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EL322near

Controlystems

Week-08

2ndApril 7thApril 2012

STABILITY

Instructor: Engr. Shiraz Latif/ Engr. Atif Fareed/ EngrAreeb Ahmed

Control System Design requirements,

Transient response covered in ch-04

Stability Ch-06 Steady-state errors ch-08

Stability:

Instructor:Shiraz Latif/ Atif Fareed/ Areeb Ahmed

BIBO Stability:

System is stable if

Every Bounded Input yields a Bounded Output.

the natural response approaches zero as time approaches infinity.

all poles in the left half-plane.

A system is marginally stable (stable) if the natural response neither decays nor grows but remains constant or oscillates

as t.

Closed-loop transfer function with Only imaginary axis poles of multiplicity 1 andother poles in the left half-plane.

Unstable System:

System is unstable if any Bounded Input yields a Unbounded Output.

The natural response grows without bound as t

l l l l l l l lose - loop ranser unc on w a leas one pole les n e rg a l-plane(positive real part) and/or poles of multiplicity greater than one on the imaginaryaxis

3Instructor:Shiraz Latif/ Atif Fareed/ Areeb Ahmed

A h sical s stem whose natural res onse rowswithout bound can cause damage to the system.

Time response plot shows the instability by transie

thus total response never approaches a steady state valu


Close loop poles LHP pure exponential decay or damped sinusoidal natural

response

Natural response decays to zero as t

System is stable

RHP ex onentiall increasin res onse +ve oles

Natural response approaches infinity to zero as t

System is unstable

Ima inar Axis oles with multi licit 1 ure sinusoidalresponse Natural response neither decay, nor grow

System is marginally stable


Figure 6.1

Closed-loop polesand response:a. stable s stem;


b. unstable system

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Marginally stable

Note that theroots are on the


Finding the stability through roots is not easy because Finding roots of higher order polynomial is difficult G(s)

Unity feedback gain ResponseC(s)=[G(s)/(1+G(s))] R(s)

Figure 6.2Common cause of problems in findingclosed-loo oles:

Routh-Hurwitz Criterion is the techni ue to find theG(s)/[1+G

a. original system;b. equivalent system

stability of a system, without finding the location of poles.Instructor:Shiraz Latif/ Atif Fareed/ Areeb Ahmed

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A method that ields stabilit information without theneed to solve for the close-loop poles.

It tells how many poles lies in the RHP (right half

Do not provide the exact location(coordinates) ofoles.

STEPS:

1. Generate Routh table

2. Interpret Routh table tell how many poles are,


Generating Routh table Generate Routh-table for the following system

# rows = O(D(s))+1,

#col = O(D(S)/2 for even order

O(D(s)+1/2 for odd order

excluding 1st col

Write s powers in every row

Fill Alternate coefficients of D(s) inst nd

Fill 3rd row by considering above 2 rows

like shown here

Fill each nth row by consider

(n-1)th row determinants


No. of sign change in 1st column shows

No. of unstable roots

.

Make Routh table for

11Instructor:Shiraz Latif/ Atif Fareed/ Areeb Ahmed 1Instructor:Shiraz Latif/ Atif Fareed/ Areeb Ahmed

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different way

st

Prob: Division by zero

Two Solutions:

replace zero by a very small value

Coefficient reversal

Make a polynomial of row above the row of zero

Derivate and get the new coefficients




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roblems




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Week-8 v1 (6slides) + HW