Control Systems Lecture Notes

Control Systems Lecture Notes for KJM597

Faculty of Mechanical Engineering, UiTM Shah Alam

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam

CHAPTER 1.0 1.1

MATHEMATICAL REVIEW

COMPLEX VARIABLES AND COMPLEX FUNCTION This chapter will outline an overview of basic mathematical formulation in solving control systems problem that you will encounter throughout the course.

1.1.1

Complex Variables Concept A complex variable denoted by s consists of two components: a real component x component x and and an imaginary axis component y . Graphically, the real component of s of s is represented by a x a x -axis -axis in the horizontal direction, and the imaginary c omponent is measured along the vertical jy vertical jy -axis. -axis. Figure 1.0 illustrates the complex s-plane.

Figure 1.0: Complex s-plane (source: http://mathworld.wolfram.com )

   −

Using notation =

1, all numbers in engineering calculations can be re-written as

   =

+

Where z is called a complex number. Note that j is j is the only imaginary quantity in the expression. The magnitude, |z | z| and angle,



of z of z can be obtained mathematically,

  

Magnitude of z of z=|z =|z|=

2

+

2,



− 

angle of z of z= = tan

1

2


CHAPTER 1.0 1.1

MATHEMATICAL REVIEW

COMPLEX VARIABLES AND COMPLEX FUNCTION This chapter will outline an overview of basic mathematical formulation in solving control systems problem that you will encounter throughout the course.

1.1.1

Complex Variables Concept A complex variable denoted by s consists of two components: a real component x component x and and an imaginary axis component y . Graphically, the real component of s of s is represented by a x a x -axis -axis in the horizontal direction, and the imaginary c omponent is measured along the vertical jy vertical jy -axis. -axis. Figure 1.0 illustrates the complex s-plane.

Figure 1.0: Complex s-plane (source: http://mathworld.wolfram.com )

   −

Using notation =

1, all numbers in engineering calculations can be re-written as

   =

+

Where z is called a complex number. Note that j is j is the only imaginary quantity in the expression. The magnitude, |z | z| and angle,



of z of z can be obtained mathematically,

  

Magnitude of z of z=|z =|z|=

2

+

2,



− 

angle of z of z= = tan

1

2

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam A complex number can be written in rectangular form or in polar form as follows: i.

        ∠        −     

Rectangular forms

= =

ii.

+ (cos (cos

+ sin )

Polar forms

=| |

=| |

In converting complex numbers to polar form from rectangular, we use

=

2

+

2,

= tan

1

To convert complex number to rectangular form from polar, we employ

= | |cos ,

Notes:

= | | sin sin

(source: Ogata)

3

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Notes:

4

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam 1.1.2

Complex Function Concept a) Complex Function A complex function F(s,) a function of s, has a real component and imaginary component, or

        =

−  

where is tan

and

1



+

2

are real quantities. The magnitude of ( )is

+

2

, and the angle

. The angle is measured counterclockwise from the positive real axis.

 

of ( )

b) Single-valued Function In complex function analysis, we are interested in Single-Valued Function that can uniquely determine the value of s. For instance, given the function

   =

 ∞ =

1

( + 1)

is mapped onto two points, s=0 and s=-1, in the s-plane

c) Poles and zeros of a Function Poles are the value of s that will make the function F(s) become infinity. In other words, poles

    −





are the roots of the denominator of F(s). If the denominator of F(s) involves k-multiple factors

( + ) , then

=

is called a multiple poles and of order or repeated pole of order . If

= 1, the pole is called a simple pole.

Zeros are the value of s that will make the function F(s) become zero. In other words, zeros are the numerator of F(s). As an illustrative example, consider the following complex function

         −  −  −   −  − ∞ =

G(s) has zeros at

+ 2 ( + 10)

+1

+ 5 ( + 15)2

= 2 and = 10, simple poles at = 0, = 1 and = 5, and a double pole (multiple pole of order 2) at = 15. Note that G(s) becomes zero at = . G(s) is therefore has 2 zeros and 5 poles. d) Singularities of a Function The singularities of a function are the points in the s-plane at which the function or its derivatives do not exist. A pole is the most common of singularities and plays a very important role in studies of classical control theory.

(source: ogata)

5

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam 1.2

REVIEW OF DIFFERENTIAL EQUATIONS, LINEAR SYSTEMS, IMPULSE RESPONSE AND LAPLACE TRANSFORMATIONS. DEFINITION OF STABILITY. INTRODUCTION TO STATE EQUATIONS AND TRANSFER FUNCTIONS.

1.2.1

Review of Differential Equations Differential equations generally involve derivatives and integrals of the dependant variables with respect to the independent variable. For instance, a shock absorber system of a car as in figure 1.2 can be represented by the differential equation,

Ri(t )  L

di(t ) dt



1 C

 i(t )dt  v(t )

Figure 1.2: RLC Circuit where R is the resistance, L the inductance, C the capacitance, i(t) the current and v(t) the applied voltage. The dependent variable i(t) is determined by solving the equation.

In general, a differential equation of nth-order is written as n

d y (t ) n

dt

 an1

n

d 1 y (t ) n 1

dt

   a1

dy(t ) dt

 a0 y(t )  f (t )

Which is also known as a linear ordinary differential equation if the coefficients a0, a1, … ,an-1 are not a function of y(t).

6


Laplace Transforms a) Laplace Transform Laplace transform is used to convert from time domain to s-domain. Working with differential equation is rather complicated. In analyzing and designing a control system it is easier to work in s-domain. Laplace transform is defined as;

Where

   =

+

ℒ   =

, a complex variable.

=

∞  −    0

Example 2.1: Let f(t) be a unit-step function that is defined as

t  0

 1, u (t )   0,

t  0

The Laplace transform of f(t) is obtained as 



F ( s)  u (t )e

1

 st

dt   e s

0

st

 0



1 s

Example 2.2: Consider the exponential function  f (t )  e , t  0 t

where α is real constant. The Laplace transform of f(t) is written as 



F ( s)  e 0

 t st

e

dt 

e

 ( s  ) t 

s  

0



1 s  

7


Table 2.1: Laplace Transform table for input responses Notes:

8

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam b) Laplace Transform Theorems The laplace transform has a set of theorems to solve a complex mathematical equations. Table 2.2 summarizes the Laplace Transform theorems

Table 2.2: Laplace Transform Theorems

9

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam c) Inverse Laplace Transformation Using Partial Fraction Method Given the Laplace transform F(s), the operation of obtaining f(t) is termed the inverse Laplace Transformation and is denoted by:

 ℒ−  1

=

[

]

Inverse Laplace Transform is used when we want to convert from s-domain to time domain. The inverse Laplace transform of rational functions are normally carried out using partialfraction expansion and the Laplace transform table. Consider a rational function

   =

( )

( )

where Q(s) and P(s) are polynomials of s. It is assume that the order of P(s) in s is greater than of Q(s). The polynomial P(s) may be written as

 an1s n1    a1s  a0

P( s)  s n

where a0, a1, … ,an-1 are real coefficients. This method will be emphasized for the cases of simple poles, multiple-order poles and complex poles. Case 1: Simple poles If all the poles of G(s) are simple and real, then G(s) can be written as

        …    ≠ ≠⋯  − − ⋯ − =

( )

( )

=

( )

+

+

1

( +

2

)

,

1

2

Applying partial-fraction expansion, the equation can be wr itten as

=

1

+

Where K si

+

1

2

+

+

2

+

+

 Q( s )   ( s  s i )  P( s )   s  s

 −

The numerator of each fraction is called the residue. for the pole = .

−

i

is called the residue of G(s)

The inverse transform is the written as

g (t )  K s1 e

 s1t

 K s e s t    K s e s t 2

2

n

n

10


Example 2.3: Consider the function

G(s) 

5s  3 ( s  1)(s  2)(s  3)

which is written in the partial-fraction expanded form: G ( s)  The coefficients

− − − 1,

2

,

3

K 1 s 1



K 2 s2



K 3 s3

are determined as follows:

5(1)  3  ( s  1)G(s)  1   1 s (1  2)(1  3) 5(2)  3 K 2  ( s  2)G ( s)  7 s  2 (2  1)(2  3) 5(3)  3 K 3  ( s  3)G ( s)   6 s  3 (3  1)(3  2) K 1

Thus,

G ( s) 

7 1 6   s 1 s  2 s  3

The inverse transform or time function is

g (t )  e t  7e 2t  63t Case 2: Multiple-order poles If r of the n poles is identical, G(s) is written as

       …  −     =

( )

( )

=

( )

+

1

+

2

+

( +

)

11

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Then G(s) can be expanded as

G ( s) 

K s1 s  s1



K s2

 

s  s2

K  s( n r )



s  s n r

A1

A2



s  si

( s  si ) 2



Ar

( s  si ) r

                  n - r terms of simple poles The (n-r) coefficients K -s1, K -s2

r terms of repeated poles

K -s(n−r) which correspond to simple poles may be evaluated as explained before. The coefficients A1 Ar are evaluated as follows: , … ,

…



Ar  ( s  si ) r G ( s )

s  s

i

Ar 1 

d ds

(s  s ) G(s) s   s r

i

i

2

Ar  2 

1 d

2 ! ds

(s  s ) G(s) s   s r

2

i

i



A1 

1

1

d r

( r  1)! ds

(s  s ) G(s)s  s r

r 1

i

i

Example 2.4: Consider the function

G (s ) 

2 ( s  1)(s  2) 2

G(s) can be written as G(s) 

K 1

( s  1)



A1

( s  2)



A2

( s  2) 2

The coefficient corresponding to the simple pole is K 1

 2   2 2 ( 2 )  s   s  1

12


and those of second order-pole are A2

 2    2   (s  1)  s  2

A1



d 

  2    2    (s  1) 2  ds  ( s  1)    s  2 s  2 2

The completed partial-fraction expansion is 2 2 2 G ( s)    s  1 s  2 ( s  2) 2 The time function is

g (t )  2e t  2e 2t  2te 2t Case 3: Simple complex-conjugate poles Suppose that G(s) contains a pair of complex poles: s    j and s  - - j The corresponding coefficients of these poles are K   j  (s    j )G(s)

s    j

K   j  ( s    j )G( s)

s    j

Example 2.5:

Considering transfer function G(s) G ( s) 



3 s(s K 0 s

2

 2s  5)





K 1 j 2 s  1  j 2

3 s( s  1  j 2)(s  1  j 2)



K 1 j 2 s  1  j 2

13


K 0

3 3    2   s  2s  5  s  0 5

K 1 j 2

  3 3   (2  j1)   s( s  1  j 2)  s  1  j 2 20

K 1 j 2

  3 3   (2  j1)   s( s  1  j 2)  s  1  j 2 20 3

G ( s) 

5 s



 2  j1 2  j1     20  s  1  j 2 s  1  j 2  3

and the time function is given as g (t ) 

3 5 3



  5

3 20 3 20

(2  j1)e  

( 1 j 2 )t

e

(2e

t

j 2 t

 (2  j1)e( 1 j 2)t 

 2e j 2t )  j (e  j 2t  e j 2t )

 e  j 2t  e j 2t e j 2t  e  j 2t    e 4 2  5 20  2 j 2  3 3 1    et   cos 2t  sin 2t  5 5  2  3

3

t

14


TUTORIAL 1: MATHEMATICAL REVIEW 1. Derive equations for a unit step, ramp, impulse and sinusoidal response in time domain. 2. In a unit step response graph, what is the relationship between final value theorem and steady state error?

 − 

3. Find the Laplace transform of time function

=5+3

2

.

4. Verify question (3) above by using MATLAB application. MATLAB hint >>syms s t; % Command to run MATLAB in s and t domains >>f=5+3*exp(-2*t) % Entering the function >>F=laplace(f,t,s) % Executing Laplace Transform command

5. Find the inverse Laplace Transform of a rational function and

       =

5

2

+3 +2

6. Find the inverse Laplace transform of a rational function

=

2

+ 1 ( + 2)2

7. Verify the result in question (6) above using MATLAB application. MATLAB hint >> syms st; >>F=2/((s+1)*(s+2)^2) >>f=ilaplace(F,s,t)

15


CHAPTER 2.0

INTRODUCTION TO CONTROL SYSTEMS

Control systems can be placed into three broad functional groups: 

Monitoring systems, such as Supervisory Control and Data Acquisition (SCADA) systems, which provide information about the process state to the operator;



Sequencing systems, used where some process must follow a pre-defined sequence of discrete events;



Closed-loop systems, which is widely taught in engineering course, are typically implemented to give some process a set of desired performance characteristics

The history of feedback control system begun as early as in 1769 when James Watt’s steam engine and governor are developed. The Watt stem engine often used to mark the beginning of the Industrial Revolution in England. The revolution of automatic control system continues in which the first ever autonomous rover vehicle, known as Sojourner was invented in 1997. In summary below is the history of feedback control system 1769

-

James Watt’s flyball governer

Figure 2.0: James Watt’s flyball governer 1868

-

J. C. Maxwell’s model of governer

1927

-

H. W. Bode’s feedback amplifiers

1932

-

H. Nyquist’s stability theory

16

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam 1954

-

George Devol’s robot design

1970

-

State-variable models and optimal control theory

1980

-

Robust control system design

1997

-

First ever autonomous rover vehicle “Sojourner”

Info: The mobile Sojourner had a mass of 10.5kg and 0.25 square meter solar array

Figure 2.1: Sojourner But before we go into further details, we have to know control systems’ terms and concepts. The frequently used terms and concepts are as follow: Automation Control system

-

Controlled variable Manipulated variable Plant

-

Processes

-

-

The control of a process by automatic means An interconnection of components forming a system configuration that will provide a desired response Quantity or condition that is measured and controller. Normally it is the output of the system Quantity or condition that is varied by the controller so as to affect the value of the controlled variable A plant is a piece of equipment, perhaps just a set of machine parts functioning together, the purpose of which to perform a particular operation. Any physical object to be controller (such as heating furnace, a chemical reactor etc) is called a plant A process can be defined as a natural, progressively continuing operation or development marked by a

17


Disturbances

-

Feedback control

-

Feedforward

-

series of gradual changes that succeed one another in a relatively fixed way and lead towards a particular result or end A disturbance is a signal which tends to adversely affect the value of the output of the system. If a disturbance is generated within the system, it is called internal; which an external disturbance is generated outside the system. Feedback control is an operation which in the presence of disturbances, tends to reduce the difference between the output of a system and the reference input and which does so on the basis of the difference. Feedforward has a reference signal which is act as an additional input. Source: AAMI, Fac of Mech Eng., UiTM

Figure 2.2: Input-output configuration of control system (souce: AAMI)

18


Figure 2.3: Input-output configuration of a closed-loop control system (source: AAMI) 2.1

OPEN LOOP AND CLOSED-LOOP SYSTEMS

2.1.1

Open Loop Control System

A system is said to be an open loop system when the system’s output has no effect on the control action. In open loop system, the output is neither measured nor fed back for comparison with the input.

Figure 2.4: Open loop control system An open loop control system utilizes an actuating device (or controller) to control the process directly without using feedback as shown in Figure 2.4. The advantages and the disadvantages of an open-loop control system is tabulated in t able 2.1 below

ADVANTAGES Simple and ease of maintenance Less expensive Stability is not a problem Convenient when output is hard to measure

DISADVANTAGES Disturbances and changes in calibration cause errors Output may be different from what is desired

19


Closed-loop control system

A system that maintains a prescribed relationship between the output and the reference input is called a closed-loop system or a feedback co ntrol system. The system uses a measurement of the output and feedback of the signal to compare it with the desired output.

Figure 2.5: Closed loop control system In a closed-loop control system, the actuating error signal, which is the difference between the input signal and the feedback signal, is fed to the controller so as to reduce the error and bring the output of the system to a desired value. 2.1.3

Comparison between open loop and closed-loop control system.

The table below shows the comparison between the two systems:

OPEN LOOP System stability is not a major problem, therefore easier to build

Use open loop only when the inputs are known ahead of time and there is no disturbances

2.2

CLOSED LOOP The use of feedback makes the system response relatively insensitive to external disturbances and internal variations in system parameters System stability is a major problem because the system tends to overcorrect errors that can cause oscillations or changing amplitude.

TRANSFER FUNCTION

The transfer function of a linear system is defined as the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable, with all initial conditions assumed to be zero. The Transfer function of a system (or element) represents the relationship describing the dynamics of t he system under consideration. A transfer function may be defined only for a linear, stationary (constant parameter) system. A non-stationary system often called a time-varying system, has one or more timevarying parameters, and the Laplace transformation may not be utilized. Furthermore, a transfer function is an input-output description of the behavior of a system. Thus the transfer function description does not include any information concerning the internal structure of the system and its behavior.

20


The Transfer function of linear systems

The transfer function of a LTI system is defined as the Laplace transform of the impulse response, with all the initial conditions set to zero.

G( s)  L[ g (t )] The transfer function is related to the Laplace transform of the input and the output through the following relation:

G ( s) 

Y ( s) R( s)

where all the initial conditions set to zero, and Y ( s) and R( s) are the Laplace transform of y (t ) and

r (t ) respectively.

Although the transfer function of a linear system is defined in terms of the impulse response, in practice, the input-output relation of a linear time-invariant system with continuous –data input is often described by the differential equation, so it is more convenient to derive the transfer function directly from the differential equation.

Let us consider that the input-output relation of a linear time-invariant system is described by the following nth-order differential equation with constant real coefficients: n

d y(t ) n

dt

d  y(t ) n 1

 an1

n 1

dt

 ......  a1

dy(t ) dt

m1

m

 a0 y(t )  bm

d r (t ) m

dt

 bm1

r (t )

d

m1

dt

 .....  b1

dr (t ) dt

 b0 r (t )

To obtain the transfer function of the linear system that is represented by Eq. (2.3), we simply take the Laplace transform on both sides of the equation and assume zero initial conditions. The result is

s  a  s     a s  a Y(s)  b n 1

n

n 1

1

m

0

sm

 bm1s m1    b1s  b0  R(s)

The transfer function between r (t ) and y (t ) is given by:

Y ( s)

 ..............  b1 s  b0  an1 s n1  ......  a1 s  a0

bm s m

G( s)   R( s) s n

21

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The transfer function is said to be strictly proper if m  n . If m  n then the transfer function is proper. It is improper if m  n .



