Aor
BE
TA(N
oot
herm
ater
ial:
text
book
s,ex
erci
ses,
calc
ulat
ors
etc.
Any
othe
rbas
icco
ntro
lboo
km
ustb
eap
prov
edby
me
befo
reth
eex
am.).
•S
eeho
mep
age
foro
ldex
ams
Lect
ure
142
EL2
620
2011
Que
stio
n1
Wha
t’son
the
exam
?
•N
onlin
earm
odel
s:eq
uilib
ria,p
hase
port
aits
,lin
eariz
atio
nan
dst
abili
ty
•Ly
apun
ovst
abili
ty(lo
cala
ndgl
obal
),La
Sal
le
•C
ircle
Crit
erio
n,S
mal
lGai
nTh
eore
m,P
assi
vity
Theo
rem
•C
ompe
nsat
ing
stat
icno
nlin
earit
ies
•D
escr
ibin
gfu
nctio
ns
•S
lidin
gm
odes
,equ
ival
entc
ontro
ls
•Ly
apun
ovba
sed
desi
gn:
back
-ste
ppin
g
•E
xact
feed
back
linea
rizat
ion,
inpu
t-out
putl
inea
rizat
ion,
zero
dyna
mic
s
•N
onlin
earc
ontro
llabi
lity
•O
ptim
alco
ntro
l
Lect
ure
143
EL2
620
2011
Que
stio
n2
Wha
tdes
ign
met
hod
shou
ldIu
sein
prac
tice?
The
answ
eris
high
lypr
oble
mde
pend
ent.
Poss
ible
(lear
ning
)ap
proa
ch:
•S
tart
with
the
sim
ples
t:
–lin
earm
etho
ds(lo
opsh
apin
g,st
ate
feed
back
,...
)
•E
valu
ate:
–st
rong
nonl
inea
ritie
s(u
nder
feed
back
!)?
–va
ryin
gop
erat
ing
cond
ition
s?
–an
alyz
ean
dsi
mul
ate
with
nonl
inea
rmod
el
•S
ome
nonl
inea
ritie
sto
com
pens
ate
for?
–sa
tura
tions
,val
ves
etc
•Is
the
syst
emge
neric
ally
nonl
inea
r?E
.g,x
=xu
Lect
ure
144
EL2
620
2011
Que
stio
n3
Can
asy
stem
bepr
oven
stab
lew
ithth
eS
mal
lGai
nTh
eore
man
dun
stab
lew
ithth
eC
ircl
eC
rite
rion
?
•N
o,th
eS
mal
lGai
nTh
eore
m,P
assi
vity
Theo
rem
and
Circ
leC
riter
ion
allp
rovi
deon
lysu
ffici
entc
ondi
tions
for
stab
ility
•B
ut,i
fone
met
hod
does
notp
rove
stab
ility
,ano
ther
one
may
.
•S
ince
they
dono
tpro
vide
nece
ssar
yco
nditi
ons
fors
tabi
lity,
none
ofth
emca
nbe
used
topr
ove
inst
abili
ty.
Lect
ure
145
EL2
620
2011
Que
stio
n4
Can
you
revi
ewth
eci
rcle
crite
rion
?W
hata
bout
k1
<0
<k
2?
Lect
ure
146
EL2
620
2011
The
Cir
cle
Cri
teri
on
yk1y
k2y
f(y
)
!1 k1
!1 k2
G(i
!)
Theo
rem
Con
side
rafe
edba
cklo
opw
ithy
=G
uan
du
=!
f(y
).A
ssum
eG
(s)
isst
able
and
that
k1
"f(y
)/y
"k
2.
Ifth
eN
yqui
stcu
rve
ofG
(s)
stay
son
the
corr
ects
ide
ofth
eci
rcle
defin
edby
the
poin
ts!
1/k
1an
d!
1/k
2,t
hen
the
clos
ed-lo
opsy
stem
isB
IBO
stab
le.
Lect
ure
147
EL2
620
2011
The
diff
eren
tcas
esS
tabl
esy
stem
G
1.0
<k
1<
k2:
Sta
you
tsid
eci
rcle
2.0
=k
1<
k2:
Sta
yto
the
right
ofth
elin
eR
es
=!
