Discrete control

1. 1
Discrete Control
State Space Analysis:
In many situations, one is able to find a difference equation that describes the relation between
several variables. We need to put this equation in a format that is suitable for discrete control.
The standard linear time invariant state space model is given as:
)()()1( kuHkXGkX 
With observations:
)()()1( kuDkXCky 
Where
)(kX = n-vector
)(ky = m-vector
)(ku = r-vector
Consider the discrete linear system, with scalar input and scalar output, described by:
)(...)1()()(...)2()1()( 1021 mkubkubkubnkyakyakyaky mn 
This could be put in several format.
(1)Controllable Canonical form:
Define )()1( 21 kxkx  ,…, )()1(1 kxkx nn 
Then )()(...)()()1( 1121 kukXakXakXakX nnnn  
)(
1
0
...
0
0
)(
)(
...
)(
)(
...
1...000
...............
0...100
0...010
)1(
)1(
...
)1(
)1(
1
2
1
121
1
2
1
ku
kX
kX
kX
kX
aaaakX
kX
kX
kX
n
n
nnnn
n










































































And

2. 2
       )(
)(
..
...
)(
)(
...)( 0
2
1
0110110 kub
kX
kX
kX
babbabbabky
n
nnnn 
















 
(2) Observable Canonical Form:
(3) Diagonal Canonical Form:
Optimization:
Define  



1
0
)()()()(
2
1
)()(
2
1 N
k
TTT
kuRkukXQkXNXSNXJ
The optimal values of u(k) that minimizes the objective function J are given by the following
equations:
)1()( 1
 
kHRku T
 , k=0,1,…,N-1
)1()()(  kGkXQk T
 , k=1,2,3,…,N-1, )()( NXSN 

3. 3
Linear Systems
The linear time invariant system is described using the delay operator 1
q as:
      )()()( 111
kqCkuqBkyqA 

where   A
AqaqaqaqA deg
deg
2
2
1
1
1
...1 

  B
BqbqbqbbqB deg
deg
2
2
1
10
1
... 

  C
CqcqcqccqC deg
deg
2
2
1
10
1
... 

Or
 
 
 
  )()()( 1
1
1
1
k
qA
qC
ku
qA
qB
ky 




    )()( 11
kqGkuqGu 


Notice that the delay operator 1
q plays the role of the z transform i.e. 11 
 zq . For example
  )1()()(1 1
1
1  
kyakykyqa , and the z transform of  )1()( 1  kyaky is   )(1 1
1 zyza 
 .
Remember that the definition of Z transform y(z) of the signal y(k) is given as:





0
)()(
k
k
zkyzy
which, in the frequency domain, is given as:






0
)()()(
k
kj
ez
j
ekyzyey j


i.e. we are evaluating the z-transform on the unit circle and thus obtain the frequency response (if
y(z)=H(z)= the transfer function).
For finite impulse response (FIR) model for data or filter we have:

4. 4


I
i
i ikyky
1
)()(  or 

I
i
i iIkyIky
1
)()( 
with known initial conditions )1((),...1(),0( Iyyy . The z transform y(z) will depend on the
initial conditions.
Example: )()1()2( 21 kykyky   . Rearrange we get: 0)()1()2( 21  kykyky  ,
after some manipulations we get
 
21
2
1
2
)1()0(
)(





zz
zyyzz
zy
Stability:
The linear time-invariant system with zero input
    0,0),()( 11
 
ku(k)kuqBkyqA , is asymptotically stable if 0)(lim 

ky
k
for all initial conditions y(0),y(-1),…,y(1-degA). Thus, to study asymptotic stability we must
first zero the input.
Example: Consider the first order system   11
1 
 qqA  which has the form
    0,0)1()()(1)( 11
 
kkykykyqkyqA  i.e. )1()(  kyky  . Thus, for
k=1, we have )0()1( yy  , for k=2, we have )0()1()2( 2
yyy   , and for any k and for any
initial conditions y(0), we have, )0()( yky k
 . The system is asymptotically stable,
0)(lim 

ky
k
, if 1 .
The general case with characteristic polynomial    
m
i
v
i
i
qqA
1
11
1


  is stable if ii  ,1
Stability of Nonlinear Systems:
For the general nonlinear system dynamics of the form:

5. 5
 ),...2(),1(),...,2(),1()(  kukukykyfky
We first zero the input and study only the system
 ,...0)2(,0)1(),...,2(),1()(  kukukykyfky . One has to convert the nonlinear
system into linear one by using Taylor series expansion around reference values )(kyr (to be
calculated) and keeping only the first order terms. To simplify the analysis we will confine
ourselves to the case where the nonlinearity is in y(k) not u(k). We shall first start with a simple
example and then generalize the approach.
Example: Consider the system dynamics given by   )()1()( kbukyfky  , we first find the
reference value, )(kyr ,by setting u(k)=0. Thus  )()( kyfky rr  and we solve to find )(kyr .
Using Taylor series expansion around )(kyr we get:
     
   )1()1(
)1(
)1(
)(
)1()1(
)1(
)1(
)1()(
)1()1(
)1()1(










kyky
ky
kyf
ky
kyky
ky
kyf
kyfky
r
kyky
r
r
kyky
r
r
r
Define  )()()( kykykx r , then we get the new difference equation:
   )1()1(
)1(
)1(
)()(
)1()1(





kyky
ky
kyf
kyky r
kyky
r
r
i.e.
  )1(
)1(
)1(
)(
)1()1(





kx
ky
kyf
kx
kyky r
It is this equation that we use in the analysis.

6. 6
Feedback System:
In the feedback system, we design a controller that has two inputs, y(k) and r(k). Where r(k) is a
reference/external input. In polynomial format we have:
      )()()( 111
kyqSkrqTkuqR 

Where   R
RqrqrqrqR deg
deg
2
2
1
1
1
...1 

  S
SqsqsqssqS deg
deg
2
2
1
10
1
... 

  S
SqtqtqttqT deg
deg
2
2
1
10
1
... 