Characteristic Equation: The characteristic equation of a LTI system is defined as the equation obtained by setting the denominator polynomial of the transfer function to zero. Thus, the characteristic equation of the system described by the Eq. (2.4) is

sn

 an1s n1    a1s  a0  0

Later, we shall show that the stability of a linear single-input single-output system is governed completely by the roots of the characteristic equation.

2.2.2

Transfer function of multivariable system

The definition of a transfer function is easily extended to a system with multiple inputs and outputs. A system of this type is often referred to as a multivariable system. Figure 2.6 shows a control system with two inputs and two outputs.

Figure 2.6: General block representation of a two-input, two-output system

Since the principle of superposition is valid for linear systems, the total effect on any output due to all the inputs acting simultaneously is obtained by adding up the outputs due to each input acting alone. Thus, using transfer function relations we can write the simultaneous equations for the output variables as

Y 1 ( s )  G11 ( s) R1 ( s )  G12 ( s) R2 ( s ) Y 2 ( s)  G21 ( s) R1 ( s )  G22 ( s) R2 ( s)

22

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam th

th

where G ij ( s ) is the transfer function relating the i output to the j input variable. Thus

Gij



Y i ( s) R j ( s)

In general, for j inputs and i outputs, we can write the simultaneous equations for the output variables as

Y 1 ( s)  G11 (s) Y ( s) G (s)  2    21         G ( s) ( ) Y s i    i1

G12 ( s)  G1 j ( s )   R1 ( s) 

            Gij ( s )   R j ( s)

G22 ( s)  G2 j ( s) R2 ( s)

 G i 2 ( s)

It is convenient to express Eq. (2.7) in a matrix-vector form

Y(s)  G(s)R(s) where

Y 1 ( s )  Y ( s) 2  Y ( s )        ( ) Y s i   is the i  1 transformed output vector; whereas

 R1 ( s )   R ( s ) 2  R( s )         R j ( s) is the j  1 transformed input vector; and

G11 (s) G (s) 21 G ( s)       Gi1 (s) is the i  j transfer-function matrix.

G12 ( s )  G1 j ( s) 

       Gij ( s) 

G22 ( s )  G2 j ( s)

 G i 2 ( s)

23


DEFINITION OF STABILTY

A stable system is defined as a system which gives a bounded output in response to a bounded input.

The concept of stability can be illustrated by considering a circular cone placed on a horizontal surface, as shown in Fig. 2.7 and Fig. 2.8.

Figure 2.7: The stability of a cone. ----------------------------------------------------------------------------------------------------

Figure 2.8: Stability in the s-plane.

The stability of a dynamic system is defined in a similar manner. Let u(t), y(t), and g(t) be the input, output, and impulse response of a linear time-invariant system, respectively. The output of the system is given by the convolution between the input and the system's impulse response. Then

y(t ) 





0

u (t   ) g ( )d 

24

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam This response is bounded (stable system) if and only if the absolute value of the impulse response, g(t), integrated over an infinite range, is finite. That is





0

g ( ) d 



Mathematically, Eq. (4.24) is satisfied when the roots of the characteristic equation, or the poles of G(s), are all located in the left-half of the s-plane.

A system is said to be unstable if any of the characteristic equation roots is located in the right-half of the s-plane. When the characteristic equation has simple roots on the j - axis and none in the right-half plane, we refer to the system as marginally stable.

The following table illustrates the stability conditions of a linear continuous system with reference to the locations of the roots of the characteristic equation.

STABILITY CONDITION Stable Marginally stable of marginally unstable

Unstable

LOCATION OF THE ROOTS All the roots are in the left-half s-plane At least one simple root and no multiple roots on the j -axis; and no roots in the right-half s-plane. At least one simple root in the right-half splane or at least one multiple-order root on the j - axis.

The following examples illustrate the stability conditions of systems with reference to the poles of the closed-loop transfer function M(s).

M ( s) 

M ( s) 

M ( s ) 

20

Stable

s  1s  2s  3 20( s  1) ( s  1)(s 2

 2s  2)

20( s  1) ( s  2)( s 2

Unstable due to the pole at s = 1

 4)

Marginally stable or marginally unstable due to s =  j 2.

25


M ( s)  2.3.1

Unstable due to the multiple-order pole at s =  j 2.

10 (s 2

 4) 2 (s  10)

Open loop and Closed loop stability

A system is open-loop stable if the poles of the loop transfer function G(s)H(s) are all in the left hand side of s-plane.

Controller

ysp

+



-

H(s)

e(s)

Plant

G(s)

y

Figure 2.9: A typical closed-loop system A system is closed0loop stable (or simply stable) if the poles of the closed-loop transfer function (or zeros of 1+G(s)H(s) are all in the left hand side of s-plane 2.4

BASIC CONTROL ACTIONS

The following six basic control actions are very common among industrial automatic controllers: 1. 2. 3. 4. 5. 6. 2.4.1

Two-position or on-off controller Proportional controller Integral controller Proportional-plus-integral controller Proportional-plus-derivative controller Proportional-plus-derivative-plus-integral controller Two-position of on-off control action

In a two-position control system, the actuating element has only two fixed positions which are, in many cases, simply on and off. Two-position or on-off control is relatively simple and inexpensive and, for this reason, is very widely used in both industrial and domestic control systems. Let the output signal from the controller be m(t) and the actuating error signal be e(t). In two position control, the signal m(t) remains at either a maximum or minimum value, depending on whether the actuating error signal is positive or negative, so that

     =

1

( )>0

=

2

( )<0

26

 




−

Where 1 and 2 , are constants. The minimum value 2 , is usually either zero or 1 . Two-position controllers are generally electrical devices, and an electric, solenoid-operated valve is widely used in such controller. Pneumatic proportional controller with very high gain act as two-position controller and are sometimes called pneumatic two-position controller. Figure 2.10 show the block diagrams for two -position controller. The range through which the actuating error signal must move before the switching occurs is called the differential gap.

Figure 2.10: Two-position controller 2.4.2

Proportional controller

For a controller with proportional control action, the relationship between the output of the controller m(t) and the actuating error signal e(t) is

      =

or, in Laplace Transform

( )

Where



( )

( )

=

, is termed the proportional sensitivity or the gain.

27

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Whatever the actual mechanism may be and whatever the form of the operating power, the proportional controller is essentially an amplifier with and adjustable gain. The proportional action has the following two properties: 1. Reduce rise time 2. Does not eliminate steady state error

Example 2.1: Given a system consist of mass-spring and damper

x

k

M

F

b a) b) c) d) e) f) g)

The second order PDE is: Taking the LT The TF is therefore: Let M=1kg, b=10N.s/m, k =20 N/m & F(s)=1, therefore X(s) / F(s): From the Transfer Function, the DC gain is: Corresponding to the steady state error of: The settling time is: Open Loop Response 0.05 0.045 0.04 0.035 ) m ( t n e m e c a l p s i D

0.03 0.025 0.02 0.015 0.01 0.005 0 0

0.2

0.4

0.6

0.8

1 Tim e (sec)

1.2

1.4

1.6

1.8

2

28

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam P control (K) reduces the rise time, increases the overshoot and reduces the steady state error. h) The closed-loop transfer function of the system with P controller is X(s)/F(s)=G/(1+G): i) Let the P gain (K) equal 300 Closed Loop Step : K = 300 1.4

1.2

1 ) m ( t n e m e c a l p s i D

0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Tim e (sec)

Rise time and ss error reduced, slightly reduced settling time but increased overshoot. 2.4.3

Integral controller

In a controller with integral control action, the value of t he controller output m(t) is changed at a rate proportional to, the actuating error signal e(t). That is

          ( )

Therefore; Where



=

=

( )

0

is an adjustable constant. The transfer function of the integral controller is

( )

( )

=

If the value of e(t) is doubled, then the value of m(t) varies twice as fast. For zero actuating error, the value of m(t) remains stationary.

29

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The integral controller has the following properties: 1. Proportional controllers often give a steady-state error. Integral controller arose from trying to add a “reset” term to the control signal to eliminate steady state error. In other words, t he integral controller “resets” the bias error from the P controller. 2. Gives large gain at low frequencies resulting in “beating down” load disturbances. 3. May make the transient response worse. 4. Controller phase starts out at -90° and increases to 0° at the break frequency. This phase lag can be compensated by derivative action. The integral controller act as “automatic reset” as shown in figure 2.11

load disturbance y sp

+



e

K

-

1 u sT i



plant

y

Figure 2.11: Automatic reset action Almost always used in conjunction with P control.

K load disturbance

y sp +



e

K

-

1 sT i

u



 

plant

y

Figure 2.12: PI control The integral term may be expressed in (i) The integral term gain. The integral term

 

and (ii)

is known as the integral time constant.

 ∞

is known as integral gain (e.g: in MATLAB)

The relationship between

  and

is as follows:

  =

=

corresponds to pure (proportional)

30


Example 2.2: a) I control reduces the rise time, increases both settling time and overshoot, and eliminates the steady-state error b) The closed-loop transfer function of the system with a PI controller is: X (s)/F (s) = ______________ . c) Let k = 30 and k i = 70. P gain (k ) was reduced because the I controller also reduces the rise time and increases the overshoot as does the P controller (double effect). Closed Loop Step : K = 30, Ki = 70 1.4

1.2


0.8

0.6

0.4

0.2

0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Time (sec)

2.4.4

Derivative controller

Introducing a derivative controller will add damping and in doing so: 1. increases system stability (add phase lead) 2. reduces overshoot 3. generally improves transient response A derivative controller may able to provide anticipative action but derivative action can make the system become noisy. Almost always used in conjunction with P control.

load disturbance y sp

c

+

 -

KT d s 1+sT d /N



 

Figure 2.12: PD control The integral term may be expressed in (i) The integral term



and (ii)

is known as the derivative time constant.

plant

y

31


The integral term



is known as derivative gain (e.g: in MATLAB)

The relationship between

  and

is as follows:

  =

Example 2.3: a) D control reduces both settling time and overshoot. b) The closed-loop transfer function of the system with a PD controller is: X (s)/F (s)=______________ c) Let k = 300 and k d = 10. Closed Loop Step : K = 300, Kd = 10 1.4

1.2


0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Tim e (sec)

d) Reduced overshoot and settling time, small effect on rise time and ss error

2

32


Closed Loop Step : K = 350, Ki = 300, Kd = 50

1.2


0.8

0.6

0.4

0.2

0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Tim e (sec)

2.4.5

PID controller

In some system the commonly implemented controller consist of the P, I and D control action. We call this type of controller as PID controller. Td s

1/(Ti s ) y sp +

-



e

u



K

G( s )

y

Figure 2.13: PID control The standard form of PID controller according to IS A (Instrument Society of America) is as follows:

Or

        =

(1 +

=

+

1

+

+

)

33


Example 2.4: a) The closed-loop transfer function of the system with a PID controller is: 2 3 2 X (s)/F (s) = (k d s +ks+k i )/(s + (10+k d )s + (20+k )s + k i ) b) Let k = 350, k i = 300 and k d = 50. Closed Loop Step : K = 350, Ki = 300, Kd = 50

1.2


0.8

0.6

0.4

0.2

0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Time (sec)

c) 2.4.6

No overshoot, fast rise and settling time and no steady-state error PID tuning

Introducing the P, I and D controller has certainly proven to contribute some effect to our system’s response. These effects are summarized as in table below.

CLOSED LOOP RESPONSE K

   =

=

RISE TIME

OVERSHOOT

Decrease

Increase

SETTLING TIME Small change

SS ERROR

Decrease

Increase

Increase

Eliminate

Small change

Decrease

Decrease

Small change

Decrease

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response. 1. 2. 3. 4. 5.

Obtain an open-loop response and determine what needs to be improved Add a proportional control to improve the rise time Add a derivative control to improve the overshoot Add an integral control to eliminate the steady-state error Adjust each of K , K i , and K d until you obtain a desired overall response referring to the table shown previously to find out which controller controls what characteristics.

34

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam 6. It is not necessary to implement all three controllers (P, I & D) into a single system. For example, if a PI controller gives a good enough response, then you don't need to add D control to the system. Simple is better. 2.5

BLOCK DIAGRAM & REDUCTION METHODS

A block diagram is used to describe the composition and interconnection of a system, or it can be used together with the transfer functions to describe the cause-and-effect relationships throughout the system. For instance, Figure 2.14 (a) shows a dc motor wiring diagram, (b) sketch, and (c) shows the block diagram with transfer function.

Figure 2.14: A dc motor: (a) wiring diagram (b) sketch

Figure 2.14 (c): A dc motor: Block diagram with transfer functions

35


Block diagram reduction method

We shall now define the block diagram elements used frequently in linear control systems and the related algebra. All component parts of a block diagram for linear time-invariant systems are shown in Figure 2.15. The characteristic of the summing junction as shown in Figure 2.15 (c) is that the output signal, C ( s) , is the algebraic sum of the input signals. The figure shows three inputs, but any number can be presented. A pickoff point, as shown in Figure 2.15 (d), distributes the input signal, R( s ) , undiminished, to several output points.

Figure 2.15: Components of a block diagram for LTI systems Figure 2.16 shows the block diagram of a linear feedback control system. The following terminology is defined with reference to the diagram.

Figure 2.16: Feedback control system

36


R( s), r (t ) = reference input (command) C ( s), c(t ) or Y ( s), y(t ) = output (controlled variable) B( s), b(t ) = feedback signal E ( s), e(t ) = actuating signal = error signal H ( s) = feedback transfer function

G( s) H ( s) = L( s ) = loop transfer function G( s) = forward-path transfer function

M (s)  C (s) R(s) or Y (s) R(s) = closed-loop transfer function or system t ransfer function.

M ( s) can be expressed as a function of G ( s) and H ( s) . From Figure 2.16, we write Y ( s)  G( s) E ( s) B( s)  H ( s)Y ( s) The actuating signal is written as

E (s)  R( s)  B(s) Thus,

Y ( s )  G ( s) R( s )  G ( s ) B( s) M ( s) 

Y ( s ) R( s )



G( s)

1  G ( s) H ( s)

The block diagram representation of a given system often can be reduced by block diagram reduction techniques to a simplified block diagram with fewer blocks than the original diagram. Table below shows some of the block diagram reduction techniques.

The block diagram reduction technique is based on the utilization of rule 6 in which eliminates feedback loops. Therefore, the other transformations are used to transform the diagram to a form ready for eliminating feedback loops.

37


For parallel subsystems as shown below in (a), the reduction technique is shown in (b).

38


Example 2.5: Block Diagram Reduction.

A block diagram of a multiple-loop feedback control system is shown in Figure 2-5. It is interesting to note that the feedback signal H1(s)Y (s) is a positive feedback signal, and the loop G3(s)G4(s)H1(s) is called a positive feedback loop. First, to eliminate the loop G3G4H1, we move H2 behind block G4 by using rule 4, and therefore obtain Figure 2-6 (a).

Multiple-loop feedback control system

39

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Eliminating the loop G3G4H1 by using rule 6, we obtain Figure 2-6 (b). Then, eliminating the inner loop containing H2/G4, we obtain Figure 2-6 (c). Finally, by reducing the loop containing H3, we obtain the closed-loop system transfer function as shown in Figure 2-6 (d).

Block diagram reduction of the system

40


Example 2.6: Reduce the system shown to a single transfer function.

Block diagram for Example 2.6

41


Steps in the block diagram reduction for Example 2.6

The block diagram representation of feedback control systems is a valuable and widely used approach. The block diagram provides the analyst with a graphical representation of the interrelationships of controlled and input variables. Furthermore, the designer can readily visualize the possibilities for adding blocks to the existing system block diagram to alter and improve the system performance. The transition from the block diagram method to a method utilizing a line path representation instead of a block representation is readily accomplished and is presented in the following section.

42


SIGNAL FLOW DIAGRAM & REDUCTION METHODS

Block diagrams are adequate for the representation of the interrelationships of controlled and input variables. However, for a system with reasonably complex interrelationships, the block diagram reduction technique is cumbersome and often quite difficult to complete. An alternative method for determining the relationship between system variables has been developed by Mason and is based on a representation of the linear system by line segments called Signal-Flow Graph (SFG). The advantage of the SFG method is the availability of a flow graph gain formula, which provides the relation between system variables without requiring any reduction procedure or manipulation of the flow graph.

2.6.1

Basic elements of SFG

When constructing a SFG, junction points or nodes are used to represent variables. The nodes are connected by line segments, called branches. A signal can transmit through a branch only in the direction of the arrow. For instance, consider that a linear system is represented by a simple algebraic equation

y2  a12 y1 where y 1 is the input, y 2 the output, and a 12 the gain between two variables. The SFG is shown in Figure 2-9. a12

Y 1

Y 2

Figure 2.17: Signal-flow graph of y2

Example 2.7: Consider the following set of algebraic equations:

y2  a12 y1  a32 y3 y3  a23 y2  a43 y4 y4  a24 y 2  a34 y3  a44 y4 y5  a25 y2  a45 y4

 a12 y1

43

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The SFG for these equations is constructed, step by step, as shown:

Step-by-step construction of the SFG of Example 2.7

44


Summary of the basic properties of SFG

The important properties of the SGF are summarized as follows: 1. SFG applies only to linear systems. 2. Nodes are used to represent variables. Normally, the nodes are arranged from left to right, from input to output. 3. Signals travel along branches only in the direction described by the arrows of the branches.

2.6.3

Definitions of the SFG terms Input Node (source) Output Node (sink)

-

Path

-

Forward Path

-

Loop

-

Path Gain

-

Loop Gain Non-touching Loops

-

-

An input node is a node that has only outgoing branches. An output node is a node that has only incoming branches. In general, we can make any non input node an output node, simply by connecting a branch with unity gain from the existing node to a new node with the same name (Example: node y 2 in Figure 2.18(b)). If we attempt to convert y 2 into input node, by using the same unity gain branch (Figure 2.18 (c)), then y2 output will differ from the original (y 2 = y2 + a12y1 + a32y3). A path is any collection of a continuous succession of branches traversed in the same direction. A forward path is a path that starts at an input node and ends at an output node, and along which no node is traversed more than once. A loop is a path that originates and terminates on the same node and along which no other node is encountered more than once. For example, there are four loops in the SFG of Example 2.7. These are shown in Figure 2.19 The product of the branch gains encountered in traversing a path is called the path gain The loop gain is the path gain of a loop Two parts of a SFG are non-touching if they do not share a common node. For example, the loop y 2-y3-y2 and y4-y4 of the SFG in Figure (d) of Example 2.7 are non-touching loops.

