System Dynamics Ch3 The State Space Equations and Their Time Domain Solution.doc

7/27/2019 System Dynamics Ch3 The State Space Equations and Their Time Domain Solution.doc

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III. The State Space Equations and Their Time Domain Solution

Introduction

This section of notes concentrates on three areas:

1. A general notation for the state variable treatment of dynamic systems (the so-called

standard state space form).

2. Techniques for putting general systems into standard form.

3. Methods for the time domain solution of the state equations, including both analytical

methods and numerical methods.

Several examples are used to illustrate these techniques and a final numerical example in

Matlab contrasts three different solution methods.

This information is organized into the following subsections:

The State Space Representation

o Example 3.1

Conversion to State Form

o Nth Order Linear Differential Equations

Example 3.2

Example 3.3

o Mixed Algebraic and Differential Equations

Example 3.4

o

Linearization of Nonlinear Systems Example 3.5

Analytic Solution to Linear Stationary Systems

o Homogeneous Systems

o Non-Homogeneous Systems

o Time-Varying Systems

Discretization of Linear Stationary Systems

Numerical Integration of the State Equations

An Example Illustrating Various Solution Schemes
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The State Space Representation (Continuous Systems)

The general state variable form for continuous dynamic systems can be written as

)(),()()( ttttdtd gxfAxx ++= (3.1)

and

)()()( ttt DuCxy += (3.2)

where ),( txf contains all the variable coefficient and nonlinear terms.

To complete this description, we note that the input, output, and state vectors do not generally

contain the same number of elements. To treat this, one can write )(tg as

)()( tt Bug =

and allow the B , C , and D matrix operators to be rectangular matrices (i.e. non-square).The system matrix, A , is required to be square.

A particular element of eqn. (3.1) can be written as

++=M

m

mim

K

k

ik

N

j

jiji tubtftxatxdt

d)(),()()( x (3.3)

and for eqn. (3.2), we have

+=M

m

mim

N

j

jiji tudtxcty )()()( (3.4)

where we have anNth order system withMinputs and a maximum ofKnonlinear terms for

any one equation.

Example 3.1 illustrates the use of eqns. (3.1 and 3.2) for a simple second order system.

The most general form of eqn. (3.1) can be written as

),,()( ttdt

duxfx = (3.5)

where the vector function f(x,u,t) contains the full inter-relationship among all the state

variables and inputs. The use of this form usually implies that only numerical simulation

techniques will be used to study the system's dynamic behavior.

Example 3.1 Standard State Space Form

Problem Statement:
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Put the following 2nd order system into standard state form:

)(2)()()()(4)(3)(7)(31211211tututxtxtxttxtxtx

dt

d++++=

)(4)(3)(5)(9)(2

2

2212tutxttxtxtx

dtd +=

and

)(2)()()( 221 tutxtxty +=

Problem Solution:

Putting this 2nd order system into standard state space form, gives

)(),()()( ttttdt

dBuxfAxx ++=

and

)()()( ttty TT udxc +=

where

=

=

)(

)(

)(

)(;)()()(

3

2

1

2

1

tu

tu

tu

ttxtxt ux

=

+=

=

040

201;

)(3

)()()(4),(;

59

372

2

211BxfA

tx

txtxtxt

t

and

[ ] [ ]020;11 ==TT

dc


The goal here is to present a recipe for the conversion of certain classes of problems into state

form. Three specific areas to be addressed are:

1. State space representation ofnth order systems,

2. Treatment of mixed algebraic and differential equations, and

3. Linearization of nonlinear systems.


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Conversion ofnth Order Linear Differential Equations (with derivatives)

Given the following nth order system,

)()()()()()()()( 11

1

1011

1

1 tubtudt

d

btudt

d

btudt

d

btyatydt

d

atydt

d

atydt

d

nnn

n

n

n

nnn

n

n

n

++++=++++

(3.6)

let

)()()( 01 tutytx =

)()()()()()( 11102 tutxdt

dtutu

dt

dty

dt

dtx ==

)()()()()()()( 22212

2

02

2

3 tutxdtdtutu

dtdtu

dtdty

dtdtx ==

or, for thejth state variable

)()()()()()( 1111

1

01

1

tutxdt

dtutu

dt

dty

dt

dtx jjjj

j

j

j

j

==

(3.7)

