Classical Control - Yazd · Classical Control Text: Feedback Control of Dynamic Systems, 4thth...
Transcript of Classical Control - Yazd · Classical Control Text: Feedback Control of Dynamic Systems, 4thth...
-
Classical Control
Topics covered:
Modeling. ODEs. Linearization. gLaplace transform. Transfer functions. Block diagrams. Mason’s Rule. Time response specifications. p pEffects of zeros and poles. Stability via Routh-Hurwitz. Feedback: Disturbance rejection, Sensitivity, Steady-state tracking.Feedback: Disturbance rejection, Sensitivity, Steady state tracking. PID controllers and Ziegler-Nichols tuning procedure. Actuator saturation and integrator wind-up.
Root locus. Frequency response--Bode and Nyquist diagrams. Stability Margins
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 1
Stability Margins.Design of dynamic compensators.
-
Classical Control
Text: Feedback Control of Dynamic Systems,4th Edition G F Franklin J D Powel and A Emami Naeini4th Edition, G.F. Franklin, J.D. Powel and A. Emami-NaeiniPrentice Hall 2002.
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 2
-
What is control?For any analysis we need a mathematical MODEL of the system
Model → Relation between gas pedal and speed:10 mph change in speed per each degree rotation of gas pedal
For any analysis we need a mathematical MODEL of the system
Block diagram for the cruise control plant:
Disturbance → Slope of road:5 mph change in speed per each degree change of slope
w
Block diagram for the cruise control plant:
Slope(degrees)
0.5 )5.0(10 wuy −=
u +
-10 y
Control(degrees)
Output speed(mph)
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 3
u + y
-
What is control?O l i t l
w
Open-loop cruise control:
ru =0.5
wuy )50(10 −=
PLANT 10u =
u +
-10 y
wrwuyol
)5.010
(10
)5.0(10
−=
=
1/10r u + yolReference
(mph)wr 5
10−=
r
wyre
wyre
ol
olol
500[%]
5
=−
=
=−=%69.7,51,65
00,65==⇒==
=⇒==
olol
ol
eewrewr
mph
OK h
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 4
rreol 500[%] == OK when:
1- Plant is known exactly2- There is no disturbance
-
What is control?Cl d l i t l
w
Closed-loop cruise control:
( )lyru −= 200.5 wuycl )5.0(10 −=
PLANT( )clyru 20
u +
-10 y
wr2015
201200
−=1/10r
+u + yclr
51
-
wyr
wryre clcl
512015
2011
+=−=
%690551165
%5.0%20110,65
+⇒
==⇒== clewr
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 5
rw
ryre clcl 201
52011[%] +=−= %69.0652012011,65 =+=⇒== clewr
-
What is control?
Feedback control can help:
• reference following (tracking)• reference following (tracking)• disturbance rejection• changing dynamic behavior
LARGE gain is essential but there is a STABILITY limit
“The issue of how to get the gain as large as possible to reduce The issue of how to get the gain as large as possible to reduce the errors due to disturbances and uncertainties without making the system become unstable is what much of feedback control design is all about”control design is all about
First step in this design process: DYNAMIC MODEL
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 6
-
Dynamic Models
MECHANICAL SYSTEMS: maF = Newton’s law
xvxbuxm
&
&&&
=−=
velocity
xva &&& == acceleration
mVub 1 T f F timbs
mUV
muv
mbv
o
oeUueVv sto
sto +
=⎯⎯⎯⎯⎯ →⎯=+==
1,
& Transfer Function
sdtd→
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 7
-
Dynamic Models
MECHANICAL SYSTEMS: αIF = Newton’s law
2 sin Tlmgml c=
+−=
θω
θθ&
&&
angular velocity
2mlI =
== θωα &&& angular acceleration
moment of inertiamlI = moment of inertia
TgTg &&&&2sin2sin ml
Tlg
mlT
lg cc =+⎯⎯ →⎯=+ ≈ θθθθ θθ Linearization
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 8
-
Dynamic Models
2mlT
lg c=+ θθ&&
θ=1x 21 xx =&Reduce to first order equations:
θ&=2x 212 mlTx
lgx c+−=&
uxgxl
Tux
x c ⎥⎦
⎤⎢⎣
⎡+
⎥⎥⎤
⎢⎢⎡
=⇒≡⎥⎦
⎤⎢⎣
⎡≡
10
010
, 21 & State Variable
Representationlmlx
⎥⎦
⎢⎣⎥⎦⎢⎣
−⎥⎦⎢⎣ 10
22
Representation
General case: GuFxx +=&
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 9
General case:
JuHxyGuFxx+=+
-
Dynamic Models
ELECTRICAL SYSTEMS:
Kirchoff’s Current Law (KCL):
Th l b i f i d i The algebraic sum of currents entering a node is zero at every instant
Kirchoff’s Voltage Law (KVL)
The algebraic sum of voltages around a loop is zero at every instant
Resistors:
iR iC iL+ + +
)()()()( tGvtitRitv RRRR =⇔=
t
Capacitors:
vR vC vL )0()(1)()()(0
C
t
CCC
C vdiCtv
dttdvCti +=⇔= ∫ ττ
Inductors:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 10
∫ +=⇔=t
LLLL
L idvLti
dttdiLtv
0
)0()(1)()()( ττ
-
Dynamic Models
ELECTRICAL SYSTEMS:
OP AMP:
+ip+
∞→−= AvvAv npO ),(
+
RIRO
A(v v )in
iO vOvp
++ 0==
= npii
vv
+-
A(vp-vn)-
invn+ 0== np ii
To work in the linear mode we need FEEDBACK!!!
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 11
To work in the linear mode we need FEEDBACK!!!
-
Dynamic Models
ELECTRICAL SYSTEMS:
R2 KCL:
R1 Cv1 vO IO
O vCR
vCRdt
dv
12
11−=
+-
( ) ( ) ∫−=⇒∞=t
IOO dvCRvtvR
01
2 )(10)( μμOC
Kv1 vO∫ RC
K 1−=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 12
Inverting integrator
-
Dynamic ModelsC O C C S S S CELECTRO-MECHANICAL SYSTEMS: DC Motor
iKT
armature currenttorque
me
at
KeiKTθ&=
=
shaft velocityemf
+−= TbJ mmm θθ &&&
0=+++− edtdiLiRv aaaa
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 13
Obtain the State Variable Representation
-
Dynamic ModelsHEAT-FLOW:
( )TTq 1=
Temperature DifferenceHeat Flow
( )
qT
TTR
q
1
21
=
−=
& qC
T =
Thermal resistanceThermal capacitance
⎞⎛ 111 ( )IoI
I TTRRCT −⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
21
111&
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 14
-
Dynamic ModelslFLUID-FLOW:
wwm −=&
Mass rate Mass Conservation law
outin wwm =
l fll fl Outlet mass flowInlet mass flow
( )outin wwAhhAm −=⇒= ρρ1&&&
A: area of the tankρ: density of fluid
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 15
ρ: density of fluidh: height of water
-
Linearization
( )uxfx ,=&Dynamic System:
( )( )oo uxf ,0 = Equilibrium
Denote uuuxxx −=−= δδDenote oo uuuxxx δδ ,
( )uuxxfx oo δδδ ++= ,&Taylor Expansion
( ) ufxfuxfx δδδ ∂+∂+≈& ( ) uu
xx
uxfxoooo uxux
oo δδδ,,
,∂
+∂
+≈
∂∂ fGfF GF δδδ&Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 16
⇒∂∂
≡∂∂
≡oooo uxux u
fGxfF
,,, uGxFx δδδ +≈&
-
Linearization
ffff ⎤⎡ ∂∂⎤⎡ ∂∂
uGxFx δδδ +≈&
mn uf
uf
ufG
xf
xf
xfF
1
1
11
1
1
, ⎥⎥⎥⎤
⎢⎢⎢⎡
∂∂
∂∂
=∂∂
≡⎥⎥⎥⎤
⎢⎢⎢⎡
∂∂
∂∂
=∂∂
≡ MM
L
MM
L
oo
oo
oo
oo
uxm
nnux
uxn
nnux
uf
ufu
xf
xfx
,1
,
,1
,
⎥⎥⎥
⎦⎢⎢⎢
⎣ ∂∂
∂∂∂
⎥⎥⎥
⎦⎢⎢⎢
⎣ ∂∂
∂∂∂
LL
Example: Pendulum with friction 0sin =++ θθθlg
mk &&&
lm
xkgx ⎥⎤
⎢⎡
=10
&
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 17
xmkx
lgx
ox⎥⎥⎦⎢
⎢⎣
−−= 1cos
-
Laplace Transform
• Function f(t) of time– Piecewise continuous and exponential order btKetf
-
Laplace transform examples• Step function – unit Heavyside Function ⎧ fStep function unit Heavyside Function
• After Oliver Heavyside (1850-1925)
1)(∞+−∞−∞∞ ωσ tjst⎩⎨⎧
≥<
=0for,10for,0
)(tt
tu
0if1)()(0
)(
000>=
+−=−===
+∞
−
−∞
−
− ∫∫ σωσ
ωσ
sje
sedtedtetusF
tjststst
• Exponential function• After Oliver Exponential (1176 BC- 1066 BC)
∞)(∫ ∫∞ ∞ ∞+−
+−−− >+
=+
−===0 0 0
)()( if1)( ασ
αα
ααα
ssedtedteesF
tstsstt
• Delta (impulse) function δ(t)sdtetsF st allfor1)()( == ∫