45


Figure 2.18 (a & b): Modification of SFG so that y 2 become output node (c): Erroneous way to make node y 2 an input node

46


Figure 2.19: Four loops in the signal-flow graph of Example 2.7

2.6.4

SFG Algebra

Based on the properties of the SFG, we can outline the following manipulation rules and algebra of SFG. 1. The value of the variable represented by a node is equal to the sum of all the signals entering the node. For the SFG of Figure 2.20 (a),

y1  a21 y2  a31 y3  a41 y4  a51 y5 2. The value of the variable represented by a node is transmitted through all branches leaving the node. In Figure 2.20 (a), we have

y6  a16 y1 y7  a17 y1 y8  a18 y1

47


Figure 2.20

3. Parallel branches in the same direction connecting two nodes can be replaced by a single branch with the gain equal to the sum of gains of the parallel branches. Example: Figure 2.20 (b). 4. A series connection of unidirectional branches can be replaced by one branch with gain equal to the product of branch gains.

48


Gain formula for SFG (Mason’s Rule)

The overall gain between the input node y in and output node y out of a SFG with N forward paths and L loops is given by

M 

yout yin

P  k



k

, k  1,2, N

k



where y in = input-node variable y out = output-node variable M = gain between y in and y out N

= total number of forward paths between y in and y out th

Pk = k forward-path gain

 = 1 – (sum of all individual loop gains) + (sum of all gain products of two non-touching loops) – (sum of all gain products of three non-touching loops) + … th

k = , which is evaluated by eliminating all loops that touch k forward-path

Procedures to solve SFG by using Mason’s rule: 1. Identify the no. of forward paths and determine the forward-path gains. 2. Identify the no. of loops and determine the loop gains. 3. Identify the non-touching loops taken two at a time, three at a time and so on. Determine the product of the non-touching loop g ains. 4. Determine  and k . 5. Substitute all of the above information into the gain formula:

M 

yout yin

P  k



k



k

, k  1,2, N

49


Care must be taken when applying the gain formula to ensure that it is applied between an input node and an output node.

Example 2.8: Consider the SFG of a closed loop control system as given in Figure below. By using the gain formula, find the transfer function Y ( s) R( s ) .

SFG of a feedback control system

1. There is only one forward path between R( s ) and Y ( s ) , and the forward-path gain is P1 = G ( s) . 2. There is only one loop; the loop gain is L1 =

 G(s) H (s) .

3. There are no non-touching loops. 4.  = 1 - L1 = 1  G( s) H (s ) and 1 = 1. 5. Thus,

Y ( s) R( s)



P11





G( s)

1  G ( s) H ( s)

50


Example 2.9: For the system shown in Figure below, determine the gain between y1 and y5.

SFG for Example 2.9

1. There are three forward paths Path 1: y 1 – y 2 – y 3 – y 4 – y 5

P1 = a12 a23 a34 a45

Path 2: y 1 – y 2 – y 4 – y 5

P2 = a12 a24 a45

Path 3: y 1 – y 2 – y 5

P3 = a12 a25

2. There are four loops Loop 1: y 2 – y 3 – y 2

L1 = a23 a32

Loop 2: y 3 – y 4 – y 3

L2 = a34 a43

Loop 3: y 2 – y 4 – y 3 – y 2

L3 = a24 a43 a32

Loop 4: y 4 – y 4

L4 = a44

3. Non-touching loops: y 2 – y 3 – y 2 and y 4 – y 4 Thus the product of the gains of the two non-touching loops: L1L4 = a23 a32 a44

4.  = 1 – (L1 + L2 + L3 + L4 ) + L1L4 = 1 – (a23 a32 + a34 a43 + a24 a43 a32 + a44 ) + a23 a32 a44

51


All the loops are in touch with forward path P1, thus 1 = 1. All the loops are in touch with forward path P2, thus 2 = 1. Two loops (y 3 – y 4 – y 3 and y 4 – y 4) are not touching with forward path P3. Thus, 3 = 1 - a34a43 – a44.

5. Thus,

 P2  2  P3  3 y1  (a a a a )  (a12 a 24 a 45 )  (a12 a 25 )(1  a34 a 43  a 44 )  12 23 34 45 1  (a 23 a32  a34 a 43  a 24 a32 a 43  a 44 )  a 23 a32 a 44

M 

y5



P1 1

Example 2.10: Consider the SFG as shown in the figure. The following input-output relation is obtained by use of the gain formula:

y7 y1



P1 1

 P2  2 G1G2 G3G4  G1G5 (1  G3 H 2 )   

where

  1  G1 H 1  G3 H 2  G1G2G3 H 3  H 4  G1G3 H 1 H 2  G1 H 1 H 4  G3 H 2 H 4  G1G2G3 H 3 H 4  G1G3 H 1 H 2 H 4

SFG for Example 2.10

52


2.7

CONVERSION FROM BLOCK DIAGRAMS TO SFG

An equivalent SFG for a block diagram can be drawn by performing the following steps: 1. Identify the input/output signals, summing junctions & pickoff points → they are replaced with nodes. 2. Interconnect the nodes & indicate the directions of signal flow by using arrows. 3. Identify the blocks - they are replaced with branches. For each negative sum, a negative sign is included with the branch. 4. Add unity branches as needed for clarity or to make connections. 5. Simplify the SFG → eliminate redundant nodes/branches (only if the node is connected to branches of a single flow in & a single flow out with unity gain). 6. Label the input/output signals and the branches accordingly.

Example 2.11: Convert the block diagram in the figure to a signal flow graph and determine the transfer function using Mason’s gain formula.

53

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The equivalent SFG:

1. There are two forward paths; the forward-path gains are: P1 = G1G2G3 P2 = G1G4

2. There are five individual loops; the loop gains ar e: L1 = −G1G2H1 L2 = −G2G3H2 L3 = −G1G2G3 L4 = −G1G4 L5 = −G4H2

3. There are no non-touching loops.

4. ∆ = 1 – (L1 + L2 + L3 + L4 + L5) = 1 + G1G2H1 + G2G3H2 + G1G2G3 + G1G4 + G4H2

All the loops are in touch with forward path P1, thus 1 = 1.

54

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam All the loops are in touch with forward path P2, thus 2 = 1

5. Thus,

Y R

 

2.8

P1 1

 P2  2  G1G2 G3

1  G1G2 H 1

 G1G4

 G2 G3 H 2  G1G2 G3  G1G4  G4 H 2

STATE SPACE EQUATIONS

State space approach is an alternative method for representing physical system. In order to use this approach, we have to limit our approach to linear, time-invariant systems or system that can be linearized by the methods we have covered previously.

In state space method, the models are constructed in the time domain. This means we can work directly with the governing differential equations to model, analyze and design a wide range of system. In contrast, classical control design practices looking at the frequency domain output to interpret system’s physical dynamics. With the arrival of space exploration, requirements for control systems increased in scope. Hence the use of classical control design seems inadequate.

Many systems do not have just a single input. Multiple-input, multiple-output systems can be compactly represented in state space with a model similar in form and complexity to that used for single-input, single-output systems. To address the multiple input and output system a convenient matrix based is used in representing the state space. In addition, the state space approach is also attractive because of the availability of numerous state-space software packages for the personal computer

55

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The table below outlines the advantages and disadvantages of state space models

ADVANTAGES Multiple input / output models are now possible

DISADVANTAGES Difficult to examine robustness (stability margins)

Possible to minimize “error critera” (optimal control) Possible to examine stability in more depth

More work than classical control for “simple” problems

Ideally suited to computer-based design and analysis

2.8.1

“optimal” systems require “optimal” error criteria

Definition of state space terms State of a system

-

State Variables

-

State Vector

-

State Space

-

State trajectory

-

A set of quantities which completely determine the evolution of the response of a system (in the absence of external inputs) Set of variables that define the state. These variables are not unique. For example x1, x2,…. The (column) vector of the nth state variables: =[ 1 ] 2 Note: system is of order n (i.e it is described by an nth order D.E. The n-dimensional space in which the components of the state vector are the co-ordinate axes. The path in state space produced by the state vector as it changes with time.

     …

Note: The selection of state variables is not unique. In the first instance, it is often reasonable to choose something with “physical meaning”, often something associated with system “energy”

56


State space model

We begin our state space equation with a state equation. A state equation consist of the state equation and output equation as follows:

   =

=

+

+

State Equation Output Equation

Now A, B, C and D are all matrices involved in a state space equation A = (n x n) state matrix that describes “internal (homogenous) motion B = (n x r) input matrix that describes how r inputs affect n states C = (m x n) output matrix that describes how n states contribute to m outputs D = (m x r) direct transmission matrix that describes how r inputs are fed through to m outputs.

LECTURER’S NOTES:

57

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam LECTURER’S NOTES:

58


CHAPTER 3.0

SYSTEM PERFORMANCE ANALYSIS

The ability to adjust the transient and steady-state response of a control system is a beneficial outcome of the design of feedback systems. Since time is used as an independent variable in most of control systems, it is usually of interest to evaluate the state and output responses with respect to time, or simply the time response.

In the analysis problem, we will use selected input signals to test the response of control systems. This response will be characterized by a selected set of response measures. In this chapter, we will strive to delineate a set of quantitative performance measures that adequately represent the performance of the control systems.

3.1

Time Response and Test Signals

The time response of a control system is usually divided into two parts: the transient response and the steady-state response. Let y(t) denote the time response of a continuous-data system; then, in general, it can be written as y(t) = y t(t) + y ss(t)

(3.1)

where y t(t) denotes the transient response and y ss(t) denotes the steady-state response.

In control systems, the transient response is defined as the part of the time response that goes to zero as time becomes very large. Thus y t(t) has the property

lim yt (t )  0 t 

(3.2)

The steady-state response is simply the part of the total response that remains after the transient has died out. All real stable systems exhibit transient phenomena to some extent before the steady state is reached.

In the design problem, specifications are usually given in terms of the transient and steady-state performance, and controllers are designed so t hat the specifications are all met by the design system.

59


Since it is difficult to design a control system that will perform satisfactorily for all possible forms of input signals, it is necessary, for the purpose of analysis and design, to assume some basic types of test signals properly for the prediction of the system's performance to other more complex inputs.

3.1.2

Step-Input Function

The step-function input represents an instantaneous change in the reference input. The mathematical representation of a step function of magnitude A is

t  0

 A r (t )   0

t  0

Mathematically, r(t) = Aus(t), where us(t) is the unit-step function. The step function is shown in Fig. 3.1(a).

3.1.3

Ramp-Input Function

The ramp function is a signal that changes constantly with time. Mathematically, a ramp function is represented by

r (t )  Atu s (t ) where A is a real constant. The ramp function is shown in Fig. 3.1(b).

3.1.4

Parabolic-Input Function

The parabolic function represents a signal that is one order faster than the ramp function. Mathematically, it is represented by

r (t )  The parabolic function is shown in Fig. 3.1(c).

2

At

2

u s (t )

60

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Fig. 3.1 shows the three time-domain test signals.

Figure 3.1: Test input signals: (a) Step, (b) Ramp, (c) Parabolic.

3.2

First & Second Order System: Transient & Steady State Response

For linear control systems, the time response is characterized by using the unit-step input. The response of the control system to the unit-step input is called the unit-step response. Fig. 3.2 illustrates a typical unit-step response of a linear control system. With reference to the unit-step response, the following performance criteria (parameters) are defined:

1. Maximum overshoot: Let y max denotes the maximum value of y(t) and y ss be the steady-state value of y(t) and y max  y ss. The maximum overshoot of y(t) is defined as, Maximum overshoot = y max − y ss

Percentage of maximum overshoot



maximum overshoot y ss

 100%

(3.3)

2. Delay time: The delay time, t d is defined as, the time required for the step response to reach 50% of its final value. 3. Rise time: The rise time, t r is defined as, the time required for the step response to rise from 10 to 90 percent of its final value. 4. Settling time: The settling time, t s is defined as, the time required for the step response to reach and stay within a specified percentage (5%) of its final value.

61


Figure 3.2: Step response of a control system.

Analytically, these quantities are difficult to establish, except for simple systems that are lower than the third order.

3.2.1

Transient Response of a Prototype of Second-Order Systems

Although it is true that second-order control systems are rare in practice, their analysis generally helps to form a basis for the understanding of analysis and design of higher-order systems, especially the ones that can be approximated by second-order systems.

Consider that a second-order control system with unity feedback is represented by the block diagram shown in Fig. 3.3. The open-loop transfer function of the system is

G( s) 

 n2 ss  2 n 

(3.5)

62

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam where ζ and  n are real constants. The closed-loop transfer function of the system is

Y ( s) R( s)



 n2 s2

(3.6)

 2 n s   n2

The characteristic equation of the prototype of the second-order system is obtained by setting the denominator of Eq. (3.6) to zero

(s)  s 2  2 n s   n2  0

(3.7)

The system is stable (Bounded output for bounded input) if the roots of the characteristic equation is located on the left half of s-plane, and marginally stable (Oscillation for a bounded input) if the characteristic equation has simple roots on the imaginary axis with all other roots on the left half of splane. For an unstable (Unbounded output for any bounded input) system, the characteristic equation has at least one root on the right half of the s-plane or it has a repeated j  roots.

Figure 3.3: A prototype of a second-order control system.

For a unit-step input, R(s) = 1/s, the output response is given as

Y ( s) 

 n s(s

2

2

 2 n s   n

2

(3.8)

)

By taking inverse Laplace transform, we obtain the unit step response of the control system

y(t )  1 

e

 nt

1  

2



sin  n 1   2 t  cos 1 



t  0

(3.9)

63

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Fig. 3.4 shows the unit-step response of the second-order system for various values of  . It may be noted that the response becomes more oscillatory with larger overshoot as  decreases.

Figure 3.4: Unit-step response of a second-order system with various ζ values.

3.2.2

Damping Ratio and Damping Factor

The effects of the system parameters ζ and  n on the step response y(t) can be studied by referring to the roots of the characteristic equation in Eq. (4.7). The roots can be expressed as

s1 , s2

  n  j n    j

1   2

(3.10)

where

 = ζ n and

(3.11)

64


   n 1   2

(3.12)

The physical significance of ζ and  is now investigated. As seen from Eq. (4.9), the factor  =  n appears as a constant multiplied by t in the exponential term of the response y(t). Therefore,  controls the rate of rise or decay of the unit-step response y(t). In other words,  controls the "damping" of the system and is called the damping factor .

The inverse of  , 1/  is proportional to the time constant of the system. When  = 1, the oscillations disappear and the system is said to be critically damped. Under this condition,  =  n. Thus, we can regard  as

Damping ratio,  

  n



actual damping factor damping factor at the critical damping

(3.13)

When  < 1, the system is under-damped and when  > 1, the system is over-damped.

3.2.3

Natural Undamped Frequency

The parameter  n is defined as the natural undamped frequency. As seen from Eq. (3.10), when  = 0, the roots of the characteristic equation are imaginary. Thus, the unit-step response of the system becomes purely oscillatory with angular frequency of  n. For 0 <  < 1, the imaginary parts of the roots have the magnitude of the actual (damped) frequency of oscillation. Thus

   n 1   2 Fig. 3.5 illustrates the relationships between the location of the roots of the characteristic equation and

 , ζ , and  n.

65


Figure 3.5: The relationships between the location of the roots of the characteristic equation and  , ζ , and  n.

The effect of the roots of the characteristic equation on the damping of the second-order system is illustrated in Fig. 3.6.

66


Figure 3.6: Step-response comparison for various locations of the roots of the characteristic equation in the s-plane.

67


Analytical Expression for Maximum Overshoot

By taking the derivative of Eq. (3.9) with respect to time t and setting the result to zero, we get

dy(t ) dt



 n 1  

2

 n 1   2 t  n

e

 nt

sin  n . 1   2 . t

(3.14)

n  0,1,2,...

From which we get

t 

n

 n 1  

n  0,1,2,...

2

(3.15)

For the unit-step responses shown in Fig. 3.4, the first overshoot is the maximum overshoot. This corresponds to n = 1 in Eq. (4.15). Thus, the time at which the maximum overshoot occurs is

t max





(3.16)

 n 1   2

With reference to Fig. 3.4, the overshoots occur at odd values of n, that is, n =1, 3, 5, …, and undershoots occur at even values of n.

The magnitude of the overshoots and undershoots can be determined by subistituting Eq. (3.14) into Eq. (3.9). This results in y(t)max or min . Therefore

maximum overshoot  y max

 1  e  / 1

2

(3.17)

and the percentage of maximum overshoot is

percentage of maximum overshoot

 100e - / 1-

2

(3.18)

The relationship between the percent maximum overshoot and the damping rat io, as given in Eq. (3.18), is plotted in Fig. 3.7.

68


Figure 3.7: The relationship between the percent maximum overshoot and the damping ratio.

3.2.5

Delay Time and Rise Time

It is more difficult to determine the exact analytical expressions of the delay time t d, rise time t r , and settling time t s. However, we can utilize the linear approximation

t d 

1  0.7

 n

0    1.0

(3.19)

The plot of  nt r versus ζ is shown in Fig. 3.8. This relation can be approximated by a straight line over a limited range of ζ .

t r 

0.60  2.16

 n

0    1

(3.20)

69


Figure 3.8: Normalized rise time versus ζ for the prototype second-order system.

From this discussion, the following conclusions can be made: 1. t r and t d are proportional to ζ and inversely proportional to  n. 2. Increasing (decreasing) the natural undamped frequency  n will reduce (increase) t r and t d .

The settling time t s can be approximated as

t s



3

 n

(3.21)

We can summarize the relationships between t s and the system parameters as follows: 1. For ζ < 0.69, the settling time is inversely proportional to ζ and  n. A practical way of reducing the settling time is to increase  n while holding ζ constant. 2. For ζ > 0.69, the settling time is proportional to ζ and inversely proportional to  n. Again, t s can be reduced by increasing  n.

70


STABILITY & PERFORMANCE SPECIFICATIONS – ROUTH-HURWITZ STABILITY TEST

The discussions in the preceding sections lead to the conclusion that the stability of a linear timeinvariant system can be determined by checking on the location of the roots of the characteristic equation. When the system parameters are all known, the roots of the characteristic equation can be solved by means of a root-finding computer program.