In these expressions, j is defined as

00 b=

0111 ab =

021122 aab =

01111 jjjjj aaab = (3.8)

Using eqns. (3.7) and (3.8), one gets the following SISO matrix representation of eqn. (3.6):

)()()( tuttdt

dbAxx += (3.9)

)()()( tdutty T += xc (3.10)

with


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[ ] 002

1

2

1

;001;; bd

x

x

x

T

nn

===

=

=

cbx (3.11)

and

=

11

100

010

010

aaa nn

A (3.12)

Example 3.2 illustrates the use of the above general conversion algorithm. Note that for the

special case where the derivatives of the forcing function are not present (i.e. for RHS = u(t)

only), we have [ ]100 =Tb and d= 0. This latter situation is actually the usualsituation, and it is much easier to visualize and implement the conversion algorithm (since all

the is are zero except n = 1).

Example 3.2 Conversion of 3rd Order System to State Form (General Case)

Problem Statement:

Given

)(7)(2)(4)(8)(6)(2

2

3

3

tutudtdtyty

dtdty

dtdty

dtd +=++

convert this 3rd order linear stationary system to state form.

Problem Solution:

Using the recipe given previously, we have

)()()( 11 tutxdt

dtx jjj =

with

yuyx == 01

dt

dyu

dt

du

dt

dyx == 102

udt

ydu

dt

du

dt

ud

dt

ydtx 2)(

2

2

212

2

02

2

3 ==

and


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011 jjjj aab =

000 ==b

with

00 b=

00111 == ab

2021122 == aab

5)2(6703122133 === aaab

Therefore

)(

5

2

0

)(

)(

)(

684

100

010

)(

)(

)(

3

2

1

3

2

1

tu

tx

tx

tx

tx

tx

tx

dt

d

+

=

and

[ ] [ ] )(0

)(

)(

)(

001)(

3

2

1

tu

tx

tx

tx

ty +

=

Example 3.3 illustrates this less general, but very common case. It also highlights the

importance of the state matrix in determining the solutions to linear stationary systems. We

will focus further on the properties of the state matrix in later subsections.

Example 3.3 Conversion of 3rd Order System to State Form (Usual Case)

Problem Statement:

Given

)()()()()( 322

2

13

3

tutyatydt

daty

dt

daty

dt

d=+++

convert this system to state form, noting that the RHS does not contain derivatives of u(t).

Also show that the eigenvalues of the resultant system matrix are identical to the roots of the

characteristic equation associated with the homogeneous solution to the given 3 rd-order

equation.

Problem Solution:


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For this case, let's make the following substitutions:

yx =1

1

1

2 xdt

dx

yx ===

22

3 xdt

dxyx ===

or, in general,

dt

tdxtx

j

j

)()(

1=

with

yx =1

Now, from the defining equation with 33 x

dt

dxy == , we have

uxaxaxatxdt

d+= 1322313 )(

Therefore, the three equations for 321 and,, xxx

can be put into matrix form, giving

)(

1

0

0

)(

)(

)(

100

010

)(

)(

)(

3

2

1

1233

2

1

tu

tx

tx

tx

aaatx

tx

tx

dt

d

+

=

Also, since the solution of interest is y(t) (a single output), the general expression for the

output equation reduces to

[ ] [ ] )(0

)(

)()(

001)(

3

2

1

tu

tx

txtx

ty +

=

This case is very similar to that given in Ex. 3.2, except now, with only u(t) on the RHS of the

defining ODE, the b vector can be written as [ ]T100 =b .

