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TIME DOMAIN ANALYSISEmail: [email protected]
URL: http://shasansaeed.yolasite.com/
1SYED HASAN SAEED
BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO & GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 2
DEFINITIONS
TIME RESPONSE: The time response of a system is the output (response) which is function of the time, when input (excitation) is applied.
Time response of a control system consists of two parts
1. Transient Response 2. Steady State Response
Mathematically,
Where, = transient response
= steady state response
SYED HASAN SAEED 3
)()()( tctctc sst
)(tct
)(tcss
TRANSIENT RESPONSE: The transient response is thepart of response which goes to zero as timeincreases. Mathematically
The transient response may be exponential oroscillatory in nature.
STEADY STATE: The steady state response is the part ofthe total response after transient has died.
STEADY STATE ERROR: If the steady state response ofthe output does not match with the input then thesystem has steady state error, denoted by .
SYED HASAN SAEED 4
0)(
tcLimit tt
sse
TEST SIGNALS FOR TIME RESPONSE:
For analysis of time response of a control system,
following input signals are used
1. STEP FUNCTION:
Consider an independent voltage source in series witha switch ‘s’. When switch open the voltage atterminal 1-2 is zero.
SYED HASAN SAEED 5
Mathematically,
;
When the switch is closed at t=0
;
Combining above two equations
;
;
A unit step function is denoted by u(t) and defined as
;
;
SYED HASAN SAEED 6
0)( tv 0 t
Ktv )( t0
Ktv
tv
)(
0)( 0 t
t0
1)(
0)(
tu
tu 0t
t0
Laplace transform:
£f(t)=
2. RAMP FUNCTION:
Ramp function starts from origin and increases ordecreases linearly with time. Let r(t) be the rampfunction then,
r(t)=0 ; t<0
=Kt ; t>0
SYED HASAN SAEED 7
ss
edtedtetu
ststst 1
.1)(
000
K>0
t
r(t)
LAPLACE TRANSFORM:
£r(t)
For unit ramp K=1
SYED HASAN SAEED 8
2
00
)(s
KdtKtedtetr stst
2)(
s
KsR
tr(t)
K<0
0
3. PARABOLIC FUNCTION:
The value of r(t) is zero for t<0 and is quadratic functionof time for t>0. The parabolic function represents asignal that is one order faster than the ramp function.
The parabolic function is defined as
For unit parabolic function K=1
SYED HASAN SAEED 9
2)(
0)(
2Kttr
tr
0
0
t
t
2)(
0)(
2ttr
tr
0
0
t
t
LAPLACE TRANSFORM:
£r(t)
SYED HASAN SAEED 10
3
3
0 0
2
)(
2)(
s
KsR
s
Kdte
Ktdtetr stst
IMPULSE RESPONSE: Consider the following fig.
The first pulse has a width T and height 1/T, area of thepulse will be 1. If we halve the duration and doublethe amplitude we get second pulse. The area underthe second pulse is also unity.
SYED HASAN SAEED 11
We can say that as the duration of the pulseapproaches zero, the amplitude approaches infinitybut area of the pulse is unity.
The pulse for which the duration tends to zero andamplitude tends to infinity is called impulse. Impulsefunction also known as delta function.Mathematically
δ(t)= 0 ; t ≠ 0
=∞ ; t = 0
Thus the impulse function has zero value
everywhere except at t=0, where the amplitude
is infinite.