For design purposes, there will be unknown or variable parameter embedded in the characteristic equation, and it will be feasible to use the root-finding programs. The method outlined below is well known for the determination of stability of a LTI system without involving root solving.

3.3.1

Routh-Hurwitz Criterion

The Routh-Hurwitz criterion represents a method of determining the location of zeros of a polynomial with constant real coefficients with respect to the left and right half of the s-plane, without actually solving for the zeros.

Consider that the characteristic equation of a linear time-invariant SISO system is of the form n F ( s)  an s

 an1s n1    a1s  a0  0

(3.25)

where all the coefficients are real. In order that Eq. (3.25) does not have roots in the right half of splane, it is necessary and insufficient that the following conditions hold: 1. All the coefficients of the equation have the same sign 2. None of the coefficients vanishes

However, these conditions are not sufficient, for it is quite possible that an equation with all its coefficients nonzero and of the same sign still will not have all the roots in the left half of the s-plane.

The first step in the Routh-Hurwitz criterion is to arrange the coefficients of the Eq. (3.25) as follows:

71


s s

n n 1

an

an 2

an 4



an 1

a n 3

a n 5



Further rows of the schedule are then completed as follows:

sn

an  2

an  4



an  3

an 5



bn 1

bn 3

bn 5



3 s n  cn 1

cn  3

cn  5









an

s n 1 an 1 s

n2





s0

hn 1

where

bn1   bn3   cn1  

1

an

a n 2

an1 an1

a n 3

an

an 4

1

an1 an1

a n 5

1 an1

an3

bn1 bn1

bn3

and so on.

Once the Routh's tabulation has been completed, we investigate the signs of the coefficients in the first column of the tabulation. The roots of the equation are all in the left half of the s-plane if all the elements of the first column of the Routh's tabulation are of the same sign. The number of changes of signs in the elements of the first column equal the number of roots with positive real parts or in the righthalf s-plane.

72


Example 3.1: Consider the equation

s  2s  1s  3  s3  4s 2  s  6  0 This equation has one negative coefficient. Thus, we know without applying Routh's test that not all the roots of the equation are in the left-half s-plane. In fact, from the factored form of the equation, we know that there are two roots in the right-half s-plane, at s = 2 and s = 3. For the purpose of illustrating, the Routh's tabulation is made as follows:

s

3

s

2

4 6

s

1

2.5

0

6

0

1

s0

1

Since there are two sign changes in the first column of the tabulation, the equation has two roots located in the right-half s-plane.

Example 3.2: Consider the equation

2s 4

 s 3  3s 2  5s  10  0

Since this equation has no missing terms and the coefficients are all of the same sign, it satisfies the necessary conditions for not having roots in the right half or on the imaginary axis of the s-plane. However, since these conditions are necessary but not sufficient, we have to check the Routh's tabulation.

s4

2

3

10

s

3

1

5

0

s

2

7

10

0

s

1

6.43

0

0

s

0

10

0

0

Since there are two changes in the first column of the tabulation, the equation has two roots in the right half of the s-plane.

73




Special Cases When Routh's Tabulation Terminates Prematurely

Depending on the coefficients of the equation, the following difficulties may occur that prevent the Routh's tabulation from completing properly: 1. The first element in any one row of Routh's tabulation is zero, but the others are not. 2. The elements in one row of Routh's tabulation are all zero.

In the first case, we replace the zero element in the first column by an arbitrary small positive number  , and then proceed with Routh's t abulation. This is illustrated by the following example:

Example 3.3: Consider the characteristic equation of a linear system:

s4

 s 3  2s 2  2s  3  0

Since all the coefficients are nonzero and of the same sign, we need to apply the Routh-Hurwitz criterion. Routh's tabulation is carried out as follows:

s

4

1

2

3

s

3

1

2

0

s2

0

3

2

1

Since the first element of the s row is zero, the element in the s row would all be infinite. To overcome 2

this difficulty, we replace the zero in the s row by a small positive number  and then proceed with the tabulation. 2



1



s s s

0

3

3 3



0 0

Since there are two sign changes in the first column of Routh's tabulation, the equation has two roots in the right-half s-plane.

74


In the second special case, when all the elements in one row of Routh's tabulation are zeros before the tabulation is properly terminated, it indicates that one or more of the following conditions may exist. 1. The equation has at least one pair of real roots with equal magnitude but opposite signs. 2. The equation has one or more pairs of imaginary roots. 3. The equation has pairs of complex-conjugate roots forming symmetry about the origin of the splane (e.g. s = -1  j1, s = 1  j1).

The situation with the entire row of zeros can be remedied by using the auxiliary equation A(s) = 0, which is formed from the coefficients of the row just above the row of zeros in Routh's tabulation. The roots of the auxiliary equation also satisfy the original equation. To continue with Routh's tabulation when a row of zeros appears, we conduct the following steps: 1. Form the auxiliary equation A(s) = 0 by use of the coefficients from the row just preceding the row of zeros. 2. Take the derivative of the auxiliary equation with respect to s; this gives dA(s)/ds = 0. 3. Replace the row of zeros with the coefficients of dA(s)/ds = 0. 4. Continue with Routh's tabulation in the usual manner.

Example 3.4: Consider the following characteristic equation of a linear control system:

s5

 4s 4  8s 3  8s 2  7s  4  0

The Routh's tabulation is

s

5

1

8

7

s

4

4

8

4

s

3

6

6

0

s

2

4

4

s

1

0

0

75


A(s) = 4s + 4 = 0 The derivative of A(s) with respect to s is dA(s)/ds = 8s = 0 From which the remaining portion of the Routh's tabulation is

s

1

8

s

0

4

0

Since there are no sign changes in the first column, the system is stable. Solving the auxiliary equation A(s) = 0, we get the two roots at s = j and s = -j , which are also two of the roots of the characteristic equation. Thus, the equation has two roots on the j - axis, and the system is marginally stable. These 1

imaginary roots caused the tabulation to have an entire row of zeros in the s row.

Example 3.5: Consider that a third-order control system has the characteristic equation

s 3  3408.3s 2

 1204 103 s  1.5 107 k  0

Determine the crucial value of k for stability.

s

3

s

2

s

1

s

0



1

1204  103

3408.3 7 3 1.5  10 k  3408  1204  10

1.5  10 k 7

0

3408 7 1.5  10

For the system to be stable, all the coefficients in the first column must have the same sign. This lead to the following conditions:



1.5  107 k  410.36  107 3408

Therefore, the condition of k for the system to be stable is

0  k  273.57

0

76

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam If we let k = 273.57 , the characteristic equation will have two roots on the j - axis. To find these roots, we substitute k = 273.57 in the auxiliary equation, as follows:

A( s)  3408.3s 2

 4.1036 109  0

which has roots at s = j1097.27 and s = -j1097.27 .

Thus if the system operate with k = 273.57 , the system response will be an undamped sinusoid with a frequency of 1097.27 rad/sec.

3.4

STEADY STATE RESPONSE – STEADY STATE ERROR

One of the objectives of most control systems is that the system output response follows a specific reference signal accurately in the steady state. Steady-state error is the difference between the output and the reference in the steady state. S teady-state errors in control systems are almost unavoidable and generally derive from the imperfections, frictions, and the natural composition of the system. In the design problem, one of the o bjectives is to keep the steady-state erro r below a certain tolerable value.

3.4.1

Definition of the Steady-State Error with respect to System Configuration

Let us refer to the closed-loop system shown in Fig. 3.11, where r(t) is the input, e(t) the actuating signal, and y(t) is the output. The error of the system may be defined as:

Figure 3.11: Closed-Loop Control System.

77


e(t )  reference signal  y(t )

(3.26)

where the reference signal is the signal that the output is to track. When the system has unity feedback (i.e. H(s) = 1), the error is simply

e(t )  r (t )  y(t )

The steady-state error is defined as

ess

 lim e(t )  lim sE (s) t 

 lim s0

s 0

sR( s)

(3.27)

1  G( s )

Clearly, ess depends on the characteristics of G(s). More specifically, ess depends on the number of poles that G(s) has at s = 0. This number is known as the system type. Fig. 3.12 shows steady state errors for different input functions.

78


Figure 3.12: Steady-state errors (a) step input, (b) ramp input

Now let us investigate the effects of the types of inputs on the steady-state error.

79


Steady-State Error of System with a Step-Input Function

When the input r(t) to a control system with unity-feedback is a step function with magnitude A, then R(s) = A/s and the steady-state error is written from Eq. (4.27),

ess

 lim s 0

sR( s)

1  G( s)

 lim s 0

A

1  G( s)



A

1  lim G( s)

(3.28)

s0

For convenience, we define

k p

 lim G( s) s 0

as the step-error constant . Then Eq. (4.28) becomes

ess



A

1  k p

We can summarize the steady-state error due to a step-function input as follows:

3.4.3



Type 0 system:



Type 1 or higher system:

ess



A

1  k p

= constant

ess = 0

Steady-State Error of System with a Ramp-Input Function

When the input to the unity-feedback control system is a ramp function with amplitude A,

r (t )  Atu s (t ) where A is a real constant, the Laplace transform of r(t) is

R( s ) 

A s2

The steady-state error is written using Eq. (4.27) as follows:

(3.29)

80


ess

A

 lim

s  sG( s)

s 0



A

lim sG( s)

(3.30)

s 0

We define the ramp-error constant as

k v

 lim sG( s ) s 0

Then Eq. (3.30) becomes

ess



A k v

(3.31)

The following conclusions may be stated with regard to the steady-state error of a system with ramp input:

3.4.4



Type 0 system:

ess = 



Type 1 system:

ess = A/k v = constant




ess = 0

Steady-State Error of System with a Parabolic Input

When the input is described by the standard parabolic form

r (t ) 

2

At

2

u s (t )

The Laplace transform of r(t) is

R( s ) 

A s

3

The steady-state error of the system is

ess 

A 2

lim s G( s) s 0

(3.32)

81

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Defining the parabolic-error constant as

k a

 lim s 2G(s)

(3.33)

s0

the steady-state error becomes

ess



A

(3.34)

k a

The following conclusions are made with regard to the steady-state error of a system with parabolic input: 

Type 0 system:

ess = 



Type 1 system:

ess = 



Type 2 system:

ess = A/k a = constant




ess = 0

Example 3.5: Find the steady state errors of the following system

G ( s) 

k ( s  3.15) s( s  1.5)( s  0.5)

H(s)



1

It is clear that this system is a type 1 system. The steady-state errors are: Step input

Step-error constant, k p = 

ess = A/1+k p = 0

Ramp input

Ramp-error constant, k v = 4.2k ess= A/k v = A/(4.2k)

Parabolic input Parabolic-error constant, k a = 0 ess = A/k a = 

3.4.5

Steady-State Error for Non-unity Feedback System

For non-unity feedback control, we usually find the equivalent unity-feedback system, as shown in Fig. 3.13.

82


Figure 3.13: Forming an equivalent unity feedback for nonunity feedback system.

We have to take into consideration, that the above steps require that input and output of the same units. The following example summarizes the concepts of steady-state error, system type, and the steady state errors.

Example 3.6: For the system shown in Fig. 4.14, find the system type and the steady state error for the unit step function. Assume input and output units are the same.

83

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Figure 3.14: Nonunity feedback control system for Example 3.6.

The first step in solving the problem is to convert the system of Fig. 3.14 into an equivalent unity feedback system. Using the equivalent forward transfer function of Fig. 3.13(e) along with

G(s) 

100 s ( s  10)

and

H ( s) 

1 s5

we find

Ge ( s) 

G( s)

1  G ( s ) H ( s)  G ( s)



100( s  5) s

3

 15s 2  50s  400

Thus, the system is type 0, and

k p

Ge ( s )   lim s

100  5

0

 400

The steady-state error is

ess



1 1  k p

 4



5 4

84


FREQUENCY RESPONSE ANALYSIS

In practice, the performance of a control system is m easured more realistically by its time-domain characteristics. The reason is that the performance of most control systems is judged based on the time response due to certain test signals.

In design problems, there are no unified methods of arriving at a designated system that meets timedomain performance specifications. On the other hand, in frequency domain, a wealth of graphical and other techniques are available that are useful for system analysis and design, irrespective of the order of the system.

It is important to realize that there are corr elating relations between the frequency- and time-domain performances in linear system so that time-domain properties of the system can be predicted based on the frequency –domain characteristics. With these in mind, we shall study the frequency response analysis of control systems.

3.5.1

Frequency Response of a System

It is well known from linear system theory that, when the input to a linear time invariant system is sinusoidal with amplitude R and frequency  o, i.e.,

r(t)  R sin ωot the steady-state output of the system, y(t), will be a sinusoid with the same frequency  o, but possibly with different amplitude and phase; i.e .

y(t)  Y sin (ωot  φ) where Y is the amplitude of the output sine wave and  is the phase shift.

Let the transfer function of a SISO system be M(s); the output Y(s) and the input R(s) are related through

Y(s)  M(s)R(s)

85

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam For sinusoidal steady-state analysis, we replace s by j , and equation (6.3) becomes

Y(j )  Μ  j ) R j ) By writing Y(j  ) and M(j  ) as (similar expression for R j ) also):

Y ( j )  Y ( j ) Y ( j ) M ( j )  M ( j )  M ( j ) Y ( j )

 M ( j ) R( j )

and the phase relation:

Y ( j )   M ( j )   R( j )

Thus, for the input and output signals described by equations (6.1) and (6.2),

Y  M ( j o ) R

   M ( j o ) Thus, by knowing the transfer function M(s), the frequency response of the system can be obtained. The frequency response of the loop transfer function G(s)H(s) [G(s) if H(s) is unity] can be plotted in several ways. The two commonly used r epresentations are: a. Bode diagram, or Logarithmic plot. b. Polar plot, or Nyquist plot.

86


Frequency Response – Bode Diagram

A Bode diagram consists of two graphs. One is a plot of the logarithm of the magnitude of a sinusoidal transfer function; the other is a plot of the phase angle; both are plotted against the frequency on a logarithmic scale.

where the base of The standard representation of the logarithmic magnitude of G(j  ) is 20 log |G(j  )|, the logarithm is 10. The unit used in this representation is the decibel (dB). The curves are drawn on a semilog paper, using the log scale for frequency and linear scale for either magnitude (in dB) or phase angle (degrees).

The main advantage of Bode diagrams is that the multiplication of magnitudes can be converted into addition. Furthermore, a simple asymptotic method is av ailable for sketching the approximate curve. Should the exact curve be desired, corrections could be made easily to these basic asymptotic plots.

In Bode diagrams, the frequency ratios are expressed in terms of octaves or decades. An octave is a frequency band from  1 to 2 1, where  1 is any frequency. A decade is a frequency band from  1 to 10 1, where  1 is any frequency.

Basic Factors of G(j )H(j ):  ) are: The basic factors that very frequently occur in an arbitrary open-loop transfer function G(j  )H(j a. Constant gain, K . n

b. Zeros and poles at the origin,  j  . c.

1

Simple zeros and poles, 1  j T  .



d. Quadratic factors, 1  2  j /  n    j /  n 

a.

Real Constant: G(s)H(s) = K

G(jω)H(jω) = K Magnitude: |G(jω)H(jω)| (dB) = 20 log 10 |K| (dB). Phase angle: G(jω)H(jω) = 0°.

2

 . 1

87

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The Bode plot for any value of K is shown in Fig. 6.1.

Figure 6.1: Bode plot for gain K .

b. Poles and zeros at the origin: G(s)H(s) = s

±n

n

For s : n

G(jω)H(jω)=(jω)

Magnitude: |G(jω)H(jω)| (dB) = 20n log10 | jω| (dB) = 20n log10 ω (dB)

(6.11)

Phase angle: G(jω)H(jω) = 90n° (a constant).

-n

For s : -n

G(jω)H(jω)=(jω)

Magnitude: |G(jω)H(jω)| (dB) = −20n log10 | jω| (dB) = −20n log10 ω (dB)

(6.12)

Phase angle: G(jω)H(jω) = −90n° (a constant).

The Bode magnitude plots are a straight line in semi log coordinate. The slope of the line is ±20 n dB/decade i.e. the magnitude change by ±20n dB for the frequency change of 10 times. The straight line 0

passes through 0 dB at  = 1. The phase angle (  ) of ±j  is constant and equal to ±90 .The Bode plots are shown in Fig. 6.2.

88


Figure 6.2: Bode diagrams for (a)G(j  ) = 1/j  (b)G(j  ) = j .

c.

Simple zeros and poles: G(s)H(s) = (1+sT)

G(jω)H(jω)= (1+ jωT)

±1

±1

Magnitude: |G(jω)H(jω)| (dB) = ±20 log10 |1 + jωT| (dB) 2 2

= ±20 log10 √*1 + ω T ] (dB)

(6.11)

89

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam To obtain asymptotic approximation we consider both very large and very small values of  . For low frequencies, such that  T << 1, the log magnitude may be approximated by

 20 log

1   2T 2

 20 log1  0 dB

1   2T 2

 20 log T dB

For high frequencies, such that  T >> 1,

 20 log

At  =1/T , log magnitude = 0 dB while at  =10/T , log magnitude = ±20 dB. Thus, the value of

 20 log T increases/decreases with 20 dB/decade. Hence, the magnitude plot can be approximated by two straight-line asymptotes, one a straight line at 0 dB for the frequency range 0 <  < 1/T and the other a straight line with slope ±20 dB/decade for the frequency range 1/T <  < . The frequency,

 =1/T , at which the two asymptotes meet is called the corner frequency or break frequency.

Phase angle: G(jω)H(jω) =

At corner frequency,

 tan 1  T .

G(jω)H(jω)

= ±45. The phase plot can be approximated by a straight line passing

through 0 at one decade below corner frequency and ±90  at one decade above corner frequency. The Bode plots are shown in Fig. 6.3 and Fig. 6.4.

An advantage of the Bode diagram is that for reciprocal factors, for example the factor 1/(1+ j T ), the log-magnitude and phase angle curves need only be changed in sign, since

20 log

1 1  j T

(1+ j T ) .

 20 log1  j T

and phase angle of 1/(1+ j T ) =

 tan 1  T =  (phase angle of

90


Figure 6.3: Bode plot for (1+ j T ).

