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3.6 State Models

EXAMPLE 3.20—BACKING A CAR

Consider backing a car close to a curb. The transfer function from steering angle to distance from the curve is non-minimum phase. This is a mechanism that is similar to the aircraft.

EXAMPLE 3.21—REVENUE FROM DEVELOPMENT

The relation between revenue development effort in a new product development is a non-minimum phase system. This means that such a system is very difficult to control tightly.

3.6 State Models

The state is a collection of variables that summarize the past of a system for the purpose of prediction the future. For an engineering system the state is composed of the variables required to account for storage of mass, momentum and energy. An key issue in modeling is to decide how accurate storage has to be represented. The state variables are gathered in a vector, the state vector x. The control variables are represented by another vector u and the measured signal by the vector y. A system can then be represented by the model

dx

f Dx, uE

 

 

dt

3.33

E

 

D

 

y nDx, uE

The dimension of the state vector is called the order of the system. The system is called time-invariant because the functions f and n do not depend explicitly on time t. It is possible to have more general timevarying systems where the functions do depend on time. The model thus consists of two functions. The function f gives the velocity of the state vector as a function of state x, control u and time t and the function n gives the measured values as functions of state x, control u and time t. The function f is called the velocity function and the function n is called the sensor function or the measurement function. A system is called linear if the functions f and n are linear in x and u. A linear system can thus be represented by

dxdt Ax Bu y Cx Du

where A, B, C and D are constant varying matrices. Such a system is said to be linear and time-invariant, or LTI for short. The matrix A is

117

Chapter 3. Dynamics

y

θ

x

Figure 3.28 An inverted pendulum. The picture should be mirrored.

called the dynamics matrix, the matrix B is called the control matrix, the matrix C is called the sensor matrix and the matrix D is called the direct term. Frequently systems will not have a direct term indicating that the control signal does not influence the output directly. We will illustrate by a few examples.

EXAMPLE 3.22—THE DOUBLE INTEGRATOR

Consider a system described by

dx

 

>

0

1

>

 

 

 

 

D

E

 

 

 

>

 

 

>

 

:

 

 

;

 

dt

 

8

0

0

9 x

 

8

0

1

9 u

3.34

y

 

:

1

0

; x

 

 

 

 

 

 

 

 

8

 

 

9

 

 

 

 

 

 

 

 

 

:

 

 

;

 

 

 

 

 

 

This is a linear time-invariant system of second order with no direct term.

EXAMPLE 3.23—THE INVERTED PENDULUM

Consider the inverted pendulum in Figure 3.28. The state variables are the angle θ x1 and the angular velocity /dt x2, the control variable is the acceleration un of the pivot, and the output is the angle θ.

Newtons law of conservation of angular momentum becomes

d2θ

J dt2 mnl sin θ mul cosθ

118

3.6 State Models

R

L

+

 

ω

u

M

M

J

e

 

 

 

i

D

 

 

 

 

 

 

(a)

 

(b)

 

Figure 3.29 Schematic diagram of an electric motor.

Introducing x1 θ and x2 /dt the state equations become

 

 

 

>

 

 

x2

>

 

 

 

J

J

dxdt

8 mnl sin x1

 

mlu cos x1

9

 

 

 

>

 

 

 

 

 

>

 

 

 

>

 

 

 

 

 

>

 

 

 

>

 

 

 

 

 

>

y

 

:

 

 

 

 

 

;

 

x1

 

 

 

 

 

 

It is convenient to normalize the equation by choosing

J

/

m

n

l as the

unit of time. The equation then becomes

p

 

 

 

 

dx

 

>

x2

>

 

 

 

 

D E

 

 

 

 

 

 

 

 

 

 

 

 

 

 

>

 

 

>

 

 

 

 

 

 

 

 

 

:

 

u cos x1

;

 

 

 

 

3.35

 

dt

8 sin x1

 

9

 

 

 

 

 

y

 

x1

 

 

 

 

 

 

 

 

This is a nonlinear time-invariant system of second order.

