stability part 1

8/6/2019 Stability Part 1

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Understand the importance of system

stability

Identify stable & unstable systems

State the characteristic function,

characteristic equation, system zeroes and

poles of a system

Analyse stability in the s-plane

Objectives


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On 7 Nov 1940, at about 10 a.m. in

TACOMA, Washington, USA something

happened .


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Think about the Tacoma story.

Find out more about the story and answer the following:

What happened in Tacoma on 7 Nov 1940? What could be the cause of that?

How is it related to what you are learning?

What have you learned ?

Your Tasks


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Concepts on System Stability

What is stability of a system?

Why is it important?

What are some examples?

Can we know if a system is stable from its

mathematical model? If so, how?


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A control system is stable if the output eventually settles

down to a finite value due to a set-point change, load change

or disturbance.

SystemInput

r(t)

Output

c(t)

SystemStable-value)finite(a)(lim!

gp

tct

SystemUnstable-value)(infinite)(lim g!gp

tct

What is Stability ?


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The following are time-response of outputs of stable systems

C(t)

t0

C(t)

t0

SystemStable-value)finite(a)(lim !gp

tct

a a


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The following are time-response of outputs of unstable systems

C(t)

t0

C(t)

t0

Systemnstable-value)(in inite)(lim g!gp

tct


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A special case is the marginally stable system where the

output oscillates continuously.

C(t)

t0


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Robotic Arm

Nuclear Power Plant

Missile

Oscillator

Tacoma Narrows Bridge **

Why is stability important?

What are some examples?


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There are many ways of determining the stability of control

systems :

a. Based on Time Response

b. Based on Roots of Characteristic Equation

c. Routh- Hurwitz Stability Criterion


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Stability based on Time Response

The procedures to determine the stability of a closed-loop

control system are:

a. Find the closed-loop transfer function

b. Find the output c(t)

c. Find c(t) as t E


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Example 1

C(s)R(s)+

- 11

s

inputstepafors

1R(s)*

2

11

2

1

)()(

2

1

)(

)(

!

v!

!

!

ss

ssRsC

ssR

sC


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1

221

)(

)(

kc

ekktct

!g

!

Since the output stabilises at k1, the system is stable

Example 1...


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Example 2

C(s)R(s) +

- )5.0)(2.0(1

ss

)5.0)(2.0(1

11

)5.0)(2.0(1

1)()(

)5.0)(2.0(1

1

)(

)(

v!

!

!

sss

ssss

sss

s


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)sin()()( 135.0

211 J!! tekksCLtc t

system is stable.1)(lim ktct

!

gp

Example 2...


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Example 3

C(s)R(s) +

- )2.0)(5.0(1

ss

)2.0)(5.0(1

11

)2.0)(5.0(1

1)()(

)2.0)(5.0(1

1

)(

)(

v!

!

!

sss

ssss

sss

s


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)94.0sin()()( 115.0

211 J!! tekksCLtc t

gp

gp )(limtc

t

system is unstable.

Example 3...


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Example 4

C(s)+R(s)

- 31

2s

4

11

4

1

)()(

4

1

)(

)(

2

2

2

v!

!

!

ss

ssRsC

ssR

sC


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Sustained Oscillation

)2sin()()( 1211 J!! tkksCLtc

p

gp

)(lim tct

Example 4...

System is marginally stable


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Stability based on Roots of Characteristic Equation

The Characteristic Equation is formed from the denominator

of the closed-loop transfer function.

Consider the closed-loop control system :

C(s)R(s)+

-G(s)

H(s)


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0)()(1

:issystemtheo( )quationsticharacteriThe

)()(1

)(

)(

)(

:issystemor thisunctiontrans erloopclosedThe

!

!

sHsG

sHsG

sG

s

s


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Determination of the Stability

The following are steps to determine the stability of a

closed-loop control system using the Characteristic Equation:

1. Find the Characteristic Equation of the system.

2. Determine the roots of the Characteristic Equation.

3. Locate the roots on the S-plane.


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Determination of the Stability (continued..)

If the characteristic equation of the system has a

pair of complex-conjugate roots, the system response

will be oscillatory.

System is marginally stable if there is

at least one pair ofcomplex-conjugate

roots on the j[ axis.

4. System is stable ifall the root are in the

Left-Hand-Side of the S-plane.

System is unstable ifone or more rootsare in the Right-Hand-Side of the S-plane.


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We will now use the CE to determine the stability of

Example 1 to Example 4 :

The CE for Example 1 is :

position.2-at thecrossabydrepresenteisThis

-2sisrootthei.e.

02

!

!s

x

-2

Since the root lies on the LHS of the s-plane the

system is stable.


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In Example 2

)5.0)(2.0(1

1

)(

)(

!

sss

s

Roots :s1 = - 0.35 + j, s2 = - 0.35 - j

j0.661.25-

0)5.0)(2(1:

s!

! ssCE


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-0.3

x

x

j

-j

Since the roots are on the LHS of the s-plane, the

system is stable.

However, as the roots are complex, the transient period

is oscillatory.


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1)5.0)(2.0(

1

)(

)(

!

sssR

sC

In Example 3

CE : (s + 0.2)(s - 0.5) + 1 = 0

Roots of CE are located on RHS System is unstable

x

x

0.94j

-0.94j

0.15

Roots : s1 = 0.15 + 0.94j, s2 =0.15 - 0.94j


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4

1

)(

)(2

!

ssR

sC

In Example 4

CE : s2 + 4 = 0

x

x

2j

-2j

Roots : s1 = 2j, s2 = - 2jRoots of CE are on the j[ axis System is marginally stable


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Click anywhere on this slide to for an interesting

program that shows the relationship between the

pole positions(roots of the Characteristic equation

and the transient response.

Have fun.

stability part 1

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