∞−δ
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 19
)()(0∫−
-
Laplace Transform Pair TablesSignal Waveform Transformg
impulse
step
)(tδ 1
s1)(tu
ramp
exponential
s
21
s1
)(ttu
)(tue tα−exponential
damped ramp
β
α+s
2)(
1
α+s
)(tue
)(tutte α−
sine
cosine
22 β
β
+s
22 β+s
s
( ) )(sin tutβ
( ) )(cos tutβ
damped sine
damped cosine α+s
22)( βα
β
++s
β+s
( ) )(sin tutte βα−
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 20
damped cosine22)( βα
α
++
+
s
s( ) )(cos tutte βα−
-
Laplace Transform Properties
{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLLLinearity: (absolutely critical property)
{ } { } { } )()()()()()( 2121 ff 21 ff( )ssFdf
t=
⎭⎬⎫
⎩⎨⎧∫0
)( ττLIntegration property:⎭⎩0
)0()()( −−=⎭⎬⎫
⎩⎨⎧ fssF
dttdf
LDifferentiation property:
)0()0()(22)(2
−′−−−=⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
fsfsFsdt
tfdL
⎭⎩
)0()0()0()()( )(21 −−−−′−−−=⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧ −− mmm
m
mffsfssF
dttfd
LmsL
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 21
⎭⎩
-
Laplace Transform Properties
)()}({ αα +=− sFtfe tL
Translation properties:
s-domain translation: )()}({ α+sFtfeL
{ } 0)()()( >=−− − asFeatuatf as forL
s domain translation:
t-domain translation:
Initial Value Property: )(lim)(lim0
ssFtfst ∞→+→
=
Final Value Property: )(lim)(lim0
ssFtfst →∞→
=0st →∞→
If all poles of F(s) are in the LHP
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 22
-
Laplace Transform Properties1 s )(1)}({
asF
aatf =LTime Scaling:
Multiplication by time:sdFttf )()}({LMultiplication by time:
dsttf )}({ −=L
Convolution: )()(})()({0
sGsFdtgft
=−∫ τττL
Time product: λλωσ
dsGsFtgtfj
∫+
= )()(1)}()({LTime product: λλπ ωσdsGsF
jtgtf
j∫ − −= )()(2)}()({L
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 23
-
Laplace Transform
Exercise: Find the Laplace transform of the following waveform
[ ] )()2cos(2)2sin(22)( tutttf += ( )24)( +ssF[ ] )()2cos(2)2sin(22)( tutttf −+= ( )( )4)( 2 += sssFExercise: Find the Laplace transform of the following waveform
( )
[ ]ddxxtuetf
t
tt
400
4 4sin5)()( − += ∫ ( )( )23
1648036)(++++
=ssssssF
[ ] ( )tudttedtuetf
tt
4040 5)(5)(
−− +=
( )24020010)(
++
=sssF
E iExercise: Find the Laplace transform of the following waveform
)2()(2)()( TtAuTtAutAutf −+−−= ( )eAsFTs 21)(−−
=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 24
)()()()(fs
)(
-
Laplace transformsLinearsystem
mai
n) Complex frequency domain(s domain)Differential
equationLaplace
transform LAlgebraicequation
ain
(tdo
m
Classicaltechniques
Algebraictechniquesm
e do
ma
q
Response Inverse Laplace
q
Response
Tim
psignal
ptransform L-1
ptransform
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 25
• The diagram commutes– Same answer whichever way you go
-
Solving LTI ODE’s via Laplace Transform
( ) ( ) ( ) ( ) ubububyayay mmm
mn
nn
01
101
1 +++=+++−
−−
− LL
Initial Conditions: ( )( ) ( ) ( )( ) ( )0,,0,0,,0 11 uuyy mn KK −−
kk tfd ∑−⎫⎧ 1)(
⎤⎡⎤⎡
Recall jj
jkkk sfsFdttfd ∑
=
−−−=⎭⎬⎫
⎩⎨⎧
0
)1( )0()()( sL
111 imin −−−
⎥⎦
⎤⎢⎣
⎡−=⎥
⎦
⎤⎢⎣
⎡−+− ∑∑∑∑∑
−
=
−−
=
−
=
−−−
=
−
=
−− ji
j
jiim
ii
ji
j
jiin
ii
n
j
jjnn susUsbsysYsasysYs1
0
)1(
0
1
0
)1(1
0
1
0
)1( )0()()0()()0()(
011
1
0
)1(
00
)1(
0
011
1
011
1
)0()0()()(
asasas
subsyasU
asasasbsbsbsbsY n
nn
j
j
ji
ii
j
j
ji
ii
nn
n
mm
mm
++++
−+
++++++++
= −−
=
−−
==
−−
=−
−
−−
∑∑∑∑LL
L
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 26
For a given rational U(s) we get Y(s)=Q(s)/P(s)
-
Laplace Transform
Exercise: Find the Laplace transform V(s)
)(4)(6)( tttdv 34)(
3)0(
)(4)(6)(
−=−
=+
v
tutvdt ( ) 6
36
4)(+
−+
=sss
sV
Exercise: Find the Laplace transform V(s)
2
( )( )( ) 12
3215)(
+−
+++=
sssssV
2)0('2)0(
5)(3)(4)( 222
=−−=−
=++ −
vv
etvdt
tdvdt
tvd t
2)0(,2)0( vv
What about v(t)?
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 27
( )
-
Transfer Functions
( ) ( ) ( ) ububyayay mmn
nn
01
101
1 ++=+++−
−−
− LL
Assume all Initial Conditions Zero:
( ) ( ) )()( 11 UbbbY mnn
)(1m sBbsbsb +++− L
( ) ( ) )()( 01110111 sUbsbsbsYasasas mmnnn +++=++++ −−−− LLInputOutput
)()(
)()()()()(
011
1
011
1
011
m
nn
nm
bsbsbsYH
sUsAsBsU
asasasbsbsbsY
+++
=++++
+++=
−
−−
−
L
L
L
)())(()()()(
21
011
1
011
m
nn
nm
zszszsK
asasasbsbsb
sUsYsH
−−−=
+++++++
== −−
−
L
L
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 28
)())(( 21 npspspsK
−−−=
L
-
Rational Functions
• We shall mostly be dealing with LTs which are rational functions – ratios of polynomials in s
1mm bbbb
)())((
)(01
11
011
1n
nn
n
mm
mm
zszszsasasasabsbsbsbsF
++++++++
= −−
−−
L
L
L
• pi are the poles and zi are the zeros of the function)())(()())((
21
21
n
m
pspspszszszsK
−−−−−−
=L
L
pi are the poles and zi are the zeros of the function• K is the scale factor or (sometimes) gain• A proper rational function has n≥m• A proper rational function has n≥m• A strictly proper rational function has n>m• An improper rational function has n
-
Residues at simple polesF i f l i bl i h i l d Functions of a complex variable with isolated, finite order poles have residues at the poles
)()()()())(()())(()( 2121 nm
psk
psk
psk
pspspszszszsKsF
−++
−+
−=
−−−−−−
= LL
L
)()()()())(( 2121 nn pspspspspsps
( ) )()()()( 21 inii pskkpskpskF −++++−+−( ))()(
)()(
)()()(
2
2
1
1
n
ini
iii ps
pkps
ppspsFps
−++++
−+
−=− LL
)()(lim sFpsk ipsi i−=
→Residue at a simple pole:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 30
-
Residues at multiple polesCompute residues at poles of order r:Compute residues at poles of order r:
rr
rm
psk
psk
psk
pszszszsKsF
)()()()()())(()(
12
1
2
1
1
1
21
−++
−+
−=
−−−−
= LL
rjsFrpsjrds
jrdpsjr
k ii
j L1,)()(lim)!(1
=⎥⎦⎤
⎢⎣⎡ −
−
−
→−=
pppp )()()()( 1111
( )31522
+
+
s
ssExample:( ) ( )31
321
11
2
+−
++
+=
sss( )
3)1(
)52()1(lim 323
3−=
+++
−→ ssss
s1
)1()52()1(lim 3
23
1=⎥⎥⎦
⎤
⎢⎢⎣
⎡+
++−→ s
sssdsd
s2
)1()52()1(lim!2
1323
22
1=⎥⎥⎦
⎤
⎢⎢⎣
⎡+
++−→ s
sssdsd
s
( ) ( )
)1(3 +→ ss )1(1 ⎥⎦⎢⎣ +→ sdss )1(!2 1 ⎥⎦⎢⎣ +→ sdss
( ) )(3231252 212
1 LssL t⎟⎞
⎜⎛
⎟⎞
⎜⎛ + −−−
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 31
( ) ( ) ( )( ) )(32
11112
321
31 tutte
sssL
sL t −+=⎟⎟
⎠⎜⎜⎝ +
−+
++
=⎟⎟⎠
⎜⎜⎝ +
-
Residues at complex poles
• Compute residues at the poles )()(lim sFasas
−→
• Bundle complex conjugate pole pairs into second-order terms if you want
• but you will need to be careful( )[ ]222 2))(( βααβαβα ++−=+−−− ssjsjs
• Inverse Laplace Transform is a sum of complex exponentials• The answer will be real
( )[ ]
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 32
-
Inverting Laplace Transforms in Practice
• We have a table of inverse LTs• Write F(s) as a partial fraction expansion
F(s) = bmsm + bm−1s
m−1 +L+ b1s+ b0ans
n + an−1sn−1 +L+ a1s+ a0
(s z )(s z ) (s z )= K (s− z1)(s− z2)L(s− zm)
(s− p1)(s− p2)L(s− pn )
=α1
( )+
α2( )
+α31
( )+
α32( )2
+α33
( )3+ ...+
αq( )
• Now appeal to linearity to invert via the table
s− p1( ) s− p2( ) (s− p3) s− p3( )2 s− p3( )3 s− pq( )
pp y– Surprise!– Nastiness: computing the partial fraction expansion is best
d b l l i h id
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 33
done by calculating the residues
-
Example 9-12)3(20 +• Find the inverse LT of
)52)(1()3(20)( 2 +++