SYED HASAN SAEED 12
An impulse function is the derivative of a step function
δ(t) = u(t)
£δ(t) = £
SYED HASAN SAEED 13
11
.)( s
studt
d
INPUT r(t) SYMBOL R(S)
UNIT STEP U(t) 1/s
UNIT RAMP r(t) 1/s2
UNIT PARABOLIC - 1/s3
UNIT IMPULSE δ(t) 1
THANK YOU FOR
ATTENTION
SYED HASAN SAEED 14
TIME RESPONSE
OF
FIRST ORDER SYSTEM
Email : [email protected]
URL: http://shasansaeed.yolasite.com/
1SYED HASAN SAEED
BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 2
RESPONSE OF FIRST ORDER SYSTEM WITH UNIT STEPINPUT:
For first order system
SYED HASAN SAEED 3
sT
T
ssC
sTssC
ssR
sRsT
sC
sTsR
sC
1
1)(
)1(
1)(
1)(
)(1
1)(
1
1
)(
)(
Input is unit step
After partial fraction
Take inverse Laplace
Where ‘T’ is known as ‘time constant’ and defined asthe time required for the signal to attain 63.2% offinal or steady state value.
Time constant indicates how fast the system reachesthe final value.
Smaller the time constant, faster is the systemresponse.
SYED HASAN SAEED 4
632.011)(
1)(
1/
/
eetc
etc
TT
Tt
When t=T
RESPONSE OF FIRST ORDER SYSTEM WITH UNIT RAMPFUNCTION:
SYED HASAN SAEED 5
Ts
Ts
T
ssC
sTssC
ssR
sRsT
sC
sTsR
sC
1
11)(
)1(
1)(
1)(
)(1
1)(
1
1
)(
)(
2
2
2
Input is unit Ramp
After partial fraction
We know that
Take inverse Laplace, we get
The steady state error is equal to ‘T’, where ‘T’ is thetime constant of the system.
For smaller time constant steady state error will besmall and speed of the response will increase.
SYED HASAN SAEED 6
TTeTLimit
eTte
TeTttte
tctrte
TeTttc
Tt
t
Tt
Tt
Tt
)(
)1()(
)(
)()()(
)(
/
/
/
/
Error signal
Steady state error
RESPONSE OF THE FIRST ORDER SYSTEM WITH UNIT IMPULSE FUNCTION:
Input is unit impulse function R(s)=1
SYED HASAN SAEED 7
TteT
tc
TsTsC
sTsC
sRsT
sC
/1)(
/1
11)(
1.1
1)(
)(1
1)(
Inverse Laplace transform
SYED HASAN SAEED 8
SYED HASAN SAEED 9
1)(
1)(
1)(
2
sR
ssR
ssRFor unit Ramp Input
For Unit Step Input
For Unit Impulse Input
TtTeTttc /)(
Ttetc /1)(
TteT
tc /1)(
It is clear that, unit step input is the derivative of unit ramp input and unit impulse input is the derivative of unit step input. This is the property of LTI system.
Compare all three responses:
THANK YOU
SYED HASAN SAEED 10
TIME RESPONSE
OF
SECOND ORDER SYSTEM
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URL: http://shasansaeed.yolasite.com/
1SYED HASAN SAEED
REFERENCE BOOKS:
1. AUTOMATIC CONTROL SYSTEM KUO & GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 2
SYED HASAN SAEED 3
Block diagram of second order system is shown in fig.
R(s) C(s)_
+)2(
2
n
n
ss
ssR
sRsssR
sC
AsssR
sC
nn
n
nn
n
1)(
)(2)(
)(
)(2)(
)(
22
2
22
2
For unit step input
SYED HASAN SAEED 4
22
22
2
2
)1(2
.1
)(
nn
nn
n
ss
ssssC
Replace by )1()( 222 nns
Break the equation by partial fraction and put )1( 222 nd
1
)3()()(
.1
2222
2
A
s
B
s
A
ss dndn
n
)2()1()(
.1
)(222
2
nn
n
sssC
SYED HASAN SAEED 5
22)( dns Multiply equation (3) by and put
)2()(
))((
)(2
2
2
nnn
nd
dndn
dnn
dn
n
n
dn
ssB
sj
jj
jB
jB
sB
js
Equation (1) can be written as
SYED HASAN SAEED 6
)4()(
.)(
1)(
)(
1)(
2222
22
dn
d
d
n
dn
n
dn
nn
ss
s
ssC
s
s
ssC
Laplace Inverse of equation (4)
)5(sin.cos.1)(
tetetc d
t
d
nd
t nn
21 ndPut
SYED HASAN SAEED 7
tte
tc
ttetc
dd
t
dd
t
n
n
sincos.11
1)(
sin.1
cos1)(
2
2
2
)sin(1
1)(
1tan
cos
sin1
2
2
2
te
tc d
tn
Put
SYED HASAN SAEED 8
)6(1
tan)1(sin1
1)(2
12
2
te
tc n
tn
Put the values of d &
)7(1
tan)1(sin1
)(
)()()(
212
2
te
te
tctrte
n
tn
Error signal for the system
The steady state value of c(t)
1)(
tcLimitet
ss
Therefore at steady state there is no error betweeninput and output.