EXAMPLE 3.24—AN ELECTRIC MOTOR

A schematic picture of an electric motor is shown in Figure 3.29 Energy stored is stored in the capacitor, and the inductor and momentum is stored in the rotor. Three state variables are needed if we are only interested in motor speed. Storage can be represented by the current I through the rotor, the voltage V across the capacitor and the angular velocity ω of the rotor. The control signal is the voltage E applied to the motor. A momentum balance for the rotor gives

J dt Dω kI

119

Chapter 3. Dynamics

Figure 3.30 A schematic picture of a water tank.

and Kirchoffs laws for the electric circuit gives

E RI L dIdt V k ddtω

I C dVdt

Introducing the state variables x1 ω , x2 V , x3 I and the control variable u E the equations for the motor can be written as

 

 

 

 

D

0

 

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

J

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

>

 

 

 

J

 

>

 

>

>

 

8

 

 

9

 

 

 

dx

 

>

 

 

 

 

 

 

 

1

 

 

>

 

> L

>

 

 

 

 

 

 

 

 

 

 

>

 

 

 

 

 

 

 

 

 

 

>

 

>

 

 

>

 

:

 

 

;

 

 

 

 

 

J L

L

 

J L

L

 

 

 

 

 

 

 

 

 

dt

 

>

kD

1

 

k2

R

>

 

>

1

 

>

 

 

 

 

D

E

>

 

 

 

C

>

8

 

9 uy

 

1

0 0

 

x

 

 

8

0

0

 

 

 

 

9 x

 

0

 

 

 

 

3.36

 

 

>

 

 

 

 

 

 

 

 

 

 

>

 

>

 

 

>

 

 

 

 

 

 

 

 

 

 

>

 

 

 

 

 

 

 

 

 

 

>

 

>

 

 

>

 

 

 

 

 

 

 

 

 

 

>

 

 

 

 

 

 

 

 

 

 

>

 

:

 

 

;

 

 

 

 

 

 

 

 

 

 

:

 

 

 

 

 

 

 

 

 

 

;

 

 

 

 

 

 

 

 

 

 

 

 

 

This is a linear time-invariant system with three state variables and one input.

EXAMPLE 3.25—THE WATER TANK

Consider a tank with water where the input is the inflow and there is free outflow, see Figure 3.30 Assuming that the density is constant a mass balance for the tank gives

dV

dt qin qout

The outflow is given by

p

 

 

qout a

2nh

120

 

 

3.6 State Models

There are several possible choices of state variables. One possibility is to characterize the storage of water by the height of the tank. We have the following relation between height h and volume

Z h

V ADxEdx

0

Simplifying the equations we find that the tank can be described by

dh

 

1

 

 

 

 

Dqin ap2nhE

dt

ADhE

p qout a 2nh

The tank is thus a nonlinear system of first order.

Equilibria

To investigate a system we will first determine the equilibria. Consider the system given by D3.33E which is assumed to be time-invariant. Let the control signal be constant u u0. The equilibria are states x0 such that the dx/dt 0. Hence

f Dx0, u0E 0 Notice that there may be several equilibria.

For second order systems the state equations can be visualized by plotting the velocities for all points in the state space. This graph is called the phase plane shows the behavior qualitative. The equilibria corresponds to points where the velocity is zero. We illustrate this with an example.

EXAMPLE 3.26—THE PHASE PLANE

 

 

 

 

 

Consider the

 

 

x2 x23

 

 

dx

 

>

>

 

 

:

 

x1

 

x22

;

 

dt

8

9

The equilibria are given by

 

>

 

 

>

x2 x23 0 x1 x22 0

There are three equilibria:

x1

1

x2

1

x1

1

x2

1

x1

0

x2 0

The phase plane is shown in Figure 3.31. The phase plane is a good visualization of solutions for second order systems. It also illustrates that nonlinear systems can be interpreted as a vector field or a flow.

121

Chapter 3. Dynamics

y

x ' = y − y 3 y ' = − x − y 2

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

−3

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

 

 

 

 

 

x

 

 

 

 

 

Figure 3.31 Phase plane for the second order system dx1/dt x2 x23 ,dx2/dt

x1 x22.

Linearization

Nonlinear systems are unfortunately difficult. It is fortunate that many aspects of control can be understood from linear models. This is particularly true for regulation problems where it is intended to keep variables close to specified values. When deviations are small the nonlinearities can be approximated by linear functions. With efficient control the deviations are small and the approximation works even better. In this section we will show how nonlinear dynamics systems are approximated. We will start with an example that shows how static systems are approximated.

EXAMPLE 3.27—LINEARIZATION OF STATIC SYSTEM

Consider the system

y nDuE

A Taylor series expansion around u u0 gives

y nDu0E nPDu0EDu u0E . . .

The linearized model is

y y0 nPDu0EDu u0E

The linearized model thus replaces the nonlinear curve by its tangent at the operating point.

