FET/BSEE/F-13 2016
CASE STUDY “ANNTENA AZIMUTH POSITION CONTROL SYSTEM”
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FET/BSEE/F-13 2016
Case study compete sout!o" #!t$ c$ae"%es &&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&CHAPTER'(&&&&&& &&&&CHAPTER'(&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&& CASE STUDY) “ANNTENA AZIMUTH POSITION CONTROL SYSTEM” A position control control system system finds widespread widespread applications applications in antenna, robot robot arms and and computer disc drives. drives. In this case study we will see how the system system works and how we will will affect changes in its its performance. SYSTEM CONCEPT) CONCEPT)
SYSTEM DIA*RAM DIA*RAM +deta!ed ayout,) ayout,)
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(Fig-1)
FET/BSEE/F-13 2016
(Fig-)
-UNCTIONAL .LOC/ DIA*RAM)
(Fig-3)
!he input command in the angular displacement into a voltage, similarly the output angular displacement is converted in to voltage by the potentiometer in the feedback path .signal and power amplifier boost the difference difference between the input and output voltages this amplified actuating signal will drive the plant, the system normally operates to derive the error to "ero when input and output voltages matches the error will be "ero then the motor will not turn. #o motor is driven only when output and input do not match, greater greater the difference between input and output larger larger E--ECT OO THE *AIN *A IN OTHE SI*NAL SI* NAL 0ITH STEADY1 STEA DY1STA STATE TE RESPONSE) RE SPONSE) the input voltage, and faster the motor will turn. $y increasing the gain of the signal there will be increase in the steady- state response because gain is proportional value that shows the relationship between the magnitudes of the input to the magnitude of the output signal at steady-state response. If the gain is increased for the given actuating system the motor will b driven harder, it turns faster towards its final 3 | P a g position e because because of increase increase in speed increased increased momentum would would cause the motor motor to overshoot the final value the motor will stop when the output match the input however the transient response will be different.
FET/BSEE/F-13 2016
IMPORTANT IMPORTANT)) Increasing gain or decreasing decreasing gain beyond the particular safety "one can cause the system to become unstable.
Fig-4
E2PLAINATION E2PLAINATION O- *RAPH) %s we know steady-state error is the difference difference between the input and the output after the transient have been effectively disappeared steady-state error decrease with the increase in gain and decreases with increase in gain this graph shows this this "ero error error in steady-state steady-state response response that is when transient transient response response disappear the the output position e&uals e&uals the commanded input position. position. in some system the steady-state steady-state error is not "ero for these types of systems a simple gain ad'ustment to regulate the transient response is either not affected or leads to trade off between desired transient response and the desired steady state accuracy solution to this problem is use of controller with dynamic response along with an
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FET/BSEE/F-13 2016
amplifier now it is possible to design the desired transient response and desired steady state accuracy without the trade off re&uired by a simple setting of gain. In this case controller is known as a compensator.
SCHYMETIC DIA*RAM) DIA*RAM)
(Fig -)
.LOC/
DIA*RAM) DIA*RAM)
Fig-
-ROM .LOC/ DIA*RAM)
3
Kp
11
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K
k s+ a
km s s + am
kg
FET/BSEE/F-13 2016
Kp
(Fig -*)
+++++++++++++++++++++++++++++++++++++%!/0+++++++++++++++++++++++++++++++++++++++++
ANNTENA ANNTENA CONTROL) CONTROL) TRANS-ER TRANS-ER -UNTIONS -UNTIONS !he physical systems can b model mathematically with transfer functions in systems are composed of subsystem of different types such as 2echanical, /lectrical and /lectromechanical. PROBLE PRO BLEM: M:
t$e t4a"s5e4 5u"ct!o" 5o4 eac$ su6system o5 a"te""a pos!t!o" co"t4o system us!"% co"5!%u4at!o" (7 Find t$e SU.SYSTEMS Input potentiometer potentiometer
INPUT %ngular rotational rotational from user
pre amplifier amplifier
3oltage oltage from potentiometer
OUTPUT i(t)
2otor 4utpu utputt po poten tentiom tiomet eteer
Pa4amete4 3 6 7 71 a 0a ; a =a 7b 7t 61 6 6@ ; A = A
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vp ( t ) vp ( t )
3oltage from pre amplifier
ea ( t )
3oltage from power amplifier %ngula gularr ro rotat tation ional fro from loa load d
CON-I*URATIONS)
v i ( t )
3oltage from power amplifier
ve ( t ) =v i ( t )− vo ( t ) ower amplifier
3oltage from pre amplifier
o(t)
3oltage 3oltage to motor
ea ( t )
%ngular rotational rotational from from load
o(t)
3oltage from pre amplifier vo ( t )
SYSTEMETIC PARAMETERS PARAMETERS Co"5!%u4at!o" ( 15 15 155 155 8 5.5 5.51 5. 5. 5 5 1 1 .oc9 d!a%4am pa4amete4s
Co"5!%u4at!o" 8 15 1 15 15 5.5 5.51 1 1 5 5 5 @
De5!"!t!o"s 3oltage across potentiometer !urns !urns of potentiometer reamplifier reamplifier gain ower amplifier gain gain ower amplifier pole 2otor resistance9ohms: 2otor inertial constant9kg-m<: 2otor damping const96 m s>rad $ack /mf constant 93-s>rad: 93-s>rad: 2otor !or !or&ue &ue constant96-m>%: constant96-m>%: ?ear teeth ?ear teeth ?ear teeth Aoad inertia constant 9kg-m<: Aoad damping constant 96-m>rad:
FET/BSEE/F-13 2016
Pa4amete4 7 p 7 71 % 7 m %m 7g
Co"5!%u4at!o" ( 5.@18 155 155 .58@ 1.*1 5.1
Co"5!%u4at!o" 8 @.18 1+ 15 15 5.8 1.@ 5.
