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Lecture 7: State Variables Analysis II

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Course Outline Control Elements: Transducer, Switches, Actuators, Valves,

Motors

Control Fundamentals: Open loop and closed loop systems,transfer function, signal flow graph, gain formula

Modeling: Mathematical modeling of linear electrical andmechanical systems, state variables, state equations andstate diagrams

Analysis and Design: Stability, controllability andobservability of systems, state variables, state transitionmatrix, transient and steady state response, root locusmethod. Nyquist criterion, PID controllers, lead lagcompensators, pole-zero cancellations.

Practical systems: Analog and microprocessor basedcontrol systems, design examples

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Recap -Example 1

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The motion of any finite dynamic system can be expressed as a set of first-order ordinary

differential equations. This is often referred to as the state-variable representation. For

example, Newtonslaw for a single mass Mmoving in one dimensionx under force Fis:

We define one state variable as the position and the other state variable as the velocity

, this equation can be written as:

Furthermore, first-order linear differential equations can be concisely expressed using matrix

notation. If we collect the state into a column vector x, and the coefficients of the state

equations into a square matrix F, and the coefficients of the input into the column vector G,

these equations can be written in matrix form as:

Where Fis the system matrix and Gis the input matrix. If we take the input to be

FxM

xx 1

xx 2

21

xx

M

Fx 2

M

F

1

0

x

x

00

10

x

x

2

1

2

1

GuFxx

1xy

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Recap

4

This to can be expressed in matrix form as:

Where His a row vector, referred to as the output matrix. Collecting these matrix notations, we

have an extremely compact notation..

Similarly, the differential-equation models of more complex systems, such as those developed in

earlier lectures on mechanical, electrical, and electromechanical systems, can be described by

state variables through selection of positions, velocities, capacitor voltages, and inductor

currents as state variables.

2

1

01 x

xy

Hxy

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Recap -Example 2

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A robot-arm drive system for one joint can be represented by the following differential equation:

where v(t) = velocity, y(t) = position and i(t) is the control-motor current. Put the equations in state variable form

and set up the matrix form for

We know that the velocity is the derivative of the position, therefore we have

and from the problem statement

This can be written in matrix form as

),()()()(

321 tiktyktvk

dt

tdv

121 kk

vdt

dy

)()()( 321 tiktyktvkdt

dv

ikv

y

kkv

y

dt

dv

312

010

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Recap - Example 3

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As an example of the formulation of equations in state-variable form, consider the two-mass

system below:

If we take the state as the position and velocity of each mass as follows:

dx

dx

yx

yx

4

3

2

1

udykdybyM )()(

0)()( ydkydbdm

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Recap

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the equations in state-variable form are:

The state-space matrices are:

43214

43

43212

21

xM

bx

M

kx

M

bx

M

kx

xx

M

ux

M

bx

M

kx

M

bx

M

kx

xx

M

b

M

k

M

b

M

k

1000

M

b

M

k

M

b-

M

k-

0010

F

0

0

1

0

MG 0001H

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Recap - Example 4

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Consider the equation of motion for the simple pendulum shown below:

where . If we choose as the state variables and , then the equations are:

These equations are nonlinear but are in state-variable form. We can linearize the equations and

hence for smallx.

2

12

2

21

sinml

Txx

xx

c

2

2 sin

ml

Tc

lg2

1x

2x

sin

22

21

sinml

Tx

xx

c

0

0F

2

10

ml

G

0

1

H

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State Diagrams

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Phase-variable format

Input feedforward format

Controllability Canonical form

Observability Canonical form

Diagonal Canonical form

Jordan Canonical form

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State Diagrams

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To illustrate the derivation of the state diagram, we consider the fourth-order transfer function:

Note that the system, is fourth order and hence we identify 4 state variables, hence we will use

four integrators. The state diagram will be:

The numerator terms represent forward-path factors in Masonsgain formula.

43

1

2

2

1

3

4

1

2

2

3

3

4

1

)(

sasasasa

sb

asasasas

b

sG

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State Diagrams II

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To illustrate the derivation of the state diagram, we consider the fourth-order transfer function:

The state diagram will be:

The numerator terms represent forward-path factors in Masonsgain formula. The general form

shown above is called thephase variable format.

