# EE1354 MODERN CONTROL SYSTEMS

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1

NPR COLLEGE OF ENGINEERING &TECHNOLOGY

BE EEE
-
III/ SEMESTER VI

EE1354

MODERN CONTROL SYSTEMS

Prepared By:

A.R.SALINIDEVI

Lect/
EEE

2

EE1354

MODERN CONTROL SYSTEMS

(Common to EEE, EIE and ICE)

L T P C

3 1 0 4

UNIT I STATE SPACE ANALYSIS OF CONTINUOUS TIME SYSTEMS 9

State variable representation

Conversion of state variable form to transfer function and vice
versa

Eigenvalues and Eigenvectors

Solution of state equation

Controllability and
o
bservability

Pole placement design

Design of state observer

UNIT II z
-
TRANSFORM AND SAMPLED DATA SYSTEMS 9

Sampled data theory

Sampling process

Sampling theorem

Signal reconstruction

Sample

and hold circuits

z
-
Transform

Theorems on z
-
Transfo
rms

Inverse z
-
Transforms

Discrete

systems and solution of difference equation using z transform

Pulse transfer
function

Response of sampled data system to step and ramp Inputs

Stability studies

Jury’s test and bilinear transformation

UNIT III S
TATE SPACE ANALYSIS OF DISCRETE TIME SYSTEMS 9

State variables

Canonical forms

Digitalization

Solution of state equations

Controllability

and Observability

Effect of sampling time on controllability

Pole
placement by state feedback

Linear obs
erver design

First order and second order
problems

UNIT IV NONLINEAR SYSTEMS 9

Types of nonlinearity

Typical examples

Phase
-
plane analysis

Singular points

Limit
cycles

Construction of phase trajectories

Describing function method

Basic conc
epts

Saturation

Relay

Backlash

Liapunov stability analysis

Stability in the
sense of Liapunov

Definiteness of scalar functions

Second method of
Liapunov

Liapunov stability analysis of linear time invariant syst
ems and non
-
linear system

UNIT V MIMO SYSTEMS 9

Models of MIMO system

Matrix representation

Transfer function representation

Poles
and

Zeros

Decoupling

Introduction to multivariable Nyquist plot and singular values
analysis

Model predictive con
trol

L: 45 T: 15 Total: 60

TEXT BOOKS

1. Gopal, M., “Digital Control and State Variable Methods”, 3rd Edition, Tata McGraw Hill,

2008.

2. Gopal
, M., “Modern Control Engineering”, New Age International, 2005.

REFERENCES

1. Richard C. Dorf and Robert H. Bishop, “Modern Control Systems”, 8th Edition, Pearson

Education, 2004.

2. Gopal, M., “Control Systems: Principles and Design”, 2nd Edition, Tata M
cGraw Hill,

2003.

3. Katsuhiko Ogata, “Discrete
-
Time Control Systems”, Pearson Education, 2002.

3

MODERN CONTROL SYSTEM

Unit I

STATE SPACE ANALYSIS OF CONTINUOUS
TIME SYSTEMS

State Variable Representation

The state variables may b
e to
tally independent of each other, leading

to diagonal or normal form or they could be derived as the derivatives of the output. If
them is no direct relationship between various
states. We

could use a suitable
trans
formation to obtain the representation in diagonal form
.

Phase Variable Representation

It is often convenient to consider the output of the system as one of the state
variable and remaining state variable

as derivatives of this state variable. The state
variables thus obtained from one of the system variables and its (n
-
1) derivatives, are
known as n
-
dimensional phase variables.

In a third
-
order mechanical
system,

the

output may be
displacement

v
x
x
x
2
.
1
,
1
and
a
x
x
3
2
.
in the c
ase of
motion
of translation
or
angular
displacement
w
x
x
x
2
.
1
,
1
1
and
.
3
.
2
w
x
x
if the motion is
rotational,

Where

v

,
,
,
a
w
v

respectively, are
velocity
, angu
lar velocity acceleration,

angular
acceleration.

Consider a SISO system desc
ribed by nth
-
order differential equation

Where

u
is, in general, a function of time.

The
nth
order transfer function of this sy
stem is

State variable representation

=
C潮癥牳楯渠潦=獴慴e=癡物r扬e=景f洠瑯t瑲a湳ne爠晵fc瑩潮oa湤n

=
䕩来b癡汵敳=a湤n䕩来b癥c瑯牳r

=
p潬畴楯渠潦=獴慴s=e煵q瑩潮o

=
C潮瑲潬oa扩b楴y=

=
m潬攠灬oce浥湴⁤m獩g渠

=
ae獩s渠潦⁳瑡te⁯扳=牶er
=
4

With the states
(each being function of

time) be defined as

Equation
becomes

Using above Eqs stat
e equations in phase satiable loan can he obtained as

Where

Physical Variable
Representation

In this representation the state variables are real physical variables, which can
b
e

measured and used

for manipulation or for control pu
rposes.

The
approach generally
d is to
break the block diagram of the transfer functio
n into subsystems in such a
way that the physical variables can he identified. The governing equations for the
subsystems can he used to identify the physical variables. To illustrate the approach
cons
id
er the block diagram of Fig.

One may represent the transfer function of this system

as

Taking
H(s) = 1, the block diagram of can be redrawn as in Fig.

physical v
ariables can
be speculated

as x
1
=y, out
put,
.
2
w
x
the angular velocity
a
i
x
3

the armature

curre
nt in a position
-
control system.

5

Where

The state space representation can be
obtained by

And

State space models from transfer functions

A simple example of system has an input and output as shown in Figure 1. This class
of system

has general form of model given in Eq.(1).

1 1
1 0 1 0
1 1
( ) ( )
n n m
n m
n n m
d y d y d u
a a y t b b u t
dt dt dt
 

Models of this form have the property of the following:

1 1 2 2 1 1 2 2
( ) ( ) ( ) y( ) ( ) ( )
u t u t u t t y t y t

(2)

where, (y
1
, u
1)

and (y
2
,u
2)

each satisfies Eq,(1).

