Control Lecture 8 Poles Performance and Stability

7/28/2019 Control Lecture 8 Poles Performance and Stability

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Lecture 7: Poles, systemperformance and system stability


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Zeros: All the values of s for which n(s) = 0

G sn s

d s

b s b s b s b

a s a s a s a

m

m

m

m

n

n

n

n( )

( )

( )= =

+ + + +

+ + + +

1

1

1

1

0

1

1

1

1

0

Poles: All the values of s for which d(s) = 0

When the system G(s) has more than one pole or one zero at the

same co-ordinates on the s-plane, we say that G(s) has multiplepoles or multiple zeros.


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0.6

0.8

1

0

0.5

1

s1 s2 s3 s4

First

Order

Time

30 400

0.2

.

Real Axis-1 -0.5 0 0.5 1

-1

-0.5

1 2 3 4

Step Response Pole-zero mapSystem


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0

1

2

p1

R

-n- - 2

n 1-2

C O

Second Order

System

Real Axis

Pole-zero map

-1.5 -1 -0.5 0 0.5 1-2

-1 p2


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Second Order Systems:

For 0 < < 1,

G ss s

n

n n

( ) =+ +

2

2 22

Pole-zero m ap

-

-3

-2

-1

0

1

2

3

4

Lines of constant

damping rat io ,

Semicirc lesof constant

( )212

1n np j = +

The angle OCP1 is tan = ? and R2 = n

2

Therefore all poles with constant value for n- will lie on a semi-circle,

radius R(=n) from the origin.For different values of, we find a number of straight lines makingdifferent angles with the real axis. We can then easily establish a

relationship between the location of poles and and n

. This is shown

here.

Real A x is

-4 -3 -2 -1 0 1 -5

- n

2 n np =


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Step response performance measures

overshoot

rise time

peak time

settling time


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Step response

Complex conjugate poles,underdamped system:

21nt

2

1 1

2

2

1( ) ( ) ( )

1 2

1nn

KY s G s U s

ss s

= =

+ +

21n

Amplitude (steady-state value): K

Rise time: from the equation above find the smallest tsuch thaty(t)=K

(a numerical solution is required)90% Rise time: from the equation above find the smallest tsuch that

y(t)=0.9K(also suitable for overdamped systems, a numerical

solution is required)

Peak time: from the equation above find the smallest tsuch that:

( )

0

dy t

dt =


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Step response performance measures

overshoot

21p

n

T

=

Peak time:

Peak value:

2/ 1

40.02n s

Ts

n

e T

=

p

2% Settling time: find time Ts such that after this time the amplitudeis within 2% of the steady-state value


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Higher order systems

( ) ( ) ( )1 2

1 2 1 2

( ) m

m m

BK B BG s

s s s s s s s s s s s s= = + + +

Partial fraction decomposition of the transfer function:

Dominant pole of the transfer function: lowest absolute

value, i.e. closest to zero. Produces the slowest response


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What do we mean by bounded signals? A time domainsignal, x(t), is assessed by the behaviour of its magnitude

over an infinite time interval. As time tends to infinity,

the absolute value of the signal magnitude can either:continuously decrease and/or increase, (or stay constant)

but remain within a bounded range.


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Decayingexponential signals have Laplace transforms with poles inthe LHP.

Growing or increasing exponential signals have Laplace transforms

with poles in the RHP.

We can generalise this observation as follows:

Poles in LHP and RHP: Si nals whose trans orms have all the oles

in the LHP are bounded. Signals whose transforms have any one polein the RHP are unbounded.

Poles on j axis: Signals whose transforms have poles in the LHPand no multiple poles on the jaxis are bounded, otherwise they areunbounded.


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What is a stable system?We call a system stable if its output signal is bounded for any

bounded input signal. We call this type of system stability

bounded-input bounded-output stability.

Half Plane.

M Using MATLAB to check the stability of a system

Enter the System transfer function using: s = tf(s); g = ..

then either run: pzmap(g). If all

the poles are located inthe LHP the system is stable. Otherwise the system is unstable.


