CD Reference

8/12/2019 CD Reference

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LabVIEW Control Design andSimulation ModuleAlgorithm References

This document contains an index of Control Design VIs that use specific

algorithms to calculate the output of the VI. These algorithms include ones

derived by National Instruments and published algorithms. Table1-1lists

the VIs of the LabVIEW Control Design and Simulation Module that use

special algorithms to calculate the VI outputs. Table1-1lists the VIs by

palette. The first column contains the name of the palette, the second

column contains the name of the VI, and the third column provides the

name of the algorithm and links to the sections that contain the algorithm

derivation.

The derivations of the algorithms are located in sections that follow

Table1-1. Each section also contains references to published algorithms,

which National Instruments used to implement the Control Design VIs.

This document also contains the derivations of theLyapunov and Sylvester

Equationsand theRiccati Equations.


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Control Design and Simulation Module Algorithm Reference 2 ni.com

Table 1-1. Algorithm References for Control Design VIs

Palette Name VI Name Reference

Dynamic Characteristics CD Covariance

Response VI

Lyapunov Equation Solver

Discrete Lyapunov Equation Solver

Staircase

CD DC Gain VI Convert State-Space Model toTransfer Function Model

Staircase

CD Norm VI Lyapunov Equation Solver

Convert State-Space Model to

Transfer Function Model

Discrete Lyapunov Equation Solver

State-Space Norm

Staircase

CD Parametric Time

Response VI



Linear Response

Staircase

Zero-Order-Hold

CD Pole-Zero Map VI Staircase

Transmission Zeros (State-Space)

CD Root-Locus VI Convert State-Space Model to


Root Locus Plots


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National Instruments Corporation 3 Control Design and Simulation Module Algorithm Reference

Frequency Response CD All Margins Convert State-Space Model to


All Margins (Linear System)

CD Bandwidth Convert State-Space Model to


CD Bode Convert State-Space Model to


Bode, Nyquist, and Nichols Analysis

CD Gain and

Phase Margins VI



Gain and Phase Margin (Linear

System)

CD Nichols VI Convert State-Space Model to



CD Nyquist VI Convert State-Space Model to



CD Singular Values VI Convert State-Space Model to


Singular Values

Staircase

Table 1-1. Algorithm References for Control Design VIs (Continued)



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Model Interconnections CD Append VI Staircase

CD Feedback VI Convert State-Space Model to


Staircase

Unit Feedback

CD Parallel VI Convert State-Space Model to


Pade Approximation of Delay for a

SISO Transfer Function Model

Staircase

CD Series VI Convert State-Space Model to

Transfer Function ModelPade Approximation of Delay for a

SISO Transfer Function Model

Staircase

Model Reduction CD Minimal Realization VI Staircase

Canonical State-Space Realization

Minimal State-Space Realization

CD Model Order

Reduction VI

Model Order Reduction




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State Feedback Design CD Ackermann Ackermann

CD Kalman Gain VI Continuous Algebraic Riccati

Equation Solver

Discrete Algebraic Riccati Equation

Solver

Hamiltonian Reduction Algorithm

for Solving Riccati Equations

Kalman Gain

Continuous Algebraic Riccati

Equation Somer via Schur and QZ


Solver via Schur and QZ

Full Rank Factorization

CD Linear Quadratic

Regulator VI


Equation Solver


Solver

Hamiltonian Reduction Algorithm

for Solving Riccati Equations


Equation Somer via Schur and QZ


Solver via Schur and QZ

Full Rank Factorization

CD Pole Placement VI Lyapunov Equation Solver

Pole Placement

CD State Estimator State Estimator and State-Space

Controller

Unit Feedback

CD State-Space Controller State Estimator and State-Space

Controller

Unit Feedback




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Implementation CD Current Observer

Corrector / Predictor

Implementation Current Observer

(Point-by-Point)

CD Predictive Observer Implementation Predictive Observer

(Point-by-Point)

CD Discrete Recursive

Kalman Corrector /

Predictor

Adjust Stochastic State-Space System

with Non-Zero-Mean Noise

Excitation

Discrete-Time Recursive Kalman

Filter

Implementation

(Simulation Module)

CD Continuous Observer Implementation Continuous

Observer (Point-by-Point)

CD Continuous Recursive

Kalman Filter

ContinuousTime Recursive

Kalman-Bucy FilterAdjust Stochastic State-Space System

with Non-Zero-Mean Noise

Excitation




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Pole Placement

Assume the pair (A, B) is controllable and consider Kthe controller gain

such that it places the poles, Eig(A BK), in the location you specify.

Then, there exists a similarity transformation Tsuch that

Where represents the user-defined poles along the diagonal. Rearrangingthe terms, you get the following equation.

AT+ BKT= T

AT T= BKT

When KTis unknown, set KTequal to G. You get the following equation.

AT T= BG

Note that Tmust be invertible. Therefore, (, G) should be observable

The CD Pole Placement VI is implemented such that Gis random but

(, G) is observable.

Reference:

Varga, A.Robust Pole Assignment via Sylvester Equation Based State

Feedback Parametrization. IEEE International Symposium on Computer

Aided Control Systems Design, CACSD' 2000. Anchorage, AK, 2000.

Singular Values

Consider a system defined in terms of the state space model, and the

equivalent transfer function model, shown below.

T1

A BK( )T =

Eig ( ) Eig A( ) 0=

x Ax Bu+=

y Cx Du+=

X s( ) Is A( )1

BU s( )=

Y s( ) CX s( ) DU s( )+=

Y s( )

U s( )----------- C

Is A( )

1B D

+=


11/114


The CD Singular Values VI calculate the eigenvalues of the following

equation:

(): [(I A)1B+ D)]

This calculation uses a singular values decomposition. To show this result

in decibels, you convert as 20log(w).

First-Order-HoldThis algorithm implements what is called the triangle-hold equivalent of a

first-order hold. The effect of this hold is to extrapolate samples and

connect them in a straight line. In discrete form, this extrapolation is

non-causal. You derive the filter that does this extrapolation as follows.

Consider the following unit impulse.

A first-order hold transfer function gives the following impulse response.

In order to construct this function in the s-domain, apply the superposition

principle of ramp functions.


12/114


As shown in the figure, the ramp has a slope of , which you can describe

in the s-domain as follows:

slope =

At time T, the ramp starts at .

At time 0, the ramp changes the sign of the slope .

At time T, the response levels are .

The composite transfer function becomes

H(s) =

Placing the first-order hold in series with the system transfer function G(s),

you get the following:

Y(s) = H(s)*G(s) *U(s)= *G(s)*U(s)

Define the unit impulse function as

and the auxiliary input variables as

By applying the Inverse Laplace Transform to the definitions above and

expressing G(s) in state-space form, you get the following:

= Ax+ Bv

= u(t + T)(t + T) 2u(t)(t) + u(t T)(t T) =

You can place these expressions in one state-space model, as illustrated in

the following expression:

1

T---

1

s2

----

eTs

s2

------1

T---

2s

2---- 1T---

eTs

s2

---------1

T---

eTs

2 eTs

+

s2T

-------------------------------

eTs

2 eTs

+

s2T

-------------------------------

U s( ) eTs

2 eTs

+( )U s( )=

sW s( ) U s( )=

sT V s( ) W s( )=

x

v wT----=

x w u


13/114


where represents the unit impulse function. This model is shown in the

following figure.

Looking at theZero-Order-Holdsolution for the equation above,

in which case the impulse response becomes

= u[(K 1)T](t (K 1)T) 2u(KT)(t KT) +

u[(K + 1)T](t (K + 1)T)

because only consists of impulses

x

v

w

t( )

A B 0

0 0 I

T---

0 0 0

M

x

v

w

t( )

0

0

1

u

B

+=

u

t( ) M t( ) Bu+=

eAt

t( ) eMt

M t( ) eMt

Bu=

deMt

t( )dt

----------------------- eMt

Bu=

K 1+( )T[ ] eMt KT[ ] e M t K 1+( )T[ ] BudtKT

K 1+( )T

+=

u

u


14/114


Similarly for (t (K + 1)T)

Using Taylor Series expansion to evaluate the exponential of a matrix, you

get the following:

eMT= I + MT + +... =

In the case of matrix M,

M0= I; ; ;

; ... ;

Substituting of such terms into the Taylor Series Expansion results in the

following:

To simplify the expression for eMT, define the following matrices,

, ,

f t( ) t KT( ) f t( ) 0=KT

KT

=KTK 1+( )T

K 1+( )T[ ] eMT

KT[ ]=

MT( )2

2!---------------- MT( )

i

i!---------------

i 0=

M1

A B 0

0 0 1 T0 0 0

= M2

A2

AB B T

0 0 00 0 0

=

M3

A3

A2B ABT

0 0 0

0 0 0

= Mi

Ai

Ai 1

B Ai 2

BT

0 0 0

0 0 0

=

eMT MT( )

i

i!---------------

i 0=

AT( )

i

i 0=

AT( )i 1

B

i 1=

AT( )i 2

B

i 2=

0 1 1

0 0 1

= =

AT( )i

i 0=

= 1 AT( )i 1 B

i 1=

= 2 AT( )i 2 B

i 2=

=


15/114


which give

Therefore

x(k + 1) = x(k) + 1v(k) + 2w(k)

Based on Franklin Powell

v(k) = u(k)

w(k) = u(k + 1) u(k)

substituting into the state-space model results in the following:

x(k + 1) = x(k) + 1u(k) + 2[u(k + 1) u(k)]

= x(k) + [1 2]u(k) + 2u(k + 1)

Because all k + 1 terms need to be on the right-hand side of the equation,

you need the following definition:

(k) =x(k) 2u(k)

Evaluating the previous equation at k + 1 results in the following:

(k + 1) =x(k + 1) 2u(k + 1) =

= [x(k) 2u(k)] + 2u(k) + (1 2)u(k)

= [x(k) 2u(k)] + [1 2+ 2]u(k)

= (k) + [1 2+ 2]u(k)

eMT

12

0 I 0

0 0 I

=

x k 1+( )

v k 1+( )

w k 1+( )

12

0 1 0

0 0 1

x k( )

v k( )

w k( )

=

x k( )

1 2( )u k( )

1u k( ) 2u k( )+


16/114


From this expression you can recognize the following system matrix, which

is equivalent for the state dynamics,

= , = 1 2+ 2

and for the output calculation,

y(k) = Cx(k) + Du(k)

= C[x(k) 2u(k)] + C2u(k) + Du(k)

= C(k) + [C2+ D]u(k)

= C, = C2+ D

Notice that because the change of variables are made to have all k + 1 terms

on the left-hand side of the system equation, the initial conditions are

affected by the inputs at time t = 0 in the following manner:

Initial Conditions: (0) = (0) =x(0) 2u(0)

, ICMAT = [I 2]

where ICMAT is used to calculate the initial conditions based on the new

set of states, .