= natural frequency of oscillation or undampednatural frequency.
= damped frequency of oscillation.
= damping factor or actual damping ordamping coefficient.
For equation (A) two poles (for ) are
SYED HASAN SAEED 9
n
d
n
2
2
1
1
nn
nn
j
j
10
Depending upon the value of , there are four cases
UNDERDAMPED ( ): When the system has two complex conjugate poles.
SYED HASAN SAEED 10
10
From equation (6):
Time constant is
Response having damped oscillation with overshoot andunder shoot. This response is known as under-dampedresponse.
SYED HASAN SAEED 11
n/1
UNDAMPED ( ): when the system has two imaginary poles.
SYED HASAN SAEED 12
0
From equation (6)
Thus at the system will oscillate.
The damped frequency always less than the undampedfrequency ( ) because of . The response is shown infig.
SYED HASAN SAEED 13
ttc
ttc
n
n
cos1)(
)2/sin(1)(
0
For
n
n
SYED HASAN SAEED 14
CRITICALLY DAMPED ( ): When the system has two real and equal poles. Location of poles for critically damped is shown in fig.
1
SYED HASAN SAEED 15
)(
11
)(
)()(
2.
1)(
1
2
2
2
2
22
2
n
n
nn
n
n
n
nn
n
ssssss
sssC
ssssC
For
After partial fraction
Take the inverse Laplace
)8()1(1)(
1)(
tetc
etetc
n
t
t
n
t
n
nn
SYED HASAN SAEED 16
From equation (6) it is clear that is the actual damping. For , actual damping = . This actual damping is known as CRITICAL DAMPING.The ratio of actual damping to the critical damping is known as damping ratio . From equation (8) time constant = . Response is shown in fig.
n
1 n
n/1
OVERDAMPED ( ): when the system has two realand distinct poles.
SYED HASAN SAEED 17
1
Response of the system
From equation (2)
SYED HASAN SAEED 18
)9()1()(
.1
)(222
2
nn
n
sssC
)1( 222 ndPut
)10()(
.1
)(22
2
dn
n
sssC
We get
Equation (10) can be written as
)11())((
)(2
dndn
n
ssssC
After partial fraction of equation (11) we get
SYED HASAN SAEED 19
Put the value of d
)12(
112
1
112
11)(
22
22
dn
dn
s
sssC
)13(
)1(112
1
)1(112
11)(
222
222
nn
nn
s
sssC
Inverse Laplace of equation (13)
From equation (14) we get two time constants
SYED HASAN SAEED 20
)14()1(12)1(12
1)(22
)1(
22
)1( 22
tt nn eetc
n
n
T
T
)1(
1
)1(
1
22
21
SYED HASAN SAEED 21
)15()1(12
1)(22
)1( 2
tnetc
From equation (14) it is clear that when is greater thanone there are two exponential terms, first term has timeconstant T1 and second term has a time constant T2 . T1 <T2 . In other words we can say that first exponential termdecaying much faster than the other exponential term.So for time response we neglect it, then
)16()1(
1
22
n
T
THANK YOU
SYED HASAN SAEED 22
TIME DOMAIN SPECIFICATIONS OF SECOND ORDER SYSTEM
Email : [email protected]
URL: http://shasansaeed.yolasite.com/
1SYED HASAN SAEED
SYED HASAN SAEED 2
BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 3
Consider a second order system with unit step input andall initial conditions are zero. The response is shown in fig.