+=
sssssF
*kkk21211
)( 221js
kjs
ksksF
+++
−++
+=
)3(20
π5)3(20
10152
2)3(20)()1(
1lim1
j
sss
ssFss
k =−=++
+=+
−→=
π42555
21)21)(1()3(20)()21(
21lim2
jej
jsjssssFjs
jsk =−−=
+−=++++
=−++−→
=
55 ⎤⎡)(252510)( 4
5)21(45)21(
tueeetfjtjjtjt
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡++=
−−−++−− ππ
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 34
)()4
52cos(21010 tutee tt ⎥⎦⎤
⎢⎣⎡ ++= −−
π
-
Not Strictly Proper Laplace Transforms8126 23 +++• Find the inverse LT of
348126)( 2
23
++
+++=
ssssssF
– Convert to polynomial plus strictly proper rational function• Use polynomial division 22)( 2
+++=
sssF
35.0
15.02
34)( 2
++
+++=
++
sss
ss
• Invert as normal
31 ++ ss
)(5.05.0)(2)()( 3 tueetdt
tdtf tt ⎥⎦⎤
⎢⎣⎡ +++= −−δδ
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 35
-
Block Diagrams
Series:1G 2G
21GGG =
1G +Parallel:
2G+
21 GGG +=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 36
-
Block DiagramsNegative Feedback:Negative Feedback:
G+)(sR )(sE )(sC BRER
−= Error signalReference input
G-
H)(sB HCBGEC
BRE
==
−= Error signal
Output
Feedback signalH HCB = Feedback signal
GC)1(
)1(GHG
RCGRCGHGHCGRC
+=⇒=+⇒−=
1E)1(
1)1(GHR
EREGHHGERE+
=⇒=+⇒−=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 37
Rule: Transfer Function=Forward Gain/(1+Loop Gain)
-
Block DiagramsPositive Feedback:Positive Feedback:
G+)(sR )(sE )(sC BRER
+= Error signalReference input
G+
H)(sB HCBGEC
BRE
==
+= Error signal
Output
Feedback signalH HCB = Feedback signal
GC)1(
)1(GHG
RCGRCGHGHCGRC
−=⇒=−⇒+=
1E)1(
1)1(GHR
EREGHHGERE−
=⇒=−⇒+=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 38
Rule: Transfer Function=Forward Gain/(1-Loop Gain)
-
Block Diagrams
Moving through a branching point:
G)(sR )(sC
)(B
G)(sR )(sC
)(B≡
G/1)(sB)(sB
Moving through a summing point:Moving through a summing point:
G+)(sR )(sC G +)(sR )(sC≡G
+
)(sB
G+
G
≡
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 39
)(sB
-
Block Diagrams
H
Example:
1G+)(sR )(sC
2G 3G+
1H
-
1G- 2G 3G
2H2
212321
321
1 GGHGGHGGG++
)(sC)(sR
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 40
-
Mason’s RuleH
H
4H
1H+)(sU )(sY
2H 3H+
6H
++ +
+
1+ 2 3
5H
+
7H
4H
5
nodes branches
Signal Flow Graph
)(sU )(sY1H 2H 3H
6H
1 2 3 4
b a c es
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 41
)(sU )(
5H 7H
-
Mason’s RuleP th f t d b h i th di ti f th Path: a sequence of connected branches in the direction of the signal flow without repetitionLoop: a closed path that returns to its starting nodeForward path: connects input and output
∑ ΔΔ== iiGUsYsG 1)()()(
Forward path: connects input and output
∑Δ i iisU )()(
path forward ith the of gain =iG
touch)notdothatloopstwopossibleallofproductsgain
gains) loop (alltdeterminan system the
+
=
=Δ
∑∑-
(
1
touch not do that loops three possible all of products gain
touch)notdo thatloopstwopossibleallofproducts gain
+
−
+
∑∑
L
)(
(
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 42
pathforwardiththetouchnot doesthatgraphtheofpartthe for of value Δ=Δi
-
Mason’s Rule
4H
6H
)(sU )(sY1H 2H 3H
H 7H
1 2 3 4
5H 7H
73515674736251
6244321
1)()(
HHHHHHHHHHHHHHHHHHHHH
sUsY
+−−−−−+
=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 43
-
Impulse Response
)()()(0
tudtu =−∫∞
ττδτDirac’s delta:
Integration is a limit of a sum ⇓
u(t) is represented as a sum of impulsesu(t) is represented as a sum of impulses
By superposition principle, we only need unit impulse response
h )(t)(tSystem Response:
( )τ−th Response at t to an impulse applied at τ
h )(ty)(tu
τττ dthuty )()()( ∫∞
System Response:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 44
τττ dthuty )()()(0∫ −=
-
Impulse Response
h )(ty)(tut-domain:Impulse response
τττ dthuty )()()(0∫∞
−=
The system response is obtained by convolving the input with
)()()()( thtyttu =⇒=δ
Convolution: )()(})()({ sUsHdthu =−∫∞
τττL
The system response is obtained by convolving the input with the impulse response of the system.
Convolution: )()(})()({0
sUsHdthu∫ τττL
H )(sY)(sUs-domain:
)()()( sUsHsY = )()(1)()()( sHsYsUttu =⇒=⇒=δImpulse response
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 45
The system response is obtained by multiplying the transfer function and the Laplace transform of the input.
-
Time Response vs. Polest1Real pole: teth
ssH σ
σ−=⇒
+= )(1)( Impulse
Response
00
<>σ Stable
Unstable0
-
Time Response vs. Poles
Real pole:
teths
sH σσσ
σ −=⇒+
= )()(
1
Impulse Response
στ 1= Time Constant
tetyss
sY σσ
σ −−=⇒+
= 1)(1)( Step Response
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 47
-
Time Response vs. Poles2ω I l Complex poles:
2
22 2)(
ω
ωζωω
++=
nn
n
sssH Impulse
Response
( ) ( )222 1 ζωζωω
−++=
nn
n
s
::
ζωn Undamped natural frequency
Damping ratio2
( )( )2
)(dd
n
jsjssH
ωσωσω
−+++=
( ) 222
d
n
s ωσω
++=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 48
21, ζωωζωσ −== ndn
-
Time Response vs. PolesComplex poles:Complex poles:
( ) ( ) ( )tethssH dtn
nn
n ωζ
ωζωζω
ω σ sin1
)(1
)(2222
2−
−=→
−++=( ) ( )s nn ζζωζω 11++
Impulse Response
00
<>
σσ Stable
Unstable0
-
Time Response vs. PolesComplex poles:Complex poles:
( ) ( ) ( )tethssH dtn
nn
n ωζ
ωζωζω
ω σ sin1
)(1
)(2222
2−
−=→
−++=( ) ( )nn ζζζ
Impulse Response
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 50
-
Time Response vs. PolesComplex poles:Complex poles:
( ) ( ) ( ) ( )⎥⎦⎤
⎢⎣
⎡+−=→
−++= − ttety
sssY d
dd
t
nn
n ωωσω
ζωζωω σ sincos1)(1
1)( 222
2
( )
Step ResponseResponse
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 51
-
Time Response vs. Poles2
Complex poles:
2
22
2
2)(
ωζωω
++=
nn
n
sssH
( ) ( )2222
1 ζωζωω
−++=
nn
n
sCASES:
ns ωζ :022 +=
CASES:
Undamped
( ) ( )( )
nn
n
s
s
ωζ
ζωζωζ
ζ
:1
1:12
222
+
−++<
p
Underdamped
Critically damped( )( )[ ] ( )[ ]nn
n
ss
s
ωζζωζζζ
ωζ
11:1
:122 −−+−++>
+= Critically damped
Overdamped
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 52
-
Time Response vs. Poles
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 53
-
Time Domain Specifications
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 54
-
Time Domain Specifications
1- The rise time tr is the time it takes the system to reach the vicinity of its new set point2- The settling time t is the time it takes the system 2 The settling time ts is the time it takes the system transients to decay3- The overshoot Mp is the maximum amount the
t h t it fi l l di id d b it fi l system overshoot its final value divided by its final value4- The peak time tp is the time it takes the system to preach the maximum overshoot point
tt π 8.1≅n
rn
p tt ωζω
ζ
ζπ 64
121
2≅
−=
−
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 55n
sp teM ζωζ 6.421 == −
-
Time Domain Specifications
Design specification are given in terms of
tMtt sppr tMtt ,,,
These specifications give the position of the polesp g p p
dn ωσζω ,, ⇒
Example: Find the pole positions that guarantee
sec3%,10sec,6.0 ≤
-
Effect of Zeros and Additional poles
Additional poles:1- can be neglected if they are sufficiently to the left of the dominant onesof the dominant ones.2- can increase the rise time if the extra pole is within a factor of 4 of the real part of the complex poles.
