Acciones básicas de control y técnicas de compensación
Carlos Alberto Cruz [email protected]
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
16.06 Principles of Automatic Control Lecture 10
PID Control
A�common�way�to�design�a�control�system�is�to�use�PID�control.�
PID =�proportional-integral-derivative�
Will�consider�each�in�turn,�using�an�example�transfer�function�
A Gpsq “
s2 ` a1s ` a2
Proportional (P) control
In�proportional�control,�the�control�aw�is�simply�a�gain,�to�that�u is�proportional�to�e:�
-+ k
pG(s) y
r e u
u “ kpe
For�our�example,�the�characteristic�equation�is�
0 “1 ` kpGpsqkpA“1 `
s2 ` a1s ` a2
ñ 0 “s 2 ` a1s ` a2 ` kpA
1� �
16.06 Principles of Automatic Control Lecture 10
PID Control
A�common�way�to�design�a�control�system�is�to�use�PID�control.�
PID =�proportional-integral-derivative�
Will�consider�each�in�turn,�using�an�example�transfer�function�
A Gpsq “
s2 ` a1s ` a2
Proportional (P) control
In�proportional�control,�the�control�aw�is�simply�a�gain,�to�that�u is�proportional�to�e:�
-+ k
pG(s) y
r e u
u “ kpe
For�our�example,�the�characteristic�equation�is�
0 “1 ` kpGpsqkpA“1 `
s2 ` a1s ` a2
ñ 0 “s 2 ` a1s ` a2 ` kpA
1� �
16.06 Principles of Automatic Control Lecture 10
PID Control
A�common�way�to�design�a�control�system�is�to�use�PID�control.�
PID =�proportional-integral-derivative�
Will�consider�each�in�turn,�using�an�example�transfer�function�
A Gpsq “
s2 ` a1s ` a2
Proportional (P) control
In�proportional�control,�the�control�aw�is�simply�a�gain,�to�that�u is�proportional�to�e:�
-+ k
pG(s) y
r e u
u “ kpe
For�our�example,�the�characteristic�equation�is�
0 “1 ` kpGpsqkpA“1 `
s2 ` a1s ` a2
ñ 0 “s 2 ` a1s ` a2 ` kpA
1� �
Ecuación característica
The�resulting�natural�frequency�is� �
a !n “ a2 ` kpA
So�in�the�example,�increasing�kp increases�the�natural�frequency,�but�reduces�the�damping� �
ratio.� �
Plot�of�pole�location�vs�kp:� �
Im(s)
Re(s)increasing k
p
-a1/2
open looppole location
closed-looppole location
Derivative (D) control
To�add�damping�to�a�system,�it�is�often�useful�to�add�a�derivative�term�to�the�control,�
uptq “kpeptq ` kDe9ptq or�
vpsq “kpEpsq ` kDsEpsq“pkp ` kDsqEpsq“KpsqEpsq
What�is�the�characteristic�equation?�
0 “1 ` KpsqGpsq pkp ` kDsqA“1 ` s2 ` a1s ` a2
0 “s 2 ` pa1 ` kDAqs ` pa2 ` kpq
2� �
La frecuencia natural resulta:
Al incrementar kp, se incrementa la frecuencia natural, pero se reduce el amortiguamiento.
Acciones básicas de control
The�resulting�natural�frequency�is� �
a !n “ a2 ` kpA
So�in�the�example,�increasing�kp increases�the�natural�frequency,�but�reduces�the�damping� �
ratio.� �
Plot�of�pole�location�vs�kp:� �
Im(s)
Re(s)increasing k
p
-a1/2
open looppole location
closed-looppole location
Derivative (D) control
To�add�damping�to�a�system,�it�is�often�useful�to�add�a�derivative�term�to�the�control,�
uptq “kpeptq ` kDe9ptq or�
vpsq “kpEpsq ` kDsEpsq“pkp ` kDsqEpsq“KpsqEpsq
What�is�the�characteristic�equation?�
0 “1 ` KpsqGpsq pkp ` kDsqA“1 ` s2 ` a1s ` a2
0 “s 2 ` pa1 ` kDAqs ` pa2 ` kpq
2� �
Acciones básicas de control
El término derivativo se suma al término proporcional para (idealmente) tomar en cuenta los valores futuros del error.