91


Figure 6.4: Log-magnitude curve (with asymptotes) for 1/(1+ j T ).



d. Quadratic factors: G ( s) H ( s)  1  2 s /  n   s /  n 

2



G ( j ) H ( j )  1  2  j /  n    j /  n 

2



1



1

Magnitude: 2

|G(jω)H(jω)| (dB) = ±20 log10 |1 + 2ζ(jω/ωn )+ (jω/ω n ) | (dB) 2

= ±20 log10

  2     1  2    2    n    n 

2

(dB)

(6.12)

92

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam If ζ > 1, this quadratic factor can be expressed as a product of two first-order factors with r eal zeros/poles. If 0 < ζ < 1, this quadratic factor is the product of two complex-conjugate zeros/poles. The asymptotic frequency response curves can be obtained as follows.

For low frequencies such that  / n << 1, the log magnitude becomes ±20 log 1 = 0 dB. The low frequency asymptote is thus a horizontal line at 0 dB. For high frequencies such that  / n >> 1, the logmagnitude becomes

 20 log

 2 2 n



 40 log

  n

dB. The equation for the high frequency asymptote is a

straight line with a slope of ±40 dB/decade.

The frequency  n is the corner frequency. The two asymptotes just derived are independent of the value of ζ . Fig. 6.5 shows exact curves with the straight-line asymptotes and the exact phase angle curves.

Phase angle of the quadratic factor is:

    2   n  1     tan  2     1        n    

(6.13)

93


Figure 6.5: Bode plot for Eqn. (6.12) and (6.13).

94


Example 6.1: Sketch the Bode plot for the following function: G ( s) 

Solution:

G( s) 

1 s2

,

1 s2

    1 1 1    G ( j )  j  j  2 2  1   2 

Magnitude:

 1  20 log G ( j )  20 log  - 20log  2 

G( j )   tan 1

   1    2 

2

 2

-5 -10 ) -15 B d ( e d -20 u t i n g a -25 M ; ) 0 g e d ( e -20 s a ) h 1 P ( Y-40 : o T

-60 -80 -1 10

0

10

Frequency (rad/sec)

Figure 6.6: Bode plots for the system in Example 6.1.

1

10

95


General Procedure for Plotting the Bode Diagrams: 

Rewrite the sinusoidal transfer function as a product of the basic factors discussed above.



Identify the corner frequencies associated with these basic fac tors.



Draw the asymptotic log-magnitude curves with proper slopes between the corner frequencies considering all the basic factors together. The exact curve, which lies very close to the asymptotic curve, can be obtained by adding contributions from all the factors and proper corrections.



3.5.3

Phase-angle curve can be drawn by adding the phase-angle curves of individual factors.

Polar Plot (Nyquist Plot)

The polar plot of a sinusoidal transfer function G(j  ) is a plot of the magnitude of G(j  ) versus the phase angle of G(j  ) on polar coordinates as  is varied from zero to infinity. Note that, in polar plots, a positive (negative) phase angle is measured counterclockwise (clockwise) from the positive real axis. The polar plot is very often called the Nyquist plot in control system engineering. An example of such a plot is shown in Fig. 6.7. Each point on the polar plot of G(j  ) represents the terminal point of a vector at a particular value of  . In the polar plot, it is important to show the frequency graduation of the locus.

96


Example of a Polar Plot.

a. Poles and zeros at the origin: G(s)H(s) = s±n

1

G ( j ) H ( j )   j 

The polar plot of G(j  )H(j  ) = 1/ j  is the negative imaginary axis since

G  j  H ( j ) 

1 j



The polar plot of G(j  )H(j  ) = j  is the positive imaginary axis.

j





1



  900

(6.14)

97


b. Simple zeros and poles: G(s)H(s) = (1+sT) ±1

1

G ( j ) H ( j )  1  j T 

For the sinusoidal transfer function

G j  H ( j ) 

1 1  j T



1 1   T 2

2

  tan 1  T

(6.15)

the values of G(j  )H(j  ) at  = 0 and at  = 1/T are, respectively,

 

G j 0 H ( j 0)  100 and G j

1   1  1   450  H  j   T   T  2

0

If  approaches infinity, the magnitude approaches 0 and the phase angle approaches –90 . The polar plot of this transfer function is a semicircle as the frequency is varied from 0 to

.

It is shown in Fig. 6.8.

The center is located at 0.5 in the real axis and t he radius is equal to 0.5. The lower semicircle corresponds to 0     , and the upper semicircle corresponds to

     0 .

Figure 6.8: (a) Polar plot of 1/(1+j T) ; (b)Same plot in X -Y plane.

98

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The polar plot of the transfer function 1+ j T is simply the upper half of the straight line passing through the point (1, 0) in the complex plane and parallel to the imaginary axis as shown in Fig. 6.7.

. Figure 6.7: Polar plot of 1+ j T



c. Quadratic factors: G ( s) H ( s)  1  2 s /  n   s /  n 



2

G ( j ) H ( j )  1  2  j /  n    j /  n 

2



1



1

The low and high frequency portions of the polar plot of the following transfer function

G j  H ( j ) 

1 1  2  j /  n    j /  n 

2

are given, respectively, by

lim G j  H ( j )  100 and lim G j  H ( j )  0  1800  0

 

Thus, the high frequency portion is tangent to the negative real axis. The polar plots are shown in Fig. 6.8.

*Phase angle of the quadratic factor is the same as Eqn. (6.13):

99


    2     1 n    tan   2    1         n    

Figure 6.8: Polar plots of

1 1  2  j /  n    j /  n 

Next, consider the following transfer function:

G j  H ( j )  1  2  j /  n    j /  n 

  2   2    1  2   j   n    n  The low-frequency portion of the curve is:

lim G j  H ( j )  100

 0

and the high-frequency portion of t he curve is:

2

2

for ζ > 0.

100


lim G j  H ( j )  1800

 

The general shape of the polar plot is shown in Fig. 6.9.

Figure 6.9: Polar plot of 1  2  j /  n    j /  n  for ζ > 0. 2

Example 6.2: Draw polar plot of G ( s) 

1 s2

Solution: First substitute s = j  in G(s).

G ( j ) 

G ( j )

1 j  2



1

 2  4

G( j )  tan 1

 2

101


Figure 6.10: Polar plot of G(s) in Example 6.2.

Example 6.3: Draw polar plot for the system with G ( s ) 

Solution:

G ( j ) 

10 s( s  1)( s  2)

10 j ( j  1)( j  2)

G ( j )



10

 ( 2  1) ( 2  2)

G( j )  90  tan1

 2

 tan1 

102

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The plot is shown in Fig. 6.11.

Figure 6.11: Polar plot for the system in Example 6.3.

NYQUIST STABILITY TEST – THE CAUCHY CRITERION

The Cauchy criterion (from complex analysis) states that when taking a closed contour in the complex plane, and mapping it through a complex function G(s), the number of times, N, that the plot of G(s) encircles the origin is equal to the number of zeros, Z , of G(s) enclosed by the frequency contour minus the number of poles, P, of G(s) enclosed by the frequency contour. N = Z – P Encirclements of the origin are counted as positive if they are in the same direction as the original closed contour or negative if they are in the opposite direction.

When studying feedback control, we are not as interested in G(s)H(s) as in the closed-loop transfer function G(s)H(s)/[1+G(s)H(s)]

If 1+G(s)H(s) encircles the origin, then G(s)H(s) will enclose the point -1. Since we are interested in the closed-loop stability, we want to know if there are any closed-loop poles (zeros of 1+ G(s)H(s)) in the right-half plane.

103


The Nyquist Stability Criterion usually written as Z = P + N, where >

Z is the number of right hand plane poles for the closed loop system (or zeros of 1+G(s)H(s))

>

P is the number of open-loop poles (in the RH side of the s-plane) of G(s)H(s) (or poles of 1+G(s)H(s)), and

>

N is the number of clockwise encirclements of (-1,0)

“A feedback control system is stable if and only if the number of counter-clockwise encirclements of the critical point (-1,0) by the GH polar plot is equal to the number of poles of GH with positive real parts.” (Nyquist Stability Criterion Definition)

Example: •

Consider the unity feedback applied to the following system G(s)=K /[s(s+3)(s+5)]

•

The loop transfer function is G( j )H( j )=K /[s(s+3)(s+5)]|K =1,s= j 

•

The number of open-loop poles in the RH side of the s-plane, P = __

104

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam •

For stability Z = 0, therefore N must also be __

•

From Nyquist diagram it can be seen that K can be increased by ____________ before the Nyquist diagram encircles -1.

•

For marginal stability, K = _____

RELATIVE STABILITY – GAIN AND PHASE MARGIN

•

K is a variable (constant) gain

•

G(s) is the plant under consideration

Gain margin is defined as the change in open loop gain required to make the system unstable. Systems with greater gain margins can withstand greater changes in system parameters before becoming unstable in closed loop. Phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable.

105


Stability Analysis with Bode Plot

Bode plot is a very useful graphical tool for the analysis and design of linear control systems.

Advantages of the Bode Plot over Nyquist plot: 1. Gain crossover, phase crossover, gain margin, and phase margin are more easily determined on the Bode plot. 2. For design purposes, the effects of adding controllers and their parameters are more easily visualized on the Bode plot.

Identifying Marginal Values from Bode Plot

The gain margin is the difference between the magnitude curve and 0dB at the point corresponding to the frequency that gives us a phase of −180° (the phase cross over frequency, ω p).

The phase margin is the difference in phase between the phase curve and −180° at the point corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, ωg).

106


Figure 6.12: Determination of GM and PM from the Bode plot.

Example 6.4: Consider the loop transfer function given as

L( s ) 

2500 s( s  5)( s  50)

From the provided Bode diagram, find the GM and PM and corresponding frequencies.

107


Solution:

Figure 6.13: Bode diagram for Example 6.4.

Fig. 6.14 illustrates PM and GM of a stable and unstable system in Bode diagrams.

108


Figure 6.14: Phase and Gain margins for stable and unstable systems.

Stability Analysis with Nyquist Plot OPEN LOOP Gain Margin, GM

Phase Margin, M

Gain margin and phase margin

CLOSED LOOP The change in open-loop gain, expressed in dB, required at -180° to make the closedloop system unstable. A good range is 2
109


A gain of a will move the system response to the critical point If a phase shift of  degrees occurs, then the system will become unstable

110

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam CHAPTER 4.0

PID CONTROLLER

There are three broad categories of PID tuning techniques: i)

Feature-based techniques

ii) Techniques that require an analytical model iii) Optimisation (minimisation of an error criterion)

But before looking at PID tuning, we need to look at modeling of simple process dynamics. There are two common approaches:

transient response methods,

which look at the time domain characteristics of the system

response to a step or impulse

frequency response methods, which look at the response to an impulse, white noise or one or more sinusoids

Stick to transient response models and very simple frequency response for the moment.

111


Transient Response Method of Modelling

Step response modeling probably the most common approach.

Peak Overshoot

1.5

Unit Step Input

e1.05 d u1 t i 0.95 l p0.9 m A

0.5

Rise Time 0.1 0

0 Time Delay

Time

Settling Time

Step Response i)

Peak Overshoot 

(peak - final value)/final value*100%



Measure of maximum value of response



Indication of the largest error between input and output



Increases as damping decreased



Well designed systems generally have overshoot less than 30%

ii) Rise Time 

Measure of the speed of response



Time necessary for the response to rise from 10% to 90% o f its final steady state error

iii) Time Delay 

Time for system to show any response

112

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam iv) Delay Time 

Time necessary for the step response to reach some value (often 50%) of the steady state value. Not to be confused with Dead time = Time Delay

v) Settling Time 

The time taken for the step response to decrease and stay within a specified range of the final value.



Often 1%, 2% or 5%

vi) Decay Ratio 

Defined as the ratio between two consecutive maxima of the error for a step change in the set-point



The value d=1/4, which is called quarter amplitude damping, is used traditionally but is often too high



For a second order system given by

CL( s)

G( s)

  

1  G ( s)  n2 s

2

 2 n s   n2 1

1  2 s /  n  ( s /  n ) 2

The decay ration is given by

d  e

 2 / 1 2

113


Step Response Model 1 – 2 parameter model

18 16 G(s) =

14

a -sL e Ls

12 e d u10 t i l p m8 a

6 4 1 step response for e-2s s

2 2

a

4

6

8

L

10 12 time

Model of Integrator with time delay (first order response) 

Response rises linearly over time



characterised by two parameters



L - essentially the “dead time”



a - where a/L is the slope.

14

16

18

20

114


Step Response Model 2 – 3 parameter model (first order)

T

2

k

1.5

G(s) =

e d u t i l p1 m a

k -sL e 1+sT

0.5 step response for 0 0

1

2

3

L

4

2 -2s e 1+s

5 6 time

7

8

9

10

Response rises smoothly and is stable: 

characterised by three parameters, gain k, time constant T and time delay L.



Most common model for PID tuning



Tangent to step response must be drawn at the location of the largest slope



Need some alternatives that are more robust

Problem:

115


Step Response Model 3 – Alternative 3 parameter model (first order)

T k

2

1.5

G(s) = .63k

e d u t i l p1 m a

k -sL e 1+sT

0.5 step response for 0 0

1 L

2

3

4

5 time

2 -2s e 1+s 6



Response is 63% of final value at t = T



63% =1-e



Still some sensitivity to high frequency noise

-1

7

8

9

10

116


Step Response Model 4 – Another alternative 3 parameter model (first ord er)

L+T

2

k

1.5

G(s) =

e d u t i l p1 m a

A1

k e -sL 1+sT

0.5 A2 0 0

1

2

3

4

L 3 Parameter Model Alternative: 

Effective at removing noise



Let (L+T) = A 1/k



Can show that T = (A2/k)e

1

5 time

6

7

8

9

10

117


Step Response Model 5 – Second order response

Second order response generally has the following features: 

Oscillatory



3 parameters - need k, w, z

Tp 3.5 d = e2 / e1

e1-e2=o(1-d)

3 2.5 e d 2 u t i l p 1.5 m a

k

G(s) = 1

k2 s2 + 2s + 2

0.5 0 0

2

4

6

8

10 12 time

14

16

18

20

-2 2 1/2 d = e (1- )

Tp =

or

2

 (1-2)1/2

= or

1 2 1/2

(1 + (2/ log d) ) 2 = Tp(1-2)1/2

Time delay can be added to this model and determined as done previously in the tutorial.

118


Frequency Response Modelling

Some tuning formulas are based upon the frequency response of the plant

Parameters of interest are the ultimate gain K u and the ultimate period T u

Find these by first closing the loop and then disabling the integral and derivative parts of the controller (T d=0, T i =very large), and increasing the proportional gain until the system begins to oscillate. Gain at this point = K u and period of oscillation = T u

Controller

ysp

+

 -

K

Plant

G(s)

y

e(s)

Problems with looking for ultimate gain in this way 

Often time consuming in practice, requiring several trials



Process can be detrimental to plant equipment and product quality



Can easily mistake other responses for t he ultimate gain a.

small amplitude “limit cycles” due to valve friction or hysteresis

b. large amplitude oscillations due to actuator saturation

Alternative method for finding ultimate gain and period: 

for the closed-loop system, find controller gain that produces a 1/4 decay ratio (overshoot of one peak is 25% of the peak before it). Let this be K 25%



K u = 2 K 25%



Period of oscillation, T 25%, will be approximately T u (a little longer in practice, but close enough)

119


Nyquist Plot Frequency Reponse Modelling

On a Nyquist plot: 

Look at open-loop response



static gain K p = point on plot where w=0



K u = -1 divided by ultimate point



T u = 2p divided by w at ultimate point

ultimate point -1 +

Im G(j)

=0

Re G(j)

120


Simple Tuning Law

There are several tuning law can be implemented in order to obtain the desired response. These tuning law are as follows: i)

Ziegler – Nichols 

Step Response



Ultimate Gain method



Generalised ZN

ii) Chien, Hrones and Reswick Method iii) Cohen Coon Method

4.3.1

Ziegler – Nichols Method

Ziegler-Nichols rule was first presented in 1942. This tuning law was developed empirically based on large number of cases. It can be said as a standard starting point.

There are some drawbacks of using this rule mainly because it needs additional manual tuning and not particularly robust.

Now we will employ this method for both step response and frequency response modeling.

121

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam 4.3.1.1 Z-N method for Step Response modeling

Uses 2 parameter model “a” and “L”

CONTROLLER TYPE P

1/a

PI PID

0.9/a 1.2/a

K

Ti

Consider the plant G(s) = (s+1)

a= 0.218

From step response a = _____ L = _____ PID controller

-3

L= 0.806

Tp

4L 3L 2L

Tp is the estimate of the period of the closed loop system

Example:

Td

L/2

5.7L 3.4L

122

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam K = ____, T i = ____ and T d = ____ Overshoot in setpoint response is too large

123

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam 4.3.1.2 Z-N Method for Frequency Response Modelling

PID settings based upon ultimate gain, K u, and ultimate period, T u

Aims to achieve effective disturbance rejection, and acceptable set point following

CONTROLLER TYPE P

0.5Ku

PI PID

0.4Ku 0.6Ku

K

Example: Consider the plant G(s) = (s+1)-3

Ti

Td

Tp

Tu 0.8Tu 0.5Tu

0.125Tu

1.4Tu 0.85Tu

124

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam From the Nyquist diagram Ultimate Gain, K u = _____ Ultimate Period, T u = _____ = _____ PID controller K = ____ T i = ____ and T d = ____

4.3.2

The Chien, Hrones and Reswick Method – Improved Z-N step response modeling

The CHN method is a modified Z-N step response rules which can gives better damped closed loop response.

The tuning method gives you two options depending on the desired response and they can be either: 

quickest response without overshoot, or



quickest response with 20% overshoot

125


CONTROLLER TYPE P PI PID

K

Ti

Td

0.3/a 0.6/a 0.95/a

4L 2.4L

0.42L

Option for 0% overshoot

CONTROLLER TYPE P PI PID

K

Ti

Td

0.7/a 0.7/a 1.2/a

2.3L 2L

0.42L

Option for 20% overshoot 4.3.3

Cohen-Coon Method Based on plant model

G(s) 

K 0

1  sT

e

sL

Attempts to position dominant poles that give quarter amplitude decay ratio by employing: 

Method same as the ZN rules



This minimises the SS error due to load disturbances.

For PID control, 3 poles are assigned, two complex conjugate poles and the third real pole is positioned at the same distance from the origin as the other 2 poles.