1. DELAY TIME (td): The delay time is the time requiredfor the response to reach 50% of the final value infirst time.
2. RISE TIME (tr): It is time required for the response torise from 10% to 90% of its final value for over-damped systems and 0 to 100% for under-dampedsystems.
We know that:
SYED HASAN SAEED 4
21
2
2
1tan
1sin1
1)(
te
tc n
tn
Where,
Let response reaches 100% of desired value. Put c(t)=1
SYED HASAN SAEED 5
01sin1
1sin1
11
2
2
2
2
te
te
n
t
n
t
n
n
0 tneSince,
)sin())1sin((
0))1sin((
2
2
nt
t
n
n
Or,
Put n=1
SYED HASAN SAEED 6
2
2
1
)1(
n
r
rn
t
t
2
21
1
1tan
n
rt
Or,
Or,
3. PEAK TIME (tp): The peak time is the time requiredfor the response to reach the first peak of the timeresponse or first peak overshoot.
For maximum
SYED HASAN SAEED 7
te
tc n
tn
2
21sin
11)(Since
)1(01
1sin
11cos1
)(
0)(
2
2
22
2
tnn
nn
t
n
n
et
te
dt
tdc
dt
tdc
Since,
Equation can be written as
Equation (2) becomes
SYED HASAN SAEED 8
0 tne
sin1
1sin11cos
2
222
tt nn
Put cosand
cos1sinsin1cos 22 tt nn
cos
sin
))1cos((
))1sin((
2
2
t
t
n
n
SYED HASAN SAEED 9
nt
nt
pn
n
)1(
))1tan((
2
2
The time to various peak
Where n=1,2,3,…….
Peak time to first overshoot, put n=1
21
n
pt
First minimum (undershoot) occurs at n=2
2min
1
2
n
t
4. MAXIMUM OVERSHOOT (MP):
Maximum overshoot occur at peak time, t=tp
in above equation
SYED HASAN SAEED 10
te
tc n
tn
2
21sin
11)(
21
n
ptPut,
2
2
2
1
1.1sin
11)(
2
n
n
n
n
etc
SYED HASAN SAEED 11
2
2
1
2
2
1
2
1
11
1)(
1sin
sin1
1)(
)sin(1
1)(
2
2
2
etc
etc
etc
Put,
sin)sin(
SYED HASAN SAEED 12
2
2
2
1
1
1
11
1)(
1)(
eM
eM
tcM
etc
p
p
p
100*%21
eM p
5. SETTLING TIME (ts):
The settling time is defined as the time required for thetransient response to reach and stay within theprescribed percentage error.
SYED HASAN SAEED 13
SYED HASAN SAEED 14
Time consumed in exponential decay up to 98% of the input. The settling time for a second order system is approximately four times the time constant ‘T’.
6. STEADY STATE ERROR (ess): It is difference betweenactual output and desired output as time ‘t’ tends toinfinity.
n
s Tt
44
)()( tctrLimitet
ss
EXAMPLE 1: The open loop transfer function of a servosystem with unity feedback is given by
Determine the damping ratio, undamped natural frequencyof oscillation. What is the percentage overshoot of theresponse to a unit step input.