Reference:

Franklin, G.F., J. D. Powell, and M. L. Workman.Digital Control of

Dynamic Systems, 3th ed., p. 204. Reading, MA: Addison-Wesley, 1997.

Linear Response

Linear Response for continuous systems: After applying a zero-order-hold

conversion from continuous to discrete, the following assumptions are

made to the input profile:

In the case of a unit step, the input is hold at 1

In the case of an impulse response, you look at the effect of the impulse

before the system has been converted to discrete form. Therefore, in a

continuous system,

= Ax+ Bu

y = Cx+ Du

AdF.O.H

BdF.O.H

CdF.O.H

DdF.O.H

xdF.O.H

I 2[ ] x 0( )

u 0( )=

x


17/114


Taking the Laplace transform,

X(s)s x(0) = AX(s) + Bu(s)

X(s) = (Is A)1[x(0) + Bu(s)]

Because u(s) of impulse (0) is 1, you get

X(s) = (Is A)

1

[x(0) + B]

Note Impulse enter to the system as if you were using or shifting initial conditions.

Therefore, for impulse in input i, =x(0) + bi, where biis the ithcolumn of B.

Therefore, the system is discretized and the impulse enters as an initial condition

equivalent to bi.

Linear Simulation

Discrete System:

x(k + 1) = Ax(k) + Bu(k)

y(k) = Cx(k) + Du(k)

which makes integration straightforward after initial conditions are

defined.

Continuous System:

The system must be discretized, and the initial conditions must be adjusted

properly depending on the discretization method.

For Zero-Order-Hold: (Ac, Bc, Cc, Dc) (Ad, Bd, Cd, Dd)

IC not a function of u0. Integration is identical to Discrete Systems.

First-Order-Hold: Based on the algorithm in the First-Order-Holdsection,

the initial conditions needs to be re-evaluated based on the change of

variables.

which you can evaluate at t = 0.

xi*

0( )

k( ) x k( ) 2u k( )

0( ) I 2[ ] x 0( )

u 0( )=


18/114


Unit Feedback

Consider the following continuous state-space model.

for unit feedback in the SISO case, as shown in the following figure.

Notice there is no one-to-one correspondence between randyin the MIMO

case, as shown in the following figure.

Therefore, a Matrix describes connectivity for the feedback path. Forexample, in the figure above, you obtain that:

where u = r +y'= r + y

x

Ax Bu+=

y Cx Du+=

y1'

y2'

0 0 1

1 0 0

y1

y2

y3

=


19/114


Substituting, you get the following:

y= Cx+ D(r + y)

y= (I D)1Cx+ (I D)1Dr

or , where , .

Substituting the previous expressions in the state-space model dynamics

expression gives:

= Ax+ Bu

= Ax+ B(r + y)

where y is a function of the states and the reference, r:

= Ax+ B(r + + )

grouping common terms for the states and the reference:

= (A + )x+ B(I + )r

In this manner the equivalence of A and B matrices are identified:

= A + B(I D)1C

= B(I + (I D)1D)

In summary,

y Cx Dr+= C I D( )1C= D I D( )

1CD=

x

Cx Dr

BC D

A

B


20/114


is equivalent to

If (I D) is not invertible, there is at least one unfeasible ri, as thefollowing example illustrates.

A = 0, B = 0, C = 0,

and connectivity

from G

y = u

Therefore r is not an independent input to the system but identical to the

null vector r = 0.

r is not chosen but fixed. In this case, you cannot calculate the unit

feedback transfer function.

D1 0

0 1=

u1 r1 y1=

u2 r2 y2=u r y+=


21/114


Grammians

Controller Continuous: By definition the Controllable Grammians is given

by the following expressions

and

The easier way to calculate the grammians is by pre and post multiplying

Wcby A and AT, and adding the two terms, as illustrated below:

Notice that the right-hand side of the equation can be grouped in the

following way:

Substituting the end values into the expression above and considering thatfor stable systems the exponential goes to zero at infinite, you get the

following expression:

AWc+ WcAT+ BBT= 0

which solved using Lyapunov.

Observer Continuous: By definition the Observable Grammians is given by

the following expressions:

,

Wc 0 tf( , ) eA

BBTe

AT

d

0

tf

Wc Wc 0 tf( , )tf lim

AWc WcAT

+ A eA

BBTe

AT

d

0

eA

BBTe

ATdA

T

0

+=

d eA

BBTe

AT

( )

d-------------------------------------d

0

=

eA

BBTe

AT

0

BBT

==

W0 0 tf( , ) eAT

CTCeAd0

tf

= W0 W0 0 tf( , )

tf lim


22/114


Following an identical reasoning than for the Controller Grammians, you

obtain the following expression:

which also is solved using Lyapunov.

Controller Discrete: The following expression defines the Controller

Grammians for discrete systems:

Multiply each side by A, ATrespectively

If you define J = k + 1, you get the following:

To reinitialize the index to start at J = 0 and make it identical to Wcthe term

BBT must be added and subtracted giving:

Therefore, AWcAT= Wc BBTis also solved using Discrete Lyapunov.

ATW0 W0A C

TC+ + 0=

Wc AkBB

TA

T( )

k

k 0=

n

Wc Wcn lim A

kBB

TA

T( )

k

k 0=

=

AWcAT

Ak 1+

BBT

AT

( )k 1+

k 0=

=

AJ

BBT

AT

( )J

J 1=

=

AWcAT

AJBB

TA

T( )

J

J 0=

BBT

=


23/114


Observer Discrete: The following expression defines the Observer

Grammians for discrete systems:

for which, following a similar reasoning in the Controller Grammians for

discrete systems, you get ATW0A = W0 CTC.

The previous expression is also solved using Discrete Lyapunov.

Implementation:

B is decomposed using SVD

B = USVTBBT= Us2UT

UUT= UTU = I

For Controller Continuous, substitution of the previous expression gives:

AWc+ WcAT+ Us2UT= 0

which could be pre- and post-multiplied by UT and U, respectively.

UTAWcU + UTWcATU + s2= 0

because Wcis symmetric.

Continuous Lyapunov

For Controller Discrete

AWcAT Wc+ Us

2UT= 0

W0 AT

( )kC

TCA

k

k 0=

n

W0 W0n lim AT( )kCTCAk

k 0=

=

UTAU

A

UTWcU

Wc

UTWcU

Wc

UTA

TU

AT

s2

+ + 0=

AWc WcAT

s2

+ + 0=


24/114


which could be pre- and post-multiplied by UT and U, respectively.

UTAWcATU UTWcU + s

2= 0

Discrete Lyapunov

Note: Wc= UWcUT

Reference:

Kailath, T.Linear Systems, 1st ed., p. 609. Englewood, NJ: Prentice-Hall,

1980.

Balancing

The CD Balance State-Space Models (Grammians) VI has only one entry

per column or row in the transformation matrix T.

The tolerance is defined as follows.

Rearranging the terms, you get the following relationship.

If TRUE, Tis not adjusted. If FALSE, set K equal to the following:

UTAU

A

UTWcU

Wc

UTA

TU

A( )T

UTWcU

Wc s

2+ 0=

T

1

2

3

4

=

ma xmi n----------- tolerance

ma xmi n-----------

1

tolerance---------------------- 1 0

Kma xmi n-----------

old

ma xmi n-----------

new

tolerance=,


25/114


The new maxand the new minare as follows:

This way, the new set of maxand min satisfies the tolerance.

Tis rescaled accordingly to the modifications made, and the procedure is

repeated until one of the following conditions are met.

1. K min= 0, Tis not full rank.2. K max= 0, Tis not full rank.

3. .

4. The number of iterations is equal to the dimension of T.

Reference:

Alan J. Laub, Michael T. Heath, Chris C. Paige, and Robert C. Ward,

Computation of System Balancing Transformations and Other

Applications of Simultaneous Diagonalization Algorithms. IEEETransactions on Automatic Control, vol. AC-21, num. 2, 1987.

