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state space control - 21.11

XXXXXXXXXX Example of learning parameter estimator to obtain desired responses.

21.4SUPPLEMENTAL OBSERVERS

•In many cases some state variables can be measured directly, but others cannot. In these cases it is more efficient to only estimate the unmeasurable variable values [Eronini].

•This can be

W = KX

–1

X =

= M

–1

M =

(Note: the inverse must exist)

---

----

---

= MX

----

= MAX + MBU = MAM

–1

+ MBU

----

= ----MX = M( AX + BU)

----

The matrices can be partitioned to isolate the unmeasurable state variables. Note that the Eigenvalues of Hw must be relatively small.


MAM–1	=	… …
		Hy Hw
d ˆ
d ˆ			ˆ
----W =	HyY + HwW + BwU
dt

…

MB =

21.5REGULATED CONTROL WITH OBSERVERS

•For a regulated system the control output ’u’ can be estimated using the desired

state space control - 21.12

system state.

u = Kx = Ky

• The system state is often difficult to measure directly, so an estimator can be

used. .

u = –Kx = –Kx

• A regulator can use an estimate of the system state to

			L
			CL
+	-	d	ˆ	ˆ	u	d	x	y
+	-	----x		ˆ	u	----x	x	y
		dt	1	x	B	dt	1	C
			--	–K	B		--	C
		+	s				s
+		+
+

A					A

system

estimator and regulator

state space control - 21.13

For a regulator assume the state variables go to zero, so....

u = –Kx

A closed loop estimator

----x =

Ax + Bu + L( y – y)

y = Cx

The state model for the system is,

Ax + Bu

----x =

y = Cx

These can be combined to obtain,

----x =

Ax + Bu + L( y – y) =

x( A – BK – LC) + x( LC)

Ax + Bu =

----x =

x( A) + x( –BK)

Resulting in the matrix,

–BK

----

LC A – BK – LC

state space control - 21.14

An error variable can be introduced, and written in matrix form,

x = x – x

I 0

= Tx

≈ x˜

T =

I 0

= T–1

I –I

An error variable can be introduced, and written in matrix form,

= TA

T–1 ≈

The previous matrix can be revised

–BK

I 0

Acl

I –I

LC A – BK – LC

I –I

I 0

A – BK

Acl

I –I

A – BK – A + BK + LC

A – BK

Acl

A – LC

The new poles for this system can be written,

= ( A – BK) ( A – LC)

– BK( 0) = A2 – ALC – ABK + BKLC

sI –

•The estimators and regulators can be designed separately and then combined. This is the separation principle.

•The simplified block diagram for the system is,

d	ˆ	ˆ		d
----x		ˆ		----x
dt		x	+	dt	x
	1	–BK	+	1	x
	--	–BK		--
+	s			s
+				+
+				+
A – BK – LC				A
				A
		LC

state space control - 21.15

• The compensator,

The previous estimator equations can be combined with a compensator

u = –Kx

d	ˆ	ˆ	ˆ	ˆ	ˆ	ˆ	ˆ
----x = Ax + Bu + L( y – y)				= Ax – BKx + L( y – Cx)			= ( A – BK – LC) x + ( L) y
dt

These can be used to define a new compenator state,

x = xc

u = –Cc xc

----xc = Acxc + Bcy dt

where,

Ac = A – BK – LC

Bc = L

Cc = K

• This can be converted to a transfer function for the regulator.

These can be used to define a new compenator state, sxc = Acxc + Bcy

xc( sI – Ac) = Bcy

		xc	Bc
		---- =	---------------
		y	sI – Ac
u = –Cc xc
		u
		---- = –Cc
		xc	–CcBc
u	=	u xc	–CcBc
--	=	---- ----	= ---------------
y		xc y	sI – Ac

• The system can also be set up for a compensator by assuming the setpoint input is not zero,

state space control - 21.16

First a system error is defined.