1/k
2
3.k
1<
0<
k2:
Sta
yin
side
the
circ
le
Oth
erca
ses:
Mul
tiply
fan
dG
with
!1.
Onl
yC
ase
1an
d2
stud
ied
inle
ctur
es.
Onl
yG
stab
lest
udie
d.
Lect
ure
148
EL2
620
2011
Que
stio
n5
Ple
ase
repe
atan
tiwin
dup
Lect
ure
149
EL2
620
2011
Trac
king
PID
∑
∑1 s 1 T
t
e s
−y
vu
e
−+
∑∑
−y
eu
v
−+
e s
1 Tt1 s
Act
uato
r m
od
el
Act
uato
r
Act
uato
r
∑
K
KT
dsK
K Ti
KT
ds
K Ti
∑
(a)
(b)
Lect
ure
1410
EL2
620
2011
Ant
iwin
dup—
Gen
eral
Sta
te-S
pace
Mod
el
yu
∑∑
∑∑
sat
u G
−K
Dy
vs
−1 K
F−
KC
CD
G
F s−1
CD
xc
xc
Cho
ose
Ksu
chth
atF
!K
Cha
sst
able
eige
nval
ues.
Lect
ure
1411
EL2
620
2011
Que
stio
n6
Ple
ase
repe
atLy
apun
ovth
eory
Lect
ure
1412
EL2
620
2011
Sta
bilit
yD
efini
tions
An
equi
libriu
mpo
intx
=0
ofx
=f(x
)is
loca
llyst
able
,iff
orev
eryR
>0
ther
eex
ists
r>
0,su
chth
at
#x(0
)#<
r$
#x(t
)#<
R,
t%
0
loca
llyas
ympt
otic
ally
stab
le,i
floc
ally
stab
lean
d
#x(0
)#<
r$
lim
t!"
x(t
)=
0
glob
ally
asym
ptot
ical
lyst
able
,ifa
sym
ptot
ical
lyst
able
fora
llx(0
)&
Rn
.
Lect
ure
1413
EL2
620
2011
Lyap
unov
Theo
rem
for
Loca
lSta
bilit
yTh
eore
mLe
tx=
f(x
),f(0
)=
0,an
d0
&!
'R
n.
Ass
ume
that
V:!
(R
isa
C1
func
tion.
If
•V
(0)
=0
•V
(x)>
0,fo
rallx
&!
,x)=
0
•V
(x)"
0al
ong
allt
raje
ctor
ies
in!
then
x=
0is
loca
llyst
able
.Fu
rthe
rmor
e,if
•V
(x)<
0fo
rallx
&!
,x)=
0
then
x=
0is
loca
llyas
ympt
otic
ally
stab
le.
Lect
ure
1414
EL2
620
2011
Lyap
unov
Theo
rem
for
Glo
balS
tabi
lity
Theo
rem
Letx
=f(x
)an
df(0
)=
0.A
ssum
eth
atV
:R
n(
Ris
aC
1fu
nctio
n.If
•V
(0)
=0
•V
(x)>
0,fo
rallx
)=0
•V
(x)<
0fo
rallx
)=0
•V
(x)(
*as
#x#
(*
then
x=
0is
glob
ally
asym
ptot
ical
lyst
able
.
Lect
ure
1415
EL2
620
2011
LaS
alle
’sTh
eore
mfo
rG
loba
lAsy
mpt
otic
Sta
bilit
yTh
eore
m:
Letx
=f(x
)an
df(0
)=
0.If
ther
eex
ists
aC
1fu
nctio
nV
:R
n(
Rsu
chth
at
(1)
V(0
)=
0
(2)
V(x
)>
0fo
rallx
)=0
(3)
V(x
)"
0fo
rallx
(4)
V(x
)(
*as
#x#
(*
(5)
The
only
solu
tion
ofx
=f(x
)su
chth
atV
(x)
=0
isx(t
)=
0fo
rallt
then
x=
0is
glob
ally
asym
ptot
ical
lyst
able
.