Thus,
 
 
 
  )()()( 1
1
1
1
ky
qR
qS
kr
qR
qT
ku 




    )()( 11
kyqGkrqG yr


Substituting, we get the closed loop output y(k):
   
       
   
        )()()( 1111
11
1111
11
k
qSqBqRqA
qCqR
kr
qSqBqRqA
qTqB
ky 







The polynomial          11111 
 qSqBqRqAqAc
is called the characteristic polynomial.
Stationary and Transient Response:
It is usually of interest to find the transient response for a unit step input, sinusoidal input, or
others. For example, administering the patient with steady level of glucose (step function) might

7. 7
cause overshoot in the patient’s temperature, blood pressure, or others. Thus, one has to design
the control such that the transient response is within acceptable limits. In this analysis, we shall
use the table of the inverse Z transform, the final and initial value theorems.
Some Inverse Z Transform for Zero Initial Conditions:
1
1
1

 az
has the inverse ,...2,1,0, kak
,
azaz
z


 

1
1 1
1
has the inverse ,...3,2,1,1

kak
,
Final Value Theorem:
For a stable system,   )(1lim)(lim 1
1
zyzky
zk


 , where y(z) is the Z transform of y(k).
Initial Value Theorem:
If )(lim zy
z 
exists, then the initial value y(0) of y(k) is given by )0()(lim yzy
z


.
Consider the simple stable system:
          0,1,1,1),()( 11111
 
kaqB-azqAkuqBkyqA -
i.e.
  )(
1
1
)( 1
zu
-az
zy -

which could be rewritten as:
)1()1()(  kukayky or )1()1()(  kukayky or )()()1( kukayky 
We need to find the transient and steady state response to a unit step function u(k)=1.

8. 8
Remember that the Z transform of y(k+n),  )( nkyZ  , depends on the initial conditions as
follows:
  )1()...2()1()0()()( 21
 
nzyyzyzyzzyznkyZ nnnn
The Z transform of a unit step function, u(k)=1, is given as:
 1
1
1
)( -
-z
zu  .
The Z transform of the output, y(k+1), is given as:
  )0()()1( zyzzykyZ 
The Z transform of the RHS of the equation, )()()1( kukayky  , is given as:
  )()()()( zuzaykukayZ 
Equating both sides we get:
   )()()1( kukayZkyZ 
i.e. )()()0()( zuzayzyzzy 
Rearrange we get:
  )()0()( zuzyzyaz 
Which yields:
   az
zu
az
zy
zy




)()0(
)(
If )1()()1(  kukayky

9. 9
Then, for zero initial conditions and unit step function as input, y(z) is obtained as:
      111
11
1
)(
1
1)(
)( 




z-az
zu
-azaz
zzu
zy --
Using the method of partial fraction expansion, order of numerator is less or equal to order of
denominator, we get:
         1111
1
)( 21
2
11







 
zazzaz
z
z-az
zy -

Where    
    11
)( 2
1













 a
a
zzaz
z
az
z
zy
az
azaz

and    
    azzaz
z
z
z
zy
z
zz 












 1
1
1
1
)(
1
1
2
1
2
Using the inverse Z transform we get:
  ,...3,2,1,1)( 2121  kaaky kkk

For a stable system, the steady state (final value) is:
 
)(
1
1
lim
1
1
)(lim 11
2 zy
za
ky
zk









 

The transient part that will die out is:
   
,...2,1,
11
1
1 





k
a
a
a
a
a
a
k
kk

Structure of Control Systems:
The general feedback system has the following relations:
    )()()( 11
kqGkuqGky u 



10. 10
    )()()( 11
kyqGkrqGku yr


We shall first give some examples of simple linear feedback systems and their applications.
Consider the case where   01

qG ,   11

qGr . In terms of the z transform, the transfer
function becomes:
 
   zGzG
zG
zr
zy
yu
u


1)(
)(
Example1, Inverse system design:
In some applications one is interested in the synthesis of the inverse of a given discrete time
system with transfer function p(z). Assuming that   KzGu  , and setting the feedback
  )(zpzGy  , then
 zpK
K
zr
zy
1)(
)(


If the gain is sufficiently large such that   1zpK , then:
     zpzpK
K
zpK
K
zr
zy 1
1)(
)(



We could obtain this by using operational amplifiers.
Example 2, stabilization of unstable systems:
Consider a first order system with   1, 

 a
az
b
zGu

11. 11
The system is unstable and we assume that the feedback is a constant gain i.e.   KzGy  , the
closed loop transfer function becomes:
 
   
 
    Kbaz
b
KzG
zG
zGzG
zG
zr
zy
u
u
yu
u






11)(
)(
The closed loop system will be stable if the pole is placed inside the unit circle by choosing:
1 aKb which leads to
b
a
K


1
Thus, we stabilize the system with a constant gain in the feedback loop. This is called a
proportional feedback system.
As another example, consider the second order unstable system:
 
   1,2




 a
azaz
b
az
b
zGu
We propose to use a proportional plus derivative feedback system;
  zKKzGy 21 
Which yields
 
   
 
       bKazbKz
b
azbzKK
azb
zGzG
zG
zr
zy
yu
u
12
22
21
2
/1
/
1)(
)(







The close loop system will be stable as long as the poles are inside the unit circle.
Example 3, tracking system:

12. 12
One of major applications of feedback is to have the output follows/tracks a reference input. As
an example consider the system where we choose:
     111 
 qGqGqG eyr
in the equation     )()()( 11
kyqGkrqGku yr


i.e.   )()()( 1
kykrqGku e  
then )(
)()(1
)()(
)( zr
zGzG
zGzG
zy
ue
ue


Define )()()( kykrke 
Then )()()()( zezGzGzy ue
and )(
)()(1
1
)( zr
zGzG
ze
ue

For good tracking system, we would like 
 j
ez
ze )( to be close to zero for the desired frequency
range. Thus, in this range, we need )()( zGzG ue to be large. This means that for good tracking we
need large gain.

