Feedback Control Systems (FCS) Dr. Imtiaz Hussain email: [email protected]@faculty.muet.edu.pk URL :

Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: [email protected]

URL :http://imtiazhussainkalwar.weebly.com/

Lecture-34-35Modern Control Theory

Introduction• The transition from simple approximate models, which are

easy to work with, to more realistic models produces two effects.

– First, a large number of variables must be included in the models.

– Second, a more realistic model is more likely to contain nonlinearities and time-varying parameters.

– Previously ignored aspects of the system, such as interactions with feedback through the environment, are more likely to be included.

Introduction• Most classical control techniques were developed for linear constant

coefficient systems with one input and one output(perhaps a few inputs and outputs).

• The language of classical techniques is the Laplace or Z-transform and transfer functions.

• When nonlinearities and time variations are present, the very basis for these classical techniques is removed.

• Some successful techniques such as phase-plane methods, describing functions, and other methods, have been developed to alleviate this shortcoming.

Introduction• The state variable approach of modern control theory provides a

uniform and powerful methods of representing systems of arbitrary order, linear or nonlinear, with time-varying or constant coefficients.

• It provides an ideal formulation for computer implementation and is responsible for much of the progress in optimization theory.

• The advantages of using matrices when dealing with simultaneous equations of various kinds have long been appreciated in applied mathematics.

• The field of linear algebra also contributes heavily to modern control theory.

Introduction• Conventional control theory is based on the input–output

relationship, or transfer function approach.

• Modern control theory is based on the description of system equations in terms of n first-order differential equations, which may be combined into a first-order vector-matrix differential equation.

• The use of vector-matrix notation greatly simplifies the a mathematical representation of systems of equations.

• The increase in the number of state variables, the number of inputs, or the number of outputs does not increase the complexity of the equations.

6

State Space Representation

• State of a system: We define the state of a system at time t0 as the amount of information that must be provided at time t0, which, together with the input signal u(t) for t t0, uniquely determine the output of the system for all t t0.

• This representation transforms an nth order difference equation into a set of n 1st order difference equations.

• State Space representation is not unique.

• Provides complete information about all the internal signals of a system.

7

State Space Representation

• Suitable for both linear and non-linear systems.

• Software/hardware implementation is easy.

• A time domain approach.

• Suitable for systems with non-zero initial conditions.

• Transformation From Time domain to Frequency domain and Vice Versa is possible.

8

Definitions

• State Variable: The state variables of a dynamic system are the smallest set of variables that determine the state of the dynamic system.

• State Vector: If n variables are needed to completely describe the behaviour of the dynamic system then n variables can be considered as n components of a vector x, such a vector is called state vector.

• State Space: The state space is defined as the n-dimensional space in which the components of the state vector represents its coordinate axes.

9

Definitions

• Let x1 and x2 are two states variables that define the state of the system completely .

1x

2x

Two Dimensional State space

State (t=t1)

StateVector x

dt

dx

State space of a Vehicle

Velocity

Position

State (t=t1)

State Space Representation

• An electrical network is given in following figure, find a state-space representation if the output is the current through the resistor.

State Space Representation• Step-1: Select the state variables.

L

c

i

v

Step-2: Apply network theory, such as Kirchhoff's voltage and current laws, to obtain ic and vL in terms of the state variables, vc and iL.

CRL iii

LRC iii

LCC iR

v

dt

dvC

Applying KCL at Node-1

(1)

State Space RepresentationStep-2: Apply network theory, such as Kirchhoff's voltage and current laws, to obtain ic and vL in terms of the state variables, vc and iL.

RL vdt

diLtv )(

Applying KVL at input loop

)(tvvdt

diL C

L

Step-3: Write equation (1) & (2) in standard form.

(2)

LCC i

Cv

RCdt

dv 11

)(tvL

vLdt

diC

L 11

State Equations

State Space Representation

LCC i

Cv

RCdt

dv 11 )(tv

Lv

Ldt

diC

L 11

)(tvL

i

v

L

CRCi

v

L

c

L

c

10

01

11

)(tvL

i

v

L

CRCi

v

dt

d

L

c

L

c

10

01

11

State Space RepresentationStep-4: The output is current through the resistor therefore, the output equation is

CR vR

i1

L

cR i

v

Ri 0

1

State Space Representation

L

cR i

v

Ri 0

1

)(tvL

i

v

L

CRCi

v

L

c

L

c

10

01

11

)()()( tButAxtx

Where,x(t) --------------- State Vector A (nxn) ---------------- System MatrixB (nxp) ----------------- Input Matrixu(t) --------------- Input Vector

)()()( tDutCxty

Where,y(t) -------------- Output VectorC (qxn) ---------------- Output MatrixD ----------------- Feed forward Matrix

Example-1• Consider RLC Circuit Represent the system in Sate Space and find

(if L=1H, R=3Ω and C=0.5 F):– State Vector– System Matrix– Input Matrix & Input Vector– Output Matrix & Output Vector

Vc

+

-

+

-Vo

iL

Lc itudt

dvC )(

cLL vRidt

diL Lo RiV

)(tuC

iCdt

dvL

c 11 Lc

L iL

RvLdt

di

1• Choosing vc and iL as state variables

Example-1 (cont...)