The�resulting�natural�frequency�is� �
a !n “ a2 ` kpA
So�in�the�example,�increasing�kp increases�the�natural�frequency,�but�reduces�the�damping� �
ratio.� �
Plot�of�pole�location�vs�kp:� �
Im(s)
Re(s)increasing k
p
-a1/2
open looppole location
closed-looppole location
Derivative (D) control
To�add�damping�to�a�system,�it�is�often�useful�to�add�a�derivative�term�to�the�control,�
uptq “kpeptq ` kDe9ptq or�
vpsq “kpEpsq ` kDsEpsq“pkp ` kDsqEpsq“KpsqEpsq
What�is�the�characteristic�equation?�
0 “1 ` KpsqGpsq pkp ` kDsqA“1 ` s2 ` a1s ` a2
0 “s 2 ` pa1 ` kDAqs ` pa2 ` kpq
2� �
Acciones básicas de control
The�resulting�natural�frequency�is� �
a !n “ a2 ` kpA
So�in�the�example,�increasing�kp increases�the�natural�frequency,�but�reduces�the�damping� �
ratio.� �
Plot�of�pole�location�vs�kp:� �
Im(s)
Re(s)increasing k
p
-a1/2
open looppole location
closed-looppole location
Derivative (D) control
To�add�damping�to�a�system,�it�is�often�useful�to�add�a�derivative�term�to�the�control,�
uptq “kpeptq ` kDe9ptq or�
vpsq “kpEpsq ` kDsEpsq“pkp ` kDsqEpsq“KpsqEpsq
What�is�the�characteristic�equation?�
0 “1 ` KpsqGpsq pkp ` kDsqA“1 ` s2 ` a1s ` a2
0 “s 2 ` pa1 ` kDAqs ` pa2 ` kpq
2� �
The�resulting�natural�frequency�is� �
a !n “ a2 ` kpA
So�in�the�example,�increasing�kp increases�the�natural�frequency,�but�reduces�the�damping� �
ratio.� �
Plot�of�pole�location�vs�kp:� �
Im(s)
Re(s)increasing k
p
-a1/2
open looppole location
closed-looppole location
Derivative (D) control
To�add�damping�to�a�system,�it�is�often�useful�to�add�a�derivative�term�to�the�control,�
uptq “kpeptq ` kDe9ptq or�
vpsq “kpEpsq ` kDsEpsq“pkp ` kDsqEpsq“KpsqEpsq
What�is�the�characteristic�equation?�
0 “1 ` KpsqGpsq pkp ` kDsqA“1 ` s2 ` a1s ` a2
0 “s 2 ` pa1 ` kDAqs ` pa2 ` kpq
2� �
Por lo que la ecuacióncaracterística es:
Observamos que al incrementar kD se incrementa el factor de amortiguamiento sin modificar el valor de la frecuencia natural del sistema.
Por ello, se dice que la acción derivativa “introduce” amortiguamiento al sistema.