13. 13
Design of Discrete time Control Systems
One of the objectives of digital control is to design a system that has some basic properties; (1)
Stability, (2) Steady state and transient response should be within acceptable limits, (3) Good
tracking capabilities.
In many applications, the plant or the system dynamics are given in terms of continuous time
differential equations. One has to find the plant transfer functions in the z domain not in the S
domain. Through sampling and hold unit, one is able to obtain the desired result.
Consider the plant transfer function G(s). The equivalent z transform of the plant G(z) is given
as:
  )()( 1
sGzG 
 LZ
Where  )(1
sG
L is the inverse Laplace transform of G(s).
Example:
as
sG


1
)( , then   akTat
eesG 
)(1
L .
     1
1
1
1
)()( 



ze
esGzG aT
akT
ZLZ
End of example.
Consider the transfer function G(s) where there is a sampling and hold unit before the plant. The
sampling and hold unit has the transfer function
s
e Ts
1
where T is the sampling interval. The
new transfer function X(s) becomes:

14. 14
)(
1
)( sG
s
e
sX
Ts


After some manipulations we get the z transform of the system dynamics X(z) as:
   














 
s
sG
z
s
sG
zzX
)(
1
)(
1)( 111
LZZ
Where










s
sG )(1
LZ is the Z transform of the inverse Laplace transform of 



s
sG )(
i.e. we get
the inverse Laplace transform of 



s
sG )(
and then take the Z transform of the obtained time-
domain signal.
Example: Assume that:
1
1
)(


s
sG . Thus
1
11
)(
1
)(






ss
e
sG
s
e
sX
TsTs
.
   1
11
1
1)(







sssss
sG
,
and
 
kTt
ekTstepetstep
sss
sG 

















)()(
1
11)( 111
LLL
where )(tstep is the step function starting at t=0.
   
  11
1
11
1
11
1
1
1
1
1
)(
)(





















zez
ze
zez
ekTstep
s
sG
T
T
T
kT
ZLZ
And    














 
s
sG
z
s
sG
zzX
)(
1
)(
1)( 111
LZZ    
  11
1
1
11
1
1 





zez
ze
z T
T
 
 1
1
1
1





ze
ze
T
T

15. 15
End of example.
A simple, yet popular and effective, controller is the proportional plus integral plus derivative
(PID) controller. If we have one reference, r(k), and the output, y(k), is fed back to the system,
then we define the error signal e(k) as:
e(k)=r(k)-y(k)
and )(
)(
)(
zG
ze
zu
PID
The z transform of the control u(k) input to the plant or system is now given as:
     
 1
211
1
1
1
11
1
1
)(
)(
)(









z
zKKzK
zK
z
K
KzG
ze
zu DIP
D
I
PPID
      
 1
21
1
2





z
zKzKKKKK DDPDIP
Where PK , IK , and DK are to be determinedfromthe desiredsystem response. The above formula
is named positional form of PID. There is, however, another popular form named velocity form. The z
transform of the input to the plant u(z) is given as:
       )(1
1
)()( 1
1
zyzKzyzr
z
K
zyKzu D
I
P





Steady State Error:
We need to find the limit value of the tracking error e(k). We use the limit value theory as:
  )(1)( 1
1
limlim zezke
zk




16. 16
Example: PID controller and simple plant. Assume that the plant transfer function G(s) is given
as:
as
sG


1
)( , then   akTat
eesG 
)(1
L ,      1
1
1
1
)()( 



ze
esGzG aT
akT
ZLZ . The
new transfer function between the error and the output is now )()(
)(
)(
)(
)(
)(
)(
zGzG
zu
zy
ze
zu
ze
zy
PID ,
and the input output relation is now:   )()()()()( zGzGzyzrzy PID which yields
    )()()()()(1)( zGzGzrzGzGzy PIDPID  i.e.
)()(1
)()(
)(
)(
zGzG
zGzG
zr
zy
PID
PID

 . To find a relation
between the error e(k) and the reference input r(k) we substitute e(k)=r(k)-y(k) which yields:
)()(1
1
)(
)()(1
)()(
)()()(
zGzG
zr
zGzG
zGzG
zrzrze
PIDPID
PID



 . We now need to find the limits as:
     






 


 )()(1
1
)(1)(1)( 1
1
1
1
limlimlim zGzG
zrzzezke
PIDzzk
.
For reference step function 1
1
1
)( 


z
zr , we get:
  

















 )()(1
1
)()(1
1
1
1
1)( limlimlim 1
1
1
1 zGzGzGzGz
zke
PIDzPIDzk
      
  


















11
21
1
1
1
1
2
1
1
lim
zez
zKzKKKKK
aT
DDPDIPz

17. 17
  
         







 


2111
11
1 211
11
lim zKzKKKKKzez
zez
DDPDIP
aT
aT
z
= 0
For any values of PID controller, the limit of the error is zero.
End of example.
The Diophantine Equation for Pole Placement Design:
As we mentioned before, in the feedback system we design a controller that has two inputs, y(k)
and r(k). Where r(k) is a reference/external input. In polynomial format we have:
      )()()( 111
kyqSkrqTkuqR 

Where   R
RqrqrqrqR deg
deg
2
2
1
1
1
...1 

  S
SqsqsqssqS deg
deg
2
2
1
10
1
... 

  S
SqtqtqttqT deg
deg
2
2
1
10
1
... 