De5!"!t!o" otentiometer gain re amplifier amplifier ower amplifier gain gain ower amplifier pole pole 2otor and load gain 2otor and load pole ?ear ratio
BLOCK DIAGRAM:
(Fig-8) $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 1D E%ACA%!/ !0%6#F/0 FC6!I46 4F !/ #B#!/2
TRANSFER FUNCTION OF SYSTEM: INPUT POTENTIOMETER & OUTPUT POTENTIOMTER: SUBSYSTEM 1 & SUBSYSTEM 5:
Input potentiometer & Output potentiometer are configured in the same way, their transfer function will b same to neglect the dynamics for the potentiometers and simply find the relationships between output oltage and input angular displacement! In center position the output oltage is "ero! Fie turns towards either the positie control or the negatie #$ olts yield a oltage change of ten olt so transfer function is %ccording to fig-
m(s)
!ransfer !ransfer function isD
Vi ( s ) = kp θi ( s )
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GGGGGG.
/&uation-.1
Kp
3i(s)
(Fig-H)
FET/BSEE/F-13 2016
kp =
V nπ
V 10 10 = = =0.318 nπ ( 10 ) π 10 ( 3.147 ) Vp ( s )
Ve ( s )
PRE AMPLIFIER: SUBSYSTEM 2:
K %ccording to fig-#$
Vp ( s ) = k GGGGGGGGG..G.. /&uation- . Ve ( s )
(Fig #$) #$)
PPWER AMPLIFIER: SUBSYSTEM 3:
'ransfer function of amplifier are gien in problem statement two phenomena are neglected! First, we assume that saturation is neer reached! econdly the dynamics of pre a mplifiers are neglected since its speed of response is typically much greater than that of the power amplifier
Vp ( s )
%ccording to fig-##
Ea ( s )
K1/ Ea ( s ) K 1 = Vp ( s ) s + a Ea ( s ) 100 = Vp ( s ) s + 100
GGGGGGG
MOTOR AND LOAD: SUBSYSTEM 4: 4:
%ccording to fig-#
m ( s) *ow we hae to calculate calculate transfer function ( Ea ( s ) ) ∅
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(Fig-11)
/&uation-.@
FET/BSEE/F-13 2016
(Fig-12)
+loc
diagram of subsystem
G(n) = Ea(s)
m(s)
km
(Fig-13)
CIRCUIT DIAGRAM:
'his is an electro-mechanical system which we can obsere from schematic diagram as we now that in electromechanical system dc motors are used!
(Fig-14)
e are concerned to input oltage so the e.uation is
RiIa ( s ) + La s Ia ( s )+ Vb ( s )= Ea ( s ) Tm ( s )= Kt Ia ( s )
/////! 0.uation-!4
///////////// 0.uation-!1
here
Ia ( s )=
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Tm ( s ) Kt
///////////!!0.uation-!2
FET/BSEE/F-13 2016
Vb ( s )= Kbθm ( s )
///////////// 0.uation-!