Zeros:1- a zero near a pole reduces the effect of that pole in th ti the time response.2- a zero in the LHP will increase the overshoot if the zero is within a factor of 4 of the real part of the complex poles (due to differentiation).3- a zero in the RHP (nonminimum phase zero) will depress the overshoot and may cause the step
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 57
depress the overshoot and may cause the step response to start out in the wrong direction.
-
Stability1)( bbbb mm
011
1
011
1
)()(
asasasbsbsbsb
sRsY
nn
n
mm
mm
++++++++
= −−
−−
L
L
)())(()(Y)())(()())((
)()(
21
21
n
m
pspspszszszsK
sRsY
−−−−−−
=L
L
)( kkksY)()()()(
)(2
2
1
1
n
n
psk
psk
psk
sRsY
−++
−+
−= L
Impulse response:
)()()()(1)( 21 nkkksYsR +++=⇒= L
)()()( 21 npspsps −−−
tpn
tptp nekekekty +++= L21 21)(
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 58
-
Stabilityttt
We want:
tpn
tptp nekekekty +++= L21 21)(
nie tpi 10 ∀→We want: nie tpi K10 =∀⎯⎯→⎯ ∞→
Definition: A system is asymptotically stable (a.s.) if Definition: A system is asymptotically stable (a.s.) if
ipi ∀< 0}Re{pi}{
1
0)(
)( 011
1 ++++=−
− asasassan
nn LCharacteristic polynomial:
Characteristic equation:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 59
0)( =saCharacteristic equation:
-
Stability
Necessary condition for asymptotical stability (a.s.):
iai ∀> 0
Use this as the first test!
If any a
-
Routh’s Criterion
Necessary and sufficient conditionDo not have to find the roots pi!pi
Routh’s Array:
n
n
aaasaas
5311
421L
L− n
a Depends on whether n is even or odd
n
n
cccsbbbs
3213
3212
531
−
−
L,,1
7613
1
5412
1
3211 a
aaaba
aaaba
aaab −=−=−=n dds
0
214
M
−
L
L
,,
,,
31312
21211
1
31512
1
21311
111
cbbcdcbbcd
bbaabc
bbaabc
aaa
−=
−=
−=
−=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 61
nas0
MM
,,1
21
1 cd
cd
-
Routh’s Criterion
How to remember this?
Routh’s Array:
Lmmmsn 221 = am jjL
2232221
1131211
mmmsmmmsmmms
n
n
−
− ,
,
12,2
22,1
= −
−
am
am
jj
jj
MMMM
L
L
4342413
333231
mmmsmmms
n−
3,1,11,11,21,2
≥∀−= +−−+−−
immmm
m jiijii
jiMMMM 3,1,1
, ≥∀−
im
mi
ji
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 62
-
Routh’s Criterion
The criterion:
• The system is asymptotically stable if and only if all the elements in the first column of the Routh’s array are positivecolumn of the Routh’s array are positive
• The number of roots with positive real parts is equal to the number of sign changes in the first column of the Routh arrayarray
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 63
-
Routh’s Criterion - Examples
Example 1: 0212 =++ asas
0322
13 =+++ asasasExample 2:
Example 3: 044234 23456 =++++++ ssssss
0)6(5 23 =+−++ kskssExample 4:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 64
-
Routh’s Criterion - Examples
Example: Determine the range of K over which the system is stable
K+)(sR )(sY( )( )1+sK
- ( )( )61 +− sss
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 65
-
Routh’s Criterion
Special Case I: Zero in the first columnWe replace the zero with a small positive constant ε>0 and proceed as before. We then apply the stability criterion by taking the limit as ε→0
Example: 010842 234 =++++ ssss
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 66
-
Routh’s Criterion
Special Case II: Entire row is zeroThis indicates that there are complex conjugate pairs. If the ith row is zero, we form an auxiliary equation from the previous nonzero row:
L+++= −−+ 331
21
11 )(iii ssssa βββ
Where βi are the coefficients of the (i+1)th row in the βi ( )array. We then replace the ith row by the coefficients of the derivative of the auxiliary polynomial.
Example: 02010842 2345 =+++++ sssss
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 67
-
Properties of feedback
Disturbance Rejection:
Open loop
Kr +A+ y
wOpen loop
oK A y
wArKy o +=
K+r ++ y
wClosed loop
cK-+r +A y
wAK
rAK
AKycc
c
++
+=
11
1
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 68
cc
-
Properties of feedback
Disturbance Rejection:
Ch t l t f 0Choose control s.t. for w=0,y≈r
wryK +=⇒= 1Open loop:
rwryA
Kc =+≈⇒>> 01
wryA
Ko +⇒Open loop:
Closed loop:
Feedback allows attenuation of disturbance without h ( h )having access to it (without measuring it)!!!
IMPORTANT: High gain is dangerous for dynamic response!!!
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 69
IMPORTANT: High gain is dangerous for dynamic response!!!
-
Properties of feedback
Sensitivity to Gain Plant Changes
Open loop
Kr +A+ y
wOpen loop
oK A y
oo
o AKryT =⎟⎠⎞
⎜⎝⎛=
K+r ++ y
wClosed loopo⎠⎝
cK-+r +A y
c
c
cc AK
AKryT
+=⎟
⎠⎞
⎜⎝⎛=
1
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 70
cc⎠⎝
-
Properties of feedback
Sensitivity to Gain Plant Changes
Let the plant gain be AA δ+Let the plant gain be
AA
TTo
o δδ =
AA δ+
Open loop:o
o
o
cc
c
TT
AA
AKAA
TT δδδδ
=
-
PID ControllerPID: Proportional – Integral – DerivativePID: Proportional Integral Derivative
P Controller:+)(sR )(sY)(sE )(sU)()()( sGsCsY
pKsC =)(-
+)(sR )(sY)(sG)(sE )(sU
.)()(1
1)()(
,)()(1
)()()()(
sGsCsRsE
sGsCsGsC
sRsY
+=
+=
)()(1)( sGsCsR +
)()(),()( sEKsUteKtu pp ==
1111Step Reference:
)0(111
)(11lim)(lim1)(
00 GKssGKsssEe
ssR
ppssss +
=+
==⇒=→→
)0(0 GK •Proportional gain is high∞→⇔= )0(0 GKe pss•Proportional gain is high•Plant has a pole at the origin
True when:
High gain proportional feedback (needed for good tracking)
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 72
High gain proportional feedback (needed for good tracking) results in underdamped (or even unstable) transients.
-
PID ControllerP Controller: Example (lecture06_a.m)
K+)(sR )(sYA)(sE )(sUpK- 1
2 ++ ss
)1()(1)(
)()(
2 AKssAK
sGKsGK
sRsY
p
p
p
p
+++=
+=
)()()( pp
12 += pn AK
ζ
ω011 ⎯⎯ →⎯==⇒ ∞→Kζ
Underdamped transient for large proportional gain
12 =nζω 122 + ∞→pKpn AKωζ
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 73
Steady state error for small proportional gain
-
PID ControllerPI Controller:PI Controller:
sKKsC Ip +=)(
-
+)(sR )(sY)(sG)(sE )(sU,)()(1
)()()()(
sGsCsGsC
sRsY
+=
)()()()()( sEKKsUdeKteKtu It
⎟⎞
⎜⎛ ++ ∫ ττ
.)()(1
1)()(
sGsCsRsE
+=
)()(,)()()(0
sEs
KsUdeKteKtu IpIp ⎟⎠
⎜⎝
+=+= ∫ ττ
01li11li)(li1)( ER
Step Reference:
0)(1
lim)(1
lim)(lim)(000
=⎟⎠⎞
⎜⎝⎛ ++
=⎟⎠⎞
⎜⎝⎛ ++
==⇒=→→→
sGs
KKssGs
KKsssEe
ssR
Ip
sIp
ssss
• It does not matter the value of the proportional gain• Plant does not need to have a pole at the origin. The controller has it!
