state space representation part-1

8/22/2019 State space representation Part-1

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Lecture -6: State-space Representation -2

hours

By Mr S Wijewardana

PhD student QMUL 21-04-

2013

Learning Objectives:1. Reduction of higher order equations to first-

order.

2. State Equations and Stare Variables.3. State-space Analysis & State-space Model.

4. From State-space to Transfer Functions.


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Solution of higher order differential equations are

much more difficult to handle than first order

differential equations.

It is easy to handle mathematically when the order

of the differential equation is of first order.

include

Consider the differential equation given below:

The easiest way to solve the

above equation is to do a

substitution:

Now we can write the equation:

xyy ...

wy .

xww .

This equation is much

easier to handle than the

second derivative equation -1

--------eq1


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From our Maths knowledge we can see that we have to use an integral factor to

solve the equation now:

Where p is the coefficient in front of the independent variable x:Nothing will happen to the equation if we multiply each term by an

integrating factor:

xdxpdx eee

xewewexxx

.

21

2

1

1

1

.2

1

)1(1

)(

)(

cecxx

dxecxwyecxw

cexe

dxexe

dxxewe

xewedx

d

x

x

x

xx

xx

xx

xx


4/47

Consider two differential equations as shown below:

Reduce them to first order differential equations:

033

0343

.......

.......

vvvuu

vvvuuu -----------eq- 1

------------eq-2

Assume:

54

.

43

.

2

.

1

.

..

5

.

43

.

21

:

;;;;

xx

xx

xux

Hence

vxvxvxuxux

Eq-1 &2 can now be

written:

033

0343

345

.

22

.

345

.

12

.

2

xxxxx

xxxxxx


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By adding and subtracting the last two equations:

402

302223

4315

.

34122

.

eqxxxx

eqxxxxx

We can write all equations as shown below:

431

.

5

.

4

.

4321

.

.

25

4

3

22322

21

xxxx

xx

xx

xxxxx

xx

We can write the above

equations in Matrix form:


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5

4

3

2

1

5

.4

. 3

.2

.1

.

01102

10000

01000

02232

00010

x

x

x

x

x

x

xx

x

x

State Equations and State Variables

)()()()(

.)(

11

1

1 tftyadt

tdya

dt

tyda

dt

tydon

n

nn

n

______________________________________________________________________

_

Consider a differential equation as shown

below:

---Eq-1


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If we substitute:

1

1

2

1

)()(

.

.

)()(

)()(

n

n

ndt

tydtx

dt

tdytx

tytx

Then we can get:

)()(

)()(

32

21

txdt

tdx

txdt

tdx

Hence we can write:

)()()()()()( 1122110 tftxatxatxatxadt

tdx nnnnn

)()()()(

.)(

11

1

1 tftyadt

tdya

dt

tyda

dt

tydon

n

nn

n

Eq-2:

State

equations


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Eq-2: as you can see clearly now a first order

differential equation with different set of variables.

(However, original equation is the nth order

equation.)The variables x1, x2, x3 are called statevariables


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State-space method is a generalized mathematical modelling technique

which can be used to analyse control systems.(also called time-domainapproach)

This has become more popular as it provides more flexibility of modelling

simple to complex nonlinear systems.

State space model can also be applied to other areas like economics,

agriculture, weather etc.

The model is simply a set of first order differential equations written in

Matrix form.

The normal classicalfrequency domain method is limited only to linear,time invariant systems(LTI systems)(But, this method gives the stability and

transient response information very rapidly)

However, state-space method can be applied to time varying systems like

missiles etc.

Other advantage of the state-space model is the flexibility of choosingstate variables.


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Consider a system described by an nth orderdifferential equation:

radtda

dtda

dtd

n

n

nn

n

121

1

.

We are free to choose our own state variable in order to bring the differential

equation into first order that we can handle easily.

Convenient way to select state variables is to choose the output, (t), and itsn-1 derivatives as state variables: this choice is normally called the Phase variable

choice.

Let us now select the state variables:

))(1(...

4

..

3

.

2

1

,

,

dotsn

nxx

xx

x

------------------eq1

------------------eq2

A dot above omega signifiesdifferentiation w.r.t. time.


11/47

From equations 1 to 3 we can get

nnxxxx

xx

1

.

32

.

2

.

,

1

Substituting the above relationship to our original nth order

differential equation: ( for convenience look at from left to right)

))(1(...

4

..

3

.

2

1

,

,

dotsn

nxx

xx

x

nnnnn

nnnnn

xaxaxaxarx

rxaxaxaxax

112211

.