Tustins Transformations

The Model Conversion VIs use Tustins methods to convert models from

discrete to continuous and from continuous to discrete. The following

section describes the algorithm used to implement the discrete to

continuous conversion.

maxnew maxoldtolerance

K----------------------=

minnew minoldK

tolerance----------------------=

ma xmi n-----------

new

ma xmi n-----------

old

tolerance

K----------------------

K

tolerance----------------------

-------------------------- tolerance==

ma xm in-----------

tolerance


26/114


Discrete to Continuous

*

*

*

*

I AT2

--------+ I AT

2--------

1

=

AT

2------------

I AT2

--------+=

I( ) I+( )T

2---A=

A I+( )1

I( )2

T---=

IAT

2--------

1

B T=

B1

T------- I

AT

2--------

=

1

T------- I I+( )

1 I( )( )=

H TC IAT

2--------

1

=

C 1T

-------H I AT2

-------- 1T

-------H I I+( ) 1 I( )( )==

J D C IAT

2--------

1 BT

2-------+=

D J C I I+( )1

I( )( )1 BT

2-------=

D J

1

T---

T

2---H I I+( )1

I( )( )

2

=

D JH

2---- I I+( )

1 I( )( )=


27/114


Reference:

Franklin, G.F., J. D. Powell, and M. L. Workman.Digital Control of

Dynamic Systems, 3th ed., p. 200. Reading, MA: Addison-Wesley, 1997.

Note The continuous to discrete conversion for Tustins method uses the algorithms

presented inDigital Control of Dynamic Systems.

Model Order Reduction

Step 1: Find a similarity transformation that place the state to eliminate at

the end.

For example,

Eliminate state 1 and 2. (Place them at the bottom of the state vector.) The

similarity transformation is given by the following similarity matrix T:

,

,

This way: ,

The states have been reordered. The following procedure explains how to

obtain T.

x

x1

x2

x3

=

T

0 0 1

1 0 0

0 1 0

= A

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

TA

a31 a32 a33

a11 a12 a13

a21 a22 a23

= TATT

a33 a31 a32

a13 a11 a12

a23 a21 a22

=

x

x3

x1

x2

= A TAT1

=


28/114


Consider the number of states to eliminate in descending order.

where

e = number of states to eliminate,

n = total number of states

Consider Tinitialized as the identity matrix, and counter index iinitialized

at 1.

1. Take Jirow from T, denoted as .

2. Take n irow from T, denoted as tni.

3. Swap them in T. In other words, substitute in row niand tniin

row Ji.

4. Make inew= i+ 1 and continue to point 1 until inew> e.

The Match Gain OptionMatch Gain allows you to maintain a steady-state gain while eliminating

states.

Consider the following continuous system description:

Where sstands for selected and estands for eliminated.

Sets the dynamics of the eliminated states to zero, , which is a

quasi-steady state assumption, where represents fast dynamics.

J

J1

J2

Je

=

tJi

tJi

x s

xe

Ass Ase

Aes Aee

xs

xe

Bs

Be

u+=

y Cs Cexs

xe

Du+=

xe 0=

x e

0 Aesxs Aeexess

Beu+ +=


29/114


Solving for

Substituting in the selected set of states dynamics, :

where now a new set of A, B, C and D matrices are identified, as illustrated

below.

Substituting fory, you get the following equation:

y= Csxs+ Ce[Aesxs + Beu] + Du

Note In the case of Discrete Systems, = (I A)1.

Canonical State-Space Realization

A state-space model can be transformed into a controllable canonical formusing the transformation matrix T defined as follows:

T = QW

where Q is the controllability matrix

Q = [B: AB: ... An1B]

where |SI A| = sn+ a1sn1+ ... + an1z + an= 0

x ess

xess

A ee1

Aesxs Beu+[ ] Aesxs Beu+[ ]= =

xs

xs Assxs Ase Aesxs Beu+[ ] Bsu+ +=

Ass AseAes+[ ]xs

Ass

Bs AseBe+[ ] u

Bs+=

Cs CeAes+[ ] xs

Cs

D CeBe+[ ] u

D+=

W

an 1 an 2 a1 1

an 2

an 3

1 0

a1 1 0 0

1 0 0 0

=


30/114


Reference:

Ogata, K.,Discrete-Time Control Systems. Englewood, NJ: Prentice-Hall,

1987.

Minimal State-Space Realization

The CD Minimal State Realization VI eliminates states that do not have anyeffect on the control problem. For example, this function eliminates states

that are unobservable and/or uncontrollable.

The controllability matrix is defined as,

Q= [BAB... An1B]

The observability matrix is defined as,

Each row of Cand OTrefers to the controllability and observability of the

corresponding state.

Let Cirefers to row iand Cjrefers to columnjin the matrix.

If CiCiT, the state is uncontrollable.

If OjTOj, the state is unobservable.

is a small constant that represents tolerance.

The CD Minimal State Realization VI eliminates all states that are either

uncontrollable or unobservable and returns a reduced model.

O

C

CA

:

CAn 1

=


31/114


Zero-Order-Hold

Continuous to Discrete: Consider the state dynamics equation:

Pre-multiply the expression above by the exponential term

The right-hand side of the equation can be grouped in the following

derivative term:

Integration of the previous expression between K and K + 1 gives

Evaluation of the left-hand side of the equation provides the following

terms:

Set = t KT. Substituting this change of variables in the equation abovegive the following result:

u(+ KT) = u(t)

d= dt

Which leads to

x Ax Bu+=

eAt

x eAt

Ax eAt

Bu+=

eAt

x eAt

Ax eAt

Bu=

d eAt

x( )dt

-------------------- eAt

Bu=

eAt

xKT

K 1+( )T

eAt

BudtKT

K 1+( )T

=

eA K 1+( )T

x K 1+( )T[ ] eAK T

x KT[ ]

eAt

eA KT+( )

=

eA KT+( )

Bu KT+( )d0

T

e

AKTe

ABu KT+( )d

0

T

==

eAT

x K 1+( )T[ ] x KT[ ] eA

Bu KT+( )d0

T

=


32/114


Solving forx[(K + 1)T]

Zero-Order-Hold

Consider the following equation where the input u(t) remains constant in

the [K K+1] interval:

Substituting this assumption in the previous expression gives:

You can take the input out of the integral.

and you can evaluate the integral accordingly.

This last expression allows you to identify the system matrices A and B for

the zero-order-hold conversion.

,

Code Implementation using the Taylor Series Expansion

x K 1+( )T[ ] eAT

x KT[ ] eA T( )

Bu KT+( ) d0

T

+=

eAT

x KT[ ] eAT

eA t KT( )

Bu t( )dtKT

K 1+( )T

+=

eAT

x KT[ ] eA t K 1+( )T[ ]

Bu t( )dtKT

K 1+( )T

+=

u t( )KT

K 1+( )Tcons t tan u KT( )= =

eAT

x KT[ ] eA t K 1+( )T[ ]

dtBu KT( )KT

K 1+( )T

+=

eAT

x KT[ ] eA t K 1+( )T[ ]

A1

KT

K 1+( )T

Bu KT( )=

eAT

x KT[ ] A1

eAT

A1

[ ]Bu KT( )=

eAT

x KT[ ] I eAT

( )A1Bu KT( )=

AdZOH

eAT

= BdZOH

eAT

I( )A1B=

eMT

I MTMT( )

2

2!---------------- + + + MT

( )i

i!---------------

i 0=

= =


33/114


Consider that matrices A and B can be appended together. Therefore, the

exponential of the appended matrix is given by:

The addition of the infinite number of terms in the series gives,

Because the matrix summation is equivalent to the summation of each of its

elements, you get the following expression:

where the definition of the exponential of the A matrix can be substituted,

and therefore, the system matrices can be identified:

Note:

y = Cx+ Duy(K) = Cx(K) + Du(K)

MA B

0 0M

2 A B

0 0

A B

0 0

A2

AB

0 0= = =

M3 A3

A2

B0 0

= Mi Ai

Ai 1

B0 0

=

eMT A

iA

i 1B

0 0i 1=

1

i!---=

I 00 0

+

eMT e

AT Ai 1

B

i!----------------

i 1=

0 0

eAT A

i

i!-----

i 1=

A1B

0 0

= =

eAT

eAT

I( )A1B

0 0=

eA B T Ad

ZOHBd

ZOH

=


34/114


Initial Conditions: For the initial condition, as the zero order hold makes the

response causal, you get the following:

and the initial conditions are not a function of the inputs at t = 0. In

summary you have:

When going from discrete to continuous, you solve for the continuous set

of system matrices, as shown in the following:

State-Space Norm

The CD Norm VI calculates two types of normsthe 2-norm and the

infinity-norm.

2-normFor stable systems, the following procedure outlines the steps for

computing the 2-norm.

For continuous systems, when you solve AX+ XAT+ BBT= 0 for X, you

get the following norm and frequency.

Frequency = NaN

xdZOH

0( ) x 0( ) Ou 0( )+ ICMAT I O==

BdZOH

eAcT

I( )Ac1Bc=

AdZOH

eAcT

=

CdZOH

C=

DdZOH

D=

AclnAd

ZOH

T------------------=

Bc Ac eAT

I( )1Bd

ZOH=

Ac AdZOH

I( )1Bd

ZOH=


35/114


where Gis the transfer function matrix.

For discrete systems, when you solve AXAT X+ BBT= 0 for X, you get

the following norm and frequency.

Frequency = NaN

If the system is unstable, the CD Norm VI produces an error. In this case,

the norm is and frequency is NaN. Systems with also have asthe norm and NaNas the frequency.

Infinity-normThe procedure for computing the infinity-norm is different for SISO

systems and MIMO systems.

SISO SystemsFor SISO systems, the CD Norm VI converts a state-space model to a

transfer function model.

Create frequency vector. For example, a vector of i.

Compute the norm according to the following formula.

MIMO SystemsCreate frequency vector. For example, a vector of i.

For each i, compute the USVT decomposition of H(ji)

Use this to compute the norm according to the formula,

Marginally stable systems have -norm of infinity.