Model of the form of Eq.(1) is known as linear time invariant (abbr.
LTI
) system. Assume
the system is at rest prior to the time t0=0, and, the input u(t) (0
t <
)

produces the output
y(t) (0
t <
), the model of Eq.(1) can be represented by a transfer function in
term

of
Laplace transform
variables
, i.e.:

S

y(t)

u(t)

6

1
1 0
1
1 0
( ) ( )
m m
m m
n n
n n
b s b s b
y s u s
a s a s a

(3)

Then applying the same input shifted by any amount
 of time produces the same output
shifted by the same amount q of time. The representation of this fact is given by the following
transfer function:

1
1 0
1
1 0
( ) ( )
m m
s
m m
n n
n n
b s b s b
y s e u s
a s a s a

(4)

Models of Eq.(1) having all
0 ( 0)
i
b i
, a state space description arose out of a reduction
to a system of first order differential equations. This technique is quite general. First, Eq.(1)
is written as:

( ) ( 1)
( 1)
0 1 1
,( ),,.,,;
with initial conditions: y(0)=y,(0) (0),
,(0) (0)
n n
n
n
y f t u t y y y y
y y y y
 

(5)

Consider the vector
n
x R
with
( 1)
1 2 3
,,,,
n
n
x y x y x y x y
 

, Eq.(5) becomes:

2
3
( 1)
,( ),,.,,
n
n
x
x
d
X
dt
x
f t u t y y y y

 

(6)

In case of linear system, Eq.(6) becomes:

0 1 n-1
0 1 0 0 0
0 0 1 0 0 0
0 ( ); y(t)= 1 0 0 0
0 0 1
-a -a -a 1
d
X X u t X
dt
 

 
  

(
7)

It can be shown that the general form of Eq.(1) can be written as

0 1 m
0 1 n-1
0 1 0 0 0
0 0 1 0 0 0
0 ( ); y(t)= b b b 0 0
0 0 1
-a -a -a 1
d
X X u t X
dt
 

  
  

(8)

and, will be represented in an abbreviation form:

7

; ; D=
X AX Bu y CX Du

0

(9)

Eq.(9) is known as the controller canonical form of the system
.

Transfer function from state space models

We have just showed that a transfer function model can be expressed as a state space
system of controller canonical form. In the reverse direction, it also easy to see that each
linear state space system of Eq.(9
) cab be expressed as a LTI transfer function. The procedure
is to take laplace transformation of the both sides of Eq,(9) to give:

( ) ( ) ( ) ; ( ) ( ) ( )
sX s AX s Bu s y s CX s Du s

(10)

So that

1
( )
( ) ( ) ( ) ( )
( )
n s
y s C sI A B D u s G s u s
d s

(11)

An algorithm to compute the transfer func
tion from state space matrices is given by the
Leverrier
-
-
Frame formula of the following:

1
1 2
0 1 2 1
1
1 1
0 1 0
1 0 1 2 1
( )
( )
( )
( )
,

( )
1/2 ( )

n n
n n
n n
n n
N s
sI A
d s
N s s N s N sN N
d s s d s d s d
where
N I d trace AN
N AN d I d trace AN

1 2 1 1 2
-1 1

1
( )
1
1
0 A
n n n n n
n n n n
N AN d I d trace AN
n
N d I d trace AR
n
 

(12)

Therefore, according to the algorithm mentioned, the transfer function becomes:

( ) ( )
n s CN s B CD

(13)

o
r
,
( )
( )
( )
CN s B CD
G s
d s

(14)

Eigen
Values

Consider an equation AX = Y which ind
icates the transformation of
1
n

vector
matrix X into 'n x 1' vector matrix Y by 'n x n' matr
ix operator A.

8

If there exists such a vector X such that A transforms it to a vector XX then X is
called

the solution of the equation
,

T
he set of homogeneous equations
(1) have a nont
rivial solution only under
the

condition,

The determinant | X I
-

A | is called characteristic polynomial while the equation (2)
is called the characteristic equation.

After expanding,

we get the characteristic equation as,

The 'n' roots o
f the equation (3) i.e. the values of X satisfying the above equation
(3)

are called eigen values of the matrix A.

The equation (2) is similar to
| sI
-

A |
=0, which is the characteristic

equation of the
system. Hence values of X satisfying characteristic e
quation arc the closed loop
poles of

the system. Thus eigen values are the closed loop poles of the system.

Eigen Vectors

Any nonzero vector
i
X
such that
i
i
i
X
AX
is said to be eig
en vector associated
with eigenvalue
i
.
Thus let
i

satisfies the equation

Then solution of this equation is called eigen vector of A

associated with eigen
value
i
and is denoted as M
i
.

If the rank of
the matrix [
i

I
-

A] is r, the
n there are (n
-

r) independent
Eigen

vectors. Similarly another impor
tant point is that if the eigen
values of
matrix

A
are all distinct, then the rank r of
matrix

A is (n
-

1) where n is order of the
sys
tem.

9

Mathematically, the
Eigen

vector can be
calculated by

taking cofactors of
matrix

(
i
I

-

A) along any row.

Where

C
ki
is cofactor of
matrix

(
i
I
-

A) of k
th

row.

Key Point: If the
cofactor along a particul
ar row gives

null solution i.e. all elements

of
corresponding eigen vectors are zero then cofactors along any other row must he
obtained.
Otherwise inverse of modal matrix M cannot exist.

Example

1

Obtain the Eigen values, Eigen vectors for the matrix

Solution

Eigen values are roots of

Eigen values are

To find Eigen vector,

Let

10

Where C = cofactor

For
2
=
-
2

For
3
=
-
3

Example 2

For a system with state model matrices

Obtain the system with state model matrices

11

Solution

The T
.F. is given by,

12

Solution of State Equations

Consider the state equation n of linear time invariant system as,

)
(
)
(
)
(
.
t
BU
t
AX
t
X

The matrices A and B are constant matrices. This state equation can be of two types,

1. Homogeneous and

2.

Nonhomogeneous

Homogeneous Equation

If A is a constant matrix and input control fo
rces are zero then the e
quation
takes the
form,

Such an equation is called homogeneous equation. The obvious equation is if input is
zero, In such systems, the driving force is provided by the initial
conditions of the
system to produce the ou
tpu
t. For example, consider a ser
ies RC circuit in which
capacitor is initially charged to V volts. The current is the output. Now there is
no
input control force i.e.

external voltage applied to the system. But the initial voltage on
the capacitor drives the

current through the system and capacitor starts
discharging
through the resistance R. Such a system which works on the initial
conditions without any input applied to it is called homogeneous system
.

Nonhomogeneous Equation

If A is a constant matrix and
m
atrix

U(t) is
non
-
zero

vector i.e. the input
control forces are applied to the system then the equation takes normal form as,

Such an equation is called nonhomogeneous equation. Most of the
practical

systems require inputs to

dive them. Such systems ar
c nonhomogeneous linear
systems.