Lect
ure
1416
EL2
620
2011
LaS
alle
’sIn
vari
antS
etTh
eore
mTh
eore
mLe
t!&
Rn
bea
boun
ded
and
clos
edse
ttha
tis
inva
riant
with
resp
ectt
ox
=f(x
).
LetV
:R
n(
Rbe
aC
1fu
nctio
nsu
chth
atV
(x)"
0fo
rx&
!.
LetE
beth
ese
tofp
oint
sin
!w
here
V(x
)=
0.If
Mis
the
larg
est
inva
riant
seti
nE
,the
nev
ery
solu
tion
with
x(0
)&
!ap
proa
ches
Mas
t(
*
Rem
ark:
aco
mpa
ctse
t(bo
unde
dan
dcl
osed
)is
obta
ined
ifw
ee.
g.,
cons
ider
!=
{x&
Rn|V
(x)"
c}an
dV
isa
posi
tive
defin
itefu
nctio
n
Lect
ure
1417
EL2
620
2011
Rel
atio
nto
Poin
care
-Ben
dixs
onTh
eore
mPo
inca
re-B
endi
xson
Any
orbi
tofa
cont
inuo
us2n
dor
ders
yste
mth
atst
ays
ina
com
pact
regi
onof
the
phas
epl
ane
appr
oach
esits
!-li
mit
set,
whi
chis
eith
era
fixed
poin
t,a
perio
dic
orbi
t,or
seve
ralfi
xed
poin
tsco
nnec
ted
thro
ugh
hom
oclin
icor
hete
rocl
inic
orbi
ts
Inpa
rtic
ular
,ift
heco
mpa
ctre
gion
does
notc
onta
inan
yfix
edpo
int
then
the
!-li
mit
seti
sa
limit
cycl
e
Lect
ure
1418
EL2
620
2011
Exa
mpl
e:P
endu
lum
with
fric
tion
x1
=x
2,
x2
=!
g lsi
nx
1!
k mx
2
V(x
)=
g l(1
!co
sx
1)+
1 2x
2 2$
V=
!k m
x2 2
•W
eca
nno
tpro
vegl
obal
asym
ptot
icst
abili
ty;w
hy?
•Th
ese
tE=
{(x
1,x
2)|V
=0}
isE
={(
x1,x
2)|x
2=
0}•
The
inva
riant
poin
tsin
Ear
egi
ven
byx
1=
x2
=0
and
x2
=0.
Thus
,the
larg
esti
nvar
iant
seti
nE
is
M=
{(x
1,x
2)|x
1=
k",x
2=
0}
•Th
edo
mai
nis
com
pact
ifw
eco
nsid
er!
={(
x1,x
2)&
R2|V
(x)"
c}
Lect
ure
1419
EL2
620
2011
•If
we
e.g.
,con
side
r!:x
2 1+
x2 2
"1
then
M=
{(x
1,x
2)|x
1=
0,x
2=
0}an
dw
eha
vepr
oven
asym
ptot
icst
abili
tyof
the
orig
in.
Lect
ure
1420
EL2
620
2011
Que
stio
n7
Ple
ase
repe
atth
em
osti
mpo
rtan
tfac
tsab
outs
lidin
gm
odes
.
Ther
ear
e3
esse
ntia
lpar
tsyo
une
edto
unde
rsta
nd:
1.Th
esl
idin
gm
anifo
ld
2.Th
esl
idin
gco
ntro
l
3.Th
eeq
uiva
lent
cont
rol
Lect
ure
1421
EL2
620
2011
Ste
p1.
The
Slid
ing
Man
ifold
SA
im:
we
wan
tto
stab
ilize
the
equi
libriu
mof
the
dyna
mic
syst
em
x=
f(x
)+
g(x
)u,
x&
Rn,
u&
R1
Idea
:us
eu
tofo
rce
the
syst
emon
toa
slid
ing
man
ifold
Sof
dim
ensi
onn
!1
infin
itetim
e
S=
{x&
Rn|#
(x)
=0}
#&
R1
and
mak
eS
inva
riant
Ifx
&R
2th
enS
isR
1,i
.e.,
acu
rve
inth
est
ate-
plan
e(p
hase
plan
e).