G 2 Trace CX CT( )=

G 2 Trace CXCT DD T+( )=

D 0

G arg maxi

arg iI A( )( )=

G arg maxi

ma x S H ji( )( )( )( )=


36/114


For the generation of the frequency vector, the continuous function is

approximated by multiple evaluation at grid points. The grid is a geometric

sequence, adjusted based on whether you are dealing with the continuous

domain (j) or discrete domain (ejT).

References:

Doyle, J. C., K. Glover, P. P. Khargonekar, and B. A. Francis. State space

solutions to standard H2 and Hcontrol problems. IEEE Transaction onAutomatic Control, vol. 34, no. 8, p. 83, 1989.

Robel, A. On computing the infinity norm. IEEE Transaction on Automatic

Control, vol. 34, no. 8, p. 882, 1989.

Lyapunov Equation Solver

The Lyapunov equation is defined by the following equation.

AX+ XB= C

The implementation of a solver for this algorithm is based on the Bartels

Stewart algorithm.

The following procedure outlines the steps for computing the Lyapunov

algorithm.

1. Compute the real Schur decomposition of A. For example, UTAU= R.

2. Compute the decomposition for B. For example, VTBTV= S.

3. Construct a matrix F= UTCV.

4. Solve RY+ YST= Fas a Sylvester equation.

5. Compute the Lyapunov solution X= UYVT.

References:

R. H. Bartels and G. W. Stewart.Algorithm 432: Solution of the matrix

equation AX + XB = C. Communications of the ACM, September 1972.

G.H. Golub, S. Nash, and C. Van Loan.A Hessenberg Schur Method for the

problem AX + XB = C. IEEE Transaction on Automatic Control, December

1979.


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Discrete Lyapunov Equation SolverThe Continuous Lyapunov equation is

where

If (I A1)1exists and then Ais defined as follows.

A= (I+ A1)1(I+ A1)

and

A1= (A I) (A+ I)1

Rewrite the continuous Lyapunov equation as follows:

(A+ I)H(AH I)X+ X(A I)(A+ I)1= C1

Rearranging the equation results in the following equation:

Let

This equation yields

AHXA X= C

This equation is the discrete Lyapunov equation.

Therefore, to solve the discrete equation, use the continuous solver and set

A1= (A I) (A+ I)1

The resulting equation is

A1H

X XA 1+ C1=

C1 C1H=

AH

XA X1

2--- A I+( )

HC1 A I+( )=

C 12--- A I+( )HC1 A I+( )=

C1

2--- A I+( )

HC1 A 1+( )=


38/114


Reference:

S. J. Hammarling.Numerical solution of the stable, non-negative definite

Lyapunov equation. IMA J. Numerical Analysis, vol. 2, pp. 303323, July

1982.

Hamiltonian Reduction Algorithm for

Solving Riccati EquationsThe Riccati equation is defined as follows:

ATX+ XA XBR-1BTX+ Q= 0

To solve this, in the context of optimal control, begin with the Hamiltonian

equation.

H(x(t), u(t), (t), t) = [xT(t)Qx(t) + UT(t)Ru(t)] + T(t) [Ax(t) + Bu(t)]

From the Euler LaGrange equations ,

This system matrix is symplectic, which means, if is an eigenvalue, isalso an eigenvalue.

Let Vbe a modal matrix of the Hamiltonian system.

The Riccati solution is as follows:

S(t) = (t)X1(t)

J+and Jare Jordan blocks containing unstable and stable subpartitions ofthe Hamiltonian matrix.

As tf, 0, and ,

X(t) = V12 Z

1

2---

H

x-------= H

x------- 0=

x t( )

t( )

A B R1

BT

Q AT

x t( )

t( )=

X t( )

t( )

V11 V12

V21 V22

eJ+ t tf( )

0

0 eJ t tf( )

V11 V12

V21 V22

1

I

Sf

=

eJ+ t tf( )

eJ t tf( )

eJ t tf( )


39/114


(t) = V22 Z

where

As tf, S(t) becomes

S(t) = (t) X1(t) = V22 Z[V12 Z]1

Therefore S0= V22V121

Numerical solutions involve computing this algorithm on the modal matrix

of Hfor stable eigenvalues.

References:

A.J. Laub.A Schur method for solving algebraic riccati equation. IEEE

Transaction on Automatic Control, vol. AC24, no. 6, pp. 913921,

December 1979.

Uy Loi Ly.Linear Multivariable Control, Course Notes, U. Washington,

Dec 31, 1996

Continuous Algebraic Riccati Equation Solver

The continuous Algebraic Riccati Equation is defined as follows:

ATX+ XA (XB+ S)R1(BTX+ ST) + Q= 0

Using the Hamiltonian reduction algorithm to solve this equation, the

above equation is equivalent to the following equation.

By applying the following transformations,

A= A BR1ST

M= BR1BT

Q= Q SR1ST

eJ t tf( )

*

z

V11 V12

V21 V22

1

I

Sf

=

eJ t tf( )

eJ t tf( )

AT

X XA Q+ S RT

BT

X XBR 1

ST

XBR1

BT

X SR1

ST

+ 0=


40/114


the continuous Algebraic Riccati Equation reduces to the following form:

ATX+ XA XMX+ Q= 0

This form of the continuous Algebraic Riccati Equation can be solved by

the Hamiltonian reduction procedure, as inHamiltonian Reduction

Algorithm for Solving Riccati Equationssection.

Discrete Algebraic Riccati Equation SolverThe equation that is being solved is

(A BR1ST)TX(A BR1ST) (A BR1ST)XB(BTXB+ R)1BTX(A BR1ST) + Q SR1ST= X

By making the following two transformations

F= A BR1S, G1= B, G2= R

and

H= Q SR1S,

then the equation reduces to

FTXF X FTXG1(G2+ XG1)1 XF+ H= 0

As seen in the references, this equation has the symplectic matrix, similar

to Hamiltonian matrix in the Continuous Algebraic Riccati Equation Solver

section.

Then this algorithm uses a basis for the stable eigenspace of Zto compute

the desired Riccati solution, using the same procedure as in the

Hamiltonian Reduction Algorithm for Solving Riccati Equationssection.

References:

T. Pappas, A. J. Laub, N. R. Sandell. On the numerical solution of the

discrete time algebraic riccati equation, IEEE Transaction on Automatic

Control, Vol AC25, No. 4, pp. 631641, Aug 1980.

G G1G21

G1T

=

G1T

G1T

Z F GF

TH+ G F

T

FT

H FT

=


41/114


A. J. Laub.A Schur method for solving algebraic riccati equations.IEEE

Transaction on Automatic Control, Vol AC24, No. 6, pp. 913921,

Dec 1979.

Convert State-Space Model to Transfer Function Model

Consider the following transfer function matrix.

H(s) = C(sI A)1B+ D

If you begin with

you obtain

Transposing this equation gives you H(s). The expression is then evaluated

for each input-output pair and the result is transposed.

Reference:

C. Chen.Linear System Theory and Design. pp. 90 91. Oxford University

Press.

Covert Transfer Function Model to State-Space ModelConsider the following strictly proper MIMO transfer function model.

To convert the whole transfer function model into a state-space model,consider the first outputy1.

sI A BC+( ) sI A( ) DT sI A( )+sI A

-----------------------------------------------------------------------------------------------

BC

sI A--------------- DT+

y1

y2

:

ym

G11 G12 G1l

G21 G2l

:

:

Gm1 Gml

u1

u2

:

ul

=

y1 G1iui

i 1=

l

y1 ii 1=

l

= =


42/114


43/114


Transmission Zeros (State-Space)

Consider the following state-space model

and the following transfer function matrix

H(s) = C(sI A)1B+ D

This algorithm allows you to compute the zeros of this transfer function

matrix.

To get the zeros, solve a generalized eigenvalue problem of the form

Mx = Nx where

When Mand Nare square matrices and the dimensions match, this problem

is solved directly by using the Analysis VIs. When the dimensions of the

matrices do not match, the algorithm creates matrices M1and M2by using

random numbers to pad the matrix with a smaller dimension.

Then, if a transmission zero corresponds to original system, it has to be

(with probability close to 1) a zero of both M1and M2.

The algorithm picks the zeros based on the following criteria where

z1= set of zeros of M1

z2= set of zeros of M2

For each element in z1, select the closest element in z2and set distance as d.

If d < threshold, meaning d is 1020times the norm of M, then the element

is a transmission zero of M.

References:

A.J. Laub and B. C. Moore. Calculation of transmission zeros using QZ

techniques. Tech report ESL-P-802.

A. S. Hodel. Computation of system zeros with Balancing. Linear Algebra

and its Applications. 188, 189: 423436, 1993.

x Ax Bu+=

y Cx Du+=

M A B

C D= N

In

0

0 0=


44/114


Pade Approximation of Delayfor a SISO Transfer Function Model

The objective of the CD Convert Delay with Pade Approximation VI is to

approximate a delay of by using a polynomial approximation. To

approximate the exponential, use the following rational polynomial.

Rpq(z) = Dpq(z)1Npq(z)

where

and

When approximating time delays,

where

k = 0 N = 1

k = 1 N =

k= 2 N =

etc.

is defined as the order of the polynomial.

The function sequentially computes the terms and takes the sum of products

(from k = 0 to p). Finally, the denominator is made monic and incorporated

back into the model.

e std

Npq z( ) p q k+( )!p!

p q+( )!k! p k( )!-------------------------------------------

k 0=

p

zk

=

Dpq z( )p q k+( )!q!

p q+( )!k! q k( )!-------------------------------------------

k 0=

q

z( )k

=

Npq Ts s( )p q k+( )!p!

p q+( )!k! p k( )!-------------------------------------------

k 0=

p

Ts( )ks

k=

1

2--- Tss

12 2 2 1( )------------------------------- T( )s

2s

2( )


45/114


Reference:

G.H. Golub and C.F. Van Loan.Matrix Computations, 3rd ed.,

pp. 572573. Baltimore, MD: Johns Hopkins University Press, 1996.

