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10.4 Linear analysis as an aid to simulation

Proper choice of the step size t is essential for successfully

applying numerical simulation methods. When there is

insufficient physical insight to choose an appropriate value fort , we can use sometimes use a linear analysis of the model to

estimate the required value of t . The analysis techniques for

linear and linearized models can be used when:

1. The model to be simulated is linear, but the order is high

enough or the input functions are complicated enough to

discourage a closedform solution.

!. The model to be simulated is nonlinear. "n this case, the

model can be linearized for nominal values of the system#svariables that represent the simulation conditions.

"n either case, a linear analysis can be used to estimate the

smallest time constant and the smallest oscillation period of the

model. The initial choice for the time step t is then ta$en to be

a small fraction of the smallest of these. The situation is

illustrated in %igure 1&.'. "n 1&.'a the smallest oscillation periodPis less than the smallest system time constant . "n this case,a reasonable initial choice for 1&&(Pt= , assuming that 1&&

points are sufficient to identify the form of the oscillatory

function for one period. "n 1&.'b, the opposite is true, and we

might choose 1&(=t , since a decaying curve with a time

constant can be plotted well with 1& points. The reason for

these choices is that the time step t must be small compared to

the time span over which significant changes occur in the

system variables.

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%igure 1&.' )sing the linear response to choose the step size t .

*a+ The oscillation period Pdetermines t . *b+ The time

constant determines t .

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Example 10.4

)se simulation to compute the time it will ta$e to fill a tan$ with

liquid to a height of ft if the input flow rate is tq -= ft-(min.

The initial height is 1+&* =h ft.

olution:

The hight h of liquid in a tan$ with an orifice as in the

following figure can be described by a model of the form:

gAChAqAhd !,

1

11==

uppose the given parameter values are ==11A .

%or the given values, the model is:

+,* qhfhqh ==

*1&.'1+

%rom the linearized analysis of the /0ample, the tan$#s time

constant is != when 1=h and = when =h . We can chooset to be 1&(1 of the smallest 2 that is, !.&+!*1.& ==t . The

results using the 3unge4utta method are shown in %igure 1&.5,

and they were chec$ed for accuracy by using onehalf the

original value of t . The results were essentially the same, so

we can ta$e !.&=t to be an appropriate value. The time to raise

the height to ft can be seen from %igure 1&.5 to be

appro0imately !. min.

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%igure 1&.5 3amp response of the nonlinear tan$ model.

10.5 Simulation and step inputs

The step function models a sudden change in the system#s input

and as such represents perhaps the most severe test of system

performance. "t is commonly used for that reason, but because it

signifies an instantaneous change in the input, we must be sure

that our system model is valid for such an input.

6ere are four ways of finding the step response of a model withinput derivatives. The first two will wor$ only with linear

models.

1. %ind the response with 7aplace transform. This is the most

direct approach, but care must be used to interpret the results

realistically from a physical viewpoint.

!. )se artificial initial conditions to allow elimination of the

input derivatives from the model. The resulting model can

then be handled by any conventional solution method.

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-. 3edefine the system#s state variables to convert the model

into one without input derivatives.

'. 8odel the step input with a more realistic function that

does not possess discontinuities.

Laplace transform solution

9onsider the following model:

vvbyy +=+

1& *1&.51+&+&* =y *1&.5!+

"f +*tv is a unitstep input and 1=b , then

1&

.&1.&1

1&

1+*

++=

+

+=

ssss

ssY *1&.5-+

and

tety 1&.&1.&+* += *1&.5'+

This is plotted in %igure 1&.. otice that *1&.5'+ implies that1+&* =+y . This represents an instantaneous ;ump in the value of

y at &=t , which is impossible if y represents a physical

variable.

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%igure 1&. )nitstep response of equation *1&.51+.