Integral control achieves perfect steady state reference tracking!!!
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 74
Integral control achieves perfect steady state reference tracking!!!Note that this is valid even for Kp=0 as long as Ki≠0
-
PID ControllerPI Controller: Example (lecture06_b.m)
KK I++)(sR )(sYA)(sE )(sU
sK p +
- 12 ++ ss
( )AKsKsGsKK
sY IpI
p +⎟⎠⎞
⎜⎝⎛ + )(
)( ( )AKsAKsssG
sKK
ssR Ip
Ip
Ip
++++=
⎟⎠⎞
⎜⎝⎛ ++
⎠⎝=)1()(1)(
)(23
DANGER: for large Ki the characteristic equation has roots in the RHP
0)1(23 =++++ AKsAKss
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 75
0)1( =++++ AKsAKss IpAnalysis by Routh’s Criterion
-
PID ControllerPI Controller: Example (lecture06_b.m)
0)1(23 =++++ AKsAKss IpNecessary Conditions:
)( Ip
0,01 >>+ AKAK IpThis is satisfied because 0,0,0 >>> Ip KKA
Routh’s Conditions:
AKAKs p
2
3
111 +
⇓
>−+ 01 AKAK Ip
AKAKsAKs
Ip
I
0
1
2
11−+
⇓
AKK PI
1+<
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 76
AKs I0 A
-
PID ControllerPD Controller:PD Controller:
sKKsC Dp +=)(-
+)(sR )(sY)(sG)(sE )(sU,)()(1
)()()()(
sGsCsGsC
sRsY
+=
( ) )()()()()( EKKUtdeKtKt ++
.)()(1
1)()(
sGsCsRsE
+=
( ) )()(,)()()( sEsKKsUdt
KteKtu DpDp +=+=
1111
Step Reference:
( ) )0(111
)(11lim)(lim1)(
00 GKssGsKKsssEe
ssR
pDpssss +
=++
==⇒=→→
P ti l i i hi h
PD controller fixes problems with stability and damping by adding
∞→⇔= )0(0 GKe pss•Proportional gain is high•Plant has a pole at the origin
True when:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 77
PD controller fixes problems with stability and damping by adding “anticipative” action
-
PID ControllerPD Controller: Example (lecture06_c.m)
+)(sR )(sYA)(sE )(sUsKKsC Dp +=)(- 1
2 ++ sssKKsC Dp +)(
( )( )
( )( ) )1(1)(1
)()()(
2 AKsAKssKKA
sGsKKsGsKK
sRsY
pD
Dp
Dp
Dp
+++++
=++
+= ( ) ( ) )()()( pDDp
AK pn +=12
ζ
ω AKAK DD +=+=⇒ 11ζ
The damping can be increased now independently of Kp
AKDn +=12ζω AK pn +122ωζ
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 78
pThe steady state error can be minimized by a large Kp
-
PID ControllerPD Controller:PD Controller:
sKKsC Dp +=)(-
+)(sR )(sY)(sG)(sE )(sU,)()(1
)()()()(
sGsCsGsC
sRsY
+=
( ) )()()()()( EKKUtdeKtKt ++
.)()(1
1)()(
sGsCsRsE
+=
( ) )()(,)()()( sEsKKsUdt
KteKtu DpDp +=+=
NOTE: cannot apply pure differentiation.pp y pIn practice,
sKDsKDis implemented as
sKD
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 79
1+sDτ
-
PID Controller
PID: Proportional – Integral – Derivative
⎞⎛ 1)(sR )(sY)(E )(U⎟⎟⎠
⎞⎜⎜⎝
⎛++ sT
sTK D
Ip
11-
+)(sR )(sY)(sG)(sE )(sU
⎥⎤
⎢⎡
++= ∫ dtdeTde
TteKtu D
d
p)()(1)()( ττ DpD
pI TKKT
KK == ,
⎟⎟⎞
⎜⎜⎛
++=
⎥⎦
⎢⎣
∫
sTKsU
dtT DIp
11)(
)()()(0
DpDI
I T,
⎟⎟⎠
⎜⎜⎝
++= sTsT
KsE DI
p 1)(
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 80
PID Controller: Example (lecture06_d.m)
-
PID Controller: Ziegler-Nichols Tuning
• Empirical method (no proof that it works well but it works well for simple systems)• Only for stable plants• Only for stable plants• You do not need a model to apply the method
)( −KY std
Class of plants:
1)()(
+=
sKe
sUsY std
τ
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 81
-
PID Controller: Ziegler-Nichols Tuning
METHOD 1: Based on step response, tuning to decay ratio of 0.25.
KP τTuning Table:
dIp
dp
tTt
KPD
tKP
30,9.0:
:
==
=
τ
dDdId
p
Id
p
tTtTt
KPID
t
5.0,2,2.1:
3.0
===τ
d
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 82
-
PID Controller: Ziegler-Nichols Tuning
METHOD 2: Based on limit of stability, ultimate sensitivity method.
uK-
+)(sR )(sY)(sG)(sE )(sU-
• Increase the constant gain Ku until the response becomes purely oscillatory (no decay – marginally p y y ( y g ystable – pure imaginary poles)• Measure the period of oscillation Pu
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 83
-
PID Controller: Ziegler-Nichols Tuning
METHOD 2: Based on limit of stability, ultimate sensitivity method.
Tuning Table:
21,45.0:
5.0:u
Iup
up
PTKKPD
KKP
==
=
8,
2,6.0:
2.1u
Du
IupPTPTKKPID ===
τThe Tuning Tables are the same if you make:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 84
dud
u tPtK 4,2 == τ
-
PID Controller: Integrator Windup
Actuator Saturates: - valve (fully open)- aircraft rudder (fully deflected)aircraft rudder (fully deflected)
u(Input of the plant)
cu(Output of the controller)
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 85
-
PID Controller: Integrator Windup
sKK Ip +
-
+)(sR )(sY)(sG)(sE )(sUc )(sU
Wh t h ? What happens? - large step input in r- large elarge e- large uc → u saturates- eventually e becomes small- uc still large because the integrator is “charged”- u still at maximum
h t f l ti
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 86
- y overshoots for a long time
-
PID Controller: Anti-WindupPlant without Anti-Windup:a ou dup
Plant with Anti-Windup:
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 87
-
PID Controller: Anti-WindupI t ti th l t b h In saturation, the plant behaves as:
For large K this is a system with very low gain and
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 88
For large Ka, this is a system with very low gain and very fast decay rate, i.e., the integration is turned off.
-
Steady State Tracking
The Unity Feedback Case
)(sC-
+)(sR )(sY)(sG)(sE )(sU
1)(sE=
)()(1)( sGsCsR +
kt
1)(
)(1!
)(
=
=
sR
tkttrTest Inputs: k=0: step (position)
k=1: ramp (velocity)
k 2 b l ( l ti )
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 89
1)( += kssR k=2: parabola (acceleration)
-
Steady State Tracking
The Unity Feedback Case
)(sR )(sY)(E )(U sG )(
Type nSystem
)(sC-
+)(sR )(sY)(sG)(sE )(sUn
o
ssG )(
11)(),()(
1)(,)()()( +=== kno
ssRsRsGsEs
sGsGsC )(1+ no ss
sGsSteady State Error:
Fin l V l e
)0(lim1
)(lim1)(
1lim)(lim)(lim 1 nkn
kn
n
kss Gs
Gs
GsssEtee =====−
+
Final Value Theorem
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 90
)0()()(1)()(
00100o
nsko
nsk
nosst
ss GsssGsss
sG +++ →→+→→∞→
-
Steady State TrackingThe Unity Feedback Case Type nThe Unity Feedback Case
k+)(sR )(sY)(sE o sG )(
ypSystem
)0(lim
0o
n
kn
sss Gsse+
=−
→
-ns
Steady State Error:Input (k)
Type (n)
Type 0
Step (k=0) Ramp (k=1) Parabola (k=2)
po KsGsCG +=
+=
+ 11
)()(lim11
)0(11
∞ ∞
Type 1
Type 2
psosGsCG
→)()()0(
0
vsoKsGssCG1
)()(lim1
)0(1
0
==→
111
0 ∞
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 91
Type 2aso
KsGsCsG )()(lim)0( 20
==→
0 0
-
Steady State Tracking
1)()(lim0)()(lim
0
0==
==
→
→nsGssCKnsGsCK
sv
spPosition Constant
Velocity Constant
2)()(lim 20
0
==→
→
nsGsCsKsa
sAcceleration Constant
n: Degree of the poles of CG(s) at the origin (the number of n: Degree of the poles of CG(s) at the origin (the number of integrators in the loop with unity gain feedback)
• Applying integral control to a plant with no zeros at the • Applying integral control to a plant with no zeros at the origin makes the system type ≥ I• All this is true ONLY for unity feedback systems• Since in Type I systems e =0 for any CG(s) we say that • Since in Type I systems ess=0 for any CG(s), we say that the system type is a robust property.