Tm ( s )= ( Js ² + Dms ) θm ( s ) ////////////0.uation-! ubstituting the alues of 'm(s) in aboe e.uation5
( Js ² + Dms) ( Ra+ La s ) θm ( s ) Kt
… … … … … … … … … … … … … 0.uation-!
ubstituting the alues of Ia(s) In e.uation 45
( Js ² + Dms) ( Ra+ La s ) θm( s ) Kt
[
+ kb sθm( s )=¿ )=¿
( J s + Dms) ( Ra + La s ) + kb s 2
Kt
]
Ea ( s )
θm ( s )= Ea ( s )
%s we suppose that 6a 77 8a so we can neglect 8a
[
( J s + Dms ) ( Ra ) + kb s 2
Kt
]
θm ( s )= Ea ( s ) ///////////!0.uation-!#$
θm ( s )= Ea ( s )
1
[
( J s + Dms ) ( Ra ) + kbkt s 2
θm ( s )= Ea ( s )
Kt
Kt
[ ( J s + Dms ) ( Ra Ra ) + kbkts ]
θm ( s )= Ea ( s )
10 | P a g e
]
2
//////// 0.uation-!##
Kt Ra
[
( J s + Dms ) + kbkt s 2
Ra
]
/////////!!0.uation-!#
FET/BSEE/F-13 2016
Dm Ra kbkt 1 s + (¿ (¿ )+ Jm Ra ( ¿¿ ]
s¿
θm ( s )= Ea ( s )
/////////0.uation-!#3
Kt Ra
¿
'he damping and inertial alues of system are connected to motor through a set of gears 9ear ratio =
INERTIA:
N 1 1 kg = N 2 2
'he e.uialent inertia :m is
Jm= Ja + JL
( ) N 1 N 2
;
/////////////// 0.uation-!#3
+y putting their alues we hae
:m
Jm
( ) 25
¿ 0 . 02+ ¿
(#)
250
;
< $!$3g-m;
:ISCOUS DAMPER)
=m =a J= A
( ) N 1 N 2 2
;
//////////////! 0.uation-!#4
+y substituting alues of =g, >8 and >a we hae 25
5.51J 1(
250
=m5.5 6-m-s>rad
SPRIN*) SPRIN*)
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) ;
FET/BSEE/F-13 2016
(
7m
Kt JmRa
)
GGGGGGGGGGGGGG
0.uation-!#1
+y substituting alues
7m
(
0 .5 0.03 x 8
)
7m.58@@ 6-m- >rad
MOTOR AND LOAD LOAD POLE)
DmRa + KbKa am JmRa
GGGGGGGGGGGGGG..... 0.uation-!#1
0.02 ( 8 )+ .5 ( .5)
am
( 0.03 ) 8
am1.*581.*1
θm ( s ) 2.083 = Ea ( s ) s ( s + 1.71)
θo ( s ) θm ( s ) 0.1 ( 2.083 ) = 0.1 = Ea ( s ) Ea ( s ) s ( s +1.71 )
θo ( s ) 0.2083 = Ea ( s ) s ( s + 1.71 ) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&CHALLEN*E'(&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& “TRANS-ER -UNTION” .Y USIN* USIN* CON-I*URAT CON-I*URATION ION 8)
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6ow for the same subsystems subsystems by using the above above e&uations we can solve solve the challenge by only only changing the values .the values are taken from the table configuration configuration .
INPUT POTENTIOMETER POTENTIOMETER ; OUTPUT POTENTIOMTER) POTENTIOMTER) SU.SYSTEM ( ; SU.SYSTEM SU.SYSTEM <) %ccording %ccording to fig-Ha
m(s)
!ransfer !ransfer function isD
Vi ( s ) = kp θi ( s )
Kp
3i(s)
GGGGGGGGG. /
[email protected]
kp =
(Fig-Ha)
V nπ
V 10 10 = = =3.18 nπ ( 1 ) π 1 ( 3.147 ) Vp ( s )
Ve ( s )
PRE AMPLI-IE AMPLI-IER) R) SU.SYSTEM SU.SYSTEM 8)
K
%ccording %ccording to fig-15 a
Vp ( s ) = k Ve ( s )
GGGGGGGG.. /&uation-@.
15a) (Fig 15a)
PO0ER AMPLI-IER) SU.SYSTEM AMPLI-IER) SU.SYSTEM =)
Vp ( s )
%ccording %ccording to fig-11a fig-11a
Ea ( s )
K1/ Ea ( s ) K 1 = Vp ( s ) s + a
Fig-
11a 11a Ea ( s ) = 150 Vp ( s ) s + 150
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GGGGGGGGGG /&uation-@.@
FET/BSEE/F-13 2016
MOTOR AND LOAD) LOAD) SU.SYSTEM SU.SYSTEM >) >)
m(s)
Ea+s,
G(n)= %ccording %ccording to fig-1a
km
m ( s) 6ow we have to calculate transfer transfer function ( Ea ( s ) ) ∅
INERTIA)
N 1 1 kg = N 2 2
?ear ratio ?