Root Locus Plots

Consider the closed-loop transfer function of a linear time-invariantsystem.

The characteristic equation is

1 + D(s)G(s)H(s) = 0

You can rewrite this equation as

1 + KL(s) = 0 where .

K is some parameter that enters the characteristic equation linearly.

The following are some alternate ways to represent the same equation.

(s) = D(s) + K(s)N(s) = 0

Then, because , D'(s)N(s) D(s)N'(s) = 0.

The roots sof this equation would solve the equation

such that if for any s,

then that value of sis also a root of (s).

Each of these properties are used in various ways in this algorithm.

T s( ) Y s( )

R s( )-----------

D s( )G s( )1 D s( )G s( )H s( )+---------------------------------------------= =

L s( ) N s( )

D s( )-----------=

ds K( )dk

--------------dK s( )

ds-------------- 1=

K D s( )

N s( )-----------=

K D s( )

N s( )-----------=


46/114


The high level structure of the algorithm is as follows:

1. Determine bifurcation points, for example, the roots of the equation

D'(s)N(s) D(s)N'(s) = 0.

Verify that these points are indeed bifurcation points by checking the

equation

to make sure it is real, which means the bifurcation point lies on the

real axis, etc.

2. Perform pole-zero cancellations to derive the minimal system for

analysis.

3. Determine the maximum gain required for plotting (and some

additional parameters).

To get the maximum gain, notice that the nature of the root locus is

such that roots move from poles to zeros. If you evaluate the following

equation at the zero,

you get an idea of the final gain required to visually depict this path. If

you look at this value for each zero of the system, you get an idea of

what the appropriate, final gain Kf, for the entire plot. Use a default Kf

for systems with no zeros. The largest pole/zero/bifurcation point also

is used to determine the extent of the axes for the plots.

4. Once you know the final gain value, you create a grid of gain values

based on the geometric series, and iteratively refine this grid, until theplot is estimated to be visually correct. Here you use some specialized

heuristics based on interpoint distances.

Lastly, you fill in the appropriate infinity and special points for the

plots.

References:

A.M. Krall and R. Fornaro.An Algorithm for Generating Root Locus

Diagrams. Communications of the ACM, Vol. 10, No. 3, pp. 186188,

Mar 1967.

T. O. e Silva.Automatic Generation of Root Locus Plots. Revista DoDetua, Vol 2, No. 3, pp. 273278, Sept 1998.

K D s( )

N s( )

-----------=

D s( )N s( )-----------


47/114



Consider the transfer functionH(s).

When you apply a sinusoidal input as

x(t) =Xsin(t)

For linear time-invariant systems, the output is

y(t) = Ysin(t + )

Define

as the magnitude or input ratio and

as the phase shift of the output sinusoid with respect to the input sinusoid.

If you repeat this operation for all the possible frequencies, you obtain a

magnitude and phase shift for each frequency, creating the frequency

response of the system.

The frequency response is obtained by transforming from the s-domain to

thejdomain:

H(s) =H(j) =M(j)ej(j)

where

and

H s( ) Y s( )

x s( )----------=

G j( ) Y

X---=

/G j( )imag G j( )( )real G j( )( )

--------------------------------atan= =

M j( ) Y j( )X j( )---------------=

j( ) /Y j( ) /X j( )=


48/114


For discrete systems, the frequency response is obtained by the

transformationz= ejT.

H(z) =H(ejT) =M(ejT)ej(j)T

If the system is a multiple-input multiple-output (MIMO) system, this

process is repeated for each input-output pair.

It is possible to visualize the frequency response in three types of plots.

The Bode plot is composed by two plots that shows the magnitude (in linear

or dB scale) against frequency and phase (degree) against frequency.

Nichols Plot is a graph that shows the magnitude (in dB) of the frequency

response of the input system plotted against its phase (in degrees).

Nyquist Plot is a graph that shows the imaginary part of the complex

frequency response of the input system plotted against its real part.

The algorithm used to evaluate the frequency response of a transfer

function is based on following steps:

1. Create a vector of frequencies that capture the relevant dynamics of the

system. This vector spans a frequency band that includes three decades

around the relevant natural frequencies of the system.

2. Evaluate the transfer function at each frequency and obtain the

magnitude and phase of the signal for each frequency.

For continuous systems:

where:

n = number of zeros at zero in the complex plane

m = number of poles at zero in the complex plane

num1 = numerator polynomial without the roots at zeroden1 = denominator polynomial without the roots at zero

This algorithm corrects the initial frequency of the Bode response to

start at the correct angle.

M num j( )

den j( )----------------------=

0num1 j( )den1 j( )-------------------------

jn

jm

---------+atan=


49/114


For discrete systems:

3. Correct phase for wrapping and transport delay:

For continuous:

= 0 delay= 0Tdelay

For discrete:

= 0delay= (0TdelayTd)

4. Repeat the operation for each input-output pair of the MIMO system.

5. Correct the result for units.

The frequency, magnitude, and phase are calculated in radians/seconds,

linear, and radians, respectively. If you need the result in hertz, decibels, or

degrees, use one of the following conversions:

= 2f

Mdb = 20log (M)

degree= 180/

All Margins (Linear System)The algorithm used for the CD All Margins VI is based on the calculation

of the frequency response, where it computes the frequency at which the

phase crosses 180, gm, for a system,H(s):

The corresponding gain margin at gmis calculated as:

gain margin (dB) = 0 20log10|H(jgm)

The algorithm is based on the following steps:

1. Generate a frequency response for system using the Bode algorithm.

Refer to theBode, Nyquist, and Nichols Analysissection for

information about the Bode algorithm.

M num jd( )den jd( )-------------------------=

0num1 jd( )den1 jd( )----------------------------

atan=

180

--------- H j( )

gm= 180=


50/114


2. In the magnitude response, calculate all crossovers for phase 180

(2k+ 1). Using bisection algorithm, obtain the crossover frequency

and gain. This will define all the Frequencies and Gain Margins of the

system.

3. In the phase response, calculate all crossovers for magnitude equal to

1 (0 dB) and using bisection algorithm obtain the crossover frequency

and phase. This will define all the Frequencies and Phase Margins of

the system.

Gain and Phase Margin (Linear System)

The CD Gain and Phase Margin VI is based on the CD All Margin VI

algorithm. The response of the CD Gain and Phase Margin VI is defined as:

Gain margin (dB) is smallest gain margin of the system.

Frequency (radians/seconds) is the nearest crossover frequency of 180

(2k + 1), where k = 0, 1, 2 corresponding to the smallest gain margin.

Phase margin (degrees) is the smallest phase margin of the system.

Frequency (radians/seconds) is the 0 dB crossover frequency

corresponding to the smallest phase margin.

Staircase

Reference:

Web Site of Math Department of Escuela Politecnica Nacional, Ecuador,Staircase Algorithm. http://epn.edu.ec/~fc/dpto_mat/pdf/

staircase.pdf

Kalman Gain

The CD Kalman Gain VI uses the derivation found in the following

references.

Kwakernaak, H. & Sivan, R.,Linear Optimal Control Systems.

Wiley-Interscience, 1972.


1987.


51/114


Ackermann

The CD Ackermann VI uses the derivation found in the following

reference.


1995.

State Estimator and State-Space Controller

Refer to theLabVIEW Control Design User Manualfor information about

the implementation of the CD State Estimator VI and CD State-Space

Controller VI.

Generating Correlated Gaussian Random Vectors

Problem Statement

We need to generate random samples (denoted as x andy) of the random

vectorsxRnandyRmsuch that the random vectorsxandyareGaussian-distributed and are characterized by the second-order statistics:

E{x} =mxCov{x} = Cxx=E{xxT} mxmxT

E[y} =my

Cov{y} = Cyy=E{yyT} mymy

T

Cov {x,y} = Cxy=E{xyT} mxmyT

SolutionLets first focus on the problem of generating random samples of a single

Gaussian random vectorxwith mean mx and covariance matrix Cxx. Then,we will generalize the solution to generating random samples of the two

vectorsXand Y.

Single Vector CaseSay we are able to generate one sample of the Gaussian random variable z

(denoted as ) such that z is characterized by the statistics z ~ N (mz = 0,z

2 = 1).

Now, lets generate n-independent samples {1, 2, ..., n}of the randomvariables {z1, z2, ..., zn} that are identically distributed as z~N (0, 1).


52/114


Define

Czz= Cov{z}

=E{zzT} mzmzT

z~N(0, I)

Define the affine transformation:

X=Az+b

whereARnn, b Rn

Since affine transformation of a Gaussian random vectors yields a Gaussian

random vector xis Gaussian-distributed.

We will find the matrixAand the vector b that will generatexsuch thatx~ N(mx, Cxx).

z

z1

z2

zn

=

mz

00

0

=

1 0 0

0 1 0

0 0 1

I==

E x{ } E Az b+{ }=

A=E z{ }

0=

b+

E x{ } b=


53/114


recall, given a real-symmetric matrixMRnnwith Eigenvalues{1, 2,..., n}, then M is similar to a diagonal matrix given by

under the transformation:

V1MV=

where, V= [v1, v2, ..., vn]

with {v1, v2, ..., vn} being the orthonormal Eigenvectors corresponding tothe Eigenvalues {1, 2, ..., n}

but, since {v1, v2, ..., vn} for an orthonormal basis for V

Vis unitary

VTV=I

VT= V1

VTMV=

M= VVT

now, if we factor = 1/21/2, and since = T

1/2= (1/2)T

M = 1/21/2VT

Cov x{ } Cov Az b+{ }=

E Az b+[ ] Az b+[ ]T

{ } E Az b+{ } ET

Az b+{ }=

AE zzT

{ }AT

AE z{ }bT

bE ZT

{ }AT

bbT

++ +=

AE z{ } b+[ ] AE z{ } b+[ ]T

A Czz

I

AT

bbT

bbT

+=

Cxx A AT

=

1 0 0

0 2

0

0 0 n

=


54/114


55/114


transformation w=Az+b. Finally, taking the first n-elements of the randomvector sample wgives the random vector sample x and taking the lastm-elements of the random vector sample wyields the random vectorsampley.