CONTROLLER TYPE

K

P

(1/a)*[1+0.35/(1-)]

PI

(0.9/a)*[1+0.92/(1-)]

PD

(1.24/a)*[1+0.13/(1-)]

PID

(1.35/a)*[1+0.18/(1-)]

Ti

Td

[(3.3-3)/(1+1.2)]*L [(0.27-0.36) / (10.87)]*L [(2.5-2)/(1-0.39)]*L

[(0.37-0.37)/(10.81)]*L

Where a = K 0L/T and  = L / (L+T ) and K 0 is the Open Loop DC Gain.

126


PID Tuning Rule of Thumb

PARAMETER K increases

SPEED

Increases

STABILITY Decreases

Ti increases

Decreases

Increases

Td increases

Does not really change

Increases

to much (Increase)

4.5

PID Tuning – A Summary

CONTROLLER TYPE P PI

Z-N Step

Z-N Nyquist

K=1/a

0.5Ku

K=0.9/a

PID

K=1.2/a

e d u t i l p m a

K=0.4Ku

Ti=3L Ti=2L

Td=L/2

K=0.6ku

Ti=0.5Pu

open loop step response

time

a L

Ti=0.8Pu Td=0.125Pu

127


step response with closed loop proportional control with gain Ku

Pu

e d u t i l p m a

time 4.5.1

Interpretation of ZN Ultimate Gain Approach

The ZN ultimate gain approach can be interpreted as shifting a point in the Nyquist curve. The technique is based around finding the “ultimate point”, where the Nyquist curve intercepts the r eal axis. 

P moves in direction of G( j w), or radially out.



I moves in direction of G( j w )/j w , or at -90 degrees to P



D moves in direction of j wG( j w), or at 90 degrees to P

ultimate point = -1/Ku -1

Im G(j )

=0 Re G(j )

+ I

P D A point on the Nyquist curve can be moved to an arbitrary position using PI, PD or PID control.

128

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam CHAPTER 5.0

ANALYSIS OF CONTROL SYSTEM STATE SPACE

The introduction of the state space representation has been discussed earlier in chapter 2.0. Please refer to the chapter for basic overview of state space representation.

In this chapter, we will cover InsyaAllah the extension of state space representation for example the conversion between the transfer function and state space equation. Also in this chapter we will gonna look at how to solve the time invariant state equation, controllability and observability.

5.1

State Space Representation Extended

The transfer function of any system can be converted to state space equation and vice versa. Consider a transfer function given by:

         −         −        −     − −   −−     −−    ( )

( )

=

( )

This system may also be represented in state space as:

=

+

=

+

Where x is the state vector, u is the input and y is the output. The Laplace Transform of the equations:

0 =

=

+

+

( )

( )

Assuming the initial conditions are zero

=

Or

( )

=

By premultiplying (

)

1

( )

to both sides of this equation, we obtain 1

=

( )

Substitute into output equation;

=

1

+

( )

129

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Comparing with the above equation, we see that

  − −     … …      … …           = (

5.1.1

Transfer Matrix

For MIMO system that the r inputs

1,

2,

1

)

+

and m outputs

.,

1

=

1, 2,

.,

define as:

1

2

2

. .

. .

=

The transfer matrix G(s) relates the output Y(s) to the input U(s), or

=

( )

Since the input vector u is r dimensional r dimensional and the ouput vector y is m dimensional, the transfer matrix is an m x r matrix. r matrix.

5.2

Converting State Space to transfer function

A modern complex system may have many inputs and outputs. Let say we have a state space representations of the following:

  − ⋯ −      − ⋯ −       ……  ……    −−  − −− −− …… −  −    −   − − −  …  −    −      +

1

1

+

+

1

+

=

+

0

1

1

+

+

+

1

Controllable canonical form 1

2

. .

=

1

0 0 . . 0

1 0 . . 0

0 1 . . 0

1

0 0 . . 1

2

1

2

. .

1

1

0 0 . + . 0 1

1

2

=

1 0

1

1 0

.

1

1 0

. .

+

1

0

130


Observable canonical form

    …… −−−  − −− −  …… −−    −−   …… −  −    −       …  −    0 1 . = . 0 0

1

2

. .

1

0 0 . . 0 0

0 0 . . 0 1

1

1 0

2

1

1

. .

2

. .

+

1

1

1 0

. . .

1

1 0

1

2

= 0

0

0

. .

1

+

0

1

Diagonal Canonical Form

Consider the transfer function system defined by equation below. In this case the denominator polynomial involves only distinct roots only.

     − ⋯ … −        ⋯    − − ……  ……   −−   …… − −        … −  −      ( )

( )

0

= =

0

+

1

+

1

+

1

1

+ 1 1

0 . . 0 0

2

. .

=

1

1

+

+

+

2

+ 2

0

+

2

+

1

+

( +

)

+

+

0 0 . . 0 0

2

. . 0 0

0 0 . . .

1

2

. .

1

1

2

=

1

2

1

. .

+

1

0

1 1 . + . . 1

131

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Jordan Canonical Form

Consider the case where the denominator polynomial involves multiple roots.

     − ⋯ −…                ⋯     −− − − ………   − …      … −        … −  −⋮      ( ) ( )

=

0

+

+

3

1

1

1

+

+

4

+

+

+

1

( +

5

)

The partial fraction expansion becomes

( )

( )

=

0

+

1

( +

1

+

)3

2

( +

1

)2

+

3

( +

1)

+

4

( +

4)

+

+

( +

)

A state space representation of this system in the Jordan canonical form is given by: 1

1

2 3 4

. . .

=

0 0 0 . . . 0

1

1

0 0

0

0 1

1

0 0 0

0

0

0 0 0 0 . . .

4

. . . 0

1

2

=

1

2

+

1

1

0

1 2 3 4

. . .

0 0 1 1 + . . . 1

132

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam LECTURER’S NOTE

133


Controllability and Observability

5.3.1

Controllability

Controllability is a test of the ability of the actuators. A system is controllable if it is possible to transfer any state with any set of initial conditions to any final state in some finite time period. Alternatively, a system is only controllable if every mode (or state) is connected to the control input.

A system is referred to as “stabilizable” so long as we can state control all unstable modes. This might mean that there are some stable uncontrollable states. Strictly speaking the dynamical system described by the pair (A;B) is said to be (state-feedback) stabilizable if there exists a state feedback u=-Kx such that

A+BK is stable.

In order to test the controllability of a LTI system, the “Controllability Matrix” must be of full rank. The Controllability matrix,

  …− =

i.e the controllability matrix must be invertible. Note

the difference between rank and determinant. Often in uncontrollable systems, part of the system is unconnected from input.

Additional tests are to show that the co ntrollability Gramian P is positive definite, where P may be found by the solution to the Lyapunov equation:

Alternatively;

  − +

=

.

∞    ≡    0

134

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Example 1:

Investigate the controllability of:

 − −       −   −    − ≠ =

1 1

0 , 2

1 2

=

Finding the Controllability Matrix,

0=

0=

Look at the determinant,

1 2

=

1 3

1

0

The rank of this matrix is 2, hence the system is controllable.

Example 2:


 − −     =

1 0

0 , 1

=

1 3

Example 3:


         ( )

( )

=

=

1

3

+

2

2

+

1

+

0

135


Observability

Observability is a test of the ability of the sensors. A system is observable if every initial state x(0) can be determined by observing the system output over some finite time period. A system is referred to as “detectable” if all unstable modes are state observable. This may mean the system has unobservable states which are stable. Strictly speaking the pair (C;A) is said to be detectable if there exists a matrix L such that A+LC is stable.

  ⋮−  

In order to test the observability of an LTI system, the “Observability Matrix” must be of full rank. The

observability matrix,

=

i.e the observability matrix must be invertible.

Additional tests are to show that the controllability Gramian Q is positive definite, where Q may be

  −

found by the solution to the Lyapunov equation:

Alternatively;

+

=

.

∞    ≡    0

Example 1:

Investigate the observability of:

Finding the observability matrix,

 − −         −     −  =

1 1

0 , 2

= 1

= 1

0 and

0

=

1

0

=

1 1

0 0

= 0

1

136

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Look at the determinant,



=0

The rank of this matrix is 1, hence the system is non observable.

What about the other case?

Example 2: Investigate the observability of:

 − −     =

1 1

0 , 1

= 1

3

Example 3: Investigate the observability of:

         ( )

( )

=

=

1

3

+

2

2

+

1

+

0

137


Solving the time invariant State equation

      

The homogeneous State Equations is

Matrix



which gives:

=

=

(0)

is a matrix exponential. This matrix is also known as the state transition matrix. It is

sometimes labeled as

  =

.

The state transition matrix is difficult to calculate. Hence there are two common ways of expressing it:

1. Expansion:

           ⋯  −  − Φ −  −    − 2 2

=

2. Inversion:

= +

=

1

+

=

2!

+

1

The forced response (in-homogeneous solution) is given by (assuming

          Φ     − −   Φ−  =

0 +

0

The output is therefore given by:

=

Example 1:

Obtain the state transition matrix

+

( )

( ) of the following system. 0 2

1 3

Obtain also the inverse of the state transition matrix,

1

1 2

For this sytem

=

 − −  =

The state transition matrix is given by

0 2

1 3

1

2

+

!

1

0

= 0):

138


Since

 −  −−  Φ   −  −− −  − −  − −   − −   −            Φ  −  −−  − −− −−−  −−− −−−  Φ −   Φ  Φ−  − − −− −−−  −−− −−−  0

=

0 2

0

1 = 3

(sI

A)

=

=

2

1 +3

1

1

1

[

1

s+3 2 s + 1 (s+2)

=

+3

Hence

=

Noting that

=

1

=

1

1

]

1

+ 1 ( + 2) 2

+ 1 ( + 2)

+ 1 ( + 2)

+ 1 ( + 2)

2

2 2

=

1 s

2

2

+2

+2

2

( )

1

=

2

2 2

=

2

2

+2

+2

2

Example 2: EM/APR 2008/KJM597/MEC522

a) An electro-hydraulic car suspension system can be modeled by the following state matrix

− −    −          

equation

1

  1

and

2

1

=

2

Where

1

1

1

0

=

1

2

2

+

1

2

1

2

are suspension displacements, u is an electrical actuating signal and k is the

suspension stiffness.

    

i.

Determine the condition for the system to be controllable

ii.

If y is a single output displacement, (given as

1

2

where

1

and

2

are constants),

establish the conditions which must be avoided if the system is to remain observable.

b) If the values of k =2N/m,

  1 =1

and

2 =0,

determine

i.

The eigenvalue of the system

ii.

The state transition matrix

iii.

The variation of x(t) response to a step change in u(t) at time t=0 from u=0 to u=1N for initial conditions

  1(

) and

2(

) equal to zero.

139


CHAPTER 6.0

CONTROL SYSTEM DESIGN

We have discussed so far the importance of the closed-loop system poles on the dynamic performance of the system. The transient-response specifications can be translated into desired locations for dominant closed-loop poles. The roots of the characteristic equation, which are the poles of the closedloop system, determine the absolute and relative stability of the system. Therefore, an important study in linear control systems is the investigation of the trajectories of the roots of the characteristic equation, or simply, the root loci when a certain system parameter varies. The basic properties and construction of root loci are first due to W.R. Evans (1948).

In this chapter, we will discuss the construction of root loci using simple rules. For plotting the root loci accurately, one can always use standard computer program packages like MATLAB. The basics of root loci should be thoroughly understood so that the engineers may be able to interpret the data provided by root loci for system analysis and design.

6.1

Root locus technique

Consider the second-order system shown in Fig. 6.1, which represents a typical position control system. The plant consists of a servomotor and load, driven by power amplifier with gain K . The open-loop transfer function of the system is

G( s) 

K s ( s  2)

(6.1)

140


Figure 6.1: A position control system

Figure 6.2: Root locus for Eq. (6.3)

The open-loop poles, marked



in Fig. 5.2, are at s = 0 and s = -2. The closed-loop transfer function of the

system is

Y ( s) R( s)



G( s )

1  G ( s)



K s

2

 2s  K

(6.2)

141

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The characteristic equation is

(s)  s 2  2s  K  0

(6.3)

This second order system is always stable for positive values of K . The relative stability of the system depends upon the location of the closed-loop poles

s1, 2

 1 

1  K

(6.4)

and hence on the choice of the parameter K .

As K is varied from zero to infinity, the closed-loop poles move in the s-plane as shown in Fig. 6.2. At K = 0, the root s1 is equal to the open-loop pole at s = 0, and root s2 is equal to the open-loop pole at s = −2. As K increases, the roots move toward each other. The two roots meet at s = −1 for K = 1. As K is increased further, the roots breakaway from the real axis, become complex conjugate, and since the r eal part of both roots remains fixed at s = −1, the roots move along the line  = −1.

A root locus of a system is a plot of the roots of the system characteristic equation (poles of the closedloop transfer function) as some parameters of the system are varied.

The two branches A-C-E and B-C-D of the plot of Fig. 6.2 are thus two root loci of the system of Fig. 6.1. Each root locus starts at an open-loop pole with K = 0 and terminates at infinity as K   . Each root locus gives one characteristic root (closed-loop pole) for a specific value of K .

The root locus plot gives us considerable information about the transient behavior of the system as gain K is varied. From Fig. 6.2:



For 0 < K < 1, the roots are real and distinct and the system is overdamped.



For K = 1, the roots are real and repeated. Thus, the system is critically damped.



For K > 1, the roots are complex conjugate and the system is underdamped with the value of  decreasing as K increases.

142

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Thus, by choosing appropriate value of K , we can cause a characteristic root at any point on root locus. For example, the dashed lines in Fig. 5.2 correspond to  = 0.707 . The points where the root loci cross the dashed lines have been marked

6.1.1

. These points corresponds to the closed loop poles for K = 2.

Basic properties of root loci

KG(s)

Figure 6.3: Typical control system

Consider the control system shown in Fig. 6.3. The closed-loop transfer function is

Y ( s) R( s)



KG( s)

1  KG( s) H ( s)

(6.5)

Let K be a positive quantity. The roots of the characteristics equation must satisfy the expression

or,

1 + KG(s)H(s) = 0

(6.6)

G(s)H(s) = -1/K

(6.7)

143

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Thus, any point s is a closed-loop pole or a root of the characteristic equation, if it satisfies the following conditions (K > 0): Magnitude condition: |G(s)H(s)| = 1/K

(6.8)

Angle condition:  G(s)H(s) = (2q+1)  = (2q+1) 180

o

where q = 0, ±1, ±2, …

(6.9)



The angle condition is used to determine the trajectory of the loci in the s-plane.



Once the root loci are drawn, the values of K on the loci are determined by using the magnitude condition.

Graphical Interpretation

Let KG( s) H ( s) 

K ( s  z1 )(s  z2 ) ( s  zm )

(6.10)

( s  p1 )(s  p2 ) ( s  pn )

The magnitude condition becomes m

 s  z  i

G ( s) H ( s)



i 1 n

 s  p 



1 K

(6.11)

j

j 1

The angle condition becomes m

n

i 1

j 1

G(s) H (s)   (s  zi )   (s  p j )  (2q  1)180o

where q = 0, 1, 2, …

(6.12)

144


Figure 6.4: Zero (z1) and pole ( p1) on a complex plane

In Fig 6.4, let us assume a complex pole and real zero: s+p1 and s+z1 represent the respective vectors in the complex plane. A and B are magnitudes of vectors (s+z1) and (s+p1) and  2 and  1 are angles of (s+z1) and (s+p1), respectively.

The graphical interpretation is:

The difference between the sums of the angles of the vectors drawn from the zeros and those from the poles of G(s)H(s) to s is an odd multiple of 180 .

Once the root loci are constructed, the values of K along the loci can be determined. Thus, the construction of root loci involves: 1. A search for all the points in the s-plane. 2. Find the magnitude of K on the root loci.

145


146

Properties and construction of root loci

The purpose of root locus is to show in graphical form the general trend of the roots of t he characteristic equation

(s) = 1+KG(s)H(s) = 1+F(s) = 0

(6.13)

m

 s  z  i

where

G ( s) H ( s) 

i 1 n

;

 s  p 

mn

(6.14)

j

j 1

as the parameter K is varied from zero to infinity. Every point s =  + j  in the complex plane that satisfies the angle criterion m

n

i 1

j 1

G(s) H ( s)   (s  zi )   (s  p j )  (2q  1)180o ;

q = 0, 1, 2, ….

is on the root locus. The value of the parameter K corresponding to a point on the root locus can be obtained from the magnitude criterion m

 s  z  i

G ( s) H ( s )



i 1 n

 s  p 



1 K

j

j 1

In principle, the root locus for a given F(s) can be sketched by measuring  F(s) at all the points of the 0

complex plane and marking down those places where we find  F(s) equal to an odd multiple of 180 . However, this trial-and-error method would be a very tedious task. Therefore, certain rules have been developed for making a quick approximate sketch of the root locus. This approximate sketch provides a guide for the selection of trial points such that a more accurate root locus can be obtained by a few trials. Further, the approximate root locus sketch is very useful in visualizing the effects of variation of the parameter K , the effects of shifting of pole-zero locations and of bringing a new set of poles and zeros.


Rules for Construction of Root Loci

The root locus for a given F(s) is to be sketched. F(s) has m zeros at s = -zi and n poles at s = -p j (refer to Eq. (6.14)) where m  n. These m zeros and n poles of F(s) are referred to as open-loop zeros and openloop poles, respectively.

Rule 1: Number of Root Loci (Branches) The root locus plot consists of n root loci (branches) as K varies from 0 to  . The loci are symmetric with respect to the real axis.

The characteristic equation can be written as: n

m

(s)   (s  p j ) K  (s  zi )  0 j 1

(6.15)

i 1

This equation has degree n. Thus, for each real K , there are n roots. As the roots are continuous function of the coefficients of equation, the n roots form n continuous loci as K varies from 0 to

.

Since the

complex roots occur in complex conjugate pairs, the root loci must be symmetrical about the real axis.

Rule 2: Starting and Ending Points of Root Loci As K increases from 0 to  , each root locus starts from an open-loop pole with K = 0 and ends on an open-loop zero or on  with K =  . The number of root loci ending at  equals the number of open-loop poles minus zeros.

Refer to Eq. (6.15). When K = 0, the equation has roots at -p j (j = 1, , n), which are open-loop poles. Thus, the root loci start at open-loop poles. Eq. (6.15) can be rearranged as

1

n

m

 (s  p )  (s  z )  0

K j 1

j

i

i 1

When K =  , the equation has roots at – zi (i = 1,  , m), which are open-loop zeros. Therefore, m root loci end on the open-loop zeros.