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 92
-
Steady State Tracking)(1)( =
k
tttwThe Unity Feedback Case
)(sW 11)(
)(1!
)(
+=
=
kssW
tk
tw
)(sC-
+)(sR )(sY)(sG)(sE )(sU ++
Set r=0.
)()( sGsY
Set r 0. Want Y(s)/W(s)=0.
)()()()(1
)()()( sTssT
sGsCsG
sWsY
on==
+=
nssTssTssYtyye )(lim1)(lim)(lim)(lim =====
Steady State Error: e=r-y=-y Final Value Theorem
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 93
kosksstssss ssT
sssTssYtyye )(lim)(lim)(lim)(lim
0100 →+→→∞→=====−
-
Steady State TrackingThe Unity Feedback CaseThe Unity Feedback Case
)(sC+)(sR )(sY)(sG)(sE )(sU
)(sW
+
)(sC-
)(sG+
Steady State Output:Disturbance (k)
Type (n)
Type 0
Step (k=0) Ramp (k=1) Parabola (k=2)
* ∞ ∞
Type 1
Type 2
*0 ∞
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 94
Type 2 *0 0
∞
-
Steady State TrackingExample:
)(sW
sKK Ip +
-
+)(sR )(sY( )1+ss
Aτ
)(sE )(sU+
+
wKKwK
IP
I
to 0 type to 1 type
⇒=≠⇒≠
0,00
IP
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 95
-
Root LocusController
+)(sR )(sY)(sE )(sU
Plant
)(sC-
+)(sR )(sY)(sG)(sE )(sU
)(H )(sH
Sensor
)(1)()(
)()()(1)()(
)()(
sKLsGsC
sHsGsCsGsC
sRsY
+=
+=⇒= )()( sKDsC
Writing the loop gain as KL(s) we are interested in tracking the closed-loop poles as “gain” K varies
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 96
-
Root LocusCharacteristic Equation:
0)(1 =+ sKL
The roots (zeros) of the characteristic equation are the closed-loop poles of the feedback system!!!
The closed-loop poles are a function of the “gain” K
Writing the loop gain as
nnnn
mmmm
asasasbsbsbs
sasbsL
++++++++
==−
−−
−
11
1
11
1
)()()(
L
L
The closed loop poles are given indistinctly by the solution of:
sLsKbsasbKsKL 1)(0)()(0)(10)(1 −==+=+=+
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 97
KsLsKbsa
saKsKL )(,0)()(,0
)(1,0)(1 ==+=+=+
-
Root Locus
{ } { })()()(1 LKLsKL numdenrootszerosRL +=+=when K varies from 0 to ∞ (positive Root Locus) or
from 0 to -∞ (negative Root Locus)from 0 to ∞ (negative Root Locus)
⎧ 1
⎪⎩
⎪⎨⎧
=∠
=⇔−=>o180)(
1)(1)(:0sL
KsL
KsLK
Phase condition
Magnitude condition
⎪⎧1)(1 sL Magnitude condition
⎪⎩
⎪⎨
=∠
−=⇔−=<o0)(
)(1)(:0sL
KsL
KsLK
Phase condition
Magnitude condition
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 98
-
Root Locus by Characteristic Equation Solution
Example: K-
+)(sR )(sY( )( )110
1++ ss
)(sE )(sU
)( KsY)10(11)(
)(2 KsssR +++
=
Closed-loop poles: 0)10(110)(1 2 =+++⇔=+ KsssLClosed loop poles: 0)10(110)(1 =+++⇔=+ KsssL
10,1 −−=s 0=K
24815.5 Ks −±−=
5.52
4815.5
−=
−±−=
s
Ks
04810481=−>−
KK
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 9928145.5 −±−= Kis 0481
0481
-
Root Locus by Characteristic Equation Solution
We need a systematic approach to plot the closed loop poles
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 100
We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS
-
Root Locus by Phase Condition
Example: K-
+)(sR )(sY( )( )845
12 ++++
sssss)(sE )(sU
( )ssL 1)( += ( )( )
( )( )( )isissss
sssssL
222251
845)( 2
−+++++
=
+++=
( )( )( )isisss 22225 ++++
iso 31+−=
belongs to the locus?
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 101
-
Root Locus by Phase Condition
o43.108o87.36
o45o90
o70.78
[ ] oooooo 18070.784587.3643.10890 −≈+++− iso 31+−=⇒ belongs to the locus!
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 102
[ ] o belongs to the locus!Note: Check code lecture09_a.m
-
Root Locus by Phase Condition
iso 31+−=
We need a systematic approach to plot the closed loop poles
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 103
We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS
-
Root Locus{ } { })()()(1 LKLsKL numdenrootszerosRL ++{ } { })()()(1 LKLsKL numdenrootszerosRL +=+=
when K varies from 0 to ∞ (positive Root Locus) or from 0 to -∞ (negative Root Locus)
0)()(1)(0)(1 =+⇔−=⇔=+ sKbsaK
sLsKL
Basic Properties:
• Number of branches = number of open-loop poles • RL begins at open-loop poles
RL ends at open loop e os o as mptotes
0)(0 =⇒= saK
• RL ends at open-loop zeros or asymptotes
⎩⎨⎧
>−∞→=
⇔=⇒∞=)0(
0)(0)(
mnssb
sLK
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 104
• RL symmetrical about Re-axis⎩ )(
-
Root Locus
Rule 1: The n branches of the locus start at the poles of L(s)and m of these branches end on the zeros of L(s).n: order of the denominator of L(s)n: order of the denominator of L(s)m: order of the numerator of L(s)
Rule 2: The locus is on the real axis to the left of and odd number of poles and zeros.In other words, an interval on the real axis belongs to the In other words, an interval on the real axis belongs to the root locus if the total number of poles and zeros to the right is odd.This rule comes from the phase condition!!!
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 105
-
Root Locus
Rule 3: As K→∞, m of the closed-loop poles approach the open-loop zeros, and n-m of them approach n-m asymptotes with angleswith angles
1,,1,0,)12( −−=−
+= mnlmn
ll K πφ−mn
and centered at
1,,1,0,11 −−=−−
=−−
= ∑∑ mnlmnmn
abK
zerospolesα
mnmn
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 106
-
Root Locus
Rule 4: The locus crosses the jω axis (looses stability) where the Routh criterion shows a transition from roots in the left half-plane to roots in the right-half plane.
Example:
5+)54(
5)( 2 +++
=sss
ssG
5,20 jsK ±==
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 107
-
Root Locus1
Example:4673
1)( 234 +++++
=ssss
ssG
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 108
-
Root Locus
Design dangers revealed by the Root Locus:
• High relative degree: For n m≥3 we have closed loop • High relative degree: For n-m≥3 we have closed loop instability due to asymptotes.