Implementation Current Observer (Point-by-Point)

@ time k = 0: is already available.

Calculate thepredictedstate :

@ time k: is already available.

A new measurement becomes available.

Calculate the corrected/currentstate

Calculate thepredicted state :

Relationship between Lpand Lc:

The Current Gainmatrix Lcis related to the Predicted Gainmatrix Lp

through the relationship.

this means that Ad needs to be full-rank.

Algorithm adapted from:

Franklin, Powell, Workman, Digital Control of Dynamic Systems, 3rd.

Ed., pp. 295

x 0( )

x 1( )

x 1( ) Ad x 0( ) Bd u 0( )+=

x 1( ) Ad x 0( ) Bd u 0( )+=

x k( )

y k( )

x k( )

x k( ) x k( ) Lc y k( ) y k( )[ ]+=

y k( ) Cdx k( ) Ddy k( )+=

x k 1+( )

x k 1+( ) Ad x k( ) Bd u k( )+=

Lp Ad Lc=


56/114


57/114


The observer gain (L) can be computed through pole-placement or

Ackermann.

Algorithm adapted from:

John Bay, Fundamentals of Linear State Space Systems, McGraw Hill,

1999.

ContinuousTime Recursive Kalman-Bucy Filter

Note The following derivation assumes the familiarity of the reader with basic concepts

of Estimation Theory and Linear Algebra. Much of the derivations herein follows the

developments outlined in the following references:

1. I. Rhodes, A Tutorial Introduction to Estimation and Filtering, IEEE

Transactions on Automatic Control, Vol. A C-16, No. 6, December

1971, pp. 688-706.

2. Jerry Mendel, Lessons in Estimation Theory for Signal Processing,

communications, and Control, Prentice Hall, 1995.

3. John Bay, Fundamentals of Linear State-Space Systems, McGraw

Hill, 1999.

Problem StatementGiven the continuous-Time (C.T.) Linear Time-Varying (LTV) stochastic

state-space system.

The control input vector is deterministic and known .

The matrices , , ,

, , are deterministic and

known .

The disturbance vectors and are assumed to be white (i.e.

zero-mean and temporally uncorrelated) random processes with the

following first- and second-order statistics.

x t( ) A t( )x t( ) B+ t( )u t( ) G t( )w t( )+=

y t( ) C t( )x t( ) D+ t( )u t( ) H t( )w t( ) v t( )+ +=

u t( ) IRr

t

A t( ) IRnx n

B t( ) IRnx r

G t( ) IRnx q

C t( ) IR

mx n D t( ) IR

mx r H t( ) IR

mx q

t

w t( ) v t( )

E w t( ){ } o= t

cov w t( ) w t( ),{ } E w t( ) wT

t( ){ } Q t( )s t t( ) Q t( ) QT

t( ) 0=,==

E v t( ){ } o= t


58/114


where is the Dirac delta function.

Also, it is assumed that the initial state vector is uncorrelated w/ the

random processes and

We need to find the best linear estimator in terms of

the measurements

SolutionThe best linear estimator of the state vector in terms of the

measurements has the following dynamical structure:

Kalman-Bucy Filter Derivation:

Estimator dynamics structure:

Estimator Error:

Error Dynamics:

cov v t( ) v t( ),{ } E v t( ) vT

t( ){ } R t( )s t t( ) R t( ) RT

t( ) 0>=,==

cov w t( ) v t( ),{ } E w t( ) vT

t( ){ } N t( )s t t( )==

S t( )

x to( )w t( ) v t( ) t

cov w t( ) x to( ),{ } 0 cov v t( ) x to( ),{ } 0=,=

x t( ) E*

x t( ) yt{ }=Yt y s( ) S;( )S to t),[{ , }=

x t( ) x t( )Yt

x t( ) A t( )x t( ) B t( )u t( ) L t( ) y t( ) y t( )[ ]+ +=

y t( ) C t( )x t( ) D t( )u t( )+=

L t( ) P t( )C

Tt( ) G t( )Q t( )H

Tt( ) G t( )N t( )+ +[ ]

. H t( )Q t( )HT

t( ) H t( )N t( ) NT

t( )HT

t( ) R t( )+ + +[ ]1

=

P

t( ) A= t( )P t( ) P t( )AT

t( ) G t( )Q t( )GT

t( )+ +

P t( )C t( ). H t( )Q t( )HT

t( ) H t( )N t( ) NT

t( )HT

t( ) R t( )+ + +[ ]1.C t( )P t( )

G t( )Q t( )HT

t( ) G t( )N t( )+[ ]. H t( )Q t( )HT

t( ) HtN t( ) NT

t( )HT

t( ) R t( )+ + +[ ]

. G t( )Q t( )HT

t( ) G t( )N t( )+[ ]T

x t( ) Ax t( ) Bu t( ) L t( ) y t( ) y t( )[ ]+ +=

y t( ) Cx t( ) Du t( )+ +

e t( ) x= t( ) x t( )


59/114


Covariance of estimation error:

e t( ) x t( ) x t( )=

Ax t( ) Bu t( ) Gw t( )+ +=

Ax t( ) Bu t( ) Lt yt Cx t( ) Du t( )[ ]+ +[ ]

e t( ) Ae t( ) Gw t( )+=

L t( ) Cx t( ) Du t( ) Hw t( ) v t( ) Cx t( ) Du t( )+ + +[ ]

Ae t( ) Gw t( ) L t C e t( ) Hw t( ) v t( )+ +[ ]+=

e t( ) A LC[ ]e t( ) G LH[ ]w t( ) Lv t( )+=

e t( ) t to,( )e to( ) t ,( ) G LH( )w ( ) Lv ( )[ ] d

to

t

+=

P t( ) E e t( ).eT ( ){ } E e t( ){ }0

. E

T

e ( ){ }0

=

E t to,( )e to( ) t ,( ) G LH( )w t( ) Lv t( )[ ] td

to

t

+

=

. t to,( )e to( ) t s,( ) G LH( )w s( ) Lv s( )[ ] sd

to

t

+

T


60/114


E t to,( )e to( )eT

to( )T

t to,( ){ }=

E t to,( )e to( ). t s,( ) G LH( )w s( ) Lv s( )[ ] sd

to

t

T

+

E t ,( ) G LH( )w ( ) Lv ( )[ ] t.e to( ).T t to,( )d

to

t

+

E t ,( ) G LH( )w ( ) Lv t( )[ ] td

to

t

. G LH( )w s( ) Lv s( )[ ]T

Tt to,( ) sd

to

t

+

t( ) t to,( )P to( )T

t to,( )=

E t ,( ) G LH( )w t( ) Lv t( )[ ] G LH( )w s( ) Lv s( )[ ]T

Tt s,( ) d d

to

t

to

t

+

t to,( )P to( )T

t to,( )=

t ,( )E G LH( )w ( ) Lv ( )[ ] wT

s( ) G LH( )T

vT

s( )LT

[ ]{ }T

t s,( ) d

to

t

to

t

+

t to,( )P to( )T

t to,( )=

t ,( ) G LH( )Q t( )s t s( ) G LH( )T

G LH( ) N ( )s s( )LT

[

to

t

to

t

+

LNT

( )s s( ) G LH( )T

LR ( )s s( )+ ]T

t s,( )d

t to,( )P to( )T

t to,( )=

t ,( ) G LH( )Q ( ) G LH( )T

G LH( ) N( )LT

[

to

t

+

LNT

( ) G LH( )T

LR LT

+ ]T

t ,( )d


61/114


recall:

denote:

but,

P t( ) t to,( )P to( )T

t to,( )=

t ,( ) G LH( )Q ( ) G LH( )T

G LH( ) N ( )LT

[

to

t

+

LNT

( ) G LH( )T

LR LT

+ ]T

t ,( )d

d

dt----- f

g t( )

h t( )

x t,( )dx

t

f x t,( ){ } x f h t ( ) t,( )[ ]+d h t( )d

dt------------

g t( )

h t( )

=

f g t( ) t,[ ] g t( )d

dt------------

Z G LH( )Q ( ) G LH( )T

G LH( )N ( )LT

LN ( ) G LH( )T

=

LR LT

+

t( )

t to,( )P to( )

Tt to,( ) t to,( ) to( )

Tto( )+=

t ,( )ZT

t ,( ) t ,( )Z t ,( )

+

d

to

t

+

t ,( )

IZ

t ,( )

IT

dt

dt-----

t to,( )ZT

t to,( )dt odt--------

0=

0=

+

t to,( )P to( )T

t to,( ) t to,( )P to( ) T

t to,( )+=

t ,( )ZT

t ,( ) t ,( )Z T

t ,( )+ d

to

t

+

Z+

e t( ) A LC[ ]e t( ) G LH[ ]w t( ) Lv t( )+=

t to,( ) state transition mtx for above D.E.=


62/114


>Different Riccati Equation.