147

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam In case m < n, the open-loop transfer function has (n - m) zeros at infinity. From the magnitude criterion, m

 s  z  i

i 1 n

 s  p 



1

K

j 

, we find that this is satisfied by s   e as

j

j 1

K   . Thus, (n - m) root loci end on infinity.

Rule 3: Asymptotes to Root Loci (Behavior at Infinity) The (n − m) root loci which tend to  do so along straight line asymptotes radiating out from a single point s= −  a on the real axis (called the centroid) where

  a  

(real part of open  loop poles ) 

 (real part of open  loop zeros)

nm

These (n − m) asymptotes have angles

 a



(2q  1)1800

;

nm

q  0,1, , (n  m  1)

This rule will be justified by referring to a pole-zero patterns shown in Fig. 6.5. For a point far away from the origin, the poles and zeros can be considered to cluster at the same point, say − a, as shown in Fig. 6.5. Thus, Eq. (6.15) can be approximated as m

 s  z 

K

1

i

i 1 n

 s  p 

 1

K

( s   a )

nm

0

j

j 1

This means that all m zeros are cancelled by poles, and only (n - m) poles are left at - a.

(6.16)

148


Figure 6.5: Asymptotes to root loci

From Eq. (6.16), n

 s  p  j

j 1

 ( s   a ) n  m

m

 s  z 

(6.17)

i

i 1

By simplifying this, we get,

s

nm

m  n     p j   z i s nm1    s nm  n  m a s nm1   i 1  j 1 

Thus, by comparison of coefficients, we get n

m

 ( p )   ( z ) j

  a 

j 1

i

i 1

nm

Moreover, for the point s0 to be on the root locus,

 (n  m)  (2q  1)1800 Thus,

 a



(2q  1)1800 nm

; ;

q  0,1,  q  0,1,, (n  m  1)

(6.18)

149


The (n - m) angles given by the above equation divide 360 equally and are symmetric with respect to real axis. The (n - m) root loci tend to  along (n - m) asymptotes radiating out from s = - a at angles  a.

Example 6.1: The pole-zero map of Fig. 6.5 corresponds to

F(s) 

K(s  2 ) (s  1  j 4 )(s  1  j 4 )(s  3 )(s  4 )

(6.19)

The root loci has four branches, each starting from an open-loop pole with K = 0. One root locus will terminate on open-loop zero with K =  . The other three loci will terminate on  as K   along the asymptotes radiating out from s = - a where

  a  0

 1  1  3  4  (2) 7  4 1 3 0

0

at angles 60 , 180 , and 300 , respectively. Fig. 6.6 shows the asymptotes.

Figure 6.6: Asymptotes for Eq. (5.19)

150


Rule 4: On-Locus Segments on the Real Axis A point on the real axis lies on the locus if the number of open-loop poles plus zeros on the real axis to the right of this point is odd.

For the system of Eq. (6.19), the open-loop pole-zeros are shown in Fig. 6.7(a). Take a point s0 on the real axis. Join this point to all the open-loop poles and zeros. It is seen that (i) poles and zeros on the real 0

axis to the right of this point contribute an angle of 180 each, (ii) poles and zeros to the left of this point 0

contribute angle of 0 each, and (iii) the net angle contribution of a complex conjugate pole or zero pair is always zero. 0

0

Thus,  F(s)=(mr – nr )180 =  (2q+1) 180 , q = 0, 1, 2, … where mr = number of open-loop zeros on the real axis to the right of s0 and nr = number of open-loop poles on the real axis to the right of s0. Thus, the angle criterion is satisfied if ( nr – mr ) or (nr + m r ) is odd and hence the rule. Thus, the real axis can be divided into segments on-locus and not-on-locus; the dividing points being the real open-loop poles and zeros. The on-locus segments of the real axis alternate as shown in Fig. 6.7(b).

Figure 6.7: On-locus segments of the real axis

151


Rule 5: On-Locus Points of the Imaginary Axis The intersections (if any) of root loci with the imaginary axis can be determined by use of the Routh criterion.

Segments of root loci can exist in the right half of s-plane. This signifies instability. The points at which the root loci cross the imaginary axis define the stability limits. The Routh Table determines the gains at the stability limit. By using this gain in the auxiliary equation, the value s = j 0 at the stability limit is computed.

Example 6.2: The characteristic equation of system in Eq. (5.19) is

s 4  9s3  43s 2  (143  K )s  204  2K  0

(6.20)

The corresponding Routh Table is shown below.

4

1

43

3

9

143+K

2

(244 – K)/9

204+2K

1

(18368 – 61K – K )/(244 – K)

0

204+2K

s s s s

204+2K

2

s

2

For stability, 244 – K > 0, 18368 – 61K – K > 0, and 204 + 2K > 0. It can be seen that these conditions are 1

satisfied if K < 108.4. For K = 108.4, all the coefficients in s row are zero. Thus, the auxiliary equation is 2

formed from the coefficients of s row and is given by

244  K 9

s 2  (204  2K )  0

For K = 108.4, the roots of the above equation lie on the j  axis and are given by s =  j 5.28. Thus, the root loci intersect the imaginary axis at s =  j5.28 and the corresponding value of K is 108.4.

152


153

Rule 6: Angle of Departure from Complex Poles 0

The angle of departure,  p , of a locus from a complex open-loop pole is given by  p = 180 + where  is the net angle contribution at this pole of all other open-loop poles and zeros.

Example 6.3: For the system of Eq. (6.19), the characteristic equation is

1  F(s)  1 

K(s  2 ) (s  1  j 4 )(s  1  j 4 )(s  3 )(s  4 )

0

(6.21)

The pole zero-map is shown in Fig. 6.8.

Figure 6.8: Angle of departure from complex poles

Let s0 be an arbitrary point on the root locus starting from s = -1+j4. The phase from this pole to s0 is  p. The net angle contribution of all other open-loop poles and zeros at s0 is

   2  ( 1   3   4 ) 0

Thus, the total phase of F(s) at s0 is  -  p. For s0 to be on the root locus, the total phase must be 180 . 0

So,  p = 180 + . This is the angle of departure from the complex open-loop pole.

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam If s If s0 is very close to the pole -1+j4, -1+j4, then the vectors drawn from all other poles and zeros to s0 can be approximated by the vector drawn to the pole at -1+j4, -1+j4, i.e., we consider s0 to be -1+j4 for measurement of angles  1,  2,  3, and  4. With this approximation, for this example,  1 = 900,  2 = 76 0,  3 = 630, and  4 = 53 . So,    2 0

 ( 1   3   4 )  1300

0

0

and  p = 180 +  = 50 . A rough sketch of the root locus for this

system is shown in Fig. 5.9.

Figure 6.9: Root locus plot for Eq. (6.21)

There are four open-loop poles, so there are four loci. One locus departs from real pole at –3 –3 and ends on the zero at – at –2 2 along the real axis. The second locus departs from real pole at –4 –4 and moves along the asymptote on the negative real axis. The third locus departs from the complex pole at – 1+j4 1+j4 with a 0

departure angle of  p = 50 and moves toward the asymptote radiating from the centroid at –7/3 –7/3 at an 0

angle of +60 ; it crosses the imaginary axis at j 5.28. 5.28. Using the symmetry property, the fourth locus is obtained immediately by reflection about the real axis.

154


Rule 7: Angle of Arrival at Complex Zeros 0

The angle of arrival,  z , of a locus at a complex zero is given by  z = 180 –  , where  is the net angle contribution at this zero of all other open-loop poles and zeros.

Example 6.4: Let us consider the characteristic equation

1  F ( s)  1 

 1) 0 s( s  2)

K ( s

2

(6.22)

The pole-zero map of this F(s) is shown in Fig. 6.10. Open loop poles: s = 0, – 2. Open-loop zeros: s =  j1. j1. Let s0 be an arbitrary point on the root locus terminating on the zero at s = j1. j1. Let the phase from this zero to s0 =  z. If the point s0 is very close to the zero at j1 at j1,, then the vectors drawn from the other zero at – j1 and poles at 0 and –2 –2 to s0 can be approximated by vectors to the zero at j1. j1. Under this approximation, the net angle contribution at s0 is given by 0 0 0 0 – 90 – 26.5 – 26.5 = – 26.5 – 26.5 .  = 90 – 90

For s0 to be on the root locus, the total phase must be 1800. Thus,  z = 1800 –  = 206.50. The complete root locus plot is shown in Fig. 6.10.

Figure 6.10: Angle of arrival at complex zero

155


Rule 8: Location of Multiple Roots Points at which multiple roots of the characteristic equation occur (breakaway points of root loci) are the solutions of

dK ds

0

(6.23)

where n

 s  p  j

K  

j 1

(6.24)

m

 s  z  i

i 1

Let us assume that the characteristic equation has a multiple root at s = s0 of multiplicity r . Then,

1  F (s)  (s  s0 )r M (s)

, r  2

(6.25)

where M(s) does not contain the factor (s ( s - s0). Thus, by differentiating Eq. (6.25), we have

dF ds

 (s  s0 )r 1rM (s)  (s  s0 ) M ' (s)

(5.26)

At s = s0, the RHS of Eq. (6.26) is zero. Thus, at s = s0,

dF ds

0

In pole-zero form, the characteristic equation is: m

 s  z 

K

1  F ( s)  1 

i

i 1 n

 s  p 

1

KB( s) A( s )

0

(6.27)

j

j 1

Thus,

dF ds

A( s) B( s)  A( s) B( s)

 K

 A(s)2

0

(6.28)

Therefore, the breakaway points are the roots of

A(s) B ' (s)  A' (s) B(s)  0

(6.29)

156

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam This equation can be equivalently represented as n

dK ds

0

, where K  

A(s) B(s)

 s  p  j



j 1

(6.30)

m

 s  z  i

i 1

Example 6.5: Consider the characteristic equation

1

K ( s  2)(s  3) s( s  1)

 1

KB( s) A( s)

0

(6.31)

Fig. 6.11 shows the open-loop poles and zeros on the complex plane. Root loci segments exist on the negative real axis between 0 and –1 –1 and between – between –2 2 and – and –3. 3. At K = 0, the roots are at s = 0 and s = – 1. As K increases, K increases, the two roots move away from poles at 0 and – and –1 1 toward each other inside the segment [1,0]. At some K , the two real roots will become repeated real roots and then break away from the real axis into two complex conjugate roots. Such a point is c alled a breakaway point .


Similarly as K approaches

,

one root will approach zero at s = – 2 along the negative real axis and

another will approach zero at s = –3 –3. As the root loci are continuous, the two complex conjugate roots will approach the real axis somewhere inside the segment [ – [ –3, 3, –2] –2] and then depart in opposite directions along the real axis. This point is also another breakaway point. Sometimes, such a point is also called as break-in point.

Applying Eq. (6.30) to this case, we get the solutions of dK/ds = 0 as s = – = –0.634 0.634 and s = – = –2.366 2.366.. Thus, the root locus has two breakaway points.

157


It is important to note that the condition for the breakaway point (as derived above) is necessary but not sufficient. In other words, all breakaway points on root locus must satisfy Eq. (6.30), but not all points that satisfy Eq. (6.30) are breakaway points.

SUMMARY OF RULES FOR ROOT LOCUS PLOTTING

The characteristic equation of the system is m

 s  z 

K

1  KG( s) H ( s)  1  F ( s )  1 

i

i 1 n

 s  p 

0

j

j 1

;

mn

;

K  0

K varies from 0 to 1. The root locus plot consists of n root loci (branches) as K varies

.

The loci are symmetric

with respect to real axis. K increases from 0 to , each root locus starts from an open-loop pole with K = 0 and ends on an 2. As K increases open-loop zero or on  with K =  . The number of root loci ending at



equals the number of open-

loop poles minus zeros. (n - m) m) root loci which tend to  do so along straight line asymptotes radiating out from a single 3. The (n point s= - a on the real axis (called the centroid) where

  a  

(real part of open  loop poles ) 

 (real part of open  loop zeros)

nm

These (n (n - m) m) asymptotes have angles

 a



(2q  1)1800 nm

;

q  0,1, , (n  m  1)

4. A point on the real axis lies on the locus if the number of open-loop poles plus zeros on the real axis to the right of this point is odd. By use of this fact, the real axis can be divided into segments onlocus and not-on-locus; the dividing points being the real open-loop poles and zeros.

5. The intersections (if any) of root loci with the imaginary axis can be determined by use of Routh criterion. 0 6. The angle of departure  p of a locus from a complex open-loop pole is given by  p = 180 +  , where

 is the net angle contribution at this pole of all other open-loop poles and zeros.

158

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam 0 7. The angle of arrival  z of a locus at a complex zero is given by  z = 180 -  , where  is the net angle

contribution at this zero of all other open-loop poles and zeros.

8. Points at which multiple roots of the characteristic equation occur (breakaway points of root loci) n

are the solutions of

dK ds

 s  p  j

0

K  

where

j 1 m

 s  z  i

i 1

6.1.3

A complete example

Question: Consider a feedback system with the characteristic equation

1

K s( s  1)(s  2)

0

;

K  0

(6.32)

Plot the root locus for this system.

Solution: The open-loop poles are located at s = 0, −1, −2. There are no finite open-loop open -loop zeros. The pole-zero configuration is shown in Fig. 5.12. K varies from 0 to . Rule 1 tells that the root locus plot consists of three root loci as K varies

Rule 2 tells that the three root loci originate from the three open loop poles with K = 0 and terminate on 

with K =  .

Rule 3 tells that the three root loci tend to

s   a







along asymptotes radiating out from

(real parts of poles ) 

 (real parts of zeros)

number of poles  number of zeros

 2 1  1 30

with angles

(2q  1)180

0

 a



number of poles

(2q  1)180

 number of zeros

0



3

; q  0,1,2

 600 ,1800 , 3000

; q  0,1,2, 

159

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The asymptotes are shown by dotted lines.


Rule 4 tells that the segments of real axis between 0 and –1, and between –2 and - lie on the root locus. On-locus segments are shown by thick lines in the Figure. From Fig. 6.12, it is seen that out of the three loci, one is a real-root locus originating from s = −2 and terminating on −. The other two loci originate from s = 0 and s = −1, and move on the real axis towards each other as K increases. Their meeting point corresponds to a double root. As K increases further, the root loci breakaway from the real axis to g ive complex conjugate pair of roots.

Rule 5 is used to calculate the intersection points on the imaginary axis by Routh Table. The characteristic equation can be written as

s3  3s 2  2s  K  0 The Routh Table is given below. s

3

1

2

s

2

3

K

s

1

(6-K)/3

s

0

K

160


161

For all roots to lie on the left half of the s-plane, the following conditions must be satisfied. K > 0, and (6 − K)/3 > 0 Therefore, the critical value of K , which corresponds to the roots on the imaginary axis, is 6. K = 6 makes all the coefficients on s1 row to be zero. The auxiliary equation is formed from the coefficients of the s2 row as:

3s 2  K  3s 2  6  0 The roots of this equation lie on the j  axis and are given by s   j 2 which are also the points where the two root loci intersect the imaginary axis and the intersection points correspond to K = 6.

Rule 6 and Rule 7 are not necessary in this case since there are no open-loop complex poles or zeros. Rule 8 is used to determine the breakaway points. From the characteristic equation of the system, 3

2

K = −(s + 3s + 2s). Thus, by differentiating K and equate it to zero,

dK ds

 (3s 2  6s  2)  0

The solutions of this equation are: s = −0.4226 and s = −1.5774 Thus, s = −0.4226 is the breakaway point and, since the other point s = −1.5774 is not on the root locus, it is not a breakaway point. 0

If two loci breakaway from a breakaway point, their tangents will be 180 apart. In general, if r loci 0

breakaway from a breakaway point, then their tangents will be 360 /r apart, i.e., the tangents will 0

equally divide 360 . The complete root loci are shown in Fig. 6.12. For K > 6, the system has two closed-loop poles in the right half s-plane.

-1

A closed-loop pole with  = 0.5 lies on a line passing through the origin and making an angle cos  = 60

0

with the negative real axis. From Fig. 6.12, the points of intersection are s = −0.33  j0.58 which are the dominant closed-loop poles. From the magnitude criterion, the corresponding K can be found.

K  s s  1 s  2 s  0.33 j 0.58

 1.04


Additional Example

Question: Sketch the root lo cus of a unity feedback system with forward path transfer function G(s) given as follows:

G(s) 

K s(s2  4s  5 )

Solution: The open-loop poles are located at s = 0, −2+j, −2−j . There are no finite open-loop zeros.

Rule 1 tells that the root locus plot consists of three root loci as K varies from 0 to . Rule 2 tells that the three root loci originate from the three open loop poles with K = 0 and terminate on 

with K =  .

Rule 3 tells that the three root loci tend to

s   a







along asymptotes radiating out from

(real parts of poles ) 

 (real parts of zeros)

number of poles  number of zeros

22  4 / 3 30

with angles

 a

 

(2q  1)1800 number of poles

(2q  1)1800 3

 number of zeros

; q  0,1,2, 

; q  0,1,2

 600 ,1800 , 3000 Rule 4 tells that the segments of real axis between 0 and – lie on the root locus. Rule 5 is used to calculate the intersection points on the imaginary axis by Routh Table. The characteristic equation can be written as s

3



4s

2

 5s 

K



0

162

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The Routh Table is given below. 3

1

5

s2

4

K

s

s

1

(20-K)/4

s

0

K

For all roots to lie on the left half of the s-plane, the following conditions must be satisfied. K > 0, and (20−K)/4 > 0 Therefore, the critical value of K , which corresponds to the roots on the imaginary axis, is 20. K = 20 1

makes all the coefficients on s row to be zero. The auxiliary equation is formed from the coefficients of 2

the s row as:

4s 2  K  4s 2  20  0 The roots of this equation lie on the j  axis and are given by s   j 5 which are also the points where the two root loci intersect the imaginary axis and the intersection points correspond to K = 20.

Rule 6 tells the angle of departure for complex poles. For pole −2+j ,

φ  153.43  90 and  p = 1800 +  = -63.430 For pole −2−j ,

 p = 63.43

0

Rule 8 is used to determine the breakaway points. From the characteristic equation of the system, K = 3

2

−(s + 4s + 5s). Thus, by differentiating K and equate it to zero,

dK ds

 ( 3s 2  8s  5 )  0

The solutions of this equation are: s = −1 and s = −1.667 Since the complete negative real axis is on the root loci, both are valid breakaway or break-in points.