1+s4673
1)( 234 +++++
=ssss
ssG
• Nonminimum phase zeros: They attract closed loop poles into the RHP
11)( 2 ++
−=
ssssG
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 109
Note: Check code lecture09_b.m
-
Root Locus
Viete’s formula:
When the relative degree n m≥2 the sum of the closed loop When the relative degree n-m≥2, the sum of the closed loop poles is constant
∑−= polesloopclosed1a
nnnn
mmmm
asasasbsbsbs
sasbsL
++++++++
==−
−−
−
11
1
11
1
)()()(
L
L
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 110
-
Phase and Magnitude of a Transfer Function1 bbbb mm
011
1
011
1)(asasas
bsbsbsbsG nn
n
mm
mm
++++++++
= −−
−−
L
L
)())((
Th f t K ( ) d ( ) l b
)())(()())(()(
21
21
n
m
pspspszszszsKsG
−−−−−−
=L
L
The factors K, (s-zj) and (s-pk) are complex numbers: zjiz
jj mjerzsφ ==− K1,)(
K
pk
i
ipkk
jj
KK
pkerpsφ
φ ==− L1,)( ieKK φ=
zm
zzK
izm
izizi ererereKsG
φφφφ L21 21)(
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 111
pn
pp ipn
ipipm
ererereKsG
φφφφ
L21 21
21)( =
-
Phase and Magnitude of a Transfer Functionzzz iziziz φφφ
( )
pn
pp
mK
ipn
ipip
izm
izizi
erererererereKsG
φφφ
φφφφ=
L
L
21
21
21
21)(
( )( )pnpp
zm
zzK
ipn
pp
izm
zzi
errrerrreK
φφφ
φφφφ
+++
+++
=L
L
L
L
21
21
21
21
( ) ( )[ ]pnppzmzzKippp
zm
zzn
errrrrrK φφφφφφφ +++−++++= LL
L
L2121
21
21
21
Now it is easy to give the phase and magnitude of the transfer function:
nrrr 21
o e a s e u o
pn
pp
zm
zz
rrrrrrKsG =
L
L
21
21 ,)(
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 112
( ) ( )pnppzmzzKsG φφφφφφφ +++−++++=∠ LL 2121)(
-
Phase and Magnitude of a Transfer Function)7356( +s
Example: ( )( ) )20(51)735.6()(+++
+=
sssssG
ppp
zrsG 1)( =iss o 57 +−==
p1φ
p2φ
p3φ
z1φ
pr3zr1 pr2
pr1 ( )pppzppp
sG
rrr
3211
321
)(
)(
φφφφ ++−=∠1φ2φ3φ 1φ
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 113
-
Root Locus- Magnitude and Phase Conditions
{ } { })()()(1 LKLsKL numdenrootszerosRL +=+=when K varies from 0 to ∞ (positive Root Locus) or
from 0 to -∞ (negative Root Locus)from 0 to ∞ (negative Root Locus)
( ) ( )[ ]pnppzmzzpKippp
zm
zz
pm
p errrrrrK
pspspszszszsKsL φφφφφφφ +++−++++=
−−−−−−
= LLL
L
L
L2121
21
21
21
21
)())(()())(()(
> LK 1)(0
nn rrrpspsps 2121 )())((
L 1)( 21 == pppz
mzz
p KrrrrrrKsL
⇔−=>K
sLK )(:0
( ) ( ) oLLL
180)( 2121
21
=+++−++++=∠ pnppz
mzzK
pn
pp
psL
Krrr
φφφφφφφ
⇔−=<K
sLK 1)(:0( ) ( )
L
L 1)(21
21 −==
K
pn
pp
zm
zz
p KrrrrrrKsL
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 114
K ( ) ( ) oLL 0)( 2121 =+++−++++=∠ pnppzmzzK psL φφφφφφφ
-
Root Locus
Selecting K for desired closed loop poles on Root Locus:
If s belongs to the root locus it must satisfies the If so belongs to the root locus, it must satisfies the characteristic equation for some value of K
1
Then we can obtain K as
KsL o
1)( −=
1K −=
)(1
)( o
sLK
sL
=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 115
)( osL
-
Root Locus
Example: ( )( )511)()(
++==
sssGsL
54314351)(
143 ++−++−=++==⇒+−= iissL
Kis ooo
( ) ( ) 204242
)(2222 =++−=
sL oooo
Using MATLAB:
sys=tf(1,poly([-1 -5]))so=-3+4i
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 116
[K,POLES]=rlocfind(sys,so)
-
Root LocusExample:
1)()( == sGsL 43 is +=Example: ( )( )51)()( ++== sssGsL
57
⇓
+−=o is
43 iso +−=
06.42)(
1==
⇓
LK
)( osL
57 iso +−=o
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 117
When we use the absolute value formula we are assuming that the point belongs to the Root Locus!
-
Root Locus - Compensators
Example: ( )( )511)()(
++==
sssGsL
Can we place the closed loop pole at so=-7+i5 only varying K?NO. We need a COMPENSATOR.
( ) 110)()()( +== ssGsDsL( )( )1)()( == sGsL ( )( )( )5110)()()( +++ ssssGsDsL( )( )51
)()(++ ss
sGsL
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 118
The zero attracts the locus!!!
-
Root Locus – Phase lead compensator
Pure derivative control is not normally practical because of the amplification of the noise due to the differentiation and must be approximated:be approximated:
zppszssD >
++
= ,)( Phase lead COMPENSATORps +
When we study frequency response we will understand why we call “Phase Lead” to this compensator.
( )( ) zpsspszssGsDsL >
++++
== ,51
1)()()(
How do we choose z and p to place the closed loop pole at so=-7+i5?
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 119
-
Root Locus – Phase lead compensatorExample: zpzssGsDsL
-
Root Locus – Phase lead compensator17356+s
Example:
Phase lead
( )( )511
20735.6)()()(
++++
==sss
ssGsDsL
Phase lead COMPENSATOR
57 i117
57=
+−=K
iso
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 121
-
Root Locus – Phase lead compensator
Selecting z and p is a trial an error procedure. In general:
• The zero is placed in the neighborhood of the closed-p gloop natural frequency, as determined by rise-time or settling time requirements.• The poles is placed at a distance 5 to 20 times the value of the zero location. The pole is fast enough to avoid modifying the dominant pole behavior.
The exact position of the pole p is a compromise between:The exact position of the pole p is a compromise between:
• Noise suppression (we want a small value for p)• Compensation effectiveness (we want large value for p)
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 122
-
Root Locus – Phase lag compensator17356+s
Example: ( )( )511
20735.6)()()(
++++
==sss
ssGsDsL
( )( )2
00010735.6
511
20735.6lim)()(lim)(lim −
→→→×=
++++
===sss
ssGsDsLKsssp
What can we do to increase Kp? Suppose we want Kp=10.
Ph l ( )( ) zpsss
spszssGsDsL <
++++
++
== ,51
120735.6)()()(
Phase lag COMPENSATOR
We choose: 48148101 3×z
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 123
We choose: 48.14810735.6
=×=p
-
Root Locus – Phase lag compensator17356148480 ++ ss
Example: ( )( )511
20735.6
001.014848.0)()()(
++++
++
==sss
ss
ssGsDsL
035946 i31.18
03.594.6=
+−=K
iso
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 124
-
Root Locus – Phase lag compensator
Selecting z and p is a trial an error procedure. In general:
• The ratio zero/pole is chosen based on the error /pconstant specification.• We pick z and p small to avoid affecting the dominant dynamic of the system (to avoid modifying the part of the locus representing the dominant dynamics)• Slow transient due to the small p is almost cancelled by an small z. The ratio zero/pole cannot be very big.
The exact position of z and p is a compromise between:
• Steady state error (we want a large value for z/p)• Steady state error (we want a large value for z/p)• The transient response (we want the pole p placed far from the origin)
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 125
-
Root Locus - Compensators
Phase lead compensator: pzpszssD <
++
= ,)(ps +
pzpszssD >
++
= ,)(Phase lag compensator:ps +
We will see why we call “phase lead” and “phase lag” to these compensators when we study frequency responsethese compensators when we study frequency response
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 126
-
Frequency Response
• We now know how to analyze and design systems via s-domain methods which yield dynamical information
The responses are described b the e ponential modes• The responses are described by the exponential modes– The modes are determined by the poles of the response Laplace
Transform
• We next will look at describing cct performance via frequency response methodsresponse methods– This guides us in specifying the system pole and zero
positions
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 127
-
Sinusoidal Steady-State ResponseC id t bl t f f ti ith
)()cos()( ωω =⇔= AsUtAtu
Consider a stable transfer function with a sinusoidal input:
22)()cos()( ωω
+=⇔=
ssUtAtu
The Laplace Transform of the response has poles
• Where the natural modes lie
–These are in the open left half plane Re(s)
-
Sinusoidal Steady-State Response
• Input φωφωφω cossinsincos)cos()( tAtAtAtu −=+=
ωs• Transform
R T f
2222 sincos)( ωωφ
ωφ
++
+−=
sA
ssAsU
• Response Transform
N
N
psk
psk
psk
jsk
jsksUsGsY
−++
−+
−+
++
−== L
2
2
1
1*
)()()(ωω
• Response SignalNpppjj 21
L21* 21)( tpN
tptptjtj Nekekekekkety +++++= − ωωforced response natural response
• Sinusoidal Steady State Response
44444 344444 2144 344 21responsenatural
21responseforced
)( N ekekekekkety +++++
∞→t
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 129
tjtjSS ekkety
ωω −+= *)( 0
-
Sinusoidal Steady-State Response
• Calculating the SSS response to• Residue calculation
[ ] [ ])()()(li)()(li UGjYjk
)cos()( φω += tAtu
[ ] [ ]
2sincos)(
))((sincos))((lim
)()()(lim)()(lim
ω
ωω
ωφωφωω
ωωφωφω
ωω
js
jsjs
jjAjG
jsjssAjssG
sUsGjssYjsk
→
→→
⎥⎦
⎤⎢⎣
⎡ −=⎥
⎦
⎤⎢⎣
⎡+−
−−=
−=−=
• Signal calculation
))(()(21
21)(
))((
ωφφ ωω jGjj ejGAejAG
jjj
∠+==
⎦⎣⎦⎣
g
jsk
jskLtyss
⎭⎬⎫
⎩⎨⎧
++
−= −
ωω)(
*1
( )Ktkeekeek tjKjtjKj
∠+=
+= −∠−∠
ω
ωω
cos2
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 130
( ))(cos)()( ωφωω jGtjGAtyss ∠++=
-
Sinusoidal Steady-State Response
• Response to• is ))(cos()(
)cos()(ωφωω
φωjGtAjGy
tAtu
ss ∠++=
+=
– Output frequency = input frequency– Output amplitude = input amplitude × |G(jω)|p p p p | (j )|– Output phase = input phase + G(jω)
Th F R f h f f i G( ) i i
∠
• The Frequency Response of the transfer function G(s) is given by its evaluation as a function of a complex variable at s=jω
• We speak of the amplitude response and of the phase response– They cannot independently be varied
» Bode’s relations of analytic function theory
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 131
-
Frequency Response
• Find the steady state output for v1(t)=Acos(ωt+φ)
+sL +
C t th d i t f f ti T( )
+_V1(s) R V2(s)
-
• Compute the s-domain transfer function T(s)– Voltage divider
• Compute the frequency responseRsL
RsT+
=)(
Compute the frequency response⎟⎠⎞
⎜⎝⎛−=∠
+= −
RLjT
LR
RjT ωωω
ω 122
tan)(,)(
)(
• Compute the steady state output( )[ ]RLt
LR
ARtv SS /tancos)(
)( 1222
ωφω −−+=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 132
LR )( 22 ω+
-
Bode Diagrams
Bode Diagram
5
0
• Log-log plot of mag(T), log-linear plot of arg(T) versus ω
)M
agni
tude
(dB)
-15
-10
-5ni
tude
(dB
)M
-25
-20
0
Mag
nse
(deg
)
-45
e (d
eg)
Phas
4 5 6-90
Phas
e
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 133
Frequency (rad/sec)
104
105
106
-
Frequency Response
)cos()( φω += tAtu )(sG ))(cos()( ωφωω jGtAjGyss ∠++=
Stable Transfer Function
• After a transient, the output settles to a sinusoid with an amplitude magnified by and phase shifted by .)( ωjG )( ωjG∠
• Since all signals can be represented by sinusoids (Fourier series and transform), the quantities and are extremely important
)( ωjG )( ωjG∠extremely important.