Acciones básicas de controlSo�increasing�kD increases�the�damping�ratio�without�changing�the�natural�frequency,� for this example.� �
For�kpfixed,�kDvarying,�plot�of�closed-loop�pole�location�is:� �
Im(s)
Re(s)
pole position for kD
= 0
pole position for kD
> 0
increasing kD
NB:�For�other�Gpsq,�results�may�vary.� �
Sometimes,�it’s�better�to�place�derivative�feedback�in�the�feedback�path:� �
-+ k
p
kD(s)
yr u+ G(s)
Why?�We�get�the�same�pole�locations,�but�no�additional�zeros�to�cause�additional�overshoot.�
Another�way�to�think�about�this�is�that�we�want�the�derivative�effect�on�y,�because�that�
adds�damping,�but�we�don’t�want�to�differentiate�the�reference.�
Integral (I) control
Especially�if�the�plant�is�a�type�0 system,�we�may�want�to�add�integrator�to�controller�to�
drive�steady-state�error�to�zero:�
`kIV psq “ pkp `kDsq EpsqÓ Ó
s Ó
P I D
3� �
Manteniendo fijo kP y variando kD, la ubicación de los polos en lazo cerrado resulta:
Acciones básicas de control
En algunas ocasiones es preferible derivar únicamente la salida, no el error:
So�increasing�kD increases�the�damping�ratio�without�changing�the�natural�frequency,� for this example.� �
For�kpfixed,�kDvarying,�plot�of�closed-loop�pole�location�is:� �
Im(s)
Re(s)
pole position for kD
= 0
pole position for kD
> 0
increasing kD
NB:�For�other�Gpsq,�results�may�vary.� �
Sometimes,�it’s�better�to�place�derivative�feedback�in�the�feedback�path:� �
-+ k
p
kD(s)
yr u+ G(s)
Why?�We�get�the�same�pole�locations,�but�no�additional�zeros�to�cause�additional�overshoot.�
Another�way�to�think�about�this�is�that�we�want�the�derivative�effect�on�y,�because�that�
adds�damping,�but�we�don’t�want�to�differentiate�the�reference.�
Integral (I) control
Especially�if�the�plant�is�a�type�0 system,�we�may�want�to�add�integrator�to�controller�to�
drive�steady-state�error�to�zero:�
`kIV psq “ pkp `kDsq EpsqÓ Ó
s Ó
P I D
3� �
¿Por qué?
Acciones básicas de control
En algunas ocasiones es preferible derivar únicamente la salida, no el error:
So�increasing�kD increases�the�damping�ratio�without�changing�the�natural�frequency,� for this example.� �
For�kpfixed,�kDvarying,�plot�of�closed-loop�pole�location�is:� �
Im(s)
Re(s)
pole position for kD
= 0
pole position for kD
> 0
increasing kD
NB:�For�other�Gpsq,�results�may�vary.� �
Sometimes,�it’s�better�to�place�derivative�feedback�in�the�feedback�path:� �
-+ k
p
kD(s)
yr u+ G(s)
Why?�We�get�the�same�pole�locations,�but�no�additional�zeros�to�cause�additional�overshoot.�
Another�way�to�think�about�this�is�that�we�want�the�derivative�effect�on�y,�because�that�
adds�damping,�but�we�don’t�want�to�differentiate�the�reference.�
Integral (I) control
Especially�if�the�plant�is�a�type�0 system,�we�may�want�to�add�integrator�to�controller�to�
drive�steady-state�error�to�zero:�
`kIV psq “ pkp `kDsq EpsqÓ Ó
s Ó
P I D
3� �
¿Por qué? Porque se obtiene la misma ubicación de los polos, pero se evita el cero, evitando así el incremento en el máximo pico.
Acciones básicas de control
El término integral adiciona un término “mientras” exista error. Este término se emplea en una clase de sistemas para llevar a cero el error de estado estacionario.