In addressing the second part of the problem, recall from Section II that the characteristic

equation associated with the homogeneous form of the given equation can be written as

0322

13

=+++ aaa


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Now, the eigenvalues of the state matrix are given by

( )

==

123

10

01

0det

aaa

IA

and expanding along row 1 gives

( ) 0)(1)(det 322

1

3

321

2 =+++=++= aaaaaa IA

This is exactly the same result as above. The eigenvalues of the state matrix for a linear

stationary system are identical to the roots of the characteristic equation of the original linear

constant coefficient ODE.

Thus, ( ) 0det = IA is also referred to as the characteristic equation.

Conversion of Mixed Algebraic and Differential Equations

The general formulation in eqn. (3.1) has only one derivative term in each equation. This is

the standard form. However, a linear system can have more than one derivative per equation.

For example, consider the following 2nd order system,

)(4)()(121

tuxtxdt

dtx

dt

d=++

04)(3)( 212 = xtxtxdt

d

Putting these expressions into matrix form gives

)(0

1

43

04

)(

)(

10

11

2

1

2

1tu

x

x

tx

tx

dt

d

+

=

Thus, a more general form for the state equations for linear stationary systems is

)()()( tttdt

d

BuAxxE += (3.13)

Now to put this into standard form, we simply multiply by E1, giving

( ) ( ) )()()( 11 tttdt

duBExAEx

+= (3.14)

Where ( )AE 1 now becomes the system matrix.

Another common situation that occurs is to have a complex dynamic system that is described

by a set of mixed algebraic and differential equations. Such a system is said to have embeddedstatics since the algebraic equations represent relationships that do not change with time.


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In general, systems with embedded statics can be reduced in order (by the number of

algebraic equations) via a series of algebraic manipulations. A general procedure for doing

this is as follows:

Step 1. Order the equations so that n1 equations contain derivative terms and the next n2

equations only have algebraic relationships

Step 2. Write the original equations using partitioned matrices (with g = Bu), giving

+

=

a

d

a

d

a

d

g

g

x

x

CC

CC

x

xE

)(

)(

)(

)(

10

1

2221

121111

t

t

t

t

dt

d

where

xd= vector containing elements that have derivative terms

xa= vector containing elements that appear without derivatives

Step 3. Expand the matrix expression in Step 2,

dddgxCxCxE ++= a

dt

d121111

aad gxCxC0 ++= 2221

Step 4. Solve the algebraic expressions forxa in terms ofxdand substitute into the differential

equation

)( 211

22 dda gxCCx +=

and

daddd ggCCxCCCxCxE += 1221221

1

22121111dt

d

Step 5. The new system written in standard form is

wAxxdd +=

dt

d(3.15A)

with

(21

1

221211

1

11 CCCCEA = (3.15B)

( )ad gCCgEw1

2212

1

11

= (3.15C)

Example 3.4 gives a simple, but illustrative, demonstration for dealing with systems withembedded statics.


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Example 3.4 Treating Mixed Algebraic and Differential Equations

Problem Statement:

Put the following system in standard state form:

Problem Solution:

Let's write these equations in matrix form and partition them accordingly,

This is now in the form

which has solution

where

and


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Linearization of Nonlinear Systems

In general, all systems are nonlinear. However, linear analysis methods can usually be usedfor systems analysis in some limited range around some reference state. The goal here is to

develop a method for linearizing the general nonlinear state variable equations.

The variable coefficient terms in the original equations are also often treated in a similar

manner as the nonlinear terms. This is done because most of the analytical techniques are

applicable only for linear stationary systems.

A general nonlinear nonstationary system can be written as

where all the nonlinear and variable coefficient terms are included within .

Expressing the dependent state variables and independent forcing function as deviations from

some normal operating state, , gives

Substituting these expressions into eqn. (3.16) gives eqn. (3.18) as follows:

The nonlinear nonstationary terms can be written in a first-order Taylor series expansion

about the operating point. For the ith component of the vector, the first-order approximation

gives

The last two summations simply represent matrix multiplication operations. Thus, eqn. (3.19)

written in matrix notation gives eqn. (3.20), or

where are referred to as Jacobian matrices (evaluated at the

reference operating point). These matrices can be represented symbolically as


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For example, written out in detail is given as

Note also that at the reference operating point, one has

Now, substitution of eqns. (3.20) and (3.22) into eqn. (3.18) gives

This latter expression represents a linearization of the original nonlinear nonstationary system.

