Dr.kourosh Kiani Linear Control Systems Lecture 11 Transient and Steady-State Analysis-Second Order...
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7/31/2019 Dr.kourosh Kiani Linear Control Systems Lecture 11 Transient and Steady-State Analysis-Second Order System
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Linear-Control-Systems
Present to:Amirkabir University of Technology (Tehran Polytechnic) &
Semnan University
Dr. Kourosh KianiEmail: [email protected]
Email: [email protected]
Web: www.kouroshkiani.com
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Lectureecture 1111ectureecture 1111
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Transient and Steady-Stateransient and Steady-StateResponse Analysesesponse Analysesof
Second-order systems
Transient and Steady-Stateransient and Steady-StateResponse Analysesesponse AnalysesofSecond-order systems
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Time response of 2-order systems
What is a second-order system?
Second-order systems are described bysecond-order differential equations.
Example
2
2 2
2( ) ( ) ( ) ( )2 n n n
d dy t y t y t u t
dt dt
+ + =
A prototype second-order differential equation:
y(t)---output respon se of the system;
u(t)---input to the system
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Using Laplace transform and assuming zero initialconditions, we get:
2 2 2( ) ( ) ( ) ( )2 n n ns Y s sY s Y s U s + + =
Transfer function of a second-order system:
2
2 2( )( )( ) 2
n
n n
Y sG sU s s s
= =+ +
, will determine how fastthe system oscillates during any transient response
- undamped natural frequencyn
, will determine how much the systemoscillates as the response decays toward steady state.
- damping ratio
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Finding nand for a second-ordersystem
G(s) = n2
s2
+ 2ns+
n
2
n2
= 22
2n
= 2.2
n
= 2
= 2.33
s2+2
ns+
n
2=2
s2=
n+
n22
s2=
n
n22
G(s)=22
s2+2.2s+22
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Step responseStep response of second-order systemsof second-order systems
A 2-order system:
2
2 2( ) 2n
n nG s s s
= + +
Case 1: underdamp( ),2 ed < including =2unda( )mped
2( ) sin( ),2 nt
ny t e t
= +2
2
where = 2
=tan ( / )
Case : overdam2 d( )pe >
2 2/ /
2 2( ) 2 t ty t k e k e = + + ,22
2
2where
2n n
=
Case 3: critica(2 lly da )mped =
/ /
2 2( ) 2 t ty t k e k e = + + where 2 n =
Input :
( ) ( );2
2( )
u t t
U ss
=
=
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Step responseStep response of second-order systemsof second-order systems
A 2-order system:2
2 2( )
2
n
n n
G s
s s
=
+ +
Case 1: underdamp( ),2 ed
2 2/ /
2 2( ) 2 t ty t k e k e
= + +
Case 3: critically dam( )d2 pe =
/ /
2 2( ) 2 t ty t k e k e = + +
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t0
1
y(t)The time response
of a control systemis usually dividedinto two parts:
Transient and steady-state response
( ) ( ) ( )t ss
y t y t y t= +
Transient response-- defined as the part of thetime response that goes tozero as time becomes very
large.
( )ty t
lim ( ) 2tt
y t
=
Steady-state response
-- the part of the timeresponse that remains afterthe transient has died out.
( )ss
y t
The steady-state response can still vary in a fixed pattern, such as a sine
wave, or a ramp function that increases with time.
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Time-domain SpecificationsTime-domain Specifications
. Maximum overshoot:
max;
p ssM y y=
4. Rise time: rt
5. Settling time: st
1. Steady-state value: ssy
Percent overshoot %222p
ss
M
y=
ssy
pt
3. Peak time: pt
st
error band
0.1
rt
0.9
pM
maxy
How to calculate ?pt
Is there overshoot inthe time response of afirst-order system?
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In practical applications, the followingcriteria are often usedRise time : evaluate the response speed of thesystem quickness)
Overshoot: evaluate the damping of thesystem (smoothness)Settling time: reflect both response speedand damping
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Relationship between the characteristic-equation roots and the step response
A second-order system:
2
2 2( ) 2n
n n
G ss s
= + +
Its characteristic equation: 2 2( ) 22n nD s s s = + + =
The value of determines the location of the roots of ( ) .2D s =
:3 3< < 2,22 2n ns j = underdamped
:2 = ,22 ns = critically damped
:2 > 2,22 2n ns = overdamped
:2 = ,22 ns j= undamped
:2 < 2,22 2n ns j = negatively damped
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:2 >
overdamped
:2 =
critically damped
:2 2< num=[9];>> den=[1 9];
>> pzmap(num,den)>> axis([- -4 4]);>> grid on;>> step(num,den)
Underdamped
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c(t)=2et(cos 2t+
2
2sin2t)
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2
2
2+s
ssR
2)( =
)(sG
)(sC
>> num=[9];>> den=[1 9];
>> pzmap(num,den)>> axis([- -4 4]);>> grid on;>> step(num,den)
Undamped
C(s)= 2s(s2+2)
c(t)=2cos2t
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Critically damped
C(s)=2
s(s2+2s+2)=2
s(s+2)2
c(t)=22te2te2t
22
2
2++ ss
ssR
2)( =
)(sG
)(sC
>> num=[9];>> den=[1 6 9];
>> pzmap(num,den)>> axis([- -4 4]);>> grid on;>> step(num,den)
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Step response for second ordersystemdamping cases
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Summary Overdampedverdamped
Poles: Two real at -Poles: Two real at - 11, -, - UnderdampednderdampedPoles: Two complex at -Poles: Two complex at - dd + j+ jdd, -, - dd - j- jdd Undampedndamped
Poles: Two imaginary at + jPoles: Two imaginary at + j11, - j, - j11 Criticallyritically dampedamped
Poles: Two real at -Poles: Two real at - 11,,
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Constant real part
Constant imaginary part
Constant damping ratio
S ggestions ?
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Suggestions ?
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