112211

.

...


12/47

Last equation in the previous slide is a set of first order differential

equations what we want:

Whole set of equations can now be represented in Matrix form as

shown below:

n

nn

x

x

x

r

x

x

x

aaax

x

x

x

2

1

2

1

221

.

3

.

2

.

1

.

001

1

.

0

.

.....

.....

0100

0010

These two equations represent the state-space model for the dynamicsystem of nth order differential equations. However, the standardnotation used in control systems is shown in the next slide:


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DuCxy

BuAxx

.

State-space Model

-----------State equation

-----------Output equation

X is the state vector:

):1(,

)(

)(

)(

)(2

1

vectormatrixn

tx

tx

tx

tx

n


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):1(,

)(

)(

)(

)( 2

1

vectormatrixq

ty

ty

ty

ty

q

Yis the output vector:

U is the input control vector:

):1(,

)(

)(

)(

)(2

1

vectormatrixp

tu

tu

tu

tu

p


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A is the system matrix:

):(,

21

22221

11211

vectormatrixnn

aaa

aaa

aaa

A

nnnn

n

n

B is the input matrix:

):(,

21

22221

11211

vectormatrixpn

bbb

bbb

bbb

B

npnn

p

p


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C = output matrix:

D = feed forward matrix:

):(,

21

22221

11211

vectormatrixnq

ccc

ccc

ccc

C

qnqq

n

n

):(,

21

22221

11211

vectormatrixpq

ddd

ddd

ddd

D

qpqq

p

p


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State Variables:A set of user defined minimum number of linearly independent variables which

are used to define the system variables such that the knowledge of these

variables at any time t 0 , and information on the input excitation subsequently

applied, are sufficient to determine the state of the system at any time t t 0 .

State Vector:

A vector whose elements are the state variables

State Space:

The n-dimensional space whose axes are the state variables.

State Equation:

A set of n simultaneous, first order differential equations with n variables, where

the n variables to be solved are the state variables.

Output Equation:

Output equation is an algebraic equation, which signifies the output variables of

a system as a linear combination of the state variables and the inputs.

Note here, the number of state variables we define is equal to the order of the

differential equation.

We cannot select a variable which is linearly combined with another state


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State Transition

Matrix

The solution to the homogeneous(In the matrix equation

AX=B if B =0 then we call homogeneous) state equationwhat we described earlier:

Axx .

Can be represented in the form of )0().()( xttx

)(tWhere is an n x n matrix and is the unique solution of .)0(),(.

ItA

)(t is called the state transition matrix.

Also note Atet )(


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Example:

In a normal LRC circuit if VR(t) is chosen as state variable, then i(t)

cannot be written as a state variable because VR(t)= R.i(t)( one variable

is a linear combination of the other)

Example:

Find the state-space representation in Phase-variable form for the

transfer function

65

5

)3)(2(

5

)(

)(:

)3)(2(

5

)(

)(

2

sssssR

sC

Solution

sssR

sC

State space model in phase-variable presentation is:

equationOutputDuCxy

EquationSBuAxx

.. -----eq1

------eq2


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Equation-1 and 2 can also be written in the form:

n

nn

x

x

x

Cy

Br

x

x

x

A

x

x

x

2

1

2

1

.

2

.

.

1

Cross multiplication of the TF yields:

(s2+5s+6) C(s) = 5 R(s)

Taking the s operator as: d/dt = s; d2/dt2 =

s2

rccc565

...


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rccc 565...

Now we have to select the state variables:

..

2

.

3

..

3

.

1

.

2

.

2

111

cxxcx

cxxcx

xxcx

Now I can write:

rxxx

xx

565 122.

21

.

In matrix form:

2

1

2

1

2

.1

.

01

50

5610

x

xy

rxx

xx


22/47

5 x2

1.

x

2

.

x

-5

rxxx

xx

565 122.

21

.

x1

-6

r

Y(t) or

c(t)

Block diagram for the

computer simulation


23/47

Example:

Find the state-space representation in phase-variable form for the transfer

function:

65

)1(5

)(

)(2

ss

s

sR

sC

Solution:

Cross multiplication of the TF yields the derivative of R(which is

the input variable).C(s)[s2+5s+6] = (5s+5)R(s)

For convenience let us eliminate R(s).