What is

The output is simply:

43

1

2

2

1

3

43

1

2

2

1

3

1

2

2

3

3

4

1

2

2

3

3

1

)(

sasasasa

sbsbsbsb

asasasas

bsbsbsb

sG

uxaxaxaxa 43322114x

4332211)( xbxbxbxbtc

433221 ,, xxxxxx

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State Diagrams II

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State equations

Then in matrix form we have:

and the output is:

uxaxaxaxa 43322114x

4332211)( xbxbxbxbtc

433221 ,, xxxxxx

ubAxx

)(

1

0

0

0

1000

0100

0010

4

3

2

1

32104

3

2

1

tu

x

x

x

x

aaaax

x

x

x

dt

d

4

3

2

1

3210 ,,,Dx)(

x

x

x

x

bbbbtc

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State Diagrams III

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The state diagram in the previous example is not a unique representation of the given transfer

function and other equally useful structures can be designed. An alternative form known as the

input feedforward format is illustrated below.

Using the state diagram we can obtain the following set of first-order differential equations:

43

1

2

2

1

3

43

1

2

2

1

3

1

2

2

3

3

4

1

2

2

3

3

1

)(

sasasasa

sbsbsbsb

asasasas

bsbsbsbsG

ubxxax 32131 ubxxax 23122

ubxxax 14113 ubxax 14

)(

000

100

010

001

0

1

2

3

4

3

2

1

0

1

2

3

4

3

2

1

tu

b

b

b

b

x

x

x

x

a

a

a

a

x

x

x

x

dt

d

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State Diagrams IV

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Consider the system represented by the equation:

The state variables, the outputs of the integrators are . In these terms,

the equations of motion are:

uyyyy 66116

uxxxx

xx

xx

661163213

32

21

yxandyxyx 321 ,

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State Diagrams V

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The closed-loop transfer function of a system is:

The first model is the phase variable state diagram as illustrated below:

Recalling Masonsgain formula, the denominator can be considered to be one minus the sum ofthe loop gains whereas the numerator of the transfer function is equal to the forward-path.

and the output is:

321

321

23

2

61681

682

6168

682

)(

)()(

sss

sss

sss

ss

sR

sCsT

)(

1

0

0

8166

100

010

3

2

1

3

2

1

tu

x

x

x

x

x

x

dt

d

3

2

1

286)(

x

x

x

tc

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State Diagrams VI

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The closed-loop transfer function of a system is:

The second model is the phase variable state diagram as illustrated below:

Recalling Masonsgain formula, the denominator can be considered to be one minus the sum ofthe loop gains whereas the numerator of the transfer function is equal to the forward-path.

and the output is:

321

321

23

2

61681

682

6168

682

)(

)()(

sss

sss

sss

ss

sR

sCsT

)(

1

0

0

8166

100

010

3

2

1

3

2

1

tu

x

x

x

x

x

x

dt

d

3

2

1

286)(

x

x

x

tc

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State Diagrams VII

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The closed-loop transfer function of a system is:

The first model is the input feedforward format as illustrated below:

Recalling Masonsgain formula, the denominator can be considered to be one minus the sum ofthe loop gains whereas the numerator of the transfer function is equal to the forward-path.

and the output is:

321

321

23

2

61681

682

6168

682

)(

)()(

sss

sss

sss

ss

sR

sCsT

)(

6

8

2

006

1016

018

3

2

1

3

2

1

tu

x

x

x

x

x

x

dt

d

)()( 1 txtc

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System Transfer Functions

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In order to relate the state-variable equations to our earlier consideration of poles and zeros,

we take the Laplace transform of:

We obtain:

which is now an algebraic equation. We collect the terms involving X(s) on the left-hand side,

we get:

If we multiply both sides by the inverse of , then:

The output of the system is:

Which means

GuFxx

)(G)FX(x(0))( sUsssX

x(0))(G)(X)FI( sUss

)FI( s

x(0))()(G)()(X -1-1 FsIsUFsIs

)()(H)( sJUsXsY

)(x(0))(H)(G)(H)( -1-1

sJUFsIsUFsIsY

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System Transfer Functions

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If we assume that the initial conditions are zero, the input-output relationship is:

JG)(H)(

)(

)( 1-

FsIsU

sY

sG

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Example 1

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Consider the system with state-variable description:

Find the corresponding state diagram and compute the transfer function.

0165F

01G

01H 0J

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Example 1 (cont.)

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Compute the transfer function form:

and compute:

Therefore:

ssFsI

165

6)5(51

6

)( 1

ss

s

s

FsI

6)5(

0

151

6)5(

0

1

51

610

)(

ss

s

ss

s

s

sG

)3)(2(

1)(

ss

sG

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State Diagram Example

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The phase variable form of a system is given by:

Draw the flow graph model.

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