The new system matrix for the linearized system becomes and the new input

matrix operator becomes . Note that these matrices are functions of the initial

reference operating point, . This is true since we have linearized about this condition.

Example 3.5 illustrates the use of the above expressions.

Note also that theperturbation form given in eqn. (3.23) is often useful even if the original

system is linear and stationary, since, by definition, the initial conditions for the new state

vector, , are zero.


The goal here is to present a recipe for the conversion of certain classes of problems into state

form. Three specific areas to be addressed are:

1. State space representation of nth order systems,

2. Treatment of mixed algebraic and differential equations, and


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3. Linearization of nonlinear systems.

Conversion of nth Order Linear Differential Equations (with derivatives)

Given the following nth order system,

let

or, for the jth state variable

or

In these expressions, is defined as


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Using eqns. (3.7) and (3.8), one gets the following SISO matrix representation of eqn. (3.6):

with

and

Example 3.2 illustrates the use of the above general conversion algorithm. Note that for the

special case where the derivatives of the forcing function are not present (i.e. for RHS = u(t)

only), we have and d = 0. This latter situation is actually the usual

situation, and it is much easier to visualize and implement the conversion algorithm (since all

the are zero except ). Example 3.3 illustrates this less general, but very common

case. It also highlights the importance of the state matrix in determining the solutions to linear

stationary systems. We will focus further on the properties of the state matrix in later

subsections.

Conversion of Mixed Algebraic and Differential Equations

The general formulation in eqn. (3.1) has only one derivative term in each equation. This is

the standard form. However, a linear system can have more than one derivative per equation.

For example, consider the following 2nd order system,
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Putting these expressions into matrix form gives

Thus, a more general form for the state equations for linear stationary systems is

Now to put this into standard form, we simply multiply by , giving

where now becomes the system matrix.

Another common situation that occurs is to have a complex dynamic system that is described

by a set of mixed algebraic and differential equations. Such a system is said to have embedded

statics since the algebraic equations represent relationships that do not change with time.

In general, systems with embedded statics can be reduced in order (by the number of

algebraic equations) via a series of algebraic manipulations. A general procedure for doing

this is as follows:

Step 1. Order the equations so that n1 equations contain derivative terms and the next n2

equations only have algebraic relationships

Step 2. Write the original equations using partitioned matrices (with ), giving

where

vector containing elements that have derivative terms

vector containing elements that appear without derivatives

Step 3. Expand the matrix expression in Step 2,


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Step 4. Solve the algebraic expressions for in terms of and substitute into the

differential equation

and

Step 5. The new system written in standard form is

with

Example 3.4 gives a simple, but illustrative, demonstration for dealing with systems with

embedded statics.


In general, all systems are nonlinear. However, linear analysis methods can usually be used

for systems analysis in some limited range around some reference state. The goal here is to


The variable coefficient terms in the original equations are also often treated in a similarmanner as the nonlinear terms. This is done because most of the analytical techniques are

applicable only for linear stationary systems.






17/32




gives

The last two summations simply represent matrix multiplication operations. Thus, eqn. (3.19)

written in matrix notation gives eqn. (3.20), or






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The new system matrix for the linearized system becomes and the new input

matrix operator becomes . Note that these matrices are functions of the initial





vector, , are zero.

Example 3.5 Linearization of a 2nd Order Nonlinear System

Problem Statement:

Write a linear approximation for the system described by

Take the reference state as the equilibrium point (i.e. ) with .

Problem Solution:

Since we must expand the system around , we first must identify the reference state. In

this case, one has

or

and

Rearranging these expressions gives


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and

or

Therefore, the two equilibrium states are

and

The Jacobian matrix associated with the state vector can be expressed as (note that for

this case)

For the reference state with , the linearized system is

and for , the linearized system is given as

These linear stationary systems represent linear approximations to the original system (about

different initial states).