The
solution

of the state equation is obtained

by considering basic method of
13

finding the solution of homogeneous
equation
.

Controllability

and Observability

More

specially, for system o
f Eq.(1
), there exists a similar transformation that will
diagonalize the system. In other words, There is a transformation matrix

Q such that

0
; ; X(0)=X
X AX Bu y CX Du

(1)

-1
or X=Q
X QX X
 

(2)

y = C
X X Bu X Du

    

(3)

Where

1
2
0 0
0 0
0
n

(4)

Notice

that by doing the diagonalizing transformation, the resulting transfer function between
u(s) and y(s) will not be altered.

Looking at Eq.(3
), if
0
k
b

, then
k
x
(t) is uncontrollable by the input u(t), since,
k
x
(t) is
characterized by the mode
k
t
e

by the equation:

( ) (0 )
k
t
k k
x t e x

The lake of controllability of the state
k
x
(t) is reflect by a zero k
th

row
of
B

, i.e.
k
b

.
Which

would cause a complete zero
rows

in the following matrix (known as the

controllability
matrix), i.e.:

C(A,b
)

2 1
1 1 1 1 1 1 1
2 1
2 2 2 2 2 2 2
2 3 n-1
2 1
k k k

A A A A

n
n
n
k k k k
b b b b
b b b b
B B B B B
b b b b
   

   

   
    
    
 
   

   
2 1
n n n

n
n n n n
b b b b

   

(5)

A C(A,b) matrix with all non
-
zero row has a rank of N
.

In fact

,
1
or
B Q B B QB
 
. Thus, a non
-
singular C(A,b) matrix implies a non
-
singular
matrix of C(A,b)of the following:

14

C(A,b)
2 -1

n
B AB A B A B

(6)

It is important to note that this result holds in t
he case of non
-
distinct eigenvalues as well.

Remark 1]

If matrix A has distinct eigenvalues and is

represented as a controller
canonical form, it
is easy to show the following identity holds, i.e.:

2 1 2 1
1 1 1 1 1 1 1
1 1
n n
A
 

f

or each i.

There
fore a transpose of so
-
called Vand
ermonde matrix V of n column
eigenvectors of A will
diagonalize A, i.e.,

2
1
1 1
1
1 2
2
1
2 2 2
2 2
2
1 2
2
n-1
n
1 1 n-1
1 2 n
1 1 1
1
1
1

T
n
n
n
T
n
n n
n n
W

   

   

(6)

and

1 1
T T
or, A= A
T T T T
W A W W W W W A

[Remark 2]

There is an alternative way to explain why C(A,b)

should have rank n for state controllable,
let us start from the solution of the state space system:

0
( )
( ) (0 ) ( )
f
t
At A t
t
X t e X e Bu d

(
7
)

The state controllability requires that for each X(t
f
)
nearby

X(t
0
), there is a finite sequence of
u(t; t
[to,t
f
]).

15

0
0
0
0
0
0
0
0
( 1)
0
0
( 1)
1
0
0
( ) ( )
( ) ( )
( )
= ( ) ( )
f
f
f
f
t
At
A
f
t
t
At
A
f
t
t k
n
A
k
t k
t k
i n
i
i
k i
t k
X t e X e Bu d
or
e Bu d e X t X
e Bu t k d
A B u t k
0
0
0
( 1)
i=n-1
0
i=0 0
1
i=n-1
2
2 n-1
i=0
= ( ) ( )
= AB A B A B
n
t k
k
i
i
k
t k
i
i
n
d
A B u t k d
w
w
A BW B
w

Thus, in order W has non
-
trival solution, we need that C(A,b)

matrix has exact rank n

There are several alternative ways of establishing the state space controllability:

The
(n) rows of
At
e B

are linearly independent over the real field for all t.

The controllability grammian

0
(,)
f
T
t
At T A
ram o f
t
G t t e BB e d

is non
-
singular for all
0
f
t t
.

[Theorem 1]

Replace B with b, (i.e. Dim{B}=n
1), a

pair [A,b] is

non
-
controllable if
and only if there exists a row vector
0
q

such that

, 0
qA q qb

(8
)

T
o prove the “if” part:

If there is such row vector, we have:

16

2 2 2 1
-1 1
0
0
0
0,,,,0
and 0

0
n
n n
qA q and qb
qAb qb
q I A
qA b qAb qb q b Ab A b A b
qb
qA b qb

Since
0
q
, we
conclude that:
2 1
,,,,
n
b Ab A b A b

is singular, and thus the

system is not controllable.

To prove the “only if’ part
:

If the pair is noncontrollable, then matrix A can be transformed into

non
-
controllable form like:

,
0

0
C
C
CC
C
A A
r
b
A b
A
n r

(9
)

Where
,
r rank
C(A,b) (Notice that Eq.(33) is
a
well
-
known

theorem in linear

system.)

Thus, one can find a row vector has the
form
[0 ]
q z

, where z can be selected

as
the eigenvector
of
C
A
, (i.e.:
C
zA z
), for then:

0 z 0
qA A z q

(10
)

Therefore, we have shown that only if [A, b] is non
-
controllable, there is a

non
-
zero row vector satisfying Eq.
(8).

In fact, according to Eq.(27),

1
1
i
k
t
At t t T T
i i
i
e Ve V Ve W v w e

and,
0
( )
At
X t e X
, we have:

( )
( )
0 0
1 1
0 0
( ) ( ) ( )
i ii
t t
k n
t t
At A t T T
i i i i
i i
X t e X e bu d v w e X v w b e u d

17

Thus, if b is orthogonal to
i
w
, then the state associated with
i

will not be controllable,
and
hence, the system
is not completely
controllable.

The

another form to test for the
control
lability of the [A,b] pair is known as the Popov
-
Belevitch
-
Hautus (abbrv. PBH) test is
to check if
rank sI A b n

for all s (not only at eigenvalues of A). This test is based on
the fact that if
sI A b

has rank n, there ca
nnot be a nonzero row vector q satisfying
Eq.(32). Thus by
Theorem

1, pair [A, b] must be controllable.