!he e&uivalent inertia ;m is
Jm= Ja + JL
( ) N 1 N 2
< GGGGGGGGGG. /
[email protected] /
[email protected]
$y putting their values we have
;m
Jm
( ) 50
¿ 0 . 05 + ¿
()
250
<
5.kg-m<
:ISCOUS DAMPER)
=m =a J= A
( ) N 1 N 2 2
;
///////////!0.uation-3!1
+y substituting alues of =g, >8 and >a we hae 50
5.51J @(
250
) ;
=m5.1@ 6-m-s-rad SPRIN*)
7m
(
Kt JmRa
+y substituting alues 14 | P a g e
)
GGGGGGGGGGG 0.uation-3!2
FET/BSEE/F-13 2016
7m
(
1 0.25 x 5
)
7m.86-m-rad MOTOR AND AND LOAD POLE) POLE)
DmRa + KbKa am JmRa
GGGGGGGGGGGG.. 0.uation-3!
0.13 ( 5 ) + 1 (1 )
am
( 0.25 ) 5
am1.@ 6-m-s>rad θm ( s ) 0.8 = Ea ( s ) s ( s + 1.32) θo ( s ) θm( s ) 0.2 ( 0.8 ) = 0.2 = Ea ( s ) Ea ( s ) s ( s + 1.32) θo ( s ) 0.16 = Ea ( s ) s ( s + 1.32 ) GGGGGGGGGGGGGGGGGGGG./
[email protected]
++++++++++++++++++++++++++++++++++++++%!/0+++++++++++++++++++++++++++++++++++++ REDUCTION REDUCTION O- ANTENNA ANTENNA CONTROL SYSTEM) 0eduction of antenna %"imuth position control control system or any other system is done by two methods. • • •
$lock diagram reduction reduction method. method. #ignal Flow ?raph design method 2%#46L# 0CA/
++++++++++++++++++++++++++++++++++++++++%0!%+++++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 1D
.LOC/ DIA*RAM DIA*RAM REDUCTION REDUCTION METHOD) METHOD)
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In block diagram method the functional block diagram is reduced into a simple block diagram. $lock diagram consists of blocks, signals, summing 'unction, and pickoff points. Mith 0(s) as input and (s) as output and ?(n) as the transfer function of the system. 0(s)
(s)
G (n)
(Fig-1)
De4!@at!o" o5 T4a"s5e4 T4a"s5e4 -u"ct!o" -u"ct!o" 5o4 a cosed oop system) system) FC6!I46 $A47 =I%?0%2 4F !/ #B#!/2D
(Fig-1)
0/=C!I46D
(Fig-1*) %s we can see that ? (s), (s), ?@ (s), ?K (s), ? (s) are in cascaded form #o it is written as ?p(s) ?(s)+?@(s)+?K(s)+ ?(s)+?@(s)+?K(s)+ ?(s) GGGGGGGGGG. /&uation-K.1 /&uation-K.1
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FET/BSEE/F-13 2016
km k 1 ?p(s) k+ s + a + s ( s + am) + kg 2.0833
100
?p(s) k+
s + 100 + s ( s + 1.71) +5 .1 20.83 k
?p (s)
s ( s + 1.71 ) ( s + 100 ) 20.83 k
?p (s)
2
s ( s + 101.7 s + 171) 20.83 k
?p (s)
3 s + 101.7 s ² + 171 s GGGGGGGGGGGGGGG./&uation-K.
(Fig-18) 20.83 k
?f (s)
/&uation-K.@ s + 101.7 s ² + 171 s + 6.63 k GGGGGGGGGG. /&uation-K.@ 3
(s)
0(s)
(Fig-1H) 20.83 k / π
? (s)
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3
s + 101.7 s ² + 171 s + 6.63 k
FET/BSEE/F-13 2016
6.63 k
? (s)
3
/&uation-K.K s + 101.7 s ² + 171 s + 6.63 k GGGGGGGGGG. /&uation-K.K
(s)
0(s) 6.63 k
(Fig-5)
+++++++++++++++++++++++++++++++++++++%0!$++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 1D
SI*NAL -LO0 *RAPH *RAPH METHOD) METHOD) #ignal flow graphs are alternative to functional block diagram. !hese are also snows as #F?. #F? consist only of branches, which represents systems and nodes, which represent signals. #B#!/2D #ystem is represented by a line with an arrow showing the direction of signal flow through the system. %d'acent to the line we write the transfer function. function. 64=/D % signal signal 6ode is given with signalLs signalLs name name written ad'acent ad'acent to the node.