SOLUTION: Given that
Characteristic equation
SYED HASAN SAEED 15
)5)(2(
10)(
sssG
1)(
)5)(2(
10)(
sH
sssG
0)()(1 sHsG
SYED HASAN SAEED 16
0207
0)5)(2(
101
2
ss
ss
Compare with 02 22 nnss We get
%92.1100*
7826.0
7472.4**2
sec/472.420
72
20
22 )7826.0(1
7826.0*
1
2
eeM
rad
p
n
n
n
%92.1
7826.0
sec/472.4
p
n
M
rad
EXAMPLE 2: A feedback system is described by thefollowing transfer function
The damping factor of the system is 0.8. determine theovershoot of the system and value of ‘K’.
SOLUTION: We know that
SYED HASAN SAEED 17
KssH
sssG
)(
164
12)(
2
016)164(
16)164(
16
)(
)(
)()(1
)(
)(
)(
2
2
sKs
sKssR
sC
sHsG
sG
sR
sC
is the characteristic eqn.
Compare with
SYED HASAN SAEED 18
K
ss
n
n
nn
1642
16
02
2
22
.sec/4radn
K1644*8.0*2 15.0K
%5.1
100*100*22 )8.0(1
8.0
1
p
p
M
eeM
EXAMPLE 3: The open loop transfer function of a unityfeedback control system is given by
By what factor the amplifier gain ‘K’ should be multiplied sothat the damping ratio is increased from 0.3 to 0.9.
SOLUTION:
SYED HASAN SAEED 19
)1()(
sTs
KsG
0
/
)(
)(
1.)1(
1
)1(
)()(1
)(
)(
)(
2
2
T
K
T
ss
T
K
T
ss
TK
sR
sC
sTs
K
sTs
K
sHsG
sG
sR
sC
Characteristic Eq.
Compare the characteristic eq. with
Given that:
SYED HASAN SAEED 20
02 22 nnss
T
K
T
n
n
2
12
We get
TT
K 12
T
Kn
KT2
1Or,
9.0
3.0
2
1
TK
TK
2
2
1
1
2
1
2
1
SYED HASAN SAEED 21
21
2
1
2
1
2
2
1
9
9
1
9.0
3.0
KK
K
K
K
K
Hence, the gain K1 at which 3.0 Should be multiplied
By 1/9 to increase the damping ratio from 0.3 to 0.9
BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 2
STEADY STATE ERROR:
The steady state error is the difference between theinput and output of the system during steady state.
For accuracy steady state error should be minimum.
We know that
The steady state error of the system is obtained by finalvalue theorem
SYED HASAN SAEED 3
)()(1
)()(
)()(1
1
)(
)(
sHsG
sRsE
sHsGsR
sE
)(.lim)(lim0
sEsteest
ss
SYED HASAN SAEED 4
)(1
)(.lim
1)(
)()(1
)(.lim
0
0
sG
sRse
sH
sHsG
sRse
sss
sss
For unity feedback
Thus, the steady state error depends on the input and open loop transfer function.
STATIC ERROR COEFFICIENTS
STATIC POSITION ERROR CONSTAN Kp: For unit step input R(s)=1/s
SYED HASAN SAEED 5
)()(lim
1
1
)()(lim1
1
)()(1
1.
1.lim
0
0
0
sHsGK
KsHsGe
sHsGsse
sp
ps
ss
sss
Where is the Static position error constantpK
Steady state error
STATIC VELOCITY ERROR CONSTANT (Kv):
Steady state error with a unit ramp input is given by
R(s)=1/s2
SYED HASAN SAEED 6
)()(1
1).(.
0 sHsGsRsLime
sss
vs
ss
ssss
KsHssGe
sHssGssHsGsse
1
)()(
1lim
)()(
1lim
)()(1
1.
1.lim
0
020
Where )()(lim0
sHssGKs
v
Static velocity error coefficient
STATIC ACCELERATION ERROR CONSTANT (Ka):
The steady state error of the system with unit parabolic input is given by
where,
SYED HASAN SAEED 7
as
ss
ssss
KsHsGse
sHsGsssHsGsse
ssR
1
)()(
1lim
)()(
1lim
)()(1
1.