Lect
ure
1422
EL2
620
2011
Exa
mpl
e
x1
=x
2(t
)
x2
=x
1(t
)x2(t
)+
u(t
)
Cho
ose
Sfo
rdes
ired
beha
vior
,e.g
.,
#(x
)=
ax
1+
x2
=0
$x
1=
!ax
1(t
)
Cho
ose
larg
ea
:fa
stco
nver
genc
eal
ong
slid
ing
man
ifold
Lect
ure
1423
EL2
620
2011
Ste
p2.
The
Slid
ing
Con
trol
ler
Use
Lyap
unov
idea
sto
desi
gnu(x
)su
chth
atS
isan
attra
ctin
gin
varia
ntse
t
Lyap
unov
func
tion
V(x
)=
0.5#
2yi
elds
V=
##
For2
ndor
ders
yste
mx
1=
x2,x
2=
f(x
)+
g(x
)uan
d#
=x
1+
x2
we
get
V=
#(x
2+
f(x
)+
g(x
)u)<
0+
u=
!f(x
)+
x2+
sgn(#
)
g(x
)
Exa
mpl
e:f(x
)=
x1x
2,
g(x
)=
1,#
=x
1+
x2,y
ield
s
u=
!x
1x
2!
x2!
sgn(x
1+
x2)
Lect
ure
1424
EL2
620
2011
Ste
p3.
The
Equ
ival
entC
ontr
olW
hen
traje
ctor
yre
ache
ssl
idin
gm
ode,
i.e.,
x&
S,t
hen
uw
illch
atte
r(h
igh
frequ
ency
switc
hing
).
How
ever
,an
equi
vale
ntco
ntro
lueq
(t)
that
keep
sx(t
)on
Sca
nbe
com
pute
dfro
m#
=0
whe
n#
=0
Exa
mpl
e:
#=
x1+
x2
=x
2+
x1x
2+
ueq
=0
$u
eq=
!x
2!
x1x
2
Thus
,the
slid
ing
cont
rolle
rwill
take
the
syst
emto
the
slid
ing
man
ifold
Sin
finite
time,
and
the
equi
vale
ntco
ntro
lwill
keep
iton
S.
Lect
ure
1425
EL2
620
2011
Not
e!P
revi
ous
year
sit
has
ofte
nbe
enas
sum
edth
atth
esl
idin
gm
ode
cont
rola
lway
sis
onth
efo
rm
u=
!sg
n(#
)
This
isO
K,b
utis
notc
ompl
etel
yge
nera
l(se
eex
ampl
e)
Lect
ure
1426
EL2
620
2011
Que
stio
n8
Can
you
repe
atba
ckst
eppi
ng?
Lect
ure
1427
EL2
620
2011
Bac
kste
ppin
gD
esig
nW
ear
eco
ncer
ned
with
findi
nga
stab
ilizi
ngco
ntro
lu(x
)fo
rthe
syst
emx
=f(x
,u)
Gen
eral
Lyap
unov
cont
rold
esig
n:de
term
ine
aC
ontro
lLya
puno
vfu
nctio
nV
(x,u
)an
dde
term
ine
u(x
)so
that
V(x
)>
0,
V(x
)<
0,x
&R
n
Inth
isco
urse
we
only
cons
ider
f(x
,u)
with
asp
ecia
lstr
uctu
re,
nam
ely
stri
ctfe
edba
ckst
ruct
ure
Lect
ure
1428
EL2
620
2011
Str
ictF
eedb
ack
Sys
tem
s
x1
=f 1
(x1)+
g 1(x
1)x
2
x2
=f 2
(x1,x
2)+
g 2(x
1,x
2)x
3
x3
=f 3
(x1,x
2,x
3)+
g 3(x
1,x
2,x
3)x
4
. . .
xn
=f n
(x1,.
..,x
n)+
g n(x
1,.
..,x
n)u
whe
reg k
)=0
Not
e:x
1,.
..,x
kdo
notd
epen
don
xk+
2,.
..,x
n.