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In general, all systems are nonlinear. However, linear analysis methods can usually be used

for systems analysis in some limited range around some reference state. The goal here is to


The variable coefficient terms in the original equations are also often treated in a similar

manner as the nonlinear terms. This is done because most of the analytical techniques areapplicable only for linear stationary systems.








gives

The last two summations simply represent matrix multiplication operations. Thus, eqn. (3.19)written in matrix notation gives eqn. (3.20), or




21/32





The new system matrix for the linearized system becomes and the new inputmatrix operator becomes . Note that these matrices are functions of the initial





vector, , are zero.


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Discretization of Linear Stationary Systems

As just discussed, the computer implementation of any potential solution scheme is usually a

prerequisite for the analysis and simulation of realistic systems. In general there are two good

approaches used in computer simulations of high order dynamic systems. These include thediscretization of the analytic solution scheme for linear stationary systems and general

numerical integration techniques that can be adapted to any system of interest (i.e. non-

stationary and non-linear).

The basic goal of the discretization process is to convert the standard continuous linear

stationary state equations to discrete form. That is, given

convert the continuous system into a difference equation of the form

where Tis the sampling period and and are constant matrices. For convenience, eqn.

(3.44) is often written simply as

These expressions represent a discrete recurrence relationship that can be evaluated quite

easily on the computer if the and matrices are known. Our goal here is to derive explicit

expressions for these matrices.

To accomplish this transformation, we must make an assumption concerning the behavior of

. In fact, the only assumption in our development is that can be written as a

piecewise constant function of time. Therefore, , where it is implied that the

forcing function is constant over the kth interval.

To derive the discrete form, recall that the solution to the continuous linear stationary problem

is

Letting t = (k+1)T and t0 = kT, one has


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Introducing a new variable such that or with

and

one has

Since

our final result is

By comparison with eqn. (3.45), explicit expressions for the and matrices are

and

Another form for may be more appropriate if one plans to use the infinite series expansionfor generating a numerical result for the matrix exponential. Recall that

If is expanded in a similar manner, one has


24/32

with

Note that is similar to and they both can be generated within the same routine with little

additional effort. Thus, can be found via eqns. (3.48) or (3.49).

In either case, once the matrices and have been generated, the numerical simulation of

the system can proceed using the recursion relation given in eqn. (3.45). Note that this method

gives an exact solution assuming is indeed constant within the sampling interval T.

The algorithm implied by eqns. (3.46) - (3.50) is relatively straightforward to implement on

the computer. Matlab's implementation of the discrete solution for LTI systems is roughly

similar to this algorithm.

Numerical Integration of the State Equations

An alternate solution technique to using the matrix exponential approach is to simply integrate

the defining state equations. Our goal will be to give a simple illustration that introduces the

proper terminology and describes the basic concept behind all one-point numerical integration

methods.

The advantage of numerical integration methods is that they can easily handle time-varying

systems and relatively complicated nonlinear problems.

As a brief introduction to numerical integration, we will look at the two simplest methods

available; the Eulerand Modified Eulermethods. Although these methods present the basic

concepts, they are quite inefficient relative to some of the more advanced techniques. The

methods presented here are intended only as an illustration of the basic methodology, and

these techniques are not normally implemented as derived.

Euler Method (Scalar Case)

As a starting point, consider the general first-order differential equation


25/32

In the Euler method, one assumes that the right hand side of eqn. (3.51) is constant over some

small time interval . With this condition, integrating eqn. (3.51) gives

or

Thus, the problem has been converted to a recurrence relation for x k+1 in terms of xkand the

function evaluated with the solution at time tk. This is a relatively crude approximation but, if

is chosen sufficiently small, this simple scheme can give reasonable results.