Referring to the systems described by Eqs.(26) and (27), the state
( )
i
x t

corresponding to
the mode
i
t
e

is unobservable at the output
1
y
, if
1
0
i
C

for any i=1,2,…,n. The lack of
observability of the state
( )
i
x t

is reflected by a complete zero (ith) column of so called
observability mat
rix of the system O
(,)
A C
 
, i.e.:

O
1
(,)
A C
 
11 12 1
1
1 11 2 12 1
1
1 1 1
1
1 2 12 1
1
n
n n
n n n
n
n n
C C C
C
C C C
C A
C C C
C A
  

  
 

   

  
 

(11
)

An observable state
( )
i
x t

corresponds to a nonzero column of O
(,)
A C
 
. In the case

of distinct
eigenvalues, each nonzero column
increases

the rank by one. Therefore, the rank of
O
(,)
A C
 
corresponding to the total number of modes that are observable at the output y(t) is
termed the observability rank of the system. As

in the case of controllability, it is not
necessary to transform a given state
-
space system to modal canonical form in order to
determine its rank. In general, the observability matrix of the system is defined as:

O
(,)
A C
=
1
n
C
CA
CA

= O
1
(,) (,)
A C Q A C V
   

With Q=V
-
1

nonsingular. There, the rank of O
(,)
A C

equals the rank of O
(,)
A C
 
. It is
important to note that this result holds in the case of non
-
distinct eigenvalue
s. Thus, a state
-
space system is said to be completely (state) observable if its observability matrix has a full
rank n. Otherwise the system is said to be unobservable

18

In particular, it is well known that a state
-
space system is observable if and only if

the
following conditions are satisfied:

The (n) column of
At
Ce
are linearly independent over R for all t.

The observability grammian of the following is nonsingular for
all
0
f
t t
:

0
,
T
t
A T A
ranm o
t
G e C Ce d

The (n+p)
n matrix
I A
C
b

has rank n at all eigenvalues
i
of A.

Pole Placement Design

The conventi
onal method of design of single input single output control system
consists of design of a suitable controller or compensator in such a way that the dominant
closed loop poles will have a desired damping ratio % and undamped natural frequency con.
The orde
r of the system in this case is increased by 1 or 2 if there are no pole zero
cancellation taking place. It is assumed in this method that the effects on the responses of non
-
dominant closed loop poles lo be negligible. Instead of specifying only the domin
ant closed
loop poles in the conventional method of design, the pole placement technique describes all the
closed loop poles which require
measurements of all state variables or inclusion of a state
observer in the system. The
system closed loop poles can
be placed at arbitrarily chosen
locations with the condition
that the system is completely stale controllable. This condition
can be proved and the
proof is given below.

Consider a control system described by following
slate equation

Here x is a state vector, u is a control signal which is scalar, A is n x n state matrix. B is n x 1
constant matrix.

Fig open loop control system

19

The system defined by above
equation represents open loop
system. The
state x is not fed back
to the control signal u.

Let us select the control signal to

be u =
-

Kx state. This indicates that
th
e control signal is obtained from
instantaneous state.
This is called state feedback. The k is a matrix of order l x n called
state feedback gain matrix.
Let us consider the control signal to be unconstrained. Substituting value of u in equation 1

The system defined by above
equation is shown in the Fig. 5.2. It
is a closed loop
control system as the
system state x is fed back to the
control system as the system stale x
is fed back to control signal u. Thus this

a system with

state feedback

The solution of equation 2 is say

x(t)
= e,
x(0) is the initial slate

…(3)

The stability and the transient response characteristics are determined by the eigen
values of matrix A
-

BK.
Depending on the selection of state feedback gain matrix K, the
matrix A
-

BK can be
made asymptotically stable and it is possible to make x(t) approaching
to zero as time t approaches to infinity provided x(0) * 0. The eigen value
s of matrix A
-

BK
arc called regulator poles. These regulator poles
when placed in left half of s plane then x(t)
approaches zero as time t approaches infinity.
The problem of placing the closed loop poles
at the desired location is called a pole
placemen
t problem.

Design
of State Observer

In case of state observer, the state variables are estimated based on the
measurements of the output and control variables. The concept of observability
plays important part here
in case of state observer.

Consider a s
ystem defined by following state equations

20

Let us consider x as the observed state vectors. The observer is basically a
subsystem
which reconstructs the state vector of the system. The mathematical
model of the ob
server
is same as that of the plant except the inclusion of additional
term consisting of estimation
error to compensate for inaccuracies in matrices A and
B and the lack of the initial error.

The estimation error or the observation error is the differenc
e between the
measured
output and the estimated output. The initial error is the difference
between the initial state
and the initial estimated state. Thus the mathematical
model of the observer can be
defined as,

Here x is the estimated state and C x is the estimated output. The observer has
inputs of output y and control input u. Matrix K^ is called the observer gain
matrix. It is nothing
but weighing matrix for the correction term which contains
the difference

between the
measured output y and the estimated output
cx

This additional term continuously corrects the model output and the performance
of
the observer is improved.

Full order state observer

The system equations arc already defined as

The mathematical model of the state observer is taken as.

To determine the observer error equation, subtracting equation of x from x wc get

21

The block diagram of the system and full ord
er state observer is shown in the Fig.

The dynamic behavior of the error vector is obtained from the Eigen values of matrix
A
-
K^C If matrix A
-
K^C is a stable matrix then the error vector will converge to
zero for
any initial error vector e(0)
. Hence x(t) will converge to x(t) irrespective of
values of x(0)
and x(0).

If the Eigen values of matrix A
-
KeC are selected in such a manner that the dynamic
behavior of the error vector is asymptotically stable and is sufficiently fast then
any of the e
rror vector will tend to zero with sufficient speed.

1/ the system is completely observable then it can be shown that it is possible to select
matrix K,. such that A
-
K^C has arbitrarily desired Eigen values, i.c. observer gain
matrix
Ke can be obtained to

get the desired matrix A
-
KCC.

22

UNIT I

STATE SPACE ANALYSIS OF CONTINUOUS TIME SYSTEMS

PAR
T A

1.

What are the advantages of state space analysis?

2.

What are the drawbacks in transfer function model analysis?

3.

What is state and state variable?

4.

What is a state vector?

5.

Write the state model of n
th
order system?

6.

What is state space

7.

What are phase variable
s?

8.

Write the solution of homogeneous state equation?

9.

Write the solution of nonhomogeneous state equation?

10. What

is resolvant matrix?

PART B

1.

Explain Kaman’s test for determining state controllab
ility?

2.

Explain Gilbert’s test for determining state controllability?

3.

Find the output of the system having state model,

and

The input U(t) is unit step and X(0)
0
10

4.