Lect
ure
1429
EL2
620
2011
The
Bac
kste
ppin
gId
eaG
iven
aC
ontro
lLya
puno
vFu
nctio
nV
1(x
1),
with
corr
espo
ndin
gco
ntro
lu=
$1(x
1),
fort
hesy
stem
x1
=f 1
(x1)+
g 1(x
1)u
find
aC
ontro
lLya
puno
vfu
nctio
nV
2(x
1,x
2),
with
corr
espo
ndin
gco
ntro
lu=
$2(x
1,x
2),
fort
hesy
stem
x1
=f 1
(x1)+
g 1(x
1)x
2
x2
=f 2
(x1,x
2)+
u
Lect
ure
1430
EL2
620
2011
The
Bac
kste
ppin
gR
esul
tLe
tV1(x
1)
bea
Con
trolL
yapu
nov
Func
tion
fort
hesy
stem
x1
=f 1
(x1)+
g 1(x
1)u
with
corr
espo
ndin
gco
ntro
lleru
=$(x
1).
Then
V2(x
1,x
2)
=V
1(x
1)+
(x2!
$(x
1))
2/2
isa
Con
trol
Lyap
unov
Func
tion
fort
hesy
stem
x1
=f 1
(x1)+
g 1(x
1)x
2
x2
=f 2
(x1,x
2)+
u
with
corr
espo
ndin
gco
ntro
ller
u(x
)=
d$
dx
1
! f(x
1)+
g(x
1)x
2
" !dV
dx
1
g(x
1)!
(xk!
$(x
1))
!f 2
(x1,x
2)
Lect
ure
1431
EL2
620
2011
Que
stio
n9
Rep
eatb
ackl
ash
com
pens
atio
n
Lect
ure
1432
EL2
620
2011
Bac
klas
hC
ompe
nsat
ion
•D
eadz
one
•Li
near
cont
rolle
rdes
ign
•B
ackl
ash
inve
rse
Line
arco
ntro
ller
desi
gn:
Pha
sele
adco
mpe
nsat
ion
! ref
!
eu
! out
! in
! in
1
1+
sT
1 sK
1+
sT
2
1+
sT
1
Lect
ure
1433
EL2
620
2011
•C
hoos
eco
mpe
nsat
ion
F(s
)su
chth
atth
ein
ters
ectio
nw
ithth
ede
scrib
ing
func
tion
isre
mov
ed
F(s
)=
K1+
sT2
1+
sT1
with
T1
=0.
5,T
2=
2.0:
Rea
l Axi
s
Imaginary Axis
Nyq
uist
Dia
gram
s
-10
-50
510
-10-8-6-4-20246810
with
filte
r
with
out f
ilter
05
1015
200
0.51
1.5
y, w
ith/w
ithou
t filt
er
05
1015
20-2-1012
u, w
ith/w
ithou
t filt
er
Osc
illat
ion
rem
oved
!
Lect
ure
1434
EL2
620
2011
Que
stio
n10
Can
you
repe
atlin
eari
zatio
nth
roug
hhi
ghga
infe
edba
ck?
Lect
ure
1435
EL2
620
2011
Inve
rtin
gN
onlin
eari
ties
Com
pens
atio
nof
stat
icno
nlin
earit
yth
roug
hin
vers
ion:
F(s
)f
#1(·)
f(·)
G(s
)!
Con
trolle
r
Sho
uld
beco
mbi
ned
with
feed
back
asin
the
figur
e!
Lect
ure
1436
EL2
620
2011
Rem
ark:
How
toO
btai
nf
!1
from
fus
ing
Feed
back
!
vu
k s f(·)
e
u
f(u
)
e=
# v!
f(u
)$
Ifk
>0
larg
ean
ddf
/du
>0,
then
e(
0an
d
0=
# v!
f(u
)$-
f(u
)=
v-
u=
f#
1(v
)
Lect
ure
1437
EL2
620
2011
Que
stio
n11
Wha
tsho
uld
we
know
abou
tinp
ut–o
utpu
tsta
bilit
y?
You
shou
ldun
ders
tand
and
beab
leto
deriv
e/ap
ply
•S
yste
mga
in%(S
)=
sup
u$L
2
%y% 2
%u% 2
•B
IBO
stab
ility
•S
mal
lGai
nTh
eore
m
•C
ircle
Crit
erio
n
•P
assi
vity
Theo
rem
Lect
ure
1438
EL2
620
2011
Que
stio
n12
Wha
tabo
utde
scri
bing
func
tions
?