(Fig-1) In the following signal signal flow graph of of integrator we have ?1(s) ?1(s) 1>N and ?(s) k to split split the power amplifier amplifier and motor
Ea ( s ) and load system. !he power amplifier Vp ( s )
=
100
s + 100
1
is drawn in three different functions 155 ¿ s + 100 as 1
cascade system weLve separated 155 ?@(s) and then further more weLve split
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s + 100 to 1>s ?K(s) and -155 is in
FET/BSEE/F-13 2016
feedback with so
-155 1(s). !he motor and and gain as
1
θm ( s ) 2.083 = Ea ( s ) s ( s + 1.71) also as .58@ ?(s) and in the
1
s ( s + 1.71) .the integrator 1>s is separated ( s +1.71 )
term
is written as a feedback system so it is separated as
1>s?*(s) and -1.*1(s). %t the last gain ratios are give on the arrow head as ?8(s) 5.1.output potentiometer @(s)
-1>N. STATE STATE EUATIONS) EUATIONS) #tate e&uations are written according to the signal flow graph we define he states as the output integrator. ence, the state vector is ẋ 1
x =[ ẋ 2 ] ėa $y using the figure figure of signal flow graph given above the the state e&uation of integrator areD areD
B( ? B8GGGGGGGGGGGGGGGGGGGG../&uation-.1 B8 ? 1(7(B8 387=eaGGGGGGGGGGGGGG /&uation-.
ea?1=7(9B(1(ea3=(7/ θi GGGGGGGGG.......
/&uation-.@
%long with the output output e&uationD
y? θo= 0.1 x 1 Mhere 1>N5.@18. In vector-matriO vector-matriO formD
O
0
1
0
0
−1.71 1.71
2.08 2.083 3
0
−100
−3.18
¿
0
: O J
0
θi
31.8
+++++++++++++++++++++++++++++++++++++++++++++%0!++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 1D 2%#46L# 0CA/D
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6ow we will apply masonLs masonLs rule on the signal signal flow graph obtained. obtained. !o apply masonLs masonLs rule we have to consider following steps we have to find Aoop gain, Forward-path Forward-path gain, gain, 6on touching loop, loop, 6on touching loop gain, gain, From fig-1 fig-1 we have
-OR0ARD1PATH *AIN) !1 ?1(s)?(s)? @(s)?K(s)?(s)? (s)? (s)?* (s)? (s)?8(s) GGGGGGGGGG. /&uation-.1 /&uation-.1
!11>N+k+155+1>s+.58@+1>s+1>s+5.1 100∗2.083 k ∗0.1
!1
3
s π 20.83 k
!1
3
s π
#o 6.63 k
!1
s
3
GGGGGGGGGGGGGGGGGGGGGGGGG. /&uation-.
-o4#a4d1Pat$ ?1(s)?(s)? @(s)?K(s)? (s)? (s)? (s)?* (s)? (s)?8(s)
-o4#a4d1Pat$ *a!" 6.63 k
s
INDU:IUAL LOOP *AIN) *AIN) ? A1?K(s) 1(s) ?1>s+(-155)
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3
FET/BSEE/F-13 2016
−100
s
GGGGGGGGGGGGGGGGGGGGGGGGG. /&uation-.@
(s) (s) ? A ? ? (s) 1>s+(-1.*1)
−1.71
?
s
GGGGGGGGGGGGGGGGGGGGGGGGG. /&uation-.K
? A@ ?(s)? @(s)?K(s)?(s)? (s)? (s)?* (s)? (s)?8(s) @(s)
k+155+1>s+.58@+1>s+1>s+5.1+-1>N
−100∗2.083 k ∗0.1
3
s π
GGGGGGGGGGGGGG..GGGGG. /&uation-.
I"d!@!dua I"d!@!dua Loop
I"d!@!dua I"d!@!dua Loop *a!" *a!"
−100
?K(s) 1(s)
s
−1.71
? (s) (s) (s)
?
s
−100∗2.083 k ∗0.1
(s)?* (s)? (s)?8(s) @(s) ?(s)?@(s)?K(s)?(s)? (s)?
3
s π
T0O NON1TOUCHIN* LOOP) From the the given loops we can see that only only two non-touching non-touching loops eOists ?A1 and ? and ? A.
No"1touc$!"% oops ? A1? A1 + ? A
No"1touc$!"% oop %a!" %a!"
−100 s
∗−1.71 s
171
s²
!hree non-touching loops does not eOists 6ow we have to find values of of ∆ ∧∆ k
∆ =1−¿ #um of loop gains Jsum of two non-touching loop gains GGGGGGGGGG. /&uation /&uation.
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FET/BSEE/F-13 2016
∆ =1−¿ 9? (s) J ? A(s) J ? A@ : J 9? ? A : A1 A1 ∆ =1 +
100
s
1.71
+
s
+
6.63 k
171
s³
s²
+
∆ =¿ s@J151.*1s J1*1sJ.@kGGGGGGGGGGGGGGGGGGGGGGGGGGG.. /&uation.* we know that !here eOists one forward-path so k1
∆
1
∆ −loo p gains terms touching touching the forward-path forward-path GGGGGGGGGGGGGGGGG
/&uation-.8
∆
1
1
MASONFs MASONFs RULE) RULE) 2asons rule formula formula
( s ) ?(s) R (s )
∑ T k∆ k k
∆
GGGGGGGGGGGGGGGGGGGGG.