)(tuCi

v

L

R

L

Ci

v

L

c

L

c

0

1

1

10

L

co i

vRV 0

Lo RiV

)(tuC

iCdt

dvL

c 11 Lc

L iL

RvLdt

di

1

State Equation

Output Equation

18

Example-2• Consider the following system

KM

Bf(t)

x(t)

Differential equation of the system is:

)()()()(

tftKxdt

tdxB

dt

txdM

2

2

Example-2

)(tfM

xM

Kv

M

B

dt

dv 1

• As we know

vdt

dx

dt

dv

dt

xd

2

2

• Choosing x and v as state variables

vdt

dx

)()()()(

tftKxdt

tdxB

dt

txdM

2

2

)(tfMv

x

M

B

M

Kv

x

1010

Example-2

• If velocity v is the out of the system then output equation is given as

)(tfMv

x

M

B

M

Kv

x

1010

v

xty 10)(

Example-3• Find the state equations of following mechanical translational

system.

0211

21

2

1 KxKxdt

dxD

dt

xdM 122

22

2 KxKxdt

xdMtf )(

• System equations are:

Example-3

02111

1 KxKxDvdt

dvM

122

2 KxKxdt

dvMtf )(

• Now 1

1 vdt

dx

dt

dv

dt

xd 121

2

22 vdt

dx

dt

dv

dt

xd 222

2

• Choosing x1, v1, x2, v2 as state variables

11 vdt

dx

22 vdt

dx

Example-3

21

11

11

1 xM

Kx

M

Kv

M

D

dt

dv

)(tfM

xM

Kx

M

K

dt

dv

21

22

2

2 1

• In Standard form

11 vdt

dx

22 vdt

dx

Example-32

11

11

1

1 xM

Kx

M

Kv

M

D

dt

dv

)(tfM

xM

Kx

M

K

dt

dv

21

22

2

2 1

• In Vector-Matrix form

11 vdt

dx 2

2 vdt

dx

)(tf

Mv

x

v

x

M

K

M

K

M

K

M

D

M

K

v

x

v

x

22

2

1

1

22

111

2

2

1

1

10

0

0

00

1000

0

0010

Example-3

• If x1 and v2 are the outputs of the system then

)(tf

Mv

x

v

x

M

K

M

K

M

K

M

D

M

K

v

x

v

x

22

2

1

1

22

111

2

2

1

1

10

0

0

00

1000

0

0010

2

2

1

1

1000

0001)(

v

x

v

x

ty

Eigenvalues & Eigen Vectors• The eigenvalues of an nxn matrix A are the roots of the

characteristic equation.

• Consider, for example, the following matrix A:

Eigen Values & Eigen Vectors

Example#4

• Find the eigenvalues if – K = 2– M=10– B=3

)(tfMv

x

M

B

M

Kv

x

1010

Frequency Domain to time Domain Conversion

• Transfer Function to State Space

KM

Bf(t)

x(t)

Differential equation of the system is:

)()()()(

2

2

tftKxdt

tdxB

dt

txdM

Taking the Laplace Transform of both sides and ignoring Initial conditions we get

30

)()()()(2 sFsKXsBSXsXMs The transfer function of the system is

KBsMssF

sX

2

1

)(

)(

State Space Representation:

MK

MB

M

sssF

sX

2

1

)(

)(

)(

)(

)(

)(2

2

2

1

sPs

sPs

sssF

sX

MK

MB

M

)()()(

)(

)(

)(21

21

sPssPssP

sPs

sF

sX

MK

MB

M

)(1

)( 2 sPsM

sX

)()()()( 21 sPsM

KsPs

M

BsPsF

……………………………. (1)

……………………………. (2)

From equation (2)

)()()()( 21 sPsM

KsPs

M

BsFsP ……………………………. (3)

Draw a simulation diagram of equation (1) and (3)

1/s 1/sF(s) X(s)

-K/M

-B/M

P(s) 1/M

1/s 1/sF(s) X(s)

-K/M

-B/M

P(s)

2x

12 xx

• Let us assume the two state variables are x1 and x2.• These state variables are represented in phase variable form as

given below.

1x

• State equations can be obtained from state diagram.

21 xx

212 )( xM

Bx

M

KsFx

• The output equation of the system is

1

1)( xM

tx

1/M

21 xx 212 )( x

M

Bx

M

KsFx

1

1)( xM

tx

)(1

010

2

1

2

1 tfx

x

M

B

M

Kx

x

2

101

)(x

x

Mtx

Example#5

• Obtain the state space representation of the following Transfer function.

END OF LECTURES-34-35

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