Thus,
 
 
 
  )()()( 1
1
1
1
ky
qR
qS
kr
qR
qT
ku 




    )()( 11
kyqGkrqG yr


Substituting, we get the closed loop output y(k):
   
       
   
        )()()( 1111
11
1111
11
k
qSqBqRqA
qCqR
kr
qSqBqRqA
qTqB
ky 








18. 18
The polynomial          11111 
 qSqBqRqAqAc
is called the characteristic polynomial. It has the shape of what is known as the Diophantine
equation.
Usually we know the desired shape of  1
qAc and we need to find the polynomials  1
qR and
 1
qS that satisfythe characteristicequation/polynomial. The polynomial  1
qT is obtained through
the impulse response and the steady state error for a reference input.
There are sufficient (not necessary) conditions to find a unique solution to the Diophantine equation;
namely: (1) deg S = deg A – 1, (2) deg R = deg B -1, (3) deg    1degdeg1

BAqAc , where deg
stands for the degree or order.
We notice that the noise term appears both in the output y(k) and the control u(k). There is a need to
reduce or eliminatethe noise effects from the output and the control. We will, extensively, talk about
thisissue andothersinthe nextsections. For now, assume that the noise, )(k , has constant value. In
order to eliminate the noise effect from the output y(k):
   
       
   
        )()()( 1111
11
1111
11
k
qSqBqRqA
qCqR
kr
qSqBqRqA
qTqB
ky 







we put a term in R(k) such that the noise is eliminated. Specifically; let      11
1
1 
 qRqRqR f ,
and choose   11
1 
 qqRf . In this case     0)1()()(1)( 11
 
kkkqkqRf  .
Sometimes we need to eliminate the noise from the input control u(k):
   
 
   
  )()()( 1
11
1
11
k
qA
qCqS
kr
qA
qTqA
ku
cc






19. 19
If the noise term )(k is oscillating i.e. )1()(  kk  , then we design      11
1
1 
 qSqSqS f ,
with   11
1 
 qqSf . Thus     0)1()()(1)( 11
 
kkkqkqSf  . Notice that the
choice of  1
qRf and  1
qSf depends on the noise.
The chosenfactors,to eliminate the noise,are includedinthe Diophantine equation as elements of the
matrices A and B. Specifically, the new Diophantine equation becomes:
         1
1
11
1
11
'' 
 qSqBqRqAqAc
where      111
' 
 qRqAqA f , and      111
' 
 qSqBqB f
Example:Assume thatthe systemisgivenas )()1()( kkbuky  ,and the noise tconsk tan)(  .
Thus, the systems dynamics       )()()( 111
kqCkuqBkyqA 
 with   11

qA ,   11 
 bqqB ,
and   11

qC . We needto eliminatethe noise effectsfromthe output.   11
1 
 qqRf . In this case
    0)1()()(1)( 11
 
kkkqkqRf  . If we need to eliminate the noise from the
controller, u(k), we could choose   11
1 
 qqSf . Instead, we will choose  1
qS to satisfy the
Diophantine equation as follows:        1111
1' 
 qqRqAqA f , and we need to solve
         111
1
11
' 
 qSqBqRqAqAc . Following the sufficient conditions rules we get: namely:
(1) deg S = deg A’ – 1= 0, (2) deg 1R = deg B -1 = 0, (3) deg    1deg'deg1

BAqAc = 1. Thus,
  0
1
sqS 
,   10
1
1 RqR 
, and choose a stable    11
1 
 qqAc  . Substituting in the equation
         111
1
11
' 
 qSqBqRqAqAc , we get:     0
1
10
11
11 sbqRqq 
  . This yields the
independent equations: 101 R , and 010 bsR   . Thus,   bs /10  and
       111
1
1
1 
 qqRqRqR f . We need to find an expression for  1
qT . In this example, we

20. 20
assume thatthe reference r(k) = 0r = constantand we needthe output,y(k),tofollow the reference r(k).
After the elimination of noise, by choosing   11
1 
 qqRf ,
   
  )()( 1
11
kr
qA
qTqB
ky
C


 . From the
final value theory         
  





 





)(1)(1)( 1
11
1
1
1
1
limlimlim zr
zA
zTzB
zzyzky
Czzk
. And since
 1
0
1
)( 


z
r
zr , then
 
 
 
  001
11
1 1
1
1
)(
0 limlim r
bT
r
z
zTbz
ky
zk
r  







 


 , then
 
b
T


1
)1( . One
couldchoose    
b
TqT

 1
0
1
. In thiscase,
 
 
 
   1
00
1
1
1
1
1
)()(
)()()( 






q
kyskrT
ky
qR
qS
kr
qR
qT
ku
And since    
00
1 1
s
b
TqT 

 
, then
 
    )(
11
)()(
)( 1
0
1
0
ke
q
s
q
kykrs
ku 




 . Thus, we have
an integral controller.
In order to study the transient response, we set the disturbance to zero and we write the system
equation with advances not delays i.e. )()1()( kkbuky  is written as: )()1( kbuky  .
Usingthe Z transformwe get: )()0()( zbuzyzzy  . This yields )()0()( zu
z
b
yzy  . For a unit step
function, u(k)=1 and
 1
1
1
)( 


z
zu . Thus,
 1
1
1
)0()( 


zz
b
yzy
 1
)0(


z
b
y . Using the
inverse Z transform we get:   ,...3,2,1,0,)()0(1)()0()(  kbkybkyky
k


21. 21

22. 22
Disturbance Rejection and Tracking
Annihilation is used to remove undesired signals or disturbances from the output. We shall first
explain the annihilation concept and then move to see its applications.
Consider the signal S(k) which is modeled as:
    )()( 11
kqNkSq 

with zero initial conditions i.e. S(-1)=0=S(-2)=…=S(-deg).
Where, as before, )(k is the pulse function i.e. 1)0(  and zero elsewhere,  1
 q and  1
qN
are polynomials in the delay operator  1
q . It is usually assumed that the degree of polynomial
N is less than the degree of polynomial i.e. Ndeg < deg.
Taking the Z transform of both sides we obtain:
       zNkZzNzSz  )()( 
Rearrange we get:
 
 z
zN
zS

)(
Define   


 deg
deg
2
2
1
1
1
...1 qfqfqfq
  N
N qnqnqnnqN deg
deg
2
2
1
10
1
... 
 Ndeg <deg
In the time domain, the equation     )()( 11
kqNkSq 
 could be written as:
    )(...)(...1 deg
deg
2
2
1
10
deg
deg
2
2
1
1 kqnqnqnnkSqfqfqf N
N 




23. 23
i.e. )deg(...)2()1()(
)deg(...)2()1()(
deg210
deg21
Nknknknkn
kSfkSfkSfkS
N 
 

We study the above equation at different time instants.
k=0: )deg(...)2()1()0(
)deg(...)2()1()0(
deg210
deg21
Nnnnn
SfSfSfS
N 
 