Using artificial initial conditions

The 7aplace transform of *1&.51+ with a step input of

magnitude m and initial condition cy =+&* gives:

cs

mbm

cs

mbscsVbssYs

++=

=++=++=+

+1*+*+1*+*+1&*

*1&.55+

%rom this, we see that the effect of the derivative term

vb for

step input can be represented by a proper choice of initial value+&*y . That is, if we pic$ cbmy +=+&* , then the following model

can be used to replace *1&.51+ as long as the input is a stepfunction.

vyy =+

1& *1&.5+

cbmy +=+&* *1&.5

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Redefining the state variables

The model *1&.51+ can be converted into the equivalent form

vbxx +1&1*1& =+

*1&.5=+

bvxy += *1&.5+

+&*+&*+&* = bvyx *1&.51&+

%irst, we solve *1&.5=+ by any method to obtain the

intermediate variable +*tx . Then we use *1&.5+ to obtain the

variable of interest +*ty . %or &+&*,1 == yb and a unitstep input,

*1&.51&+ gives &+&* =x . /quation *1&.5=+ gives:

1&

.&.&1

1&

+*

++

=

+

=

sssssX

or

tetx 1&.&.&+* +=

%rom *1&.5+ for +&t ,

tety 1&.&1.&+* ==

which agrees with *1&.5'+.

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+*+>*?

ass

matvL

+= *1&.51-+

as

ma

dt

vdL

nn

n

n

+

=

+1+1*>? *1&.51'+

The larger the constant a is in *1&.'11+, the closer +*tv

resembles a pure step function. This is illustrated in %igure 1&.

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1&

1

+*

+*

+

+=s

bs

sV

sY*1&.51+

%rom *1&.51-+,

+*+*

ass

masV

+= *1&.51

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%igure 1&.= 3esponse of the model *1&.511+ to a pure step

input, and to the practical step function for two values of the

parameter a .

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10. !ector"matri# methods

The state variable form of a linear model can be e0pressed

compactly using vectormatri0 notation. This compactness can

be e0ploited by simulation and analysis techniques.

!ector"matri# form of the state e$uations

The general form of the secondorder linear model in terms of

the state variables 1x and !x is:

!1!111!1!1111 ububxaxax +++=

*1&.1+

!!!1!1!!!1!1! ububxaxax +++=

*1&.!+

where we have assumed that two inputs +,* !1 uu act on the

system. This is the state variable form of the model. @ssume

also that there are two outputs +,* !1 yy , which are linear

combinations of the state variables and the inputs. Then

!1!111!1!1111 ududxcxcy +++= *1&.-+

!!!1!1!!!1!1! ududxcxcy +++= *1&.'+

We can form a twodimensional column vector # with two state

variables 1x and !x 2 that is,

=

!

1

x

x# *1&.5+

The vector # is the state vector. imilarly, the input vector is:

=

!

1

u

uu *1&.+

and the output vector is:

=

!

1

y

yy *1&.

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%rom the rules of matri0vector multiplication, the state

equations and the output equations *1&.1+*1&.'+ can be

written as:

+

=

!

1

!!!1

1!11

!

1

!!!1

1!11

!

1u

u

bb

bb

x

x

aa

aa

x

x

+

=

!

1

!!!1

1!11

!

1

!!!1

1!11

!

1u

u

dd

dd

x

x

cc

cc

y

y

or

%u# +=

*1&.=+'u(#y += *1&.+

where

=

!

1

x

x# *1&.1&+

ija=& *1&.

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*1&.'+. This diagram is called the state diagram, because it is

constructed from the state variable form of the model and thus

graphically shows how the state variables are generated.

Example 10.5

Araw the state diagram for the following model:

!!!

1!11

5-

'!'

uxx

uxxx

+=

++=

The output variables are

!!

1!11 -

xy

uxxy

=

++=

@ssuming zero initial conditions, we transform the model and

divide the first two equations by s to obtain:

+*+*

+*-+*+*+*

+>*5+*-?1

+*

+>*'+*!+*'?1

+*

!!

1!11

!!!

1!11

sXsY

sUsXsXsY

sUsXs

sX

sUsXsXs

sX

=

++=

+=

++=

)sing the guidelines for the inputs and outputs of the two

integrators, we obtain the diagram in %igure 1&.=.

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%igure 1&.= tate diagram for /0ample 1&.5.