Show the following system is completely state

controllable and observable.

And

5.

Obtain the homogenous solution of the equation X(t) =A X(t)

6.

Derive the transfer function of observer based controller?

23

UNIT II

Z
-
TRANSFORM AND SAMPLED DATA SYSTEMS

Sampled Data System

When the signal or information at any or some points in a system is in the form of
discrete pulses.
Then the

system is called discrete data system. In control engine
ering the
discrete data system is popularly known as sampled data system
s.

Sampling process

Sampling is the conversion of a continuous time signal into a
discrete time signal
obtained by taking sample of the continuous time
signal at discrete time instants
.

Thus if f (t) is the input to the sampler

The output is f(kT)

Where T i
s called the sampling interval

The recipro
cal of T

Sampled data theory

=
pa浰m楮i=灲潣e獳s

=
pam灬p湧=瑨敯te洠

=
pig湡氠rec潮獴牵r瑩潮o

=
pa浰me= a湤n 桯汤h c楲i畩瑳t

=
z
J

=

J
=

=
f湶e牳r= z
J

=

=
䑩ac牥瑥t sys瑥浳t a湤n獯汵s楯渠潦=摩晦e牥nce=e煵q瑩潮o畳楮g=z=瑲t湳景牭r

=
m畬獥u 瑲t湳晥爠晵湣瑩潮o

=
oe獰潮獥s 潦=獡浰me搠摡瑡t sy獴e洠瑯t 獴数s a湤=牡浰mf湰畴n=

=
p瑡扩汩ty⁳瑵摩e猠

=
Jury’s test and bilinear transformation
=

24

Let
1/T =F
s

is called the sampling rate. This type of sampling is called periodic
Sampling, since samples are obtained uniformly at intervals of T sec
onds
.

Multiple order sampling

A particular sampling pattern is repeated periodically

Multiple rate sampling
-

In this method two simultaneous sampling operation
s with
different time periods are carried out on the signal to produce the sampled output.

R
andom sampling

In this case the sampling instants are random

Sampling Theorem

A band limited continuous time signal with highest frequency f
m

hertz can be uniquely
recovered from its samples provided that the sampling rate F
s

is greater t
han or equal to 2fm
samples

per seconds

Signal Reconstruction

The signal given to the digital controller is a sampled data signal and in turn the
controller gives the controller output in digital f
orm.

But the system to be controlled needs an
25

analog control signal as input. Therefore the digital output of controllers must be converters
into analog form

This can be achieved by means of various types of hold circuits. The simplest hold

circuits are

the zero order hold (ZOH)
.

In ZOH, the reconstructed analog signal acquires the
same values as the last received sample for the entire sampling period

The high frequency
noises present in the reco
nstructed signal are

automatically
filtered out by the control system
component which behaves

like low pass filters
. In a first
order hold the last two signals for the current sampling period. Similarly higher order hold
circuit can be devised. First or hi
gher order hold circuits offer no particular advantage over
the zero order hold

Z
-

Transform

Definition of Z Transform

Let f

(
k
)

=

Discrete

time signal

F (
z
) =

Z {f (
k)} =z transform of f

(k)

The
z transforms

of a discrete time signal or sequence is define
d as the power series

k
k
z
k
f
Z
F
)
(
)
(

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

1

Where
z is a complex variable

Equation (1) is considered to be two sided and the transform is called two sided z transform.
Since the time index k is defined for both positive and negative va
lues.

The one sided z transform
of f
(k) is defined as

k
k
z
k
f
Z
F
0
)
(
)
(

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

2

26

Problem

1. Determine the z transform and their ROC of the discrete sequences f(k) ={3,2,5,7}

Given f (k) = {3,

2,

5,

7}

Where

f (
0)
=3

f (1) =2

f (2)

= 5

f (3) =

7

f (k) = 0 for k < 0 and k >3

By definition

k
k
z
k
f
z
F
k
f
Z
)
(
)
(
)}
(
{

The
given sequence is a finite duration se
quence. Hence the limits of summation can be
changed as k = 0 to k = 3

k
k
z
k
f
z
F
3
0
)
(
)
(

0
)
0
(
)
(
z
f
z
F
+
1
)
1
(
z
f
+
2
)
2
(
z
f
+
3
)
3
(
z
f

=
3
2
1
7
5
2
3
z
z
z

Here
)
(
z
F

is bounded
, expect when z =0

The

ROC is entire z
-
plane expect z = 0

2.

Determine the z transform of discrete sequences f (k) =u (k)

Given f (k) =u (k)

u (k) is a discrete unit step sequence

u (k) = 1 for k

0

=
0

for
k < 0

By definition

k
k
z
k
f
z
F
k
f
Z
)
(
)
(
)}
(
{
k
k
z
k
u
0
)
(

k
k
z
0

k
k
z
)
(
1
0

27

F (z) is an

infinite geometric series and it converges if
1
z

F (z)

1
1
1
z

z
1
1
1

1
z
z

3.

Find the one sided z transform of the discrete sequences generated by mathematically
sampling the continuous time function f (t)

kT
e
at
cos

Given

f (t)

kT
e
at
cos

B
y definition

F(z)

=

k
k
akT
z
kT
e
k
f
Z
0
cos
)}
(
{

k
k
T
k
j
T
k
j
T
k
a
z
e
e
e
0
2

0
0
1
1
2
1
2
1
k
k
k
T
j
T
a
k
T
j
T
a
z
e
e
z
e
e

WKT
c
c
k
k
1
1
0

1
1
1
1
2
1
1
1
2
1
)
(
z
e
e
z
e
e
z
F
T
w
j
T
a
T
w
j
aT

aT
T
j
aT
T
j
e
z
e
e
z
e
1
1
2
1
1
1
2
1

T
j
aT
aT
T
j
aT
aT
e
e
z
e
z
e
e
z
e
z
2
1

T
j
aT
T
j
aT
T
j
aT
aT
T
j
aT
aT
e
ze
e
ze
e
ze
ze
e
ze
ze
2
1

T
j
T
j
T
j
aT
T
j
aT
aT
T
j
aT
T
j
aT
aT
e
e
e
ze
e
ze
ze
e
ze
e
ze
ze
2
)
(
2

28

1
)
(
)
(
2
2
2
2
T
j
T
j
aT
aT
T
j
T
j
aT
aT
e
e
ze
e
z
e
e
ze
ze

1
cos
2
)
cos
(
2
2
2
T
ze
e
z
T
ze
ze
ze
aT
aT
aT
aT
aT

2
cos
j
j
e
e

Inverse z transform

Partial fraction expansion

(PFE)