Now, to optimize the Gain , the chosen criterion will be to minimize

the squared estimation error.

t to,( ) A LC[ ] t to,( )=

t( ) A LC[ ] t to,( )P to( )

Tt to,( )=

t to,( )P to( )T

t to,( ) A LC[ ]T

+

A LC[ ] t ,( )ZT

t ,( ) d

to

t

+

t ,( )ZT

t ,( ) A LC[ ]T

d

to

t

+

Z+

A LC[ ] t to,( )P t to,( )T

t to,( ) t ,( )ZT

t ,( ) d

to

t

+=

t to,( )P to( )T

t to,( ) t ,( )ZT

t ,( ) d

to

t

+

P t( )

A LC[ ]T

+

Z+

P

t( ) A LC[ ]P t( ) P t( ) A LC[ ]T

Z+ +=

t( ) A LC[ ]P t( ) P t( ) A LC[ ]

T

+=

G LH[ ]Q t( ) G LH[ ]T

G LH[ ]N t( )LT

LNT

t( ) G LH[ ]T

LR t( )LT

+ +

L t( )

Ee

Tt( ).e t( )

scalar

tr E eT

t( ).e t( ){ }[ ]=

E tr eT

t( ).e t( )[ ]{ }=

E tr e t( ).eT

t( )[ ]{ }=

tr E e t( ).eT

t( ){ }[ ]=

tr P t( )[ ]=


63/114


ma x

L t( )tr P

t( )[ ]{ }

L t( )

tr{ } P t( )[ ]

L t( )

tr{ } P

t( )[ ]L

AP LC P PAT

PCTL

T+

GQ GT

GQ HTL LHQGT LHQHTL+ +

GN LT

LHNLT

LNT

GT

+ LNTHTLT=

LR LT

+

=

.x

tr XAT

[ ]{ } A=

.x

tr XTA[ ]{ } A=

.x

tr AB AT

[ ]{ } AB ABT

+=

1L

tr LC P[ ]{ } CP( )T

PT

CT

= =

2L

tr PCTL

T[ ]{ }

L

tr LT

CPT

[ ]{ } PCT

= =

3L tr GQ H

TL

T[ ]{ }

L tr L

TGQ H

T[ ]{ } GQ H

T= =

4L

tr LHQGT

[ ]{ } HQ GT

( )T

GQTH

T= =

5L

tr LHQHTL

T[ ]{ }

LHQHT

L HQ HT

( )T

+=

LH QHT

LH QTH

T+=

6L tr GN LT[ ]{ }

L tr LTGN[ ]{ } GN= =

7L

tr LHNLT

[ ]{ } LHN L HN ( )T

+ LHN LN TH

T+= =


64/114


substituting for into :

8L

tr LNT

GT

[ ]{ } NT

GT

( )T

GN= =

9L

tr LNTH

TL

T[ ]{ }

LNTH

TL N

TH

T( )

T+=

LNTH

TLH N+=

10 L

tr LRL

T

[ ]{ } LR LRT

+=

now,L

tr P

t( )[ ]{ } 0

PCT

PCT

GQ HT

GQ HT

LHQHT

+ LHQHT

=

GN LHN LN TH

TGN LN

TH

TLH N LR LR 0+ + + + + +

2PCT

2GQ HT

2LH QHT

2GN 2LH N 2LNTH

T2LR 0+ + + +

LH QHT

LNTH

TLR LH N PC

TGQ H

TGN+ ++ +=

L HQ HT

R HN NTH

T+ + +[ ] PC

TGQ H

TGN+ +

L t( ) PCT

GQ HT

GN+ +[ ]. HQ HT

HN NTH

TR+ + +[ ]

1=

L t( ) P

t( )

t( ) AP LCP PA T PCTLT+=

GQ GT

GQ HTL

T LHQG

T LHQH

TL

T+ +

GN LT

LNT

GT

LHNLT

LNTH

TL

T+ +

LR LT

+

AP LC P PAT

PCTL

T+=

GQ GT

GQ HTL

T LHQG

T GN L

T LN

TG

T+

L HQ HT

HN NTH

TR

T+ + +[ ]+

P

t( ) AP PAT

GQ GT

+=

L CP HQ GT

NT

GT

+( )[ ]

PCT

GQ HT

GN+ +[ ]


65/114


P t( ) AP PA GQG + +=

L CP HQ GT

NT

GT

+ +[ ]

1

PCT

GQ HT

GN+ +[ ]LT

2

L HQHT HN NTHT R+ + +[ ]LT

3

+

1 PCT

GQ HT

GN+[ ] HQ HT

HN NTH

TR+ + +[ ]

1. CP HQG

TN

TG

T+ +[ ]

2 PCT

GQ HT

GN+[ ]. HQ HT

HN NTH

TR+ + +[ ]

1[ ]

T. PC

TGQ H

TGN+ +[ ]

T

PCT

GQ HT

GN+[ ] HQ HT

HN NTH

TR+ + +[ ]

T[ ]

1CP HQG

TN

TG

T+ +[ ]=

PCT

GQ HT

GN+[ ] HQ HT

HN NTH

TR+ + +[ ]

1CP HQG

TN

TG

T+ +[ ]=

3 PCT

GQ HT

GN+[ ]. HQ HT

HN NTH

TR+ + +[ ]

1

. HQ HT

HN NTH

TR+ + +[ ]

. HQ HT

HN NTH

TR+ + +[ ]

1[ ]

T. PC

TGQ H

T GN+[ ]

T

PCT

GQ HT

GN+[ ]= . HQ HT

HN NTH

TR+ + +[ ]

1CP HQG

TN

TG

T+ +[ ]

1 2 3+ + 1=

P t( ) AP PAT

GQ GT

+ +=

PCT

GQ HT

GN+ +[ ] . HQ H

THN N

TH

TR+ + +[ ]

1

. CP HQG T

NT

GT

+ +[ ]

AP PA GQ GT

PCT

GQ HT

GN+ +[ ].. HQ GT

NT

GT

+[ ]+ +=

AP PA GQ G

T

PC

T

CP GQH

T

GN+[ ].

. HQ G

T

N

T

G

T

+[ ]+ +=

P

t( ) AP PAT

PCT

[ ] HQ HT

HN NTH

TR+ + +[ ]

1. CP[ ] GQ G

T+ +=

GQ HT

GN+[ ] HQ HT

HN NTH

TR+ + +[ ]

1. GQ H

TGN+[ ]

T


66/114


n-step-Ahead Kalman Predictor

Note:

Much of the derivations herein follow the developments in the following

reference:

[1] Ian Rhodes, A Tutorial Introduction to Estimation and Filtering,

IEEE Transactions on Automatic Control, Vol. AC-16, No. 6,

December 1971, pp. 688706.

Problem statement:

Given the Discrete-Time LTV XXXX state-space system:

Need to find the n-step-ahead Kalman Predictor, i.e. find the state estimate

given the knowledge of the one-step-ahead

predicted state and assuming the knowledge of the matrices

and the central input

Solution:

Define,

Then, we may re-write the state difference equation as:

Similarly, we may expressxk+ n(n 2) as:

xk 1+ Akxk Bkuk Gkwk+ +=

yk Ckxk Dkuk Hkwk vk+ + +=

x k n+ k E*

xk n+ Yk{ },=

x k 1+ k

Ai{ }i k 1+=k n 1+

Bi{ }i k 1+=k n 1+

, ui{ }i k 1+=k n 1+

.

k j, Ak 1 Ak 2 Aj =

j j, I=

xk 1+ k 1 0,+ x0 k 1 i 1+,+ Biui( ) k 1 i 1+,+ Giwi( )+[ ]

i 0=

k

+=

xk n+ k n k 1+,+ xk 1+ k n i 1+,+ Biui( ) k n i 1+,+ Giwi( )+[ ]

i k 1+=

k n 1+

+=


67/114


Now:

but,

E*{ui| Yk} = ui, since uiis a deterministic input vector.

since the sequence Yk= {y0,y1, ,yk} depends only and linearly on

, all of which are assumed uncorrelated

with .

If the system is Time-Invariant, then the n-step-ahead Kalman predictorcan be represented more compactly.

k n k+ E*

xk n+ Yk{ }=

k n+ k 1+, E*

xk 1+ Yk{ }=

k n i 1+,+ BiE*

ui Yk{ } k n i 1+,+ GiE*

wi Yk{ }+[ ]i k 1+=

k n 1+

+

E*

wi Yk{ } 0 k 1+ i k n 1+ =

x0{ } wi{ }i 0=k, xi{ }i 0=

k vi{ }k

i 0=,,

wi{ }i k 1+=k n 1+

x k n k+ k n k 1+,+ x k 1 k+ k n i 1+,+ BiUi

i k 1+=

k n 1+

+=

xk n k+

k n+ k 1+,Ak n 1+ Ak n 2+ Ak 1+

k n 1 k 1+( ) 1+ n 1=+=

An 1

=

k n+ i 1+,

Ak n 1+ Ak n 2+ Ai 1+

k n 1 i 1+( ) 1+ k n i 1+=+=

Ak n i 1+

=


68/114


Lyapunov and Sylvester Equations

The Lyapunov and Sylvester equations are similar in form. Table1-2

defines both the continuous and discrete versions of each equation.

These equations differ in the underlying assumptions and meaning

attributed to the coefficient matricesA,B, and C. These differences ensurecertain properties of the solutionX.

You can use this set of equations to analyze the stability of systems and

related applications such as the computation of Grammians, norms, and

so on. Refer toNumerical Linear Algebra Techniques for Systems and

Control1for more information about these computations. Refer toRobust

Pole Assignment via Sylvester Equation Based State Feedback

Parametrization2for more information about the Sylvester equation.