So�increasing�kD increases�the�damping�ratio�without�changing�the�natural�frequency,� for this example.� �
For�kpfixed,�kDvarying,�plot�of�closed-loop�pole�location�is:� �
Im(s)
Re(s)
pole position for kD
= 0
pole position for kD
> 0
increasing kD
NB:�For�other�Gpsq,�results�may�vary.� �
Sometimes,�it’s�better�to�place�derivative�feedback�in�the�feedback�path:� �
-+ k
p
kD(s)
yr u+ G(s)
Why?�We�get�the�same�pole�locations,�but�no�additional�zeros�to�cause�additional�overshoot.�
Another�way�to�think�about�this�is�that�we�want�the�derivative�effect�on�y,�because�that�
adds�damping,�but�we�don’t�want�to�differentiate�the�reference.�
Integral (I) control
Especially�if�the�plant�is�a�type�0 system,�we�may�want�to�add�integrator�to�controller�to�
drive�steady-state�error�to�zero:�
`kIV psq “ pkp `kDsq EpsqÓ Ó
s Ó
P I D
3� �
Acciones básicas de control
Considere al sistemaExample:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
Suponga que se desea un sistema que:
1. Tenga tiempo de elevación menor a un segundo.2. Tenga un máximo pico menor al 5 %.3. Tenga error cero en estado estacionario a la entrada escalón
Acciones básicas de controlExample:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
1. Tenga tiempo de elevación menor a un segundo.The�rise�time�is�approximately� �
1.8 tr «
!n
The�rise time is�a�bit�faster�for�systems�with�less�damping,�a�bit�longer�for�systems�with�
more�damping,�and�sensitive�to�additional�poles�and�zeros.�
The�settling time can�be�approximated�via:�
´⇣!ntse «0.01
Ñ ts « 4.6 ⇣!n
Note�that,�in�reality,�settling�time�varies�discontinuously�with�⇣,�since�as�damping�increases,�
a�peak�may�decrease�from�just�over�1.01�to�just�under�1.01,�so�tsis�drastically�reduced.�
Desired pole locations
Given�specifications�on�tr,�Mp,�and�ts,�where�should�poles�be?�
tr §a
Ñ 1.8 §a !n
Ñ !n •1.8 “ !nmin a
Im(s)
Re(s)
x
x
ωn min
Likewise,�to�keep�Mpless�than�a�fixed�value,�must�have�⇣ • ⇣pMpq:�
3� �
Example:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
Example:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
wn=2
kp=3
Example:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
2. Tenga un máximo pico menor al 5 %.El amortiguamiento necesario para un máximo pico del 5% es:
Por lo que es necesario adicionar …
Mp “ peak�overshoot�
tr “ rise�time�(10%�to�90%)�
ts “ settling�time�(1%)�
tp “ time�of�peak�
Each�of�the�above�parameters�may�be�important�in�the�design�of�the�control�system.� For�
example,� the�designer�of�a�hard�disk�drive�may� specify�a�maximum�settling� time�of� the�
response�of�the�read/write�head�to�a�commanded�change�in�position.�
Peak�overshoot�is�important,�both�because�it�is�a�measure�(to�a�degree)�of�stability,�and�for�
practical�reasons,�overshoot�should�be�minimized�(think�of�an�elevator!).�
Rise�time�tr(and�to�a�lesser�extent�peak�time�tp)�is�a�measure�of�the�speed�of�response�of�the�
system.�Often,�a�maximum�trwill�be�specified.�
We�can�connect�⇣and�!nto�Mp, tp, tr,�with�two�important�caveats:� first,�some�of�the�rela-
tionships�are�approximate.�Second,�additional�poles�and�zeros�will�change�the�results,�so�all�
of�the�results�should�be�viewed�as�guidelines.�
The�step�response�of�Hpsq is�
˜ ¸1´⇣!nthsptq “ 1 ´ e cosp!dtq ` a sinp!dtq
1 ´ ⇣2
Using�elementary�calculus,�we�can�find�tp and�Mp(see�text):�
⇡ tp “
!d ? ́ ⇡⇣
Mp “e 1´⇣2
´⇡ tan ⇥ “e
where�⇥ “ sin ́ 1 ⇣.�
Typical�values:�
⇣ Mp
0.5� 0.16�
0.7� 0.05�
2� �
Example:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
Acciones básicas de control
Sí, amortiguamiento, así que al considerar un controlador PD la ecuación característica resulta:
Example:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
D
Mientras el polinomio deseado es:
Example:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �
Example:
1 Gpsq “
s2 ` s ` 1 Suppose�we�want�a�system�that�
1.� Has�rise�time�above�tr “ 1s�
2.� Has�peak�overshoot�of�Mp “ 0.05
3.� Has�zero�steady-state�error�to�step�command�
Let’s�do�one�piece�at�a�time:�
-+ k
pG(s)
Characteristic�equation�is�
0 “ s 2 ` s ` 1 ` kp
So�can�only�change�!n (and�indirectly,�⇣)�with�kp.� for�tr “ 1,�need�
1.8 1 “ ñ !n « 1.8
!n
So�let’s�take�kp “ 2 for�simplicity.�Then�
kpG 3 T “ “
1 ` kpG s2 ` s ` 4 ñ ⇣ “0.25, Low�
To�get�Mp “ 5%,�need�⇣ “ 0.7.�So�add�derivative�control.�Characteristic�Equation�is�
0 “ s 2 ` p1 ` kpqs ` 1 ` kp
The�desired�polynomial�is�
0 “ s 2 ` 2.8s ` 4
So�take�kD “ 1.8.�
4� �Pero, recordemos que un cero (de fase mínima) incrementa el máximo pico que al considerar el controlador PD será de 16 % y no del 5%
Acciones básicas de control
Por lo que usaremos la estructura:
If�PD�control�is�in�forward�loop,�
1.8s ` 3 T “
s2 ` 2.8s ` 4 and�the�peak�overshoot�will�be�16%,�not�5%.�So�instead,�use�control�structure�
-+ 3
1.8s
yr u+ G(s)
(*)
p˚q “ ”mirror�loop�feedback” With�this�structure,�we�have:�
tr “1.06s Mp “4.6% ess “0.25
So�let’s�add�integral�control:�
-+ k
v+k
I/s
kDs
yr+ G(s)
Take�kI “ 0.25 (trust�me!)�
Then�
Response�sort of meets�specs:�
T “ 3s ` 0.25
s3 ` 2.8s2 ` 4s ` 0.25
5� �
¿Cuánto vale el error en estado estacionario?
If�PD�control�is�in�forward�loop,�
1.8s ` 3 T “
s2 ` 2.8s ` 4 and�the�peak�overshoot�will�be�16%,�not�5%.�So�instead,�use�control�structure�
-+ 3
1.8s
yr u+ G(s)
(*)
p˚q “ ”mirror�loop�feedback” With�this�structure,�we�have:�
tr “1.06s Mp “4.6% ess “0.25
So�let’s�add�integral�control:�
-+ k
v+k
I/s
kDs
yr+ G(s)
Take�kI “ 0.25 (trust�me!)�
Then�
Response�sort of meets�specs:�
T “ 3s ` 0.25
s3 ` 2.8s2 ` 4s ` 0.25
5� �
Acciones básicas de control
If�PD�control�is�in�forward�loop,�
1.8s ` 3 T “
s2 ` 2.8s ` 4 and�the�peak�overshoot�will�be�16%,�not�5%.�So�instead,�use�control�structure�
-+ 3
1.8s
yr u+ G(s)
(*)
p˚q “ ”mirror�loop�feedback” With�this�structure,�we�have:�
tr “1.06s Mp “4.6% ess “0.25
So�let’s�add�integral�control:�
-+ k
v+k
I/s
kDs
yr+ G(s)
Take�kI “ 0.25 (trust�me!)�
Then�
Response�sort of meets�specs:�
T “ 3s ` 0.25
s3 ` 2.8s2 ` 4s ` 0.25
5� �
3. Error cero en estado estacionario
Como el error en estado estacionario NO es cero, debemos adicionar el término integral.
Supongamos kI = 0.25, lo que resulta en la función de transferencia en lazo cerrado:
If�PD�control�is�in�forward�loop,�
1.8s ` 3 T “
s2 ` 2.8s ` 4 and�the�peak�overshoot�will�be�16%,�not�5%.�So�instead,�use�control�structure�
-+ 3
1.8s
yr u+ G(s)
(*)
p˚q “ ”mirror�loop�feedback” With�this�structure,�we�have:�
tr “1.06s Mp “4.6% ess “0.25
So�let’s�add�integral�control:�
-+ k
v+k
I/s
kDs
yr+ G(s)
Take�kI “ 0.25 (trust�me!)�
Then�
Response�sort of meets�specs:�
T “ 3s ` 0.25
s3 ` 2.8s2 ` 4s ` 0.25
5� �
0 10 20 30 400
0.2
0.4
0.6
0.8
1
ty
5%
The�response�has�a�long�tail,�due�to�slow�pole�–�poles�are�at:�
s “ ´ 1.37 ˘ 1.40j s “ ´ 0.065
Ò slow�pole�causes�long�tail�
6� �
El polo real es lento, lo que provoca que el tiempo de asentamiento sea muy alto.