The state equations and output equations need not be linear to be

represented in vector form. The general vector form is2

)uf*#+# t,=

*1&.15+u)g*#+y = *1&.1+

,umerical simulation methods in vector form

The compactness of *1&.15+ and *1&.1+ ma$es it easy to

describe a numerical simulation method in a completely general

way. %or e0ample, the /uler method applied to *1&.15+ yields

>+,*+,*?+*+* 1 kkkkk tttttt u### +=+ *1&.1

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Euler predictor+,,*1 kkkkk tt u#f#- +=+ *1&.1=+

Trapezoidal corrector+>,,*+,,*?5.& 1111 ++++ ++= kkkkkkkk ttt u-fu#f## *1&.1+

where we have suppressed the kt notation by letting +* kk t## = .

The 3unge4utta algorithm can be e0pressed in vector form in

li$e manner.

Transfer function methods

The transfer function describes the effect of a given input on a

given output. "t does not describe the effects of other inputs orinitial conditions on that output. Therefore, the other inputs and

all initial conditions are set to zero when deriving a transfer

function. That is, the transfer function between the output iy

and the input ju is

kxjksUj

i

ij

kk

sU

sYsT

allfor,&+&*,,&+*+*

+*+*

==

= *1&.!&+

ote that the derivative *or integral+ of a vector or matri0 is the

vector or matri0 of the derivatives *or integrals+ of the elements.

Therefore, since the 7aplace transform is an integration

operator, the transform of a vector is the vector consisting of the

transforms of the elements2 namely,

=

+*

+*+*

!

1

sX

sXsX *1&.!1+

Therefore,

+&*+*+&*+*

+&*+*+*

!!

11#. =

=

ssxssX

xssXxL *1&.!!+

Transforming *1&.=+ and *1&.+ and using this result, we

obtain

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+*+*+&*+* ssss %U&.#. += *1&.!-++*+*+* sss 'U(./ += *1&.!'+

The first of these can be arranged as

+*+&*+*+* sss %U#.&0 += *1&.!5+

where the identity matri0 0 has been used to allow +*s. to be

factored out of the e0pression. The matri0

&0R =ss+* *1&.!+

is the resolvent matri0. "ts properties will tell us much about thesystem#s behavior.

The transfer function is defined for zero initial conditions. With&+&* =# , *1&.!5+ gives

+*+*+*1

sss %U&0. = *1&.!

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Example 10.6

)se *1&.!+ to find +*1! sT for the model

!!!

1!11

5-

'!'

uxx

uxxx

+=

++=

The output variables are

!!

1!11 -

xy

uxxy

=

++=

olution:

%or this model,

=

=

=

=

&&

&-

1&

11

5&

&'

-&

!'

'(

%&

Thus,

+

+=

=

+-*&

!+'*

-&

!'

1&

&1

s

sss &0

The inverse of *!0!+ matri0 is:

=

11!1

1!!!

!11!!!11

1 1

mm

mm

mmmm *1&.-&+

7etting &0 =s , we see that

+

+++

=

+

+

++=

-

1&

+-+*'*

!

'

1

+'*&

!+-*

+-+*'*

1

+* 1

s

sss

s

s

sss &0

and

+

+

+++

=

&&

&-

5&

&'

-

1&

+-+*'*

!

'

1

1&

11+*

s

ssssT

The transfer functions are obtained by carrying out the indicatedmultiplications.

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They are:

-

5+*

&+*

+-+*'*+*5

-5

+-+*'*1&+*

'

1--

'

'+*

!!

!1

1!

11

+=

=

++ +=++++=

+

+=+

+=

ssT

sT

sss

ssssT

s

s

ssT

The state vector notation also allows a compact description of

the system in graphical form, called the state vector diagram.

This can be either a bloc$ diagram or a signal flow graph. Both

are shown in %igure 1&. for the system described by *1&.=+and *1&.+. The reduced transfer function diagram is shown in

%igure 1&.c. "n the vector diagram, the signals *now multiple or

vector quantities+ are shown with a doublelined arrow.

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The transition matri#

The 7aplace transformation for the state equation %u# +=

is:

+*+&*+*+* sss %U#.&0 += *1&.-1+

%or the freeresponse calculation we can ta$e +*tu to be zero.