Modified Euler Method (Scalar Case)

A somewhat better estimate for integrating the defining equation over some specified is to

assume that f(x,u,t) varies linearly over the given step. This gives the following recurrence

relation,

The problem with eqn. (3.53) is that we need xk+1 to calculate fk+1. A possible solution is to use

the Euler method to get a first guess for xk+1 and fk+1 and then use this guess to get a betterestimate of xk+1. This crude Predictor-Corrector scheme can be summarized as follows:

Predictor Step

Corrector Step

where

with x'k+1 representing a first estimate (prediction) of the value for the dependent variable at

tk+1 and xk+1 being the best estimate (correction) for this time point.


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The modified Euler method is a simple, but illustrative example of a predictor-corrector

numerical integration scheme. Other refinements such as adaptive control/updating of the

integration step, , can be done by addressing the error in the predictor and corrector steps.

In most current algorithms, one only inputs a desired tolerance level and a variable integration

step is computed that will satisfy the specified error criterion.

The above schemes for the scalar problem can be easily generalized for solution of large

systems of equations. In this case we desire the solution of a general system of first order

equations,

By analogy to the scalar problem, one has the following expressions:

Euler Method (Matrix Case)

where

Modified Euler Method (Matrix Case)

Predictor Step

Corrector Step

where

The most important part of the above simple schemes is the choice of an appropriate time

interval, . For the solution of several state equations, the choice of becomes even more

important and difficult, since some state variables may vary rapidly while others are slowly

varying functions of time. The more advanced schemes usually address this concern by

automatically adjusting as needed to meet user specified accuracy requirements. Thus,


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"Adaptive Predictor-Corrector" methods are the norm for performing realistic time domain

simulations.

Example 3.6 An Example Illustrating Various Solution Schemes

A general 2nd

order linear stationary system can be given as

Using the general recipe for converting nth order systems into state form, one has

where

Also,

where x1(0) and x2(0) represent the initial conditions for the system.

As a specific example, consider the simple series RLC circuit shown in Fig. 3.1.

From Kirchoffs voltage law, we have

where ea(t) is the applied voltage and the voltage drops across the individual components are

given by


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where i(t) is the current in the loop. Substituting these relationships into the balance equation

gives

This combined differential and integral balance equation is put into standard differential form

by differentiating each term, or

Comparing this system to the general representation for a 2nd order LTI system, we have

Thus, the standard state space representation for this specific system is

where

and

and u(t) is the input voltage, ea(t), and y(t) is the resultant current, i(t), versus time.

For specificity in the numerical solutions, lets assume the following:

1. For the reference case, R = 100 ohms, L = 0.1 henry, and C = 0.001 farad.


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2. Let ea(t) be a step input of 10 volts. Thus, u(t) = 10 volts for all t > 0.

3. The sensitivity of i(t) to changes in these parameters can also be addressed.

4. To illustrate the various solution techniques, lets perform a time domain simulation for this

system using three different methods:

A. Continuous analytical solution using the closed form representation of

B. Discretization of the LTI system

C. Numerical integration of the state equations

All three solution schemes were implemented in Matlab within files LTIDEMO1.M and

SSEQN1.M. A brief overview of each method is given below.

Solution Method A

For a step input, the analytical solution for a LTI system can be written as

This expression can be simplified as follows:

or

where, in the last manipulation, we used that fact that and commute.

To implement this last expression we used Sylvesters Theorem to evaluate . This

required that the eigenvalues be computed from

All other matrix manipulations were completed directly within Matlab. The time domain

solutions for the nominal set of parameters given above and for the case where R = 10 ohms

are plotted in Figs. 3.2 and 3.3, respectively. Note that R = 100 ohms gives and overdampedresponse and an underdamped solution is obtained for R = 10 ohms. This behavior can be

explained simply by computing the eigenvalues for the two cases.
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Solution Method B

The discrete solution of the LTI state equations is incorporated as part of Matlabs Control