The Sylvester equation is the basis for solving both forms of these

equations. The solution is based on the Bartels-Stewart algorithm, which

reduces bothAandFto the real Schur form and then performs a backwardsubstitution to complete the solution. The underlying implementation

of this algorithm uses the Mathematics VIs, which are based on

BLAS/LAPACK algorithms.

Complete the following steps to solve the continuous version of the

Lyapunov equation.

4. Calculate the Schur decompositions ofAandB.

UTAU=Rand VTBTV= S

5. ConstructF= UTCV.

Table 1-2. Lyapunov and Sylvester Equations

Equation Continuous Discrete

Lyapunov AX+XAT+ Q= 0 AXATX+ Q= 0

Sylvester AX+XB+ C= 0 AXBX+ C= 0

1 Laub, Alan J., Paul Van Dooren, and R.V. Patel (eds).Numerical Linear Algebra Techniques for Systems and Control.Piscataway, NJ: IEEE Press, 1994.

2 Varga, A.Robust Pole Assignment via Sylvester Equation Based State Feedback Parametrization. IEEE InternationalSymposium on Computer Aided Control System Design, CACSD'2000. Anchorage, Alaska, 2000.

x k n k+ An 1

x k 1 k+ Ak n i 1+

B Ui

i k 1+=

k n 1+

+=


69/114


6. SolveRY+ YST=Fusing the backward-substitution procedure.

7. SolveX= UYVT.

You can apply the same methodology to the discrete version of the

Lyapunov equation if you complete the following steps:

1. Consider the continuous Lyapunov equation, .

2. If (IA1)1exists and ifA= (IA1)1(I+A1) and

A1= (AI)(A+I)1, the Lyapunov equation is as follows:

(A+I)T(ATI)X+X(AI)(A+I)1+ C1= 0

3. If you define the matrix Cwith the following equation:

4. Rearranging the terms in the above equation produces the following

discrete equation:

ATXAX+ C= 0

Riccati Equations

The symmetric n nalgebraic Riccati equation for the continuous timecase isATX+XAXMX+ Q= 0.

The solution of this equation is based on the reduction of a Hamiltonian

matrix, defined by the following equation:

If Vis the modal matrix of this Hamiltonian matrix, for stable eigenvalues,you can express Vwith the following equation:

You then can express the steady state solution of the Riccati equation with

the following equation:

A1TX XA 1 C1 0=+ +

C1

2--- A I+( )

TC1 A I+( )=

H AT

M

Q A=

V V11 V12

V21

V22

=

X V22V121

=


70/114


You can reduce a continuous algebraic Riccati equation,

ATX+XA (XB+ S)R1(BTX+ ST) + Q= 0, to the simpler form byapplying the following transformations:

A=ABR1ST

M=BR1BT

Q= Q SR1

ST

The following equation defines the discrete version of the Riccati equation.

If , the following equation defines the Hamiltonian matrix

for this equation.

You can use this Hamiltonian matrix in a reduction procedure to compute

the solution of the discrete algebraic Riccati equation.

If the discrete Riccati equation is defined as follows, you can reduce the

equation using the Hamiltonian matrix.

The following transformations reduce the discrete Riccati equation to the

simpler form based on the determined Hamiltonian matrix.

A=ABR1ST

B1=B

B2=R

C= QSR1ST

Convert Continuous Model to Discrete Model with InputDelay

Problem 1: ZoH discretization with uniform input delay

ATXA X A

TXB1 B2 B1

TXB1+( )

1B1

TXA C+ 0=

B B1B21B1

T=

H A BA

TC+ BA

T

A T

C A T=

A BR1S

T( )X A BR

1S

T( ) A BR

1S

T( )

TXB B

TXB R+( )

1B

TX A BR

1S

T( ) Q SR

1S

T+ X=


71/114


= (lm) Tl = 1,2,3...0m


72/114


which if we employ a ZOH discretization technique will result in capturing

the input sample for the previous time-step, i.e. u(k1) intoy(k).

y(k)=Cx(k)+Du(k1)

+[0]u(k)

Cd[C D}

Dd= 0

Problem 2: ZOH discretization with vectorized input delay.

li= 1, 2, 3, ...

0mi,1

y k( ) C D[ ]x k( ) k( )----------=

y k( ) Cd k( )x k( ) Ddu k( )+=

Ad T( ) 1

0 0=

Bd 2

I-----=

x t( ) Ax t( ) Bu t ( )+=

y t( ) Cx t( ) Du t ( )+=

x t( ) Ax t( ) b1 b2 br[ ]

u1 t 1( )

u2 t 2( )

ur t r( )

+=

1 l1 m1( )T=

2 l2 m2( )T=

r lr mr( )T=


73/114


We will consider the case when li1, 0mi 1, the (li1) delay will be set to the discrete model @ input channel i,

independently after the discretization is achieved.

x(k+1) = (T)x(k) + 1u(k-1) + 2u(k)

(T) = eAT

where,

where,

define the state vector (k) = u(k1)

y(k) = Cx(k) +Du(k1)

Problem 3: ZoH discretization with uniform output delay

1

1 1,

1 2,

1 r,

[ ]=

1 i, eA

d

miT

bi=

1 i, miT( ) i T miT( )=

miT( ) e mi

=

i T miT( ) eA

d

0

T miT

bi=

2 2 1, 2 2, 2 r,[ ]=

2 i, eA

d

0

mi

bi=

i miT( )= i miT( ) eA d

0

mi

bi=

x k 1+( ) k 1+( )--------------------

x k 1+( )

T( ) 1

0 0

x k( ) k( )----------

2I-----u k( )+=

y k( )CD[ ]

Cd

x k( ) k( )----------

x k( )

0[ ]u k( )+=


74/114


We will consider the case when j 1, 0 q 1, the (j1) delay will be set to the discrete model independently

after the discretization is achieved.

, ,

y(t) = Cx(t) +Du(t)

y(kT) = Cx(kT) +Du(kT)

= Cx(kT(lm)T) +Du(kT(lm)T)

= Cx[T(k(lm)] +Du[T(k(lm))]

but,

to ,

x t( ) Ax t( ) Bu t( )+=

y t +( ) Cx t( ) Du t( ) y t( ) Cx t ( ) Du t ( )+=+=

j q( )T=

j 1 2 3 , , ,=

0 q 1


75/114


but, ,

Note that, u[T(k(1q)] = u[k(1q)]= u(k1+q)

= u(k-1)

It can be noted that ~1 ZoH, the input signal u(t) will be sampled @

time-step kTand the value of the sample is led through time-step (k+1)T.

So, if we delay the signal by k-1 and then advance the signal by m s.t. weget u(k1+q) we will still be capturing the value of the signal @ time-step

k1, i.e. u(k1)

y(kt) = Cx[T(k(1q))] +Du[T(k(1q))]

y(k) = C[(qt)x(k1) + (qT)u(k1)] +Du(k1)

u t( ) u k 1( )T[ ]= k 1( )T t k 1 q+( )T kT<


76/114


y(k) = C[(qT)x(k1) + [C(qT)+D]u(k1)

y(k+1) = C(qT)x(k) + [C(qT)+D]u(k)

Note Few of the derivations here follows the development in

Franklin, Powell, Workman, Digital Control of Dynamic Systems, 3rd.

Ed., 1998, pp. 110-114.

Problem 4: ZOH discretization with vectorized output delay

x k 1+( )y k 1+( )-------------------

x k 1+( )

T 0

C qT( ) 0

Ad

x k( )y k( )----------

x k( )

C qT( ) D+-------------------------------

Bd

u k( )

u k( )= =

y k( )0 I[ ]

Cd

x k( )y k( )----------

0[ ]

Ddu k( )+=

x t( ) Ax t( ) Bu t( )+=

y t +( ) Cx t( ) Du t( )+=

y1 t 1+( )

y2 t 2+( )

ym t m+( )

C1

C2

Cm

x1 t( )

x2 t( )

xn t( )

D1

D2

Dm

u1 t( )

u2 t( )

ur t( )

+=

y1 t 1+( )

y2 t 2+( )

ym t m+( )

C11x1 t( ) C12x2 t( ) C1nxn t( ) D11u1 t( ) D12u2 t( ) D1rur t( )+ + + + + + +

C21x1 t( ) C22x2 t( ) C2nxn t( ) D21u1 t( ) D22u2 t( ) D2rur t( )+ + + + + + +

Cm1x1 t( ) Cm2x2 t( ) Cmnxn t( ) Dm1u1 t( ) Dm2u2 t( ) D2mrur t( )+ + + + + + +

=


77/114


Define,

We will only consider the case where .

Ifj1> 1, the (j11) delay will be set to the discrete model at the output

channel i, independently after the discretization is achieved.

y1 t( )

y2 t( )

ym t( )

C11x1 t 1( ) C12x2 t 1( ) C1nxn t 1( )+ + +

C21x1 t 2( ) C22x2 t 2( ) C2nxn t 2( )+ + +

Cm1x1 t m( ) Cm2x2 t m( ) Cmnxn t m( )+ + +

D11u1 t 1( ) D1ru t 1( )+ +

D21u1 t 2( ) D2ru t 2( )+ +

Dm1u1 t m( ) Dmru t m( )+ +

+=

y1 t( )

y2 t( )

ym t( )

C1x t 1( )

C2x t 2( )

Cmx t m( )

D1u t 1( )

D2u t 2( )

Dmu t m( )

+=

1 j1 q1( )T=

2 j2 q2( )T=

m jm qm( )T=

j1 1, 2, 3, ...= i 1, 2, ..., m=

0 q1 1


78/114

Control Design and Simulation Module Algorithm Reference 78 ni.c

CD Reference

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