Lect
ure
1439
EL2
620
2011
Idea
Beh
ind
Des
crib
ing
Func
tion
Met
hod
re
uy
!N
.L.
G(s
)
e(t)
=A
sin
!t
give
s
u(t
)=
" % n=
1
&a
2 n+
b2 nsi
n[n
!t+
arct
an(a
n/b
n)]
If|G
(in!)|
.|G
(i!)|
forn
%2,
then
n=
1su
ffice
s,so
that
y(t
)/
|G(i
!)|'
a2 1+
b2 1si
n[!
t+
arct
an(a
1/b
1)+
arg
G(i
!)]
Lect
ure
1440
EL2
620
2011
Defi
nitio
nof
Des
crib
ing
Func
tion
The
desc
ribi
ngfu
nctio
nis
N(A
,!)
=b 1
(!)+
ia1(!
)
A
e(t)
u(t
)N
.L.
e(t)
(u 1(t
)N
(A,!
)
IfG
islo
wpa
ssan
da
0=
0th
en
(u 1(t
)=
|N(A
,!)|A
sin[!
t+
arg
N(A
,!)]
/u(t
)
Lect
ure
1441
EL2
620
2011
Exi
sten
ceof
Per
iodi
cS
olut
ions
repl
acem
ents
0e
uy
!f(·)
G(s
)
!1/N
(A) AG
(i")
y=
G(i
!)u
=!
G(i
!)N
(A)y
$G
(i!)
=!
1
N(A
)
The
inte
rsec
tions
ofth
ecu
rves
G(i
!)
and
!1/
N(A
)gi
ve!
and
Afo
rapo
ssib
lepe
riodi
cso
lutio
n.
Lect
ure
1442
EL2
620
2011
Mor
eC
ours
esin
Con
trol
•E
L245
0H
ybrid
and
Em
bedd
edC
ontro
lSys
tem
s,pe
r3
•E
L274
5P
rinci
ples
ofW
irele
ssS
enso
rNet
wor
ks,p
er3
•E
L252
0C
ontro
lThe
ory
and
Pra
ctic
e,A
dvan
ced
Cou
rse,
per4
•E
L182
0M
odel
ling
ofD
ynam
icS
yste
ms,
per1
•E
L242
1P
roje
ctC
ours
ein
Aut
omat
icC
ontro
l,pe
r2
Lect
ure
1443
EL2
620
2011
EL2
520
Con
trol
Theo
ryan
dP
ract
ice,
Adv
ance
dC
ours
eA
im:
prov
ide
anin
trodu
ctio
nto
prin
cipl
esan
dm
etho
dsin
adva
nced
cont
rol,
espe
cial
lym
ultiv
aria
ble
feed
back
syst
ems.
•Pe
riod
4,7.
5p
•M
ultiv
aria
ble
cont
rol:
–Li
near
mul
tivar
iabl
esy
stem
s
–R
obus
tnes
san
dpe
rform
ance
–D
esig
nof
mul
tivar
iabl
eco
ntro
llers
:LQ
G,H
"-o
ptim
izat
ion
–R
ealt
ime
optim
izat
ion:
Mod
elP
redi
ctiv
eC
ontro
l(M
PC
)
•Le
ctur
es,e
xerc
ises
,lab
s,co
mpu
tere
xerc
ises
Con
tact
:M
ikae
lJoh
anss
Lect
ure
1444
EL2
620
2011
EL2
450
Hyb
rid
and
Em
bedd
edC
ontr
olS
yste
ms
Aim
:cou
rse
onan
alys
is,d
esig
nan
dim
plem
enta
tion
ofco
ntro
lal
gorit
hms
inne
twor
ked
and
embe
dded
syst
ems.
•Pe
riod
3,7.