1.lim
1)(
20
22030
3
)()(lim 2
0sHsGsK
sa
Static acceleration constant.
STEADY STATE ERROR FOR DIFFERENT TYPE OF SYSTEMS
TYPE ZERO SYSTEM WITH UNIT STEP INPUT:
Consider open loop transfer function
SYED HASAN SAEED 8
)1().........1)(1(
.).........1)(1()()( 21
ba
m sTsTs
sTsTKsHsG
Ke
KKe
KsHsGK
ssR
ss
p
ss
sp
1
1
1
1
1
1
)()(lim
1)(
0 Hence , for type zero system the static position error constant Kp is finite.
TYPE ZERO SYSTEM WITH UNIT RAMP INPUT:
TYPE ZERO SYSTEM WITH UNIT PARABOLIC INPUT:
For type ‘zero system’ the steady state error is infinite for ramp and parabolic inputs. Hence, the ramp and parabolic inputs are not acceptable.
SYED HASAN SAEED 9
v
ss
bass
v
Ke
sTsT
sTsTKssHssGK
1
0)....1)(1(
)...1)(1(.lim)()(lim 21
00
sse
a
ss
bass
a
Ke
sTsT
sTsTKssHsGsK
1
0)....1)(1(
)...1)(1(.lim)()(lim 212
0
2
0
sse
TYPE ‘ONE’ SYSTEM WITH UNIT STEP INPUT:
Put the value of G(s)H(s) from eqn.1
TYPE ‘ONE’ SYSTEM WITH UNIT RAMP INPUT:
Put the value of G(s)H(s) from eqn.1
SYED HASAN SAEED 10
)()(lim0
sHsGKs
p
01
1
p
ss
p
Ke
K
0 sse
)()(.lim0
sHsGsKs
v
KKe
KK
v
ss
v
11
Kess
1
SYED HASAN SAEED 11
TYPE ‘ONE’ SYSTEM WITH UNIT PARABOLIC INPUT:
Put the value of G(s)H(s) from eqn.1
Hence, it is clear that for type ‘one’ system step input and ramp inputs are acceptable and parabolic input is not acceptable.
)()(lim 2
0sHsGsK
sa
a
ss
a
Ke
K
1
0
sse
Similarly we can find for type ‘TWO’ system.
For type two system all three inputs (step, Ramp,Parabolic) are acceptable.
SYED HASAN SAEED 12
INPUT SIGNALS
TYPE ‘0’ SYSTEM
TYPE ‘1’ SYSTEM
TYPE ‘2’ SYSTEM
UNIT STEP INPUT 0 0
UNIT RAMP INPUT 0
UNIT PARABOLIC
INPUT
K1
1
K
1
K
1
DYNAMIC ERROR COEFFICIENT:
For the steady-state error, the static error coefficients gives the limited information.
The error function is given by
For unity feedback system
The eqn.(2) can be expressed in polynomial form (ascending power of ‘s’)
SYED HASAN SAEED 13
)1()()(1
1
)(
)(
sHsGsR
sE
)2()(1
1
)(
)(
sGsR
sE
SYED HASAN SAEED 14
)3(........111
)(
)( 2
321
sK
sKKsR
sE
)4().......(1
)(1
)(1
)( 2
321
sRsK
ssRK
sRK
sEOr,
Take inverse Laplace of eqn.(4), the error is given by
)5(.......)(1
)(1
)(1
)(321
trK
trK
trK
te
Steady state error is given by
)(lim0
ssEes
ss
Let s
sR1
)(
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1
2
3210
1
.......1
.11
..11
.1
.lim
Ke
ss
Kss
KsKse
ss
sss
Similarly, for other test signal we can find steady state error.