Since the initial conditions are zeros i.e.S(-1)=0=S(-2)=…=S(-deg), and )(k is the pulse
function i.e. 1)0(  and zero elsewhere, then the above equation is reduced to:
0)0( nS 
k=1: )1deg(...)1()0()1(
)1deg(...)1()0()1(
deg210
deg21

 
Nnnnn
SfSfSfS
N
which is reduced to:
11 )0()1( nSfS 
k=2: )deg2(...)0()1()2(
)deg2(...)0()1()2(
deg210
deg21
Nnnnn
SfSfSfS
N 
 

which is reduced to:
221 )0()1()2( nSfSfS 
k=l> deg :
)deg(...)2()1()(
)deg(...)2()1()(
deg210
deg21
Nlnlnlnln
lSflSflSflS
N 
 

which is reduced to:

24. 24
0)deg(...)2()1()( deg21   lSflSflSflS
i.e.   0)(1
 
kSq ,k > deg
This result is independent of the shape of the polynomial  1
qN as long as Ndeg <deg .Thus,
there is a polynomial  1
 q that when applied to the signal S(k), the output   0)(1
 
kSq after
a transient period k > deg .This polynomial is called the annihilation polynomial. Notice that
this polynomial is nothing but the denominator of the Z transform of the signal S(k).
Example: Assume that the signal is a constant quantity i.e. S(k)=a, and “a” is unknown. The Z
transform of S(k) is given as:    
  1
1
)( 




z
a
z
zN
kSZ . Thus, the annihilation polynomial
 1
 q is deduced as:      111
111



 
qzq
qz
.
Thus,     0)1()()(1)( 11
 
aakSkSkSqkSq .
Example: Assume that the signal is a ramp i.e. S(k)=a+b k, “a and b” are unknown. The Z
transform of S(k) is given as:    
   
 
 21
11
21
1
1
1
1
11
)( 













z
bzza
z
bz
z
a
z
zN
kSZ . Thus, the
annihilation polynomial  1
 q is deduced as:        212111
21111



 
qqqzq
qz
.
Thus,     )2()1(2)()(21)( 211
 
kSkSkSkSqqkSq
aS  )0( , k=0
)(2)()0(2)1( baabaSS  , k=1
    0)(22)0()1(2)2(  ababaSSS , k=2

25. 25
       212  kbakbakba
      2,02222  kbbkkkbaaa
In some applications, the polynomial   )(1
kSq
 is multiplied by another polynomial  1
 q with
finite number of elements. In this case, the above argument is still valid and we get:
        )()( 1111
kqNqkSqq 


The annihilation will occur now at:
      
deg,0)(11
kkSqq
Example: Assume that the signal is a constant quantity i.e. S(k)=a, and “a” is unknown. Assume
that    11
1 
 bqq which is of order 1 i.e. 1deg  . The Z transform of S(k) is given as:
   
  1
1
)( 




z
a
z
zN
kSZ . Thus, the annihilation polynomial  1
 q is deduced as:
     111
111



 
qzq
qz
, and 1deg 
Thus,     0)1()()(1)( 11
 
aakSkSkSqkSq .
          )(11)(11)( 211111
kSbqqbkSqbqkSqq 


        )(1)(1)( 1111
kbqakabqkqNq  

Equating both sides we get:
     )(1)(11 121
kbqakSbqqb 


26. 26
i.e. )1()()2()1()1()(  kabkakbSkSbkS 
k=0: )1()0()2()1()1()0(   ababSSbS
which yields aS )0(
k=1: )0()1()1()0()1()1(  ababSSbS 
which yields abSbS  )0()1()1(
k=2: )1()2()0()1()1()2(  ababSSbS 
which yields 0)0()1()1()2(  bSSbS
Thus, for   degk ,     0)(11
 
kSqq .
Binomial Expansion Formula:
In some situations, the denominator  1
 q of some transfer function is multiplied by the
polynomial   )(1
kSq
 . In this case, we expand the inverse of the denominator using the
Binomial expansion formula for n fraction (positive or negative) and 1x .
Thus,   1...,
3!2
)1(
11 32








 xx
n
x
nn
nxx
n
1,
0






 


xx
l
n
l
l
Which could be approximated as:

27. 27
  1,1
0






 
xx
l
n
x
L
l
ln
Example:   L
xxxx ....11 21


Thus,   )(1
kSq
 is multiplied by another polynomial  1
/1 
 q with finite number of elements.
In this case, the above argument is still valid and we get:
        )(
1
)(
1 1
1
1
1 kqN
q
kSq
q







The annihilation will occur now at:
     



deg,0)(
1 1
1
kkSq
q
Example: Assume that the signal is a constant quantity i.e. S(k)=a, and “a” is unknown. Assume
that     1,1 111
 
bqbqq . The inverse could be approximated as:
             1,...1
1
1
1 1
0
1211111
1











  bqbqbqbqbq
l
bq
q
L
l
Ll
which is of order L i.e. L

1
deg . The Z transform of S(k) is given as:
   
  1
1
)( 




z
a
z
zN
kSZ . Thus, the annihilation polynomial  1
 q is deduced as:
     111
111



 
qzq
qz
, and 1deg 
Thus,     0)1()()(1)( 11
 
aakSkSkSqkSq .

28. 28
            )(
1
1)(11)(
1
0
111111
1 kSbq
l
qkSqbqkSq
q
L
l
l

















        )(
1
)(1)(
1
0
1111
1 kbq
l
akabqkqN
q
L
l
l
 
















Thus, for   degk ,
    0)(
1 1
1 


 kSq
q
.
Example: Assume that     1,1 1
1
11
 


 qbqbq i
I
i
i . The inverse could be approximated as:
      1,
1
1
1 1
1 0
1
1
11
1









 



  qbqb
l
qb
q
i
I
i
L
l
i
I
i
i
i l
which is of order 
I
i
iL
1
i.e. 