Power series expansion

Partial fraction expansion

Let

f (k) =discrete sequence

F (z) =Z {f (
k)} = z transform of f (k)

F (z) =
n
n
n
n
m
m
m
m
a
z
a
z
a
z
b
z
b
z
b
z
b
..........
..........
2
2
1
1
2
0
1
0
0

where
n
m

The function
F (z) can be expressed as a series of sum terms by PFE

n
i
i
i
p
z
A
A
z
F
1
0
)
(

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

3

Where
0
A

is a constant

n
A
A
A
,........
,
2
1

are residues

n
p
p
p
,........
,
2
1

are poles

Power series expansion

Let f (k) =discrete sequences

F (z) = Z {f (k)} = z transform of f (k)

By definition

k
k
z
k
f
Z
F
)
(
)
(

On
E
xpanding

...)
..........
)
2
(
)
1
(
)
0
(
)
1
(
)
2
(
)
3
(
(.......
)
(
2
1
0
1
2
3
z
f
z
f
z
f
z
f
z
f
z
f
Z
F
-------
4

Problem

1.

Determine the inverse z transform of the following function

(i)

F (z) =
2
1
5
.
0
5
.
1
1
1
z
z

29

Given

F (z)

=

2
1
5
.
0
5
.
1
1
1
z
z

2
5
.
0
5
.
1
1
1
z
z

5
.
0
5
.
1
2
2
z
z
z

)
5
.
0
(
)
1
(
2
z
z
z

)
5
.
0
(
)
1
(
)
(
z
z
z
z
z
F

By partial fraction expansion

)
5
.
0
(
)
1
(
)
(
2
1
z
A
z
A
z
z
F

1
A
)
1
(
)
(
z
z
z
F

Put z =1

1
A

)
1
(
)
5
.
0
(
)
1
(
z
z
z
z

)
5
.
0
(
z
z

)
5
.
0
1
(
1

2

Put z =0.5

2
A
)
5
.
0
(
)
(
z
z
z
F

2
A
)
5
.
0
(
)
5
.
0
(
)
1
(
z
z
z
z

)
1
(
z
z

30

)
1
5
.
0
(
5
.
0

-
1

5
.
0
1
1
2
)
(
z
z
z
z
F

5
.
0
1
2
)
(
z
z
z
z
z
F

WKT

a
z
z
a
Z
k
}
{

and

1
)}
(
{
z
z
k
u
Z

On
taking inverse z transform

0
,
)
5
.
0
(
)
(
2
)
(
k
k
u
k
f
k

(i
i
)

F (z) =
5
.
0
2
2
z
z
z

Given

F (z) =
5
.
0
2
2
z
z
z

)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
2
j
z
j
z
z

)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
(
j
z
j
z
z
z
z
F

By partial fraction expansion

)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
(
*
j
z
A
j
z
A
z
z
F

A
)
5
.
0
5
.
0
(
)
(
j
z
z
z
F

Put z = 0.5+j0.5

A
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
j
z
j
z
j
z
z

)
5
.
0
5
.
0
(
j
z
z

31

)
5
.
0
5
.
0
5
.
0
5
.
0
(
5
.
0
5
.
0
j
j
j

5
.
0
5
.
0
j

*
A
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
j
z
j
z
j
z
z

Put z =0.5
-
j0.5

*
A
)
5
.
0
5
.
0
(
j
z
z

)
5
.
0
5
.
0
5
.
0
5
.
0
(
5
.
0
5
.
0
j
j
j

5
.
0
5
.
0
j

)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
(
j
z
j
j
z
j
z
z
F

)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
j
z
z
j
j
z
z
j

WKT

a
z
z
a
Z
k
}
{

On
taking inverse z transform

k
k
j
j
j
j
k
f
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
)(
5
.
0
5
.
0
(
)
(

k
k
j
j
j
j
j
j
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(

k
k
j
j
j
j
j
j
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
)(
5
.
0
5
.
0
(

1
1
)
5
.
0
5
.
0
(
)
5
.
0
5
.
0
(
k
k
j
j
j
j

2.

Determine

the inverse z transform
of z

domain function

F (z) =
2
3
1
2
3
2
2
z
z
z
z

Given

F (z) =
2
3
1
2
3
2
2
z
z
z
z

3

2
3
2
z
z

1
2
3
2
z
z

6
9
3
2
z
z

5
11
z

32

F (z)

=

2
3
5
11
3
2
z
z
z

)
2
(
)
1
(
5
11
3
z
z
z

By PFE

F (z) =
)
2
(
)
1
(
3
2
1
z
A
z
A

1
A

)
1
(
)
2
(
)
1
(
5
11
z
z
z
z

)
2
(
5
11
z
z
6
2
1
5
11

2
A
)
2
(
)
2
(
)
1
(
5
11
z
z
z
z

)
1
(
5
11
z
z

17
1
2
5
)
2
(
11

)
2
(
17
)
1
(
6
3
)
(
z
z
z
F

2
1
17
)
1
(
1
6
3
z
z
z
z
z
z

2
17
)
1
(
6
3
1
1
z
z
z
z
z
z

On taking inverse z transform

0
;
)
1
(
2
17
)
1
(
6
)
(
3
)
(
)
1
(
k
for
k
u
k
u
k
k
f
k

2.

Determine the inverse z transform of the
following

2
1
2
1
2
3
1
1
)
(
z
z
z
F

Where (
i) ROC
0
.
1
z

(ii) ROC
5
.
0
z

Given

2
1
2
1
2
3
1
1
)
(
z
z
z
F

(i)

0
.
1
z

33

..........
8
15
4
7
2
3
1
3
2
1
z
z
z

2
1
2
1
2
3
1
z
z

1

2
1
2
1
2
3
1
z
z

3
2
1
2
1
4
3
4
9
2
3
2
1
2
3
z
z
z
z
z

4
2
2
2
2
8
7
4
3
4
7
4
3
4
7
z
z
z
z
z

4
3
8
7
8
15
z
z

)
(
z
F
..........
8
15
4
7
2
3
1
3
2
1
z
z
z

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

-
(i)

k
k
z
k
f
Z
F
)
(
)
(

For a causal signal

k
k
z
k
f
z
F
0
)
(
)
(

.......
..........
)
2
(
)
1
(
)
0
(
)
(
3
2
1
z
f
z
f
z
f
z
F

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

(ii)