.......,, 321 KKK are known as “Dynamic error coefficients”
EXAMPLE 1: The open loop transfer function of unity feedback system is given by
Determine the static error coefficients
SOLUTION:
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)10)(1.01(
50)(
sssG
avp KKK ,,
0)10)(1.01(
50lim
)()(
0)10)(1.01(
50.lim
)()(.lim
5)10)(1.01(
50lim
)()(lim
2
0
2
0
0
0
0
sss
sHsGsK
sss
sHsGsK
ss
sHsGK
s
a
s
sv
s
sp
EXAMPLE 2: The block diagram of electronic pacemaker isshown in fig. determine the steady state error for unitramp input when K=400. Also, determine the value of Kfor which the steady state error to a unit ramp will be0.02.
Given that: K=400,
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,1
)(2s
sR 1)( sH
)20()()(
ss
KsHsG
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05.0
)20(1
1.
1.lim
)()(1
)(.lim
200
ss
Kss
sHsG
sRse
ssss
Now, 02.0sse Given
1000
)20(
20lim02.0
)20(1
1.
1.lim
0
20
K
Kss
s
ss
Ksse
s
sss
THANK YOU
SYED HASAN SAEED 19
BASIC CONTROL ACTION AND CONTROLLER CHARACTERISTICS
Email : [email protected]
URL: http://shasansaeed.yolasite.com/
1SYED HASAN SAEED
SYED HASAN SAEED 2
BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
INTRODUCTION:
The automatic controller determines the value ofcontrolled variable, compare the actual value to thedesired value, determines the deviations andproduces a control signal that will reduce thedeviation to zero or to a smallest possible value.
The method by which the automatic controllerproduces the control signal is called control action.
The control action may operate through eithermechanical, hydraulic, pneumatic or electro-mechanical means.
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ELEMENTS OF INDUSTRIAL AUTOMATIC CONTROLLER:
The controller consists of :
Error Detector
Amplifier
The measuring element, which converts the output variable to another suitable variable such as displacement, pressure or electrical signals, which can be used for comparing the output to the reference input signal.
Deviation is the difference between controlled variable and set point (reference input).
e=r-b
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CLASSIFICATION OF CONTROLLERS:
Controllers can be classified on the basis of type ofcontrolling action used. They are classified as
i. Two position or ON-OFF controllers
ii. Proportional controllers
iii. Integral controllers
iv. Proportional-plus-integral controllers
v. Proportional-plus-derivative controllers
vi. Proportional-plus-integral-plus-derivativecontrollers
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Controllers can also be classified according to the power source used for actuating mechanism, such as electrical, electronics, pneumatic and hydraulic controllers.
TWO POSITION CONTROL: This is also known as ON-OFF or bang-bang control.
In this type of control the output of the controller is quickly changed to either a maximum or minimum value depending upon whether the controlled variable (b) is greater or less than the set point.
Let m= output of the controller
M1=Maximum value of controller’s output
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M2=Minimum value of controller’s output
E= Actuating error signal or deviation
The equations for two-position control will be
m=M1 when e>0
m=M2 when e<0
The minimum value M2 is usually either zero or –M1
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BLOCK DIAGRAM OF ON-OFF CONTROLLER
Block diagram of two position controller is shown inprevious slide. In such type of controller there is anoverlap as the error increases through zero ordecreases through zero. This overlap creates a spanof error. During this span of error, there is no changein controller output. This span of error is known asdead zone or dead band.
Two position control mode are used in room airconditioners, heaters, liquid level control in largevolume tank.
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PROPORTIONAL CONTROL ACTION: In this type ofcontrol action there is a continuous linear relationbetween the output of the controller ‘m’ andactuating signal ‘e’. Mathematically
Where, Kp is known as proportional gain orproportional sensitivity.
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)(
)(
)()(
)()(
sE
sMK
sEKsM
teKtm
p
p
p
In terms of Laplace Transform
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INTEGRAL CONTROL ACTION:
In a controller with integral control action, the outputof the controller is changed at a rate which isproportional to the actuating error signal e(t).