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Ziegler-Nichols, ejemplo
UtilicelasreglasdeZiegler-Nychols parasintonizarelcontroladorPIDdelsiguientesistemadecontrol
÷ø
öçè
æ++ ss
K di
p tt11-
+ )2)(1(1
++ sss)(sC)(sR
p
p
KsssK
sRsC
+++=
23)()(
23
Solución:Comoelsistematieneunintegrador,seusaelmétododos.Seempleasoloeltérminoproporcional,conloqueseobtienelafuncióndetransferenciadelazocerrado
Acciones básicas de control
Ziegler-Nichols, ejemplo
023 23 =+++ pKsss
Delaecuacióncaracterísticaseobtieneelvalordelagananciaqueproduceoscilacionessostenidas
0)(2)(3)( 23 =+++ pKjjj www
023 23 =++-- pKjj www
6=pK 2=w
Elvalordegananciaeslagananciacrítica 6=crK
Mientrasqueelperíodocríticoseobtienede 2=w
4428.42 ==wp
crP
Acciones básicas de controlZiegler-Nichols, ejemplo
ConloquelosparámetrosdelcontroladorPIDresultan
2214.25.0 == cri Pt
55536036.0125.0 == crd Pt
6.36.0 == crp KK
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control
Acciones básicas de control y técnicas de compensación
Carlos Alberto Cruz [email protected]
Compensación
Compensación significa la modificación de la dinámica del sistemapara que este cumpla con determinadas especificaciones.
Esto puede hacerse modificando al sistema (lo cual generalmente noes posible) o manipulando adecuadamente la entrada al sistema.
El diseño de compensadores comúnmente se realiza, al adicionarpolos y ceros al sistema en lazo abierto, forzando que las raíces delsistema modificado correspondan con los polos deseados del sistemaen lazo cerrado
Compensación5Control synthesis using Bode plot
F(s)
-1
6
Question today: How can we design the controller F(s) to satisfy specifications on the closed-loop system, given a Bode plot of the initial loop-gain
¿Cómo podemos diseñar el controlador F(s) que satisfaga ciertas especificaciones del sistema en lazo cerrado, conocido el diagrama de Bode del sistema a controlar?
Compensación
Las especificaciones típicas del comportamiento de un sistema decontrol están asociadas a la respuesta subamortiguada de un sistemade segundo orden.
1. Velocidad en la respuesta, asociada al tiempo de elevación y enconsecuencia a la frecuencia natural del sistema.
2. Máximo pico, asociado al factor de amortiguamiento del sistema ytambién al margen de fase:
3. Error en estado estacionario, determinado por la ganancia del lazoen s=0
9Control synthesis using Bode plot
Typical specifications: The closed-loop system is typicallyspecified in terms of bandwidth (relates to speed), resonancepeak (relates to overshoot) and stationary control error
Bandwidth: Related to crossover frequency
Resonance peak: Related to phase margin
Stationary control error: determined by loop-gain in Z=0, e.g.
Compensación, cascada 11Control synthesis using Bode plot
What does the Bode plot look like for two systems in series, given the Bode plot of the two separate systems
G1 G2
Additive!
(this is the basis for our method to draw Bode plot approximately)
¿Cómo se obtiene el diagrama de Bode de dos sistemas en serie, si se conocen los diagramas de Bode de los sistemas separados?
Compensación, cascada12Control synthesis using Bode plot
Example: Series connection with (stable) zero
G1 G2
Conexión en serie con un cero (estable)
Compensación, cascada13Control synthesis using Bode plot
Does not influence gain significantly up to roughly Z=a, then followed by an increase of 20dB per decade
The phase is increased in all frequencies and goes towards an increase of 90º
Este cero no modifica la ganancia, hasta que w=a, a partir de ahí tiene una ganancia positiva de 20dB por década.