To eliminate R(s) let us introduce a dummy variable called U:

Then:

)65(

)1(5

. 2

ss

s

R

U

U

C

Now we can

write: )1(5)(

)( s

sU

sCand

)65(

1

)(

)(

2

sssR

sU


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Defining the state variables as before:

C(s) = 5s.U(s) +5u(s)

R(s) = s2.U(s) +5s.U(s) +U(s).6

uuc

ruuu

55

65

.

...

..

2

.

2

.

1

.

1;

ux

xuxux

12

122

.

21

.

55

5

xxcy

rxxx

xx

2

155

1

0

2

1

56

10

2

1.

.

x

xy

rx

x

x

x


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Y(t) or

c(t)

r(t)1

x2

1.

x

2

.

x

-5

x1

-6

5

5

Block diagram for the state space model in

phase variable form(Computer simulation

diagram)

Question:


26/47

22187

23

)(

)(23

2

ssS

ss

sR

sC

Question:


27/47

Stability

we say system x =Ax is stable if etA 0 as t

meaning:

state x(t) converges to 0, as t , no matter what x(0) is

all trajectories of x =Ax converge to 0 as t fact: x =Ax is stable if and only if all eigenvalues of A have negative real

part:

i < 0, i = 1, . . . , n


28/47

From State Space to Transfer Function

When analysing control systems more often we haveto learn how to transfer the systems from statespace to TF and vise-versa.

Assume the standard state-space model:

DuCxy

BuAxx

.

2)()()(

1)()()(

eqsDUsCXsY

eqsBUsAXssX

In Laplace-

domain:

From equation-1:

[sI-A] X(s) = B U(s)

X(s) = [sI-A]-1 BU(s)


29/47

Substituting in equation 2 we get:

Y(s) = C[sIA]-1 BU(s) + DU(s)

Y(s) = [C(sIA)-1

B +D] .U(s)

The matrix [C(sI-A)-1B +D] is called the transfer

function matrix because it relates the output

vector Y(s) to the input vector U(s). {if U(s) andY(s) are scalars} Then we can write:

DBAsICsU

sYsG 1)(

)(

)()(

2]([

)(

.)(

)(

eqAsI

DAsIBAsIadjCsG

DBAsI

AsIadjCsG


30/47

From equation-2, we can see that the dynamic characteristics and stability are

determined by poles of the closed-loop transfer function, that is , the roots of sI-A=0

This equation is called the characteristic equation in state space model.

0 AsIIf the coefficients of matrix A are real

then are also

real.0 AsI

Example: A control system is defined by the state space model by the followingequations. Find the transfer function of the system.

2

1

2

1

2

.1

.

01

0

1

32

10

x

x

y

ux

x

x

x

By comparison with the state space model:

]01[,0

1,

32

10

CBA


31/47

We have already derived the transfer function

matrix as shown below:

DBAsICsU

sYsG 1)(

)(

)()(

21

3

)(

0

1

2

13

21

1)()(

2

13

23

1)(

1

2

1

ss

s

sG

s

s

ss

oBAsICsG

s

s

ssAsI

Using Matlab:

A=[0 1;-2 -3];

>> B=[1;0];

>> C=[1 0];>> D=0;

>> [num,den]=ss2tf(A,B,C,D)

num =

0 1 3

den =

1 3 2


32/47

Use Matlab to find the transfer function of the state-space model:

01,2

0,

32

10

CBA

Answer:num =

0 0 2.0000

den =

1 3 2


33/47

Solution of State Variables

How to obtain a general solution to the linear time-invariant state equation.

(space vehicle: mass will change after some time: time variant systems)

Solution of scalar differential equations.

Consider a system with only initial conditions without any forcing function:

)()(.

taxtx

Assume a solution of x(t) is in the form of power series:

x(t)= b0

+b1t + b

2t + b

3t + ---------eq2

...32)( 2321.

tbtbbtx

By substituting the assumed solution to the first equation:

b1 + 2b2t + 3b3t2 + = a [b0 + b1t + b2t

2 +]

Equating the coefficients of equal powers of t:

b1 = a b0

b2 = a b1/2 = a2b0= (1/2!) a

2b0

b3

= (1/3!)a3.b0

--------------------eq1


34/47

Substituting in equation 2 : When t = 0,

x(0) = b0 .