Comparing equation (i) &(ii)

1
)
0
(
f
,
2
3
)
1
(
f
,
4
7
)
2
(
f
,
8
15
)
3
(
f

0
......}
..........
,
8
15
,
4
7
,
2
3
,
1
{
)
(
k
for
k
f

34

(i)

5
.
0
z

..........
30
11
6
2
5
4
3
2
z
z
z
z

1
2
3
2
1
1
2
z
z

1

2
2
3
1
z
z

3
2
2
6
9
3
2
3
z
z
z
z
z

4
3
2
2
14
21
7
3
6
7
z
z
z
z
z

5
4
3
4
3
30
45
15
14
15
z
z
z
z
z

)
(
z
F
..........
30
11
6
2
5
4
3
2
z
z
z
z

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

(i)

k
k
z
k
f
Z
F
)
(
)
(

For an anti
-
causal signal

1
0
)
(
)
(
z
k
f
Z
F
k

)
0
(
)
1
(
)
2
(
)
3
(
)
4
(
)
5
(
..
..........
)
(
1
2
3
4
5
f
z
f
z
f
z
f
z
f
z
f
z
F

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

(ii)

Comparing the equation i & ii

30
)
5
(
f
,
14
)
4
(
f
,
6
)
3
(
f
,
2
)
2
(
f
,
0
)
1
(
f
,
0
)
0
(
f

}
0
,
0
,
2
,
6
,
14
,
30
..
{.........
)
(
k
f

Difference equation

Discrete time systems are described by difference equation of the form

35

If the system is causal, a li
near difference equation provides an explicit relationship between
the input and output. This can be seen by rewriting.

Thus the nth value of the output can be computed from the nth input value and the N and M
past va
lues of the output and input, respectively.

Role of z transform in linear difference equations

Equation (1) gives us the form of the linear difference equation that describes the
system. Taking z transform on either side and assuming zero initial condit
ions, we have

Where H(z) is a z transform of unit sample response h(n).

Stability analysis

Jury’s stability test

B
ilinear transformation

Jury’s stability test

Jury’s stability test

is used to determine whether the roots of the characteristic
polynomial lie within a unit circle or

not. It consists of two parts.
One simple test for
necessary condition for stability and
another test for sufficient condition for stability.

Let us consider a general characteristic polynomial F (z)

0
,
......
..........
)
(
0
1
1
1
n
n
n
n
n
a
where
a
z
a
z
a
z
a
z
F

Necessary condition for stability

0
)
1
(
)
1
(
;
0
)
1
(
F
F
n

If

this necessary condition is not met, then the system is unstab
le. We need not check the
sufficient condition.

36

Sufficient condition for stability

2
0
2
0
1
0
0
........
..........
r
r
c
c
b
b
a
a
n
n
n

If the characteristic polynomial satisfies (n
-
1) conditions, then the system is stable

Jury’s test

Bilinear transform
ation

The bilinear transformation maps the interior of unit circle in the z plane into the left half of
the r
-
plane.

1
1
z
z
r

Or
r
r
z
1
1

Fig.
Mapping of unit circle in z
-
plan
e into left half of r
-
plane

Consider the characteristic equation

)
....(
..........
0
;
..
..........
0
2
2
1
1
i
a
a
a
z
a
z
a
z
a
n
z
n
n
n
n
n
n

Sub
r
r
z
1
1

in Equation (i)

37

)
..(
..........
0
)
1
1
(
..
..........
)
1
1
(
)
1
1
(
)
1
1
(
0
2
2
1
1
ii
a
r
r
a
r
r
a
r
r
a
r
r
a
n
n
n
n
n
n

Equation (ii) can be simplified

0
..
..........
0
1
2
2
1
1
b
r
b
r
b
r
b
r
b
n
n
n
n
n
n

Problem

1.

Check

for stability of the samp
led data
control system

s
represented by
characteristic equation.

0
2
2
5
)
(
2
z
z
i

Given

0
2
2
5
)
(
2
z
z
z
F

5
2
2
5
2
)
1
(
2
)
1
(
5
)
1
(
2
2
5
)
(
2
2
0
1
2
2
F
z
z
a
z
a
z
a
z
F

9
)
2
2
5
(
1
2
)
1
(
2
)
1
(
5
)
1
(
)
1
(
)
1
(
2
2
F
n

Here n=2

Since
0
)
1
(
)
1
(
;
0
)
1
(
F
F
n
, the necessary
condition for stability is

s
atisfied
.

Check for sufficient condition

It consisting of (2n
-
3) rows

n=2 (2n
-
3) =

(2
*2
-
3
)

= 1

So, it consists of only one row

Row z
0
z
1
z
2

1 a
0

a
1

a
2

5
,
2
,
2
2
1
0
a
a
a

1
0
a
a

The necessary condition to be satisfied

38

The necessary & sufficient conditions for stability are satisfied. Hence the system is stable

(ii)

0
05
.
0
25
.
0
2
.
0
)
(
2
3
z
z
z
z
F

)
(
z
F

0
1
2
2
3
3
a
z
a
z
a
z
a

0
05
.
0
25
.
0
2
.
0
2
3
z
z
z

Method 1

Check for necessary condition

0
05
.
0
25
.
0
2
.
0
)
(
2
3
z
z
z
z
F

6
.
0
05
.
0
)
1
(
25
.
0
)
1
(
2
.
0
1
)
1
(
2
3
F

9
.
0
]
05
.
0
)
1
(
25
.
0
)
1
(
2
.
0
)
1
(
[
)
1
(
)
1
(
)
1
(
2
3
3
F
n

Here n=3

Since
)
1
(
F
>0 &
)
1
(
)
1
(
F
n
>0

The necessary condition for stability is s
atisfied.

Check for sufficient condition

It consisting of (2n
-
3) row

n =3, (2n
-
3) = (2*6
-
3)

=3

So, the table consists of three rows

Row z
0
z
1
z
2
z
3

1 a
0

a
1

a
2
a
3

2

a
3

a
2

a
1

a
0

3 b
0

b
1

b
2

1
2
.
0
25
.
0
05
.
0
3
2
1
0
a
a
a
a

39

24
.
0
)
25
.
0
(
*
)
2
.
0
(
*
05
.
0
2
.
0
1
25
.
0
05
.
0
1875
.
0
)
2
.
0
*
25
.
0
(
05
.
0
25
.
0
1
2
.
0
05
.
0
9975
.
0
1
05
.
0
05
.
0
1
1
05
.
0
2
3
1
0
3
1
3
2
0
1
2
0
3
3
0
0
a
a
a
a
b
a
a
a
a
b
a
a
a
a
b

Row

z
0

z
1

z
2

z
3

1

0.05

-
0.25

-
0.2

1

2

1

-
0.2

-
0.25

1

3
-
0.9975

0.1875

0.24

The necessary condition to be satisfied

25
.
0
9975
.
0
,
1
05
.
0
,
2
0
3
0
b
b
a
a

The necessary and sufficient conditions for stability are satisfied. Hence the system is stable.