Toolbox. This toolbox has a whole series of functions for the analysis of linear stationary

systems using the same terminology developed in these notes. In Matlab 5.0 and higher a new

object-oriented approach to working with systems was implemented. In particular, Matlab

defines an LTI state-space object via knowledge of the matrices from the

standard state-space representation. For this situation the ss command is used as follows (see

Matlabs help file for function ss):

sys = ss(A,B,C,D)

This creates an LTI object named sys that can be manipulated within several other Matlab

functions. For the case of a unit step input, for example, one has

[Y,T,X] = step(sys)

or

[Y,T,X] = step(sys,T)

where Y and X contain the output and state vector time domain responses in a 3-D array

format. The number of rows in Y and X is determined by the number of time points in the

equally-spaced time vectorT. The Y array has as many columns as outputs and X has as

many columns as states (i.e. the order of the system). Finally, the number of pages (the length

of the 3rd dimension) is equal to the number of inputs to the system. Thus, there is a 2-D

matrix for each system input.

In the first form given above, the sample period and the number of time points in the T vector

is determined automatically by Matlab. In the second form, the T vector is specified by the

user and passed into Matlabs step function. Note that Matlab has similar functions, impulse

and lsim, respectively, if the time domain input function is a unit impulse or a general input

vector.

For the specific RLC circuit of interest in this example, the appropriate calling sequence is

sys = ss(A,10*B,C,D);

[Y,T,X] = step(sys,T);

since there is only a single input quantity and its strength is 10 units (and the default is for a

unit step input). The Matlab m-file that implements the step function actually call two other

m-files. The first one, c2d, converts the continuous state space system into a discrete state

space representation by computing the and matrices given in eqns. (3.47) and (3.48)

using Matlabs built-in matrix exponential routine, expm, with a sampling time determined by

the constant interval width in vectorT. It then calls ltitr (Linear Time Invariant Time

Response), which simply implements the recursive expression given in eqn. (3.45).


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As expected, the solutions using Method B are identical with those from Method A. The

student is referred to the second part ofLTIDEMO1.Mfor the details of the Matlab

implementation, and to Figs. 3.2 and 3.3 for displays of the output responses using Method B.

Solution Method C

The last method to be highlighted in this example is the numerical integration technique

implemented with Matlabs ode23 routine. This routine applies an adaptive step control

algorithm by obtaining error estimates using two Runge Kutta (RK) predictions of different

order. The ode23 function uses a 2nd and 3rd order RK set for medium accuracy. An overview

of how to use the ode23 routine in Matlab can be obtained by simply typing helpode23 at the

Matlab prompt (this procedure is the best way to learn how to use any Matlab function). In the

current case, the call to ode23 should resemble the following,

[T,X] = ode23(sseqn1,[to,tf],xo,options);

where to and tfare the initial and final integration times, xo represents the initial condition ofthe state vector, and options is an ODE options structure within Matlab that allows the user to

control several options within the ODE numerical integration routine (see function odeset for

the various options that can be adjusted). The default options are satisfactory for many cases,

except for possible control of the user-specified error tolerance. The outputs from ode23

includes a time vector containing the discrete time points where the state is evaluated, and the

state vector in a matrix whose columns are the various elements of the state versus time. Note

there that the time vector is determined automatically within ode23 and it is usually not

evenly spaced because of the adaptive step control algorithm that is used.

The function name, sseqn1, points to the function routine that evaluates that right hand side of

the state balance equation at each time point. This routine is supplied by the user, and it can

be specialized for any particular case of interest. For the current RLC series circuit simulation

with a step input, this function is quite straightforward as seen in SSEQN1.M.

The results from Method C, as expected, are identical to those from the other techniques. This

can be seen by comparing the results in Figs. 3.2 and 3.3. In the current example, Method B,

the discrete solution method, is clearly the easiest to use since the problem fits exactly into the

LTI class of problems, and Matlabs ss, step, impulse and lsim routines were designed to

easily treat LTI systems. For nonlinear or variable coefficient systems, the numerical

integration technique with the ode23 routine would be used (with some modification to

SSEQN1.M).
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