I
i
iL
1
1
deg .
Internal model principle and annihilation:
The system       )()()( 111
kqCkuqBkyqA 

has noise )(k as input. When designing the feedback loop
      )()()( 111
krqTkyqSkuqR 

with unknown R, S, and T, the closed loop dynamics become:
The output
   
 
   
  )()()( 1
11
1
11
k
qA
qRqC
kr
qA
qTqB
ky
cc






29. 29
and the input
   
 
   
  )()()( 1
11
1
11
k
qA
qCqS
kr
qA
qTqA
ku
cc





where          11111 
 qSqBqRqAqAc
Notice that the disturbance )(k appears in both the output, y(k), and the input to the system,
u(k). We need to reduce or eliminate the effect of the disturbance. If the feedback system is
stable, then the inverse of )( 1
qAc will have a finite number of elements i.e.  1
1
)(
1 

 q
qAc
,
where       1,
1
1 1
1 0
1
1
111






 
 



 qq
l
qq i
I
i
L
l
i
I
i
i
i l

Assume that the disturbance, )(k , could be modeled as:
    )()( 11
kqNkq  

then     
deg,0)(1
kkq 
We want to, asymptotically, eliminate the disturbance from the output y(k). In order to do that
we choose )( 1
qR such that it contains the polynomial )( 1
 q
.
i.e.    11
1
1
)( 
 qqRqR
Substitute the expressions for R and  1
1
)(
1 

 q
qAc
, we get:
   
          )(
1
)()( 1
1 0
11
1
1
1
11
kqq
l
qRqCkr
qA
qTqB
ky
I
i
L
l
i
c
i l
 
 









 

30. 30
For large values of k we get:
   
  )()( 1
11
kr
qA
qTqB
ky
c


 , k is very large.
Tracking:
In the tracking problem, we need the output y(k) to follow some known reference value r(k).
Since we usually have “d” steps delay in the system, we simply advance the reference value by
"d" steps. Assume that the feedback controller has the shape:
      )()()( 111
dkrqTkyqSkuqR  
Define    11 
 qBqqB d
d
Then for zero noise,
   
 
   
  )()()( 1
11
1
11
kr
qA
qTqB
dkr
qA
qTqB
ky
c
d
c





Assume that the reference r((k) is given by:     )()( 11
kqNkrq rr 

Define the tracking error e(k) as:
   
  )()()()()( 1
11
kr
qA
qTqB
krkykrke
c
d



   
  )(1 1
11
kr
qA
qTqB
c
d






 

     
  )(1
111
kr
qA
qTqBqA
c
dc





 
 


31. 31
We need to design  1
qT and  1
qAc such that          1111
. 
 qqTqBqA rdc . This way, the
error e(k) will asymptotically go to zero.
Remember that:          11111 
 qSqBqRqAqAc
where we obtain the values of R and S through Diophantie equation with:
(1) deg S = deg A – 1, (2) deg R = deg B -1, (3) deg    1degdeg1

BAqAc .
Let      11
1
1
1

 qAqTqT c ,      111
1

 qAqAqA cmc
Substitute for the expression of e(k) we get:
         
    )()( 11
11
1
111
1
11
kr
qAqA
qAqTqBqAqA
ke
cm
cdcm







 
 

     
  )(1
1
1
11
kr
qA
qTqBqA
m
dm





 
 

Assume          111
1
11 
 qqMqTqBqA rdm
Then
   
  )()( 1
11
kr
qA
qqM
ke
m
r




The error will asymptotically go to zero as we explained in the previous section.
Example: Assume that the system is described by the equation
)2()2()1()( 221  kubkyakyaky

32. 32
Thus,   2
2
1
1
1
1 
 qaqaqA ,   2
2
1 
 qbqB , the delay d=2 and     2
121
bqBqqBd  
. The
reference input r(k) is chosen to be a constant value i.e. 1
1
1
1
1
)(,
1
1
)( 






q
qr
z
zr . Since
    )()( 11
kqNkrq rr 
 , then       1,1 111
 
qNqq rr . The feedback system is chosen
such that it is of second order and the poles are inside the unit circle i.e.
    1
2
1
1
2
2
1
1
1
111 
 qqqaqaqA ccc  . From Diophantine equation, and to have a
unique solution, we choose   1
10
1 
 qrrqR ,   1
10
1 
 qssqS .
We need to design  1
qT and  1
qAc such that          1111
. 
 qqTqBqA rdc . For zero
delay,     2
11
bqBqBd  
. Choose         1
2
1
1
111
111

 qqqAqAqA cmc  . i.e.
   1
1
1
1 
 qqAm  ,    1
2
1
11

 qqAc  . Choose        1
20
11
1
1
11

 qtqAqTqT c 
Substitute into          1111
. 
 qqTqBqA rdc ,
we get:
       11
202
1
2
1
1 1.111 
 qqtbqq 
which is reduced to:          11
102
1
202
1
1
1
2 1.1111 
 qqtbqtbqq 
Choose “ 0t ” such that     
 1
1
1
1
02
1
1
102 1
1
1 








 qq
tb
qtb 

 i.e.
 
1
1
1
02



tb
which yields
2
1
0
1
b
t

 .
In summary, we choose the following values for the feedback polynomials:
  1
10
1 
 qrrqR ,   1
10
1 
 qssqS

33. 33
    1
2
1
1
2
2
1
1
1
111 
 qqqaqaqA ccc  =       1
2
1
1
11
111

 qqqAqA cm 
   1
1
1
1 
 qqAm  ,    1
2
1
11

 qqAc 
       1
20
11
1
1
11

 qtqAqTqT c  ,
2
1
0
1
b
t


Robust Tracking:
The system dynamics,  1
qA and  1
qB ,are usually estimated from real data. The feedback
signals y(k) are obtained from the real system. The tracking design, however, uses estimated
model. This results in error due to model mismatch. This error will be included in the design
using the annihilation principle as explained next.
Let the true system dynamics, assuming zero external noise, be given by the equation:
    )()( 1*1*
kuqBkyqA 

The estimated model used for the design is:
    )()()( 11
kkuqBkyqA  
Notice the presence of the noise or disturbance term )(k that is obtained by subtracting the two
above equations as:
          )()()( 1*11*1
kuqBqBkyqAqAk 