/&uation-.H %ccording %ccording to this formula formula
T 1 1∗∆ 1 ?(s) ∆ 6.63 k
∗1
?(s)
s³ s ³ + 101.71 s ² + 171 s + 6.63 k s³ 6.63 k ∗1
s ³ + 101.71 s ² + 171 s + 6.63 k 6.63 k
?(s)
s ³ + 101.71 s ² + 171 s + 6.63 k GGGGGGGGGGGGGGGGG. /&uation.15
+++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++++2%!A%$ 2%!A%$4=/++++++++++ 4=/++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ +
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2-FIA/D
M407#%/ D( when gain 71)
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#!/ 0/#46#/D
+++++++++++++++++++++++++++++++++++++++++++++%0!=++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++ ++++++++ $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 1D 6ow we have to 4epace t$e po#e4 amp!5!e4 #!t$ t$e t4a"s5e4 5u"ct!o" o5 u"!ty a"d t$e #e $a@e to e@auate pea9 t!meG pe4ce"ta%e o@e4s$ootG o@e4s$ootG a"d sett!"% t!me 5o4 9?( 6ow consider the block block diagram of antenna antenna a"imuth position position control control system and replace replace k1555
.LOC/ DIA*RAM) DIA*RAM)
(Fig-) Me have
km ? p(s) k+1+ s ( s + am) + kg 24 | P a g e
FET/BSEE/F-13 2016
2.083
? p(s) k+1+ s (s + 1.71) + 5.1 .2083 k
/&uation-*.1 s ( s + 1.71) GGGGG..GGGGGGGGGG. /&uation-*.1
$y using feedback formula formula we obtain .2083 k
?f(s) s ( s + 1.71 ) + 0.2083∗0.318 k .2083 k
/&uation-*. ?f(s) s ( s + 1.71 ) + 0.0663 k GGGGGGGGGGGG.. /&uation-*. .0663 k
/&uation-*.@ ?(s) s ( s + 1.71 ) + 0.0663 k GGGGGGGGGGGG... /&uation-*.@
%s k1555 then .0663∗1000
?(s) s ( s + 1.71 ) + 0.0663∗1000 66.3
?(s) s ( s + 1.71 ) + 66.3 66.3
?(s) s ² + 1.71 s + 66.3
GGGGGGGGGGGGGGGGGG. /&uation-*.K
!his is a second order e&uation so Me have to use the formulaD
T ( s )=
!n s ² + 2 "!n + !n ² GGGGGGGGGGGGGGG.. /&uation-*.
6ow compare the denominator of e&uation *.K with *. *. we have 6%!C0%A 6%!C0%A 0/#46#/D
25 | P a g e
Pn<.@
FET/BSEE/F-13 2016
Pn 8.1K =%2I6? 0%!I4D 0%!I4D
QPn 1.*1 1.71
Q
2∗8.14
Q5.15
PEA/ TIME) TIME) eak time is the time time given for the maOimum maOimum peak of the system or time time for first overshoot overshoot in the system. system. !o calculate the peak time standard formula is given
!p
π !# secGGGGGGGGGGGGGGGGGGGG. /&uation-*.
Mhere
!#
1
!n
√ 1− " ² ² sec. GGGGGGGGGGGGGGG. /&uation-*.*
π !p
sec
8.14 √ 1 − 0.105²
!p 5.@88 sec PERCENTA*E PERCENTA*E O:ERSHOOT O:ERSHOOT)) −π"
e
OS ?
e
e
26 | P a g e
√ 1−" ² ²
∗100
GGGGGGGGGGG. /&uation-*.8
−π ( (0.105) 1 −( √ 0.105) ²
−3.14∗0.105 √ 1−(0.105) ²
∗100
+155
FET/BSEE/F-13 2016
R4#5.*1*8+155
R4#*1.*8R
SETTLIN* TIME) TIME) !ime taken by the system to reach H8R of steady-state response is called settling time.
!sK $ sec 4
!s !s "!n secGGGGGGGGGGGGGGGGGGGG. /&uation-*.H 4
!s !s
0.105∗8.14
sec
!sK. !sK. sec
+++++++++++++++++++++++++++++++++++++++++++++%0!/++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 1D 6ow we have to de4!@e t$e system o5 pa4t d to a cosed oop step 4espo"se o5 t$e system
CLOSEDLOOP STEP RESPNSE O- THE SYSTEM) From e&uation-@5 e&uation- @5
( s ) 66.3 = ?(s) R ( s ) s ² + 1.71 s + 66.3 GGGGGGGGGGGGGGGGGGGG. /&uation-8.1
R ( s )=
1
s
27 | P a g e
(s)
¿ R
( s )∗66.3 s ² + 1.71 s + 66.3
FET/BSEE/F-13 2016
1
(s)
s
¿
∗66.3
s ² + 1.71 s + 66.3
¿
66.3
(s)
¿ (s) s ³ + 1.71 s ² + 66.3 s
s ( s ² + 1.71 s + 66.3 )
66.3
GGGGGGGGGGGGGGGGGGGG. /&uation-8.