5p
•H
owis
cont
roli
mpl
emen
ted
inre
ality
:
–C
ompu
ter-
impl
emen
tatio
nof
cont
rola
lgor
ithm
s
–S
ched
ulin
gof
real
-tim
eso
ftwar
e
–C
ontro
love
rcom
mun
icat
ion
netw
orks
•Le
ctur
es,e
xerc
ises
,hom
ewor
k,co
mpu
tere
xerc
ises
Con
tact
:D
imos
Dim
arog
onas
Lect
ure
1445
EL2
620
2011
EL2
745
Pri
ncip
les
ofW
irel
ess
Sen
sor
Net
wor
ksA
im:
prov
ide
the
part
icip
ants
with
aba
sic
know
ledg
eof
wire
less
sens
orne
twor
ks(W
SN
)
•Pe
riod
3,7.
5cr
•TH
EIN
TER
NE
TO
FTH
ING
S
–E
ssen
tialt
ools
with
inco
mm
unic
atio
n,co
ntro
l,op
timiz
atio
nan
dsi
gnal
proc
essi
ngne
eded
toco
pew
ithW
SN
–D
esig
nof
prac
tical
WS
Ns
–R
esea
rch
topi
csin
WS
Ns
Con
tact
:C
arlo
Fisc
hion
Lect
ure
1446
EL2
620
2011
EL1
820
Mod
ellin
gof
Dyn
amic
Sys
tem
sA
im:
teac
hho
wto
syst
emat
ical
lybu
ildm
athe
mat
ical
mod
els
ofte
chni
cals
yste
ms
from
phys
ical
law
san
dfro
mm
easu
red
sign
als.
•Pe
riod
1,6
p
•M
odel
dyna
mic
alsy
stem
sfro
m
–ph
ysic
s:la
gran
gian
mec
hani
cs,e
lect
rical
circ
uits
etc
–ex
perim
ents
:pa
ram
etric
iden
tifica
tion,
frequ
ency
resp
onse
•C
ompu
tert
ools
form
odel
ing,
iden
tifica
tion,
and
sim
ulat
ion
•Le
ctur
es,e
xerc
ises
,lab
s,co
mpu
tere
xerc
ises
Con
tact
:H
akan
Hja
lmar
sson
Lect
ure
1447
EL2
620
2011
EL2
421
Pro
ject
Cou
rse
inC
ontr
olA
im:
prov
ide
prac
tical
know
ledg
eab
outm
odel
ing,
anal
ysis
,des
ign,
and
impl
emen
tatio
nof
cont
rols
yste
ms.
Giv
eso
me
expe
rienc
ein
proj
ectm
anag
emen
tand
pres
enta
tion.
•Pe
riod
4,12
p
•“F
rom
star
tto
goal
...”:
appl
yth
eth
eory
from
othe
rcou
rses
•Te
amw
ork
•P
repa
ratio
nfo
rMas
tert
hesi
spr
ojec
t
•P
roje
ctm
anag
emen
t(le
ctur
ers
from
indu
stry
)
•N
ore
gula
rlec
ture
sor
labs
Con
tact
:Jo
nas
Mar
tens
Lect
ure
1448
EL2
620
2011
Doi
ngM
aste
rTh
esis
Pro
ject
atA
utom
atic
Con
trol
Lab
0Th
eory
and
prac
tice
0C
ross
-dis
cipl
inar
y
0Th
ere
sear
ched
ge
0C
olla
bora
tion
with
lead
ing
indu
stry
and
univ
ersi
ties
0G
etin
sigh
tin
rese
arch
and
deve
lopm
ent
Hin
ts:
•Th
eto
pic
and
the
resu
ltsof
your
thes
isar
eup
toyo
u
•D
iscu
ssw
ithpr
ofes
sors
,lec
ture
rs,P
hDan
dM
Sst
uden
ts
•C
heck
old
proj
ects
Lect
ure
1449
EL2
620
2011
Doi
ngP
hDTh
esis
Pro
ject
atA
utom
atic
Con
trol
•In
telle
ctua
lstim
uli
•G
etpa
idfo
rstu
dyin
g
•In
tern
atio
nalc
olla
bora
tions
and
trave
l
•C
ompe
titiv
e
•W
orld
-wid
ejo
bm
arke
t
•R
esea
rch
(50%
),co
urse
s(3
0%),
teac
hing
(20%
),fu
n(1
00%
)
•4-
5yr
’sto
PhD
(lic
afte
r2-3
yr’s
)
Lect
ure
1450