Mathematically,
Where, Ki is constant
Equation (1) can also be written as
Where m(0)=control output at t=0
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)1()()( teKtmdt
di
)2()0()()( mteKtm i
Laplace Transform of eqn. (1)
The block diagram and step response is shown in fig.
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)3()(
)(
)()(
s
K
sE
sM
sEKssM
i
i
The inverse of Ki is called integral time Ti and is defined as time of change of output caused by a unit change of actuating error signal. The step response is shown in fig.
For positive error, the output of the controller is ramp.
For zero error there is no change in the output of the controller.
For negative error the output of the controller is negative ramp.
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DERIVATIVE CONTROL ACTION:
In a controller with derivative control action the output of the controller depends on the rate of change of actuating error signal e(t). Mathematically,
Where is known as derivative gain constant.
Laplace Transform of eqn. (1)
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)1()()( tedt
dKtm d
dK
)2()(
)(
)()(
d
d
sKsE
sM
ssEKsM
Eqn. (2) is the transfer function of the controller.
From eqn.(1) it is clear that when the error is zero orconstant, the output of the controller will be zero.Therefore, this type of controller cannot be usedalone.
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BLOCK DIAGRAM OF DERIVATIVE CONTROLLER
PROPORTIONAL-PLUS-INTEGRAL CONTROL ACTION:
This is the combination of proportional and integralcontrol action. Mathematically,
Laplace Transform of both eqns.
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)2()(1
)()(
)1()()()(
0
0
t
i
pp
t
ipp
dtteT
KteKtm
dtteKKteKtm
)3(1
1)(
)(
)()()(
i
p
i
p
p
sTK
sE
sM
sEsT
KsEKsM
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Block diagram shown below
In eqn (3) both and are adjustable. Is calledintegral time. The inverse of integral time is calledreset rate. Reset rate is defined as the number oftimes per minute that the proportional part of theresponse is duplicate.
Consider the fig.(2a), the error varies at
pK iT iT
1tt
Fig. (1)
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Fig. (2a)
Fig. (2b)
Consider the diagram the output of the controllersuddenly changes to due to proportional action,after that controller output changes linearly withrespect to time at rate
For unit step (t1=0), the response shown in fig.(2b).
From equation (2) it is clear that the proportionalsensitivity affects both the proportional andintegral parts of the action.
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i
p
T
K
pm
pK
PROPORTIONAL-PLUS- DERIVATIVE CONTROL ACTION:
When a derivative control action is added in series withproportional control action, then this combination isknown as proportional-derivative control action.
Block diagram is shown in fig.
Mathematically,
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)1()()()( tedt
dTKteKtm ppp
Laplace Transform of equation(1)
This is the transfer function. is known as derivativetime. Derivative time is defined as the time intervalby which the rate action advances the effect of theproportional control action.
PD control action reduces the rise time, fasterresponse, improves the bandwidth and improves thedamping.
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)2()1()(
)(
)()()(
dp
dpp
sTKsE
sM
ssETKsEKsM
dT
Consider the diagram, if the actuating error signal e(t) is aramp function at t=t1. the derivative mode causes a stepmd at t1 and proportional mode causes a rise of mp equalto md at t2. this is for direct action PD control.
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For unit ramp input, the output of the controller isshown below.
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PROPORTIONAL-PLUS-INTEGRAL-PLUS-DERIVATIVECONTROL ACTION:
The combination of proportional, integral andderivative control action is called PID control actionand the controller is called three action controller.Block diagram is shown below.
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)2(1
1)(
)(
)()()()(
)1()()(1
)()(0
d
i
p
dp
i
p
p
dp
t
i
pp
sTsT
KsE
sM
ssETKsEsT
KsEKsM
tedt
dTKdtte
TKteKtm
Laplace transform
Equation (2) is the transfer function.
THANK YOU
FOR
ATTENTION
SYED HASAN SAEED 27