$y using partial partial fraction methodD methodD
% &s + ¿ + (s) s ( s ² + 1.71 s + 66.3 )
GGGGGGGGGGGGGGGGGGGG. /&uation-8.@
Me can get the value of % by putting value s5S 66.3 s
%T s5 s ( s ² + 1.71 s + 66.3) 66.3
%T s5 ( 0 ) ² + 1.71 ( 0 )+ 66.3 66.3
%T s5
66.3
%1 6ow we have to calculate the the value of $ and For this we can find find their values by comparing comparing coefficient method method
(s) % (
s ² + 1.71 s + 66.3 ¿
J $s+s J +sGGGGGGGGGGGGGG. /&uation-8.K
(s) % ( s ² + 1.71 s + 66.3 ¿ J $s< J s Mhere %1 #o
(s)
28 | P a g e
s ² + 1.71 s + 66.3 J $s< J s
FET/BSEE/F-13 2016
s
51 J $,
sD
$-1
51.*1 J ,
-1.*1
1
¿ −
(s) 1
(s)
¿ − s
s
s +1.71 ( s ² + 1.71 s + 66.3 )
s + 1.71 /&uation-8. ( s ² + 1.71 s + 66.3 ) GGGGGGGGGGGGGGGG.. /&uation-8.
$y completing s&uare s&uare method method
s
+sJ (1.*1>) s
6ow by putting this this value to
(s)
1
( s + 0.855)+ .855
¿ −
8.0975
( s + 0.855 )² +( 8.097 ) ²
s
1
(8.0975 )
( s + 0.855)+ 8.0975
¿ −
( 0.855 ) 8.0975
(s)
1 ( s + 0.855 ) +( 8.0975 ) 0.106 ¿ − (s) s ( s +0.855 )² +( 8.097 ) ²
(s)
(s +0.855 )² +( 8.097 ) ²
s
1
s + 0.855
s
( s + 0.855) ² +( 8.097 ) ²
¿ −
GGGGGGGGGGGGGGG. /&uation-8.*
( 8.0975 ) 0.106 J (s +0.855 ) ² +( 8.097 )²
GGG..GGGGG. /&uation-8.8
6ow by taking Inverse laplas !ransform !ransform we get c (t) 1- e 29 | P a g e
− 0.855
(cos8.5H*t J5.15sin8.5H*t) GGGGGGGG.. /&uation-8.H
FET/BSEE/F-13 2016
+++++++++++++++++++++++++++++++++++++++++++++%0!F++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 1D E-!"d!"% t$e @aue o5 9 t$at %!@es t$e spec!5!c OS” 6ow it is given that R4# is 15R and the power amplifier is unity unity so we can find the value value of k ower amplifier unity unity
+-!%18=, .0663 k
?(s) s ( s + 1.71 ) + 0.0663 k −π"
e
OS ?
√ 1−" ² ²
∗100
−π"
OS( ?
e
√ 1−" ² ²
$y taking natural log (ln) 5n both sides we have ln
'() 100
−ln
30 | P a g e
− π"
=ln e
'() 100
=
√ 1− " ² ²
π"
√ 1−"
2
FET/BSEE/F-13 2016
− ln
√
J?
'() 100
π ² + ln
'() GGGGGGGGGGGG../&uation-8.15 100
$y solving
−ln
√
J?
10 100
π ² + ln ²
10 100
2.30
J?
3.89
Q5.H
T ( s )=
!n 2
2
s + 2 "!n + ! n
($y using e&uation*.)