• Bode developed methods for quickly finding and for a given and for using them in control design
)( ωjG )( ωjG∠)(sG
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 134
for a given and for using them in control design.)(sG
-
Frequency Response
• Find the steady state output for v1(t)=Acos(ωt+φ)
+sL +
C t th d i t f f ti T( )
+_V1(s) R V2(s)
-
• Compute the s-domain transfer function T(s)– Voltage divider
• Compute the frequency responseRsL
RsT+
=)(
Compute the frequency response⎟⎠⎞
⎜⎝⎛−=∠
+= −
RLjT
LR
RjT ωωω
ω 122
tan)(,)(
)(
• Compute the steady state output( )[ ]RLt
LR
ARtv SS /tancos)(
)( 1222
ωφω −−+=
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 135
LR )( 22 ω+
-
Frequency Response - Bode Diagrams
Bode Diagram
5
0
• Log-log plot of mag(T), log-linear plot of arg(T) versus ω)
Mag
nitu
de (d
B)
-15
-10
-5
nitu
de (d
B)
M
-25
-20
0
Mag
nse
(deg
)
-45
e (d
eg)
Phas
4 5 6-90
Phas
e
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 136
Frequency (rad/sec)
104
105
106
[ ] [ ] Hzffrad === ,2sec,/ πωω
-
Bode Diagrams)())(()())(()())(()(
21
21
n
m
pspspszszszsKsG
−−−−−−
=L
L
( ) ( )[ ]pnppzmzzKiz
mzz
errrKsG φφφφφφφ +++−++++= LLL 212121)(
The magnitude and phase of G(s) when s=jω is given by:
pn
pp errrKsG
L21)(
zm
zz rrrKjG ω = L21)(
Nonlinear in the magnitudes
( ) ( )pnppzmzzKp
npp
jG
rrrKjG
φφφφφφφω
ω
+++−++++=∠ LL
L
2121
21
)(
,)(
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 137
( ) ( )Linear in the phases
-
Bode DiagramsWhy do we express in decibels?)( jG
)(log20)( ωω jGjG dB =
Why do we express in decibels?)( ωjG
?)()(21
21 =⇒= dBpn
pp
zm
zz
jGrrrrrrKjG ωω
L
L
( ) ( )pnppzmzmz rrrrrrKsG log20log20log20log20log20log20log20)(log20 211 +++−++++= LLBy properties of the logarithm we can write:
21 n
Linear in the magnitudes (dB)
The magnitude and phase of G(s) when s=jω is given by:
( ) ( )( ) ( )pppzzzK
dBp
ndBp
dBp
dBz
mdBz
dBz
dBdB
sG
rrrrrrKsG
φφφφφφφ +++++++∠
+++−++++= LL 2121
)(
)(
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 138
( ) ( )pnppmsG φφφφφφφ +++−++++=∠ LL 2121)(Linear in the phases
-
Bode DiagramsWhy do we use a logarithmic scale? Let’s have a look at our example:Why do we use a logarithmic scale? Let s have a look at our example:
222
1)(
)()(⎞⎛
==⇒+
=LR
RjTRL
RsT ω222
1)(
⎟⎠⎞
⎜⎝⎛+
++
RLLRRsL ωω
Expressing the magnitude in dB:
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+−=⎟
⎠⎞
⎜⎝⎛+−=
22
1log101log201log20)(RL
RLjT dB
ωωω⎦⎣
Asymptotic behavior:
0)(:0 →→ jT ωω
( ) ωωωωω log20log20/log20/
log20)(: −=−=⎟⎠⎞
⎜⎝⎛−→∞→
dBdB L
RLRLR
jT
0)(:0 →→ dBjT ωω
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 139
LINEAR FUNCTION in logω!!! We plot as a function of logω.dBjG )( ω
-
Bode Diagrams
A cutoff frequency occurs when the gain is reduced from its i b d l b f t /
Decade: Any frequency range whose end points have a 10:1 ratio
maximum passband value by a factor :2/1
dB3log202log20log201log20 −≈−=⎟⎞⎜⎛ TTT dB3log202log20log202log20 ≈=⎟
⎠⎜⎝ MAXMAXMAX
TTT
Bandwith: frequency range spanned by the gain passband
Let’s have a look at our example:
⎨⎧ ==
⇒=1)(01)(
ωωω
jTjT
⎩⎨ ==
⇒
⎟⎠⎞
⎜⎝⎛+
=2/1)(/
1)(
2 ωωωω
jTLR
RL
jT
Thi i l fil !!! P b d i C ff f
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 140
This is a low-pass filter!!! Passband gain=1, Cutoff frequency=R/LThe Bandwith is R/L!
-
General Transfer Function (Bode Form)q⎤⎡ ⎞⎛
2
( ) ( ) ( )nn
nmo
jjjjKjG⎥⎥⎦
⎤
⎢⎢⎣
⎡++⎟⎟
⎠
⎞⎜⎜⎝
⎛+= 121
ωωζ
ωωωτωω
The magnitude (dB) (phase) is the sum of the magnitudes (dB)(phases) of each one of the terms. We learn how to plot each term, we learn how to plot the whole magnitude and phase Bode Plot.
Classes of terms:
1- ( ) KjG =ω1-2-
3
( ) oKjG =ω( ) ( )mjjG ωω =
3-
4-
( ) ( )njjG 1+= ωτω
( )q
jjG ⎥⎤
⎢⎡
⎟⎞
⎜⎛
2ωζω
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 141
( )nn
jjjG⎥⎥⎦⎢
⎢⎣
++⎟⎟⎠
⎞⎜⎜⎝
⎛= 12
ωωζ
ωωω
-
General Transfer Function: DC gain
( ) oKjG =ω
( ) = log20d KjωGMagnitude and Phase: ( )
( )⎩⎨⎧
=∠000
log20
o
odB
KK
jG
KjωG
if if
πω
g
( )35
40
160
180
200
⎩
-
General Transfer Function: Poles/zeros at origin
( ) ( )mjjG ωω =
( ) log20 ωmjωG ⋅=Magnitude and Phase: ( )
( )2
log20
πω
ω
mjG
mjωG dB
=∠
⋅=Magnitude and Phase:
( )sGm 1,1 =−=
10
20
-20
0
2( ) ssGm ,1
( ) 01 =dBG
-10
0
nitu
de (d
B)
-60
-40
ase
(deg
)
( )dB
-30
-20Mag
n
-100
-80
Pha
decdBm 20⋅
Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 143
10-1 100 101 102-40
Frequency (rad/sec)10-1 100 101 102
-120
Frequency (rad/sec)
-
General Transfer Function: Real poles/zeros
( ) ( )njjG 1+= ωτωMagnitude and Phase:
( ) ( )τω 22 1log10 +⋅= njωG dB
Magnitude and Phase:
( ) ( )ωτω 1tan−=