As before, the output, y(k), and the input, u(k), for the closed loop system, assuming perfect
knowledge of the system, are given as:

34. 34
   
  )()( 1*
11*
kr
qA
qTqB
ky
c
d



   
  )()( 1*
11*
kr
qA
qTqA
ku
c
d



Where          11*11*1* 
 qSqBqRqAqAc
and    1*1* 
 qAqAq d
d
Substituting for the derived expressions for y(k) and u(k), the disturbance could be expressed as
function of the reference signal r(k) as:
        
          
  )()()( 1*
11*
1*1
1*
11*
1*1
kr
qA
qTqA
qBqBkr
qA
qTqB
qAqAk
c
d
c
d







              
    )(1
1*
1*1*11*1*1
krqT
qA
qAqBqBqBqAqA
c
dd 




 
    )(1
1*
1
krqT
qA
qD
c




Where                 1*1*11*1*11 
 qAqBqBqBqAqAqD dd
Assuming that there is error in the estimated system dynamics, we could find another expression
for y(k) as follows:
Since     )()()( 11
kkuqBkyqA  
The closed loop output y(k) will be:

35. 35
   
 
 
  )()()( 1
1
1
11
k
qA
qR
kr
qA
qTqB
ky
cc
d





and the input
   
 
 
  )()()( 1
1
1
11
k
qA
qS
kr
qA
qTqA
ku
cc
d





where          11111 
 qSqBqRqAqAc
Assume that the reference r(k) is given by:     )()( 11
kqNkrq rr 

Define the tracking error e(k) as:
   
  )()()()()( 1*
11*
kr
qA
qTqB
krkykrke
c
d



   
  )(1 1*
11*
kr
qA
qTqB
c
d






 

     
  )(1*
11*1*
kr
qA
qTqBqA
c
dc





 
 

This formula is function of the unknown real quantities,  1* 
qAc and  1* 
qB , and will be difficult
to use.
On the other hand, if we use the expression
   
 
 
  )()()( 1
1
1
11
k
qA
qR
kr
qA
qTqB
ky
cc
d




 , we get:
   
 
 
  )()()()()()( 1
1
1
11
k
qA
qR
kr
qA
qTqB
krkykrke
cc
d






36. 36
Substitute
 
    )()( 1
1*
1
krqT
qA
qD
k
c




We get
   
 
 
 
 
    )()()()( 1
1*
1
1
1
1
11
krqT
qA
qD
qA
qR
kr
qA
qTqB
krke
ccc
d 







               
    )(11*
111111*11*
kr
qAqA
qTqDqRqTqBqAqAqA
cc
dccc





 
 

We need to design  1
qT ,  1
qR and  1
qAc such that
                  11111111*
. 
 qqTqDqRqTqBqAqA rdcc . This way, the error e(k) will
asymptotically go to zero.
Let      111
1

 qAqAqA mcc
     1
1
11
1

 qTqAqT c
     111
1

 qqRqR r
then               1
1
111111
1

 qTqBqAqAqTqBqA dmcdc
Choose           111
1
11 
 qqMqTqBqA rdm
Substitute we get:
              
                    1111
1
11
1
1111*
1111111*
111




qqRqDqTqAqTqBqAqAqA
qTqDqRqTqBqAqA
rcdmcc
dcc
                 1111
1
11111*
111

 qqRqDqTqAqqMqAqA rcrcc

37. 37
                1111
1
111*1
111

 qqRqDqTqAqMqAqA rccc
Thus, the error becomes:
              
        )()( 1
1*11
111
1
111*1
1
111
krq
qAqAqA
qRqDqTqAqMqAqA
ke r
cmc
ccc 





            
      )(1
1*1
111
1
111*
11
krq
qAqA
qRqDqTqAqMqA
r
cm
cc 





This is the desired result.
Example: Assume that the estimated system is described by the equation
)2()2()1()( 221  kubkyakyaky . Thus,   2
2
1
1
1
1 
 qaqaqA ,   2
2
1 
 qbqB . For
zero delay, d=2,     2
11
bqBqqB d
d  
. The reference input r(k) is chosen to be a constant
value i.e. 1
1
1
1
1
)(,
1
1
)( 






q
qr
z
zr .
Since     )()( 11
kqNkrq rr 
 , then       1,1 111
 
qNqq rr .
We need to find the polynomials R, S, T such that the tracking error asymptotically goes to zero
even though the estimated values of A and B are not the correct values  1* 
qB and  1* 
qA .
The feedback system is chosen such that it is of second order and the poles are inside the unit
circle i.e.     1
2
1
1
2
2
1
1
1
111 
 qqqaqaqA ccc  . From Diophantine equation, and to
have a unique solution, we choose   1
10
1 
 qrrqR ,   1
10
1 
 qssqS .
Since         11111
10
1
111

 qqRqqRqrrqR r ,

38. 38
then choose 01 rr  . and   0
1
1
rqR 
. Thus,    1
0
1
1 
 qrqR
Let         1
2
1
1
111
111

 qqqAqAqA mcc 
e.g.    1
1
1
1 
 qqAm  ,    1
2
1
11

 qqAc 
choose        1
20
1
1
11
11

 qtqTqAqT c  , i.e.   0
1
1 tqT 
then               1
1
111111
1

 qTqBqAqAqTqBqA dmcdc
        1
012
1
0
1
1
1
2
1
11 11

 qtbqAtqbqqA cc 
Choose           111
1
11 
 qqMqTqBqA rdm
i.e.
     111
102 11 
 qqMqtb 
which yields   1
1

qM ,
 
1
1
1
02



tb
, i.e.
2
1
0
1
b
t


In summary, the values, not unique, for the R, S, T polynomials, to have asymptotic zero
tracking, are:
    1
2
1
1
1
11 
 qqqAc  Assumed
   1
0
1
1 
 qrqR Obtained from Diophantine equation
  1
10
1 
 qssqS Obtained from Diophantine equation
   1
2
2
11
1
1 





 
 q
b
qT 


39. 39

40. 40
Blind Signal Separation and the Aortic Pressure
In this problem, there is one unknown source (the Aorta u(t)) and several unknown paths or
channels )(thi that generate several known pressures )(tyi at remote sites from the Aorta. Thus
the measured output is the convolution "*" of the input with the channel,
)(*)()( thtuty ii 
Convolving channel j with )(tyi and channel i with )(tyj , we get:
)(*)(*)()(*)( ththtuthty jiji 
)(*)(*)()(*)( ththtuthty ijij 
The right hand sides of both equations are equal. Thus,
)(*)()(*)( thtythty jiij 
The channels could be estimated from this equation. This is a special case of Multi-Channel
Blind Deconvolution. For exposition purposes assume that we have two unknown sources
)(),( 21 tutu and three measurements )(),(),( 321 tytyty that are the convolution of the sources with
unknown linear time invariant finite impulse response (FIR) filters. This could be represented by
the equation:





