e = r – y

The previously defined compensator can be used here,

e	= r – y
u = –Cc xc
			u
			---- = –Cc
d			xc
d			=	A		x		+ B		e
----x		c	=	A	c	x	c	+ B	c	e
dt		c	xc		c		c	Bc	c
			xc	=				Bc
			----	=		---------------
			e			sI – Ac
u	=		u xc			=		–CcBc
--	=		---- ----			=		---------------
e			xc e					sI – Ac

The system can also be modelled,

y = Cx

-x = C

d
----x = Ax + Bu
dt		x		B
		x	=	B
		--	=	-------------
		u		sI – A
y	=	y x	=	CB
--	=	- --	=	-------------
u		x u		sI – A

The feedforward and feedback loop can be simplified,

y		u y	–CcBcCB
e- =		--e u-- = (----------------------------------------sI – A ) ( sI – A)
			c
			–CcBcCB
		----------------------------------------		–CcBcCB
y	=	( sI – Ac) ( sI – A)		–CcBcCB
-	=	-------------------------------------------------	=	(-------------------------------------------------------------------sI – Ac) ( sI – A) – CcBcCB
r			–CcBcCB
		1 + ----------------------------------------( sI – A ) ( sI – A)
			c

• Consider an example of a regulator/estimator design,

state space control - 21.17

Consider the state equations for a mass-spring-damper system.

y =

1 0

----

–Ks

–Kd

-------- ---------

----

A =

B =

C =

D =

–Ks

–Kd

1 0

--------

---------

----

The regulator will be designed to have a response time of 0.5s and damped frequency of 2 rad/s.

+ j( 2)

= – 2 ± 2j

s =

------

0.5

λi( A – BK) = – 2 ± 2j

= λi

–Ks

–Kd

–

= λi

– Ks – k1

– Kd

– k2

---------

----

---------------------

--------

--------------------

–1

s 0

–

– K

– k

– K

– k

+ k

s +

--------------------

---------------------

----------------

Kd + k2

Ks + k1

Kd + k2

Ks + k1

= s s + -----------------

----------------

= s

s + -----------------

----------------M

Kd + k2

+ k2

+ k1

–

-----------------

– 4

----------------

s =

------------------------------------------------------------------------------------------------

Kd + k2

–

-----------------

k2 = 4M – Kd

–2 =

--------------------

Kd + k2

– 4

Ks + k1

2 =

-----------------

----------------

--------------------------------------------------------------

16M = ( K

+ k

) 2

– 4( K

+ k )

4( K

+ k ) = ( K

+ 4M – K ) 2 – 16M

k1 = 4M2 – 4M – Ks

K =

– 4M – Ks 4M – Kd

state space control - 21.18

To select the regulator values, component values must be selected.

M = 1kg

Kd = 1

= 10---

------

K =

– 4M – Ks

4M – Kd

K =

( 1)

– 4( 1) – 10 4

( 1) – ( 1)

K = –10 3

The desired response time for the system will be a time response of approximately 0.2sec.

G( s) =

---

-------------------------------------

F =

+ Kds + Ks

Acl = A – BK =

–

–Ks

–Kd

k1 k2

– Ks

– k1

– Kd – k2

--------

---------

----

--------------------

---------------------

Φ e( s) =

sI – Acl

s 0

–

– Ks

– k1

– Kd

– k2

Ks + k1

+ k2

+ s

--------------------

---------------------

----------------

-----------------

Kd + k2

–Ks

– k1

= s

-----------------

-------------------

( s + 0.2)

+ s

–Ks

– k1

(

0.2)

–10.04

-------------------

Kd + k2

( 0.2)

–0.6

-----------------

state space control - 21.19

The estimator may then be designed using Ackermann’s formula.