Method 2

0
05
.
0
25
.
0
2
.
0
)
(
2
3
z
z
z
z
F

Put
r
r
z
1
1

0
05
.
0
)
1
1
(
25
.
0
)
1
1
(
2
.
0
)
1
1
(
)
(
2
3
r
r
r
r
r
r
r
F

On multiplying throughout by
3
)
1
(
r
we get

40

0
6
.
0
9
.
2
6
.
3
9
.
0
0
)
2
.
0
1
.
0
4
.
0
3
.
0
(
)
8
.
0
8
.
2
2
.
3
2
.
1
(
0
)
2
.
0
1
.
0
3
.
0
2
.
0
1
.
0
3
.
0
(
)
8
.
0
2
2
.
1
8
.
0
2
2
.
1
(
0
)
2
.
0
1
.
0
3
.
0
)(
1
(
)
8
.
0
2
2
.
1
)(
1
(
0
)
1
.
0
05
.
0
05
.
0
25
.
0
25
.
0
)(
1
(
)
2
.
0
2
.
0
2
1
)(
1
(
0
)
2
1
)(
1
(
05
.
0
)
1
)(
1
(
25
.
0
)
1
)(
1
(
2
.
0
)
2
1
)(
1
(
0
)
1
(
05
.
0
)
1
)(
1
(
25
.
0
)
1
(
)
1
(
2
.
0
)
1
(
2
3
2
3
2
3
2
3
2
2
3
2
2
2
2
2
2
2
2
2
2
2
3
2
2
3
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r

The coefficient of the new characteristic equation is positive. Hence the necessary condition
for stability is satisfied.

The suf
ficient condition for stability can be determined by constructing routh array as

1
4
.........
6
.
0
:
3
.........
75
.
2
:
2
.........
6
.
0
6
.
3
:
1
.........
9
.
2
9
.
0
:
0
1
2
3
column
row
r
row
r
row
r
row
r

75
.
2
6
.
3
)
6
.
0
*
9
.
0
(
)
9
.
2
*
6
.
3
(
1
r

6
.
0
75
.
2
)
6
.
3
*
0
(
)
6
.
0
*
75
.
2
(
0
r

There is no sign change in the elements of first column of routh array. Hen
ce the sufficient
condition for stability is satisfied.

The necessary condition and sufficient condition for stability are satisfied. Hence the system
is stable.

Pulse transfer function

It is the ratio of s transform of discrete output signal of the system

to the z
-
transform of
discrete

input signal to the system. That is

)
(
)
(
)
(
z
R
z
C
z
H

(i)

Proof

Consider the z
-
transform of the convolution sum

k
k
m
z
m
r
m
k
h
k
C
Z
0
0
)
(
)
(
)]
(
[

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

(ii)

On interchanging the order of summation, we get

41

k
k
m
z
m
k
h
m
r
z
C
0
0
)
(
.
)
(
)
(

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

(iii)

Let
m
k
l

Then
l
k
when
m
l
&
0

0
l
when

m
l
m
k
m
m
z
l
h
z
m
r
z
C
)
(
.
)
(
)
(
0

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

(iv)

l
m
k
m
m
z
l
h
z
m
r
z
C
)
(
.
)
(
)
(
0

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

(v)

)
(
).
(
)
(
z
H
z
R
z
C

The pulse transfer function

)
(
)
(
)
(
z
R
z
C
z
H

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

(vi)

The
block diagram
for pulse

transfer function

UNIT II

Z
-
TRANSFORM

AND SAMPLED DATA SYSTEMS

PART A

1. What is sampled data control system?

2. Explain the terms sampling and sampler.

3. What is meant by quantization?

4. State (shanon’s) sampling theorem

5. What is zero order
hold?

6. What is region of conv
ergence?

7. Define Z
-
transform of unit step signal?

8. Write any two properties of discrete convolution.

9. What is pulse transfer function?

10. What are the methods available for the stability analysis of sampled data

control

systems?

11. What is bili
near transformation?

42

PART B

1. (i)solve the following difference equation

2 y(k)

2 y(k
-
1) + y (k
-
2) = r(k)

y (k) = 0 for k<0 and

r(k) = {1; k= 0,1,2

{0;k<0 (8)

(ii)check if all the roots of the following characteristics equation lie within the

circle.

Z4

1.368Z3+0.4Z2+0.08Z+0.002=0 (8)

2. (i)Explain the concept of sampling process. (6)

(ii)Draw the frequency response of Zero
-
order Hold (4)

(iii)Explain any two theorems on Z
-
transform (6)

3
. The

block diagram of a
sampled
data

system is shown to Fig.
(a) Obtain discrete
-
time state
model for the system. (b) Obtain the equation for
inter s
ample

response of the system.

4. The
block diagram oils sampled
-
dat
a system is shown in Fig.

(a) Obtain discrete
-
time state model for the system

(b) Find the r
esponse of the system for a unit

step input.

(c) What is the effect on system response
(i)
when

T =0.5 sec (ii) T=1
.5 sec

43

UNIT III

STATE SPACE ANALYSIS OF DISCRETE TIME SYSTEMS

State variables

Concepts of

State and
State Variables

State

The state of a dynamic
system is the smallest set of variables (called state variables)
such
that the knowledge of these variables at
t=t
0
,

together with the know
ledge of the inputs
for
0
t
t
,

completely determine the behaviour of the system for any time

0
t
t
.

The concept of state is not limited to physical systems. It is applicable to biological systems.
economic systems, social systems, and others.

State variables

The state variables of a dynamic system are the smallest set of variables t
hat
determine the
state of the dynamic system. i.e. the state variables are the minimal set of variables
such that the
knowledge of these variables at any initial time
t = to,
together with the knowledge
of the
inputs for
0
t
t

is suffici
ent to completely determine the behaviour of the system for any