La fase se incrementa para todas las frecuencias de 0 a 90º
Compensación, cascada14Control synthesis using Bode plot
Example: Series connection with (stable) pole
G1 G2
Conexión en serie con un polo (estable)
Compensación, cascada15Control synthesis using Bode plot
Can be used to give extra gain up to roughly Z=a (and then the gain drops by 20dB per decade)
The phase is decreased in all frequencies and goes towards a decrease of 90º
Este polo no modifica la ganancia, hasta que w=a, a partir de ahí tiene una ganancia negativa de 20dB por década.
La fase decrece para todas las frecuencias de 0 a -90º
Compensación, cascada16Control synthesis using Bode plot
Example: Series connection with pole-zero, 0<b<a
G1 G2
Conexión en serie polo-cero, con 0<b<a
Compensación, cascada17Control synthesis using Bode plot
Gives extra gain from Z=b and above, but does increase it significantly after Z=a
Phase increases everywhere, largest increase between a and b
No proporciona ganancia entre 0 y b. Entre b y a hay una ganancia con pendiente positiva, después de a, la ganancia es constante.
Entre a y b el crecimiento en fase es considerable
Compensación, cascada 18Control synthesis using Bode plot
Example: Series connection with pole-zero, 0<a<b
G1 G2
Conexión en serie polo-cero, con 0<a<b
Compensación, cascada 19Control synthesis using Bode plot
Gives extra gain from Z=b and below, but does increase it significantly more before Z=a
Phase decreases everywhere, largest decrease between a and b
No proporciona ganancia entre 0 y b. Entre b y a hay una ganancia con pendiente negativa, después de a, la ganancia es constante.
Entre a y b el decrecimiento en fase es considerable
Acciones básicas de control y técnicas de compensación
Carlos Alberto Cruz [email protected]
Compensación, cascada 18Control synthesis using Bode plot
Example: Series connection with pole-zero, 0<a<b
G1 G2
Conexión en serie polo-cero, con 0<a<b
Compensación, cascada 19Control synthesis using Bode plot
Gives extra gain from Z=b and below, but does increase it significantly more before Z=a
Phase decreases everywhere, largest decrease between a and b
No proporciona ganancia entre 0 y b. Entre b y a hay una ganancia con pendiente negativa, después de a, la ganancia es constante.
Entre a y b el decrecimiento en fase es considerable
Compensación
Consideremos t1 = 1/b y t2 = 1/a, con lo que la función de transferencia del compensador resulta:
Compensación
Compensación
Compensación
La forma general de un compensador de adelanto es:
Cuyas raíces en el plano complejo resultan
aa
a
a
Compensación
Compensación
El problema de diseño consiste en encontrar losparámetros del compensador, tal que se cumpla conlas especificaciones de diseño, las cuales típicamenteestan dadas por el máximo sobrepico y el tiempo deestablecimiento. O bien, por el factor deamortiguamiento y la frecuencia natural.
Compensación
Diseño de compensadores de adelanto mediante latécnica de respuesta en frecuencia.
1. Determinar la ganancia K=Kca que satisfaga losrequisitos de error en estado estacionario
2. Dibujar el Diagrama de Bode de G1(jw)=KG(jw). Laganancia ajustada pero no compensada. Evalúe elmargen de fase.
3. Determine el ángulo de deficiencia. Adicione de 5a 12 grados al ángulo de deficiencia, ya que elcompensador cambia la frecuencia de corte a laderecha y decrementa el margen de fase.
Compensación
4. Determine el factor de atenuación a dado por
Compensación
5. Determine la frecuencia a la que ocurre fmax
las frecuencias de extremos del compensador resultan:
Polo del compensador, wp=1/T
Cero del compensador. wc=1/aT
6. Usando el valor de K obtenido en el paso 1 y el de a en elpaso 4 calcule Kc.
7. Verifique que el margen de ganancia es satisfactorio, de locontrario, modifique la ubicación de los polos y de losceros para que así ocurra.
CompensaciónEjemplo.
CompensaciónGrafiquemos ahora el diagrama de Bode
Compensación
La fase adicional que debe proporcionar el sistema es de 33grados, si consideramos 5 grados adicionales, fm=38grados, por lo que a=0.24
Si a=0.24
Compensación
Compensación
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