K s 1  2 , K s 1.66 7  1.852

163


164


Effects of addition of poles and zeros to G(s)H(s)

The controller design in control systems may be treated as an investigation of the effects to root loci when poles and zeros are added to the loop transfer function KG(s)H(s).

Addition of Poles to G(s)H(s)

Adding a pole to G(s)H(s) has the effect of pushing the root loci toward the right-half s-plane.

Example 6.6: Consider the loop transfer function KG( s) H ( s) 

K s( s  2)

The root loci are shown in Fig. 6.13(a). It is noted that the system is stable for all K . Let us introduce a pole at s = −b (b > 2). The loop transfer function G(s)H(s) becomes, with b = 3,

KG( s) H ( s) 

K s( s  2)(s  b)



K s( s  2)(s  3)

The root loci are shown in Fig. 6.13(b) where the root loci bend towards the right-half s-plane. The asymptote angles and centroid are changed from 90 to 60 and –1 to –(2+b)/3, respectively. The addition of a pole may make the system unstable if K exceeds the stability limit.

Figure 6.13(a): Root loci for

K s (s  2)

165


Figure 6.13(b): Root loci for

K s( s  2)(s  3)

Addition of Zeros to G(s)H(s)

Adding left-half plane zeros to the function G(s)H(s) generally has the effect of moving and bending the root loci toward the left-half s-plane.

Fig. 6.14 shows the root loci of G(s)H(s) with a zero added at s = −3. The complex conjugate parts of root loci of the original system are bent towards the left and form a circle. Thus, the relative stability is improved by the addition of the zero.

166


Figure 6.14: Root locus for

6.1.5

K ( s  3) s ( s  2)

.

Compensator design via root locus

The preceding chapters have shown that it is often possible to adjust the system parameters in order to provide the desired system response. However, we often find that it is not sufficient to reconsider the structure of the system and redesign the system in order to obtain a suitable one. That is, we must examine the scheme or plan of the system and obtain a new design or plan that results in a suitable system. Thus the design of a control system is concerned with the arrangement, or the plan, of the system structure and the selection of suitable components performance is called compensation.

Compensation is the adjustment of a system in order to make up for deficiencies or inadequacies. In redesigning a control system to alter the system response, an additional component is inserted within the structure of the feedback system. It is this additional component or device that equalizes or compensates for the performance deficiency.

167

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam The compensating device may be electric, mechanical, hydraulic, pneumatic, or some other type of device or network and is often called a compensator. Commonly an electric circuit serves as a compensator in many control systems. A compensator is an additional component or circuit that is inserted into a control system to compensate for a deficient performance. The transfer function of a compensator is designated as

     =

0(

)/

( ), and the compensator can be placed in a suitable

location within the structure of the system. Several types of compensation are shown in Figure 6.15 for a simple, single-loop feedback control system. The compensator placed in the feedforward path is called a cascade, or series, compensator (6.15a)

Figure 6.15: Types of compensation (a) Cascaded compensation. (b) Feedback compensation. (c) Output, or load compensation. (d) Input compensation The objectives of introducing compensator can be categorized as follows: i.

Use PI and phase-lag compensators to improve steady-state error.

ii.

Use PD and phase-lead compensators to improve transient response.

168


Ideal PI compensator design

The PI compensator’s transfer function is given by:

            =

1

+

2

1(

=

The ideal PI compensator’s transfer function is given by;

=

+

+

2

1

)

169


Example 6.7: Given a uncompensated system operating with a damping ration of 0.174. Find out the steady state error for a unit step input. Design an ideal PI compensator to r educe the steady-state error to zero without appreciably affecting transient response.

If the original OLTF is



=

    1

+1

+2 ( +10)

operating with a damping ratio of 0.174, then design a PI

compensator to reduce the steady-state error to zero for a step input without appreciably affecting transient response. The compensator has a zero at -0.1, close to the compensator pole.

To achieve these requirements, the compensated system should have a dominant closed-loop pole at

 −   1

=

0.694 + 3.926,

= 164.6

170


Thus, the steady-state error is

   =

,

1

1+

=

1 1 + 164.6/20

= 0.108

The dominant pole of the compensated system and the gain are approximately the same as for the uncompensated system

   ,

=

1

1+

=

→    1

1+lim 0

( )

=0

171


General first-order compensators

Consider the first-order compensator with t he transfer function

   

When | 0 |<|

0 |,

 −− ∠ ∠− −∠−  −  ∠ ∠− −∠−  −  =

(

0)

0

the compensator is called a phase-lead compensator, because this results in a

contribution to the angle criterion of the root locus that is always positive.

When | 0 |>|

0 |,

=

0

0

=

>0

the compensator is called a phase-lag compensator, because this results in a

contribution to the angle criterion of the root locus that is always positive.

=

0

0

=

<0

172


Phase-lead compensator design

The compensator transfer function is given by:

 −−        −   −   →                                 ∠    ∠∠           −  (

=

0)

0

and it can be re-written in a form as:

+

1

=

0

+1

1

Where,

1

=

,

1

The compensator DC gain is

Assume that the parameter 1 and

1

= lim

0

0

0

=

,

1

=

0

0

1

=

1

0

is either known or c an be determined. The design problem is to find

such that the compensated system will have a closed-loop pole at

First we express

1

and G(s)H(s) as 1=

1

1

1

=

+ 0 1 +1

1

From the characteristics equation, we get 1 +

1

( 1)

=0

Equating magnitudes and angles, we can rewrite as

+

1 1

1 1

1 1

1 1

0

+1

1

( 1) = 1

0

1

( 1)

+

+1

+

= 180 180°

Where

1

=

sin

+

0

1

1

( 1 ) sin(

1

( 1 ) sin

)

=

1.

173


   −      1

=

sin( +

)+

0

1

1

( 1 ) sin

sin

Example 6.8:

Design a phase-lead compensator such that the closed-loop compensated system has a settling t ime around 4 sec. and a percent overshoot around 4.32%. The compensator has a DC gain as 0.15.


−  − 

1 + . Because the DC gain for

1=

At

1=

1 + , we have

Also, we have

 

( ) is 0.15, so we have

      −  21

+1 ( +3)

1=

=

1+



0 =0.15.

−  ∠

2.1 + 6.3 = 6.64 108.43°

 −   ∠        −             −      1=

1 + = 2 135° 1

=

1

=

sin

+

0

1

sin( +

)+

1

( 1 ) sin(

1

( 1 ) sin

0

1

1

sin

( 1 ) sin

)

= 0.19 0.1924 24

= 0.14 0.1417 17

174


Phase-lag compensator design

The negative angle contributed by the phase-lag compensator will tend to shift the root locus to the right in the s-plane, i.e., towards the unstable region. Thus, in general, the angle contribution of the phase-lag compensator must be small, which is assured by placing the pole and the zero of the compensator very close to each other.

For convenience in the design, we assume that the compensator has a unit DC gain, i.e.,

    ( )

=0

0

=

0

=1

175

    −       0

=


<1

0

Suppose that the root locus of the point of the uncompensated system passes through the point



0.

1+

0

0

  

As we choose the value of small compared to

1,

0

and

0

so

1

1

1

=

1

( ( 1)

  −− ≈ =

(

1

0

given by

1

Since



1

< 1, so

>

0.

( 1)

=

1

=

and

0

to be

1

0

1

The compensator has been chosen to have a unity DC gain; thus the open-loop

DC gain has been increased, but the transient response appears to remain unaffected.

The steady-state error

0

for the uncompensated system is

 −  −         →∞  →  →   1

 

0)

1

Now the gain required to place a root of t he locus at approximately

=

1 for

=0

to be approximately equal, and t he magnitudes of

1



, when H(s)=1, is

= lim

= lim 0

= lim

01

( )

+ ( )

176


The sready-state error compensator.



has been improved, and this is the principal use of the phase-lag

Example 6.9: Phase-lag compensator design Design a radar tracking system of the uncompensated OLTF given by



=



. Suppose that

( +2)

the design requirements are such that a time constant of 1 second and damping coefficient of 0.707 are satisfactory and the compensator has a DC gain as 1 .

So

 −   1

=

1+

0

= 2 , is acceptable. Suppose that the system is required to track aircraft that

have essentially constant velocity, which will appear to the control system as a ramp input, i.e. the antenna must rotate at a constant velocity to remain pointed directly at the aircraft. Also, it is required that the

of 0.2° with a unit ramp input.

177


178


Ideal PD compensator

The PD compensator’s transfer function is given by:

                =

1

+

2

=

2(

+

1

)

2

The ideal PD compensator’s transfer function is given by;

=

+

If the original open-loop transfer function is

=

1

+1

+ 2 ( + 5)

Then design the PD compensator zero at -2.

=

+2

179


180


Example 6.9 Given the system below, design an ideal derivative compensator such that the closed-loop compensated system has a threefold reduction in settling time and a 16% percent overshoot.


 −   1

=

4

1.205

1.205 + 2.064

= 43.45. Thus, the uncompensated system’s settling time is

= 3.302.

The desired real part of the closed-loop pole is the closed-loop pole is





=

 4

 − 

= 3.613 tan 180

=

4 1.107



=

  4

=

= 3.613. The desired imaginary part of

120.26 = 6.193.

181


182


State Space Applications

An alternative method yet powerful tools of control system design by using a State Space representation. The concept of using state space is by placing a pole at a desired location. We call this as pole placement method.

Similar in concept to classical control system design where we have to firstly formulate desired pole locations to satisfy some performance criteria, then formulate control gains to make this happen.

In designing the control system by using state space application, we have to assume that all states can be measured and used in control implementation; this is called full state feedback.

6.2.1

Controller Pole Placement Method

The controller pole placement method mainly concerns with t he controllability matrix. It takes measurement and/or estimates of the state v ariables, multiplies them by the control gains, and produce the control signal. This can be designed by pole placement or optimal control.

Let say we have a control input, u in the form of,

u = -kx where k = vector (or matrix) of proportional control gains applied to each state given by:

  …

k=[

]

Essentially, we are trying to modify the underlying differential equations.

Now, consider the state equation:

   −  −  −  =

Taking the Laplace,

+

=

=

+

=0

183

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Poles defined by:

 −  …  −  − … −   − ⋯ −   …… ……  … − −− −− … −       −  − −−  …  −    −   −     − ⋯   −     −  − … −     − det

+

=0

Suppose we have some desired pole locations:

1,

2, 3,

.

Then the desired characteristics equation is: 1

=0

2

This can be expanded to (desired characteristics equation):

+

1

1

+

+

+

1

=0

Now suppose that the state space equations are in control canonical form:

0 0 . . 0

=

1 0 . . 0

0 1 . . 0

0 0 . . 1

1

2

1

0 0 . = . 0 1

=

1 0

1

1 0

.

1

+

2)

1 0

The poles of the feedback system are defined by the expression: det

+

=0

Poles are therefore given by det

+

=

+(

1

+

1

1)

+

+(

2

2

+

+

=0

Compare this to the “desired” characteristics equation:

+

1

1

+

2

2

+

1

+

=0

The conclusion is that when the system is in control canonical form, then control gains c an be calculated by simple comparison of coefficients:

=

184


Example 6.10

    −     − 

Consider an undamped oscillator with frequency 1

0

=

2

and a SS model given by

1 0

2

1

0 1

+

2

Find the controller that places the both CL poles of the system at

2

. In other words, you want to

double the natural frequency and increase the damping ratio from 0 to 1.

Solution: Assumption: Must be a full state feedback. We need to prove the system has a full state feedback by assessing its controllability matrix.

We have an open loop poles,

       −             −   −−                            and closed loop poles,

=

The open loop characteristics equation is:

=

2

+

=

2

+4

+4

2

+

.

+

And the desired characteristics equation:

( +2

)2 =

2

Now we have to determine the poles of closed-loop system; det

det

+

0

0

2

+

1 0 + 0 1

2

0

2

=0

2

+

+

1

Comparing with the desired characteristics equation gives: 2

1

We have now,

Hence,

=4 2

+

=4

1

=3

2

=4

=

1

= 3

2

2

2

4

2

1

=0

2

2

185


Generalised Strategy

Step 1: Transform state equations into control canonical form.

′ − ′  ′ − ′ −    … …−   −−⋮ …⋱ ⋮ ⋮    …   −  − … −   =

=

(

,

= transformed state vector)

(A’=transformed state matrix)

=

Where T is the transformation matrix:

=

= controllability matrix =

1

= the following Toeplitz matrix:

0

0

2 1

Where

1,

2,

0

1

1

0 1

are from the characteristics equation of the uncontrolled system:

+

1

1

+

2

2

+

+

1

=0

Step 2: Calculate control gains by comparison with the desired characteristics equation Step 3: Transform back to original state

Transform has the form:

    − =

Controller Pole Placement using Ackermann’s Formula

For SISO systems the control gains using Ackermann’s Formula are

Where

 …

= 0

0

0

1

0

1

( )

0 =controllability matrix (note inversion again), and

A = state matrix, Where

  …  −      −  − ⋯ 

1,

2,

=

+

1

1

+

2

2

+

= coefficient of the desired characteristics equation.

+

186


Example 6.11:

    −     − 

Consider an undamped oscillator with frequency 1

0

=

2

and a SS model given by

1 0

2

1

0 1

+

2

Find the controller that places the both CL poles of the system at

2

. In other words, you want to

double the natural frequency and increase the damping ratio from 0 to 1.

Solution: Our objective is to get the vector k. According to Ackermann’s Formula:

  …  −         −  − ⋯                    − −  −         −   = 0

0

0

1

0

1

( )

The SS model must be controllable before we can proceed. Checking the controllability matrix,

=

Where

=

+

1

1

0 1

1 0

+

2

2

+

+

From the desired characteristics equation we have,

)2 =

( +2

2

+4

+4

So we have,

2

=

Matrices

2

,4

and 4

2

+4

are given by,

2

2

=

4

4

Therefore matrix

+4

=

2

=

0

2

0

0 4

4

3

0

2

4

0

0

4

2

is given by,

=

3 4

2

3

4 3

2

2

2

187

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam Hence,

    −       = 0

0 1 1

= 3

1 0

2

3 4

2

3

4 3

2

4

Example 6.12:

Consider the SS system

  − 1

2

=

1 1

  −  −   0 2

Design a control system to move the system poles to

1

2

=

+

1±

1 2

188


Observer Pole Placement Method

The observer pole placement mainly concerns with Observability matrix. The o bjective of designing the observer is to estimate some or all of the states of t he system. This can be achieved by linear observers (pole placement) or optimal observers (Kalman filters).

Let’s say a system is described by:

           =

Assume that we know

We need to extract

+

,

=

, , , .

by constructing a second linear system (a model of the target system), using the

known parameters ( , , , ) of the target system, which predicts the (measurable) target system output. If the predicted output is acceptably close to the actual output, then we can use the estimated states in place of the actual states.

In other words, we want to minimize the difference between the actual and predicted states. This difference or error in estimate of state is given by,

 − =

    −   − −−   −     −

Which converges to zero if

is stable.

The open loop dynamics is then:

=

=

+

=

And the characteristics equation,

det

=0

The observer (model) system response is now:

=

+

+

(

)

189

KJM597 Control Systems Faculty of Mechanical Engineering UiTM Shah Alam where L = vector (or matrix) of estimator (observer) gains applied to each state given by:

ℓ  ℓℓ⋮   − − −  1

2

=

The dynamics of the error become:

=

Therefore, the characteristics equation = det by adding feedback.

) = 0 i.e we can change convergence speed

(

Suppose we have some desired pole locations:

…  −  − … −   − ⋯ −   …… −−−  ……… −−  …−  −   −  − −   −     …  1,

2, 3,

.

Then the desired characteristics equation is: 1

=0

2

This can be expanded to (desired characteristics equation):

+

1

1

+

+

1

+

=0

Now suppose that the state space equations are in observer canonical form:

0 1 . = . 0 0

0 0 . . 0 0

0 0 . . 0 1

1 2

. .

1

1 0

1

=

= 0

1 0

. . .

1

1 0

0

0

1

With the observer feedback, the poles are defined by the expression: det

− −  (

) =0

190

− − −  =


…  …… ……  …−

0

1 . . 0 0

− ℓℓ−   ℓ ℓ 

0 0 . .

. . 0 0

+ 1+ . . 2 + 2 + 1+

1

1

1

Poles are therefore given by

 −    ℓ −  ℓ − ⋯  ℓ   −  − … −  

det

+

=

+(

1

+

1

1)

+(

+

2

2

2)

+

+

+

=0

Compare this to the “desired” characteristics equation:

+

1

1

+

2

2

+

+

1

=0

The conclusion is that when the system is in control canonical form, then control gains can be c alculated by simple comparison of coefficients:

ℓ  − =

Example 6.13: Compute the estimator (observer) gain matrix which will place both estimator poles at

             −  ℓ  −−  ℓ    −  1

2

Solution:

0

2

1 0

= 1

0

=

1

2

+

0 1

1

2

The desired characteristics equation is given by;

( + 10

)2 =

2

+ 20

+ 100

Now we have to determine the poles of closed-loop system; det

det

0

0

+

0

2

=0

1 + 0

1 2

0 0

2

−  10

, given

191

 ℓℓ −   ℓ   ℓ  ℓℓ      ℓℓ    ℓℓ     + +

det

det


2

+

2

1

1

2

2

+

1

+

=0

2

Comparing with the desired characteristics equation gives:

= 20

1

2

We have now,

Hence,

2

+

= 100

1

= 20

2

= 99

2

2

1

=

2

=

20 99

2

Generalised Strategy

Step 1: Transform state equations into control canonical form.

′ − ′  ′ − ′ − −    ⋮−   …  −−⋮ …⋱ ⋮ ⋮    …   −  − … −   =

=

(

,

Where T is the transformation matrix:

= transformed state vector)

(A’=transformed state matrix)

=

=(

and

)

= observability matrix =

1

= the following Toeplitz matrix:

0

2 1

Where

1,

2,

0

0

1

1

0 1

are from the characteristics equation of the uncontrolled system:

+

1

1

+

2

2

+

1

+

=0

192

Control Systems Lecture Notes

Recommend Documents