–1

L = Φ e( A)

1 0

–Ks

–Kd

+ 0.2

--------

---------

M M

1 0

–Ks

–Kd

--------

---------

1 0

–Ks

–Kd

+ 2( 0.2)

–Ks

–Kd

+ 0.2

--------

---------

--------

---------

1 1

–Ks

–Kd

--------

---------

0.4

0.04

1 0

–0.4Ks

–0.4Kd

KsKd

–Ks

----------------

0.04

------------

--------

------

1 1

–Ks

–Kd

--------

+ 0.04

---------

+ 0.4

1 0

Kd2

KsKd

0.4Ks

–Ks

0.4Kd

1 1

0.04

------------

– -------------

--------

+ ------ –

-------------

–Ks

–Kd

--------

+ 0.04

---------

+ 0.4

Kd2

KsKd

0.4Ks

–Ks

0.4Kd

0.04

------------

– -------------

--------

+ ------ –

-------------

–Kd

---------

+ 0.4

–0.6

Kd2

–Ks

0.4Kd

–9.36

+ 0.04

--------

------ – -------------

state space control - 21.20

The compensator coefficients may then be calculated.

Ac = A – BK – LC

–

–0.6

–10 3

1 0

–10 –1

----

–9.36

–

–0.6 0

0.6 1

–10 –1

–10 3

–9.36 0

9.36 –4

= L =

–0.6

–9.36

Cc = K = –10 3

The compensator transfer function can then be written,

Gc( s) = Cc ( sI – Ac) –1Bc

–1

s 0

0.6

–0.6

s – 0.6

–1

–0.6

–10 3

–

–10 3

0 s

9.36

–4

–9.36

s – 4

–9.36

s – 0.6

–1

–0.6

s – 4

9.36

–0.6

–10 3

–9.36

s – 4

–9.36

s – 0.6

–9.36

= ( s – 0.6)

s – 0.6

–1

( s – 4) – ( –1) ( –9.36)

–9.36

s – 4

–0.6

– 0.6s – 6.96

–10 3

9.36 s – 0.6

–9.36

–9.36s

2 – 4.6s – 6.96

1s2 – 4.6s – 6.96

– 22.08s + 69.6

=------------------------------------

s2 – 4.6s – 6.96

•This system can be implemented with a C program as shown below,

state space control - 21.21

#define PERIOD 0.001 // period of the interrupt

interrupt_loop(){ // this routine is called at a regular interval PERIOD

}

double analog_in(){

}

d	ˆ	ˆ		d
----x		ˆ		----x
dt		x	+	dt	x
	1	–BK	+	1	x
	--	–BK		--
+	s			s
+				+
+				+
A – BK – LC				A
				A
		LC

state space control - 21.22

21.6 LQR

-Linear Quadratic Regulator (LQR) allows the design of a regulating system that tends to zero, but also considers the control effort.

-The basic function for evaluating the controller is defined below. The integral is evaluated to reduce the value. Ideally the state value will go to zero (it is a regulator), and the plant input will be minimized. It is assumed that D=0. Here a new state ’z’ is defined. It should reflect the system state values that should be regulated to zero.

∞
JLQR = ∫ [ ( y( t) ) Ty( t)		+ r( u( t) )	2]dt
0
( y( t) ) Ty( t)	=	state cost
( u( t) ) 2 =	control cost

r = control penalty

z = Czx

∞

JLQR = ∫ [ ( x( t) ) TCzTCzx( t) + r( u( t) ) 2]dt

- the design process involves adjusting the r value to obtain the desired result.

state space control - 21.23

Assume a simple controller form, u( t) = –KLQRx( t)

Solve using an Algebraic Riccati Equation

Consider a simple system
·	y = Cx
x = Ax + Bu	y = Cx

The equivalent transfer function can be written, where a(s) is the characteristic eqn.

C( sI – A)	–1		b( s)		b( s)
C( sI – A)		B =	---------	=	----------------
			a( s)		sI – A

This is followed by.......

UNKNOWN STUFF

The desired charactersitic equation can be used to find the controller gain values.

sI – A + BKLQR = ∏ ( s – pi)

i = 1

∆ ( s) = a( s) a( –s) + r–1b( s) b( –s)

The poles are then found using Ackermann’s method.

( u( t) ) 2 = control cost

r = control penalty

∞

JLQR = ∫ [ ( x( t) ) TCTCx( t) + r( u( t) ) 2]dt

∞

JLQR = ∫ [ ( x( t) ) TCTCx( t) + r( u( t) ) 2]dt

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