Therefore the solution x(t) can be written as:

)0()(

)0(.. .!3!2

1)(3322

xetx

xtata

attx

at

...)0(!2

1)0()0()(

2

xaaxxtx

x(0) is the initial condition of state variable

Combining equation 2

Laplace Transform Method to the solution of Homogeneous State Equations


35/47

p g q

Our system is:

)()(.

tAxtx ---------eq1

Taking Laplace transform of equation-1:Laplace transform of x: (x)= X(s)

Hence s X(s)X(0) = AX(s): Note the theorem )0()(

fssF

dt

df

s X(s)- A X(s) = X(0)

(s I - A) X(s) = X(0)

X(s)= (sIA)-1 X(0)

Hence: x(t) = -1 (s IA)-1x(0)

...3

2

2

1

s

A

s

A

s

IAsI


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-1 (s IA)-1

)0()(

)(

!3!21

3322

xetx

eAsI

etAtA

AtI

At

At

At

we can write

Hence the

solution is:

)()(

3210

2

1

2

.1

.

txtx

x

x

Example:

Find x1(t) and x2(t) given initial conditions:

2

1

)0(

)0(

2

1

x

xUsing Laplace

transformmethod.


37/47

We have shown that given the initial conditions the solution to the above

state equation can be represented in the form of x(t) = eAt x(0).

This solution is for the state equation that can be represented as follows:

)()(.

tAxtx

Taking Laplace transform of both side of the equation we can write:

sX(s) -X(0) = A X(s)

(sIA)X(s) =X(0)

X(s) = (sI -A)-1 X(0)

2

1

)0(,32

10

XAWe were giventhat

Hence we can

write:

32

1

32

10

0

0

s

s

s

sAsI


38/47

tt

tt

ee

ee

tx

tx

ss

sss

s

ss

s

s

XAsI

2

2

2

1

2

1

64

34

)(

)(

)2)(1(

22)2)(1(

5

23

2

1

2

13

)0(

We can check the solution whether it is correct by applying the initialconditions: x1(0) =1, x2(0) =2, 1 = 1 and 2 =2: which implies as correct.

32

10A

In the same example if initial conditions are not given:

The state transition matrix is defined as: Atet)( -1 (s IA)-1


39/47

32

1

32

10

0

0][

s

s

s

sAsI

The inverse of (sIA) is given by:

)2)(1()2)(1(2

)2)(1(

1

)2)(1(

3

2

13

)2)(1(

1][ 1

sss

ss

ssss

s

s

s

ssAsI

Atet)( -1 (s IA)-1Since:

)(tWe can get thestate transition

matrix:

tttt

ttt

eeee

eeee22

221

222

2


40/47

We have learned how to construct signal flow graphs (SFG) for simultaneous

equations in lecture -5: The signal flow graphs which represent the state

equations or differential equations are called state diagrams. State diagram is anextension of SFG to portray state equations and differential equations.

Given the state diagram it is much easier to construct state equations or TF than

from SFG. Construction procedures and rules are same for SFG and State

diagrams but, state diagram is in Laplace domain.

State Diagram

State Diagram for State Equations:

Consider the control system:

4954

3853

26237

12352

321

3213

.

3212

.

3211

.

eqxxxy

eqrxxxx

eqrxxxx

eqrxxxx


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R(s) sX3(s) X3(s) sX2(s) X2(s) sX1(s) X1(s)

Y(s)

Initially let us define three state variable x1, x2, x3 and let these three state

variables are represented as nodes and to the left of each node, respective

derivatives of them are also shown as below:

Let us define the input as r and the output as y.

In Laplace domain we can show the derivatives and nodes as shown below:

In the next step, we interconnect the state variables and their derivatives

with the definition of integration (1/s):

sX3(s) X3(s)

1/s

Integratio

n


42/47

12352 3211.

eqrxxxx

State diagram for equation -1:

R(s) sX3(s) X3(s) sX2(s) X2(s) sX1(s) X1(s) Y(s)-51/s

2

1/s1/s

2

3


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R(s) sX3(s) X3(s) sX2(s) X2(s) sX1(s) X1(s) Y(s)


6

1/s

21/s 1/s

-7

-3

262373212

.

eqrxxxx

1/s8

1

1/s-5

-3

1/s

3853 3213.

eqrxxxx


44/47


321 954 xxxy

State diagram for equation -1:

5

1/s

9

1/s 1/s

-4


45/47


Now we have to combine all figures to get one state diagram for all equations:

3

1

8

5

1/s

9

1/s 1/s

-4

6

2

-3

-5

2-5

2

-7-3

Homework: check the numbers and the branches


46/47

End

Characteristic Equation From State-Space Representation:


47/47

The roots of the characteristic equation are the same as the eigenvalues of

matrix A. The characteristic equation of matrix A can simply be computed

with Matlab by using the command :

p = poly(A);

or p = poly(A,v); %retun as a variable v

state space representation part-1

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