122

3.6 State Models

Linearization of dynamic systems is done in the same way. We start by determining the appropriate equilibria. The nonlinear systems are then approximated using Taylor series expansions. Consider the system

dxdt f Dx, uE y nDx, uE

Consider small deviations from the equilibrium!

x x0 δ x, u u0 δ u, y y0 δ y

Make a series expansion of the differential equation and neglect terms of second and higher order. This gives

dx

 

 

f

x

 

δ x, u

 

δ u

f

x , u

 

V f

x , u δ x

 

V f

x , u δ u

dt

 

 

 

D 0

0

E

 

D 0 0E Vx D 0 0E

Vu D 0 0E

y

 

 

x

 

δ x, u

 

δ u

y

Vn

x , u δ x

Vn

x , u δ u

 

 

nD 0

0

E

0 Vx D 0 0E

Vu D 0 0E

We have f Dx0, u0E 0 because x0 is an equilibrium and we find the following approximation for small deviations around the equilibrium.

 

dDx x0E

 

A

x

x

B u u

 

 

dt

 

 

D

 

0E

D − 0E

 

 

y y0 CDx x0E DDu u0E

where

 

 

 

 

 

 

 

A

V f

Dx0, u0E

 

 

 

 

B

V f

Dx0, u0E

Vx

 

 

 

 

Vu

C

Vn

Dx0, u0E

 

 

 

 

D

Vn

Dx0, u0E

Vx

 

 

 

 

Vu

The linearized equation is thus a linear time-invariant system, compare with D3.37E. It is common practice to relabel variables and simply let x, y and u denote deviations from the equilibrium.

We illustrate with a few examples

EXAMPLE 3.28—LINEARIZATION OF THE WATER TANK

dh

 

1

 

 

 

 

Dqin ap2nhE

dt

ADhE

p qout a 2nh

123

Chapter 3. Dynamics

To determine the equilibrium we assume that the inflow is constant qin

q0. It follows that

p qout qin q0 a 2nh0

q2

h0 n0 2

2 a

Let A0 be the cross section A at level h0, introduce the deviations. The linearized equations are

dδ h

dt

δ qout

The parameter

T 2A0h0 q0

T

 

 

a 2nh0

δ h

 

 

1

δ q

T2A0h

 

 

 

 

0

 

A0

in

 

a 2nh0

δ h

 

 

q0

δ h

 

 

 

 

 

 

h0

h0

 

2 Total water volume Fm3G Flow rate Fm3/sG

is called the time constant of the system. Notice that T/2 is the time it takes to fill the volume A0h0 with the steady state flow rate q0

EXAMPLE 3.29—LINEARIZATION OF THE INVERTED PENDULUM

Consider the inverted pendulum in Example 3.23 which is described by D3.35E. If the control signal is zero the equilibria are given by

x2 0 sin x1 0

i.e. x2 θ/dt and x1 θ 0 and x1 θ π . The first equilibrium corresponds to the pendulum standing upright and the second to the pendulum hanging straight down. We have

 

f x, 0

 

 

>

 

 

0

 

 

1

>

 

f

 

>

0

 

>

 

 

 

 

 

 

V

 

 

 

 

 

 

 

 

 

V

 

 

 

 

 

 

 

 

V D x

E

 

:

 

 

 

u sin x1

0

;

,

Vu

:

 

;1

 

 

 

 

8 cos x1

 

9

8 cos x1

9

,

0 and

Evaluating the

derivatives at the upper equilibrium u

 

0, x

 

 

 

>

 

 

 

 

 

 

 

>

 

 

 

>

 

 

>

 

 

 

 

 

x2 0 we get

 

 

A

8 1 0

9 ,

B 8 0 1 9 .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

>

0

 

1

>

 

:

;

 

 

 

 

 

 

1

 

 

 

 

 

 

 

:

 

 

 

 

;

 

 

 

 

 

 

 

 

 

0, x

 

 

π

For the equilibrium when>then

pendulum is hanging down, u

 

 

 

 

>

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and x2 0 we have instead

 

1

9 ,

 

8

 

9

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A

8 01 0

B 0 1 .

 

 

 

 

 

 

 

 

 

 

 

 

 

>

 

 

 

 

>

 

:

 

;

 

 

 

 

 

 

 

 

 

 

 

 

 

>

 

 

 

 

>

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

:

 

 

 

 

;

 

 

 

 

 

 

 

 

 

 

 

 

 

 

124

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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