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Open loop system:

transfer function analysis:

The open-loop transfer function:

K(s) G(s)

R(s) U(s) Y(s)

( ) ( )( ) ( ) ( ) G K

ol

n s n sG s G s K s= =

( ) ( )( ) , K(s)

( ) ( )

G K

G K

n s n sG s

d s d s= =

Characteristic equation of the system: find the poles of the

transfer function.

Open loop poles are the roots of:

dOL

(s) = dK

(s)dG

(s) = 0

G K


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Unity feedback system

transfer function analysis:

K(s) G(s)

+

-

R(s) U(s) Y(s)

( ) ( )( ) , K(s)

( ) ( )

G K

G K

n s n sG s

d s d s= =

The closed-loop transfer function:

Characteristic equation of the system: find the poles of the

transfer function.

Closed loop poles are the roots of:

dCL(s) = dK(s)dG(s) + nK(s)nG(s) = 0

( ) ( )( ) ( )( )

1 ( ) ( ) ( ) ( ) ( ) ( )

G Kcl

G K G K

n s n sG s K sG s

G s K s d s d s n s n s= =

+ +


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Stability necessary condition

( ) 0

1 ( ) ( ) 0( ) ( ) ( ) ( ) 0

G K G K

ChE s

G s K sd s d s n s n s

=

+ =+ =

Characteristic equation (closed loop system):

1 2 2... 0n n nc s c s c s c s c s c + + + + + + =n-th order system

1 0

2

2 1 0

3 2

3 2 1 0

0

0

0

c s c

c s c s c

c s c s c s c

+ =

+ + =

+ + + =

First order system

Second order system

Third order system

Necessary condition for stability is that all coefficients in the Characteristic

Equation have the same sign (i.e. either positive or negative).


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Stability, Hurwitz-Ruth criterion

1 2 2

1 2 2 1 0... 0n n n

n n nc s c s c s c s c s c

+ + + + + + =

Assume that the Characteristic Equation is n-th order polynomial:

Build a matrix as follows: 1 3 52 4

1 3 5

0

0

0 0

n n n

n n n

n n n

c c c

c c c

c c c

2 4

0

0

0

n n nc c c

c

Calculate determinant of M and of all sub-matrices obtained bycutting the last k rows and columns of M.

If all are positive, and cn is positive, the system is stable.


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Stability, Hurwitz-Ruth criterion, examples

1 0

2

2 1 0

3 2

3 2 1 0

4 3 2

4 3 2 1 0

0

0

0

0

c s c

c s c s c

c s c s c s c

c s c s c s c s c

+ =

+ + =

+ + + =

+ + + + =

First order system

Second order system

Third order system

4-th order system

0M c=First order system For First order system and

1

1 0

2 0

0, 0

cM c c

c c

= >

2 0

3 1 2 1 3 0

2 0

0

0 , 0

0

c c

M c c c c c c

c c

= >

( )

3 1

3 2 4 1

4 2 0

3 1 2

1 3 2 4 1 0 34 2 0

0 00

0,

0 0

00

c cc c c c

c c cM and

c c

c c c c c c cc c c

>

=

>

Second order system

Third order system

4-th order system

Second order systemnecessary condition is also

sufficient condition.


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Examples

K(s) G(s)

+

-

R(s) U(s) Y(s)

Use SIMULINK to simulate the step responses of the systems

with transfer functions selected as below. Which combinations can

( )

p

Ip

Ip D

k

kK s k

s

kk k s

s

= +

+ +

2

2

1

1

1

( )

1 21

1

nn

s

K

sT

K

sT s

G s K

s s

Ke

sT

+

+

= + +

+


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Fast pump Slow pump

How do the zeros of a transfer function model arise?

Zeros arise from the internal physical pathways of a process and

represent where these internal effects are adding together orcompeting(subtracting ) with one another.

Example: The feeder tank is used to supply a steady flow of liquid

feed to downstream processes.

OutflowFeeder Tank

Unit

Voltage Step

Level

Inflow

5s+1

1

Slow pump 0.833

10s+1

Feeder Tank

0.75s+1

0.5

Fast pump


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Poles, zeros and stability

Poles and time responses for first order andsecond order systems

transfer function

Control Lecture 8 Poles Performance and Stability

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