)(
)(
)(
)(
)(
*
)()(
)()(
)()(
)(
)(
)(
3
2
1
2
1
3231
2221
1211
3
2
1
t
t
t
tu
tu
thth
thth
thth
ty
ty
ty



Where "*" is the convolution operation, )(ti is the ith error or noise.

41. 41
The objective is to find an estimate of the source signals )(),( 21 tutu .
The above equation could be written in the Z domain as:





































)(
)(
)(
)(
)(
)()(
)()(
)()(
)(
)(
)(
3
2
1
2
1
3231
2221
1211
3
2
1
z
z
z
zu
zu
zhzh
zhzh
zhzh
zy
zy
zy



Setting the error terms to zero, we get:



























)(
)(
)()(
)()(
)()(
)(
)(
)(
2
1
3231
2221
1211
3
2
1
zu
zu
zhzh
zhzh
zhzh
zy
zy
zy
We could deal with two channels at a time. This yields a total of 3 equations as:


















)(
)(
)()(
)()(
)(
)(
2
1
2221
1211
2
1
zu
zu
zhzh
zhzh
zy
zy
which gives: 


















)(
)(
)()(
)()(
)(
)(
2
1
1
2221
1211
2
1
zy
zy
zhzh
zhzh
zu
zu


















)(
)(
)()(
)()(
)(
)(
2
1
3231
1211
3
1
zu
zu
zhzh
zhzh
zy
zy
which gives: 


















)(
)(
)()(
)()(
)(
)(
3
1
1
3231
1211
2
1
zy
zy
zhzh
zhzh
zu
zu
and 

















)(
)(
)()(
)()(
)(
)(
2
1
3231
2221
3
2
zu
zu
zhzh
zhzh
zy
zy
which gives: 


















)(
)(
)()(
)()(
)(
)(
3
2
1
3231
2221
2
1
zy
zy
zhzh
zhzh
zu
zu
Thus, we have:





































)(
)(
)()(
)()(
)(
)(
)()(
)()(
)(
)(
)()(
)()(
3
2
1
3231
2221
3
1
1
3231
1211
2
1
1
2221
1211
zy
zy
zhzh
zhzh
zy
zy
zhzh
zhzh
zy
zy
zhzh
zhzh
After some manipulations we get:
 
 
 
 zhzhzhzh
zyzhzyzh
zu
zhzhzhzh
zyzhzyzh
()()()(
)()()()(
)(
)()()()(
)()()()(
32111231
311131
2
22111221
211121





i.e.
  
  )()()()()()()()(
)()()()(()()()(
31113122111221
21112132111231
zyzhzyzhzhzhzhzh
zyzhzyzhzhzhzhzh



42. 42
Through regression analysis, one could find an estimate for  )()()()( 31222132 zhzhzhzh  ,
 )()()()( 32111231 zhzhzhzh  , and consequently an estimate for )(2 zu is obtained using, for
example,     )()()()()()()()()( 222111221211121 zuzhzhzhzhzyzhzyzh 
The above steps could be repeated for )(1 zu . Thus, the two sources are obtained.
Example: To further simplify the analysis, assume that the filters are first order i.e.
1
11)( 
 zhzh ijij i=1,2,3 , j=1,2
Thus,
      
    
     )()()(
)()()(
)()(
3
2
221111121211
1
221111121211
1
2
311221321211
1
311221211321
2
2
321111121311
1
321111311121
zyzhhhhzhhhh
zyzhhhhzhhhh
zyzhhhhzhhhh






Define:   )( 3211113111211 hhhh 
  3211111213112 hhhh 
 )()( 3112212113213 hhhh 
 3112213212114 hhhh 
 )()( 2211111212115 hhhh 
 2211111212116 hhhh 
Thus the above equation could be written compactly as:
     )2()1()2()1()2()1( 363514132221  kykykykykyky 
Through regression analysis one should be able to find an estimate for only five of the six
unknown coefficients that represent the transfer functions. The sixth coefficient
 2211111212116 hhhh  will be taken as the numeraire. Another two similar equations could

43. 43
be obtained and from which we could get estimates for all the six parameters. The filter
coefficients 1ijh are then estimated from the estimated i . Once estimated, we use these filter
coefficients to find an estimate for the unknown sources using:
 
5
2111121121
2
5
6
2
)()()1()1(
)1()(

 kyhkyhkyky
kuku




In matrix format and for N data points we have:
121212  UHY
Where
























)2()2(13()1((
.
.
.
)2()2()3()3((
)1()1()2()2((
2111121121
2111121121
2111121121
12
NyhNyhNyNy
yhyhyy
yhyhyy
Y ,
2U =,




















 )1(
.
.
.
)1(
)0(
2
2
2
Nu
u
u
, 1 =




















 )3(
.
.
.
)1(
)0(
1
1
1
N


, and 12H =















56
56
56
...0
...........
00
...0



Once the coefficients of the FIR filters are estimated, we use inverse filtering to find and
estimate 2
ˆU for the source signal 2U as follows:
  12
1
1212122
ˆˆˆˆ YHHHU TT 

Where “T” stands for transpose, and ”^” on top of the variable symbol means the estimate of that
variable.

44. 44