This gives:

+&*+*+* #.&0 = ss *1&.-!+

or+&*+*+*

1#&0. = ss *1&.--+

This says that the inverse of the resolvent matri0 can be

considered as the transfer function matri0 relating the free

response +*t# to the initial condition +&*# . ince +&*# is a

vector of constants, *1&.--+ shows the free response to have

the form

+&*+*+* #2# tt = *1&.-'+

where the matri0 +*t2 is given by:

>+?*+* 11 = &02 sLt *1&.-5+

The matri0 +*t2 is the transition matri0.

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Example 10.7

9ompute the transition matri0 given the state equations:

!1!

!1

5' xxx

xx

=

=

olution:

The resolvent matri0 is:

+

=

+5*'

1

s

ss &0

and

+

++=

+

++==

s

s

sss

s

ssss

'

1+5*

+'+*1*

1

'

1+5*

'+5*

1+*+* 1&02

*1&.-+

Thus, with a partial fraction e0pansion,

'

1

-

1

1

1

-

'

+'+*1*

+5*+*D11

+

+=

++

+=

ssss

ss

ortt

eet'

11-

1

-

'+*D

=

Performing the same procedure with the other three entries in

the matri0, we obtain:

tt

tt

tt

eet

eet

eet

=

=

=

-

1

-

'+*D

-'

-'+*D

-

1

-

1+*D

'

!!

'!1

'

1!

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Example 10.8

9ompute the transition matri0 for the system of the previous

e0ample, given the state diagram shown in %igure 1&.1&a.

%igure 1&.1& *a+ tate diagram for /0ample 1&.. *b+ Partially

reduced diagram with &+&*! =x .

This approach considers the elements of +*s2 to be the transfer

functions between the two outputs +*1 s. and +*! s. , and the twoinputs +&*1x and +&*!x . et &+&*! =x temporarily, and reduce the

diagram to find the transfer functions between +*1 sX , +*! sX and+&*1x . %rom %igure 1&.1&b, the result is:

'+5*

'

+&*

+*+*D

'+5*

5

+&*

+*+*D

1

!

!1

1

1

11

++

==

++

+==

ssx

sXs

ss

s

x

sXs

These are the entries in the first column of +*s2 , as given by

*1&.-+. imilarly, with &+&*1 =x , we can reduce the diagram to

obtain +*D1! s and +*D!! s . The rest of the procedure follows as in

/0ample 1&.< in order to obtain +*t2 .

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(alculating the forced response from +*t2

"f &+* tu , the resulting response can be found from *1&.!5+ as

follows.

+*+*+&*+*+*11

ssss %U&0#&0. += *1&.-*+*? %u%U2 *1&.-+

ince >+?*+* 11 = &0sLt , *1&.-+ must be the forced response.

Therefore, the complete response from *1&.-

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3roblems

1&.1 )se the /uler method to solve the equation 1! +=

yy , with&+&* =y for 1& t using *a+ 1.&=t and *b+ &1.&=t . 9ompare

the results with the e0act solution tty tan+* = .

1&.! 3epeat Problem 1&.1 using the modified /uler method.

1&.- 3epeat Problem 1&.1 using a version of the 3unge4utta

algorithm.

1&.' )se a numerical method to find the response of the model

ttyy =+

! for !& t if &+&* =y .

1&.5 Civen the following transfer functions, find a state variable

model for each. @lso, obtain the e0pressions relating the initial

values +&*ix of the state variables to the initial values of y and

its derivatives.

*a+ -'

!

+*

+*

!++

+=

ss

s

sV

sY

*b+-'

1-

+*

+*!

!

++

++=

ss

ss

sV

sY

1&. )se a numerical method to compute the time it will ta$e to

fill a tan$ with liquid to a height of 1& feet, if the input flow rate

is ==q ft-(minute. The initial liquid height is 1+&* =h ft. The

tan$#s model is hqh !=

.

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1&.< Civen the following model:

!!!

1!11

'

!-5

uxx

uxxx

+=

++=

and the output equations

!!

1!11 !-

xy

uxxy

=

++=

Araw the state diagram.