6ow by comparing ?(s) with !(s) we have
Damp!"% 4at!o 5o4mua) 5o4mua) QPn 1.*1 1.71
Pn
2∗0.59
Pn1.KKH
A"d "atu4a 4espo"se 4espo"se !n is e&ual to formula Pn<5.5@k from here we can calculate the value of kD
(1.KKH)<5.5@k
(1.449 )² k
31 | P a g e
0.0663
FET/BSEE/F-13 2016
k @1.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&CHALLEN*E'8&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& & $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 D
+PART1A,
6ow by using configuration configuration we have to 4educe t$e 6oc9 d!a%4am
0(s)
(s)
G (n)
(Fig-K)
De4!@at!o" o5 T4a"s5e4 T4a"s5e4 -u"ct!o" -u"ct!o" 5o4 a cosed oop system) system) FC6!I46 $A47 =I%?0%2 =I%?0%2 4F !/ #B#!/2D #B#!/2D
(Fig-) 0/=C!I46D
(Fig-)
32 | P a g e
FET/BSEE/F-13 2016
%s we can see that ? (n), (n), ?@ (n), ?K (n), ? (n) are in in cascaded form #o it is written as
?p (n) ? (n)+?@ (n)+?K (n)+ ? (n) GGGGGGGGGGGGG.. /&uation-H.1
km k 1 ?p (n) k+ s + a + s (s + am) + kg GGGGGGGGGGGGG.. GGGGGGGGGGGGG.. /&uation-H. 150
0.8
?p (n) k+ s + 150 + s (s + 1.32) +5. 24 k
?p (n) s ( s + 1.32 ) ( s + 150 ) 24 k
?p (n) s (s + 151.32 s + 198 ) 2
24 k
?p (n) s + 151.32 s ² +198 s 3
GGGGGGGGGGGGG.. /&uation-H.@
(Fig-) 24 k
?f (n) s + 151 151.. 32 s ² + 198 s + 24 k ∗10 / π 3
33 | P a g e
FET/BSEE/F-13 2016
(s)
0(s) 24 k ∗10 / π 3
? (n) s + 151.32 s ² + 198 s + 76.4 k 76.4 k
? (n) s
3
+151.32 s ² + 198 s +76.4 k
0(s)
GGGGGGGGGGGGG.. /&uation-H.K
(s)
+++++++++++++++++++++++++++++++++++++%0!$++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 D
SI*NAL -LO0 *RAPH *RAPH METHOD) METHOD) #ignal flow graphs are alternative to functional block diagram. !hese are also snows as #F?. #F? consist only of branches, which represents systems and nodes, which represent signals. #B#!/2D #ystem is represented by a line with an arrow showing the direction of signal flow through the system. %d'acent to the line we write the transfer function. function. 64=/D % signal signal 6ode is given with signalLs signalLs name name written ad'acent ad'acent to the node.
34 | P a g e
FET/BSEE/F-13 2016
Fig-8 In the following signal signal flow graph of of integrator we have ?1(s) ?1(s) 15>N and ?(s) k to to split the power amplifier amplifier and motor
Ea ( s ) 1 150 ¿ = and load system. !he power amplifier Vp ( s ) s + 150 is drawn in three different functions 155 s + 150 as 1
cascade system weLve separated 15?@(s) and then further more weLve split
feedback with so
-155 1(s). !he motor and and gain as
1
s + 150 to 1>s ?K(s) and -15 is in
θm ( s ) 0.8 = Ea ( s ) s ( s + 1.32) also as 5.8 ?(s) and in the term
1
s ( s + 1.32) .the integrator 1>s is separated ( s +1.32 )
is written as a feedback system so it is separated as
1>s?*(s) and -1.@(s). %t the last gain ratios are give on the arrow head as ?8(s) 5. and output potentiometer @(s) -15>N.
STATE STATE EUATIONS) EUATIONS) #tate e&uations are written according to the signal flow graph we define he states as the output integrator. ence, the state vector is ẋ 1
x =[ ẋ 2 ] ėa $y using the figure figure of signal flow graph given above the the state e&uation of integrator areD areD
B( ? B8GGGGGGGGGGGGGGGGGGGG../&uation-15.1 B8 ? 1(7=8B8 37eaGGGGGGGGGGGGGG
35 | P a g e
/&uation-15.
FET/BSEE/F-13 2016
ea?1K7>K9B(1(<ea3>7>/ θi GGGGGGGGG......./&uation-15.@
%long with the output output e&uationD
y? θo= 0.2 x 1 where 15>N@.18. In vector-matriO vector-matriO formD
O
0
1
0
0
−1.32
0.8
−97.49
0
−150
¿
0
: O J
0
θi
477.46
+++++++++++++++++++++++++++++++++++++++++++++%0!++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 D 2%#46L# 0CA/D 6ow we will apply apply masonLs masonLs rule rule on the signal flow flow graph obtained. obtained. !o apply masonLs masonLs rule rule we have to consider following steps we have to find Aoop gain, Forward-path gain, 6on touching loop, 6on touching loop gain,
-OR0ARD1PATH *AIN) !1 ?1(s)+ ?(s)+?@(s)+?K(s)+?(s)+? (s)? (s)?* (s)+? (s)+?8(s)GGGGGGGGGGGGGG../&uation-11.1
!115>N+k+15+1>s+5.8+1>s+1>s+5. 150∗0.8 k ∗0.2
!1
36 | P a g e
3
s π
11. GGGGGGGGGGGGGGGGGGGGG /&uation 11.
FET/BSEE/F-13 2016
240 k
!1
3
s π
#o 76.39 k
!1
s
3
GGGGGGGGGGGGGGGGGGGG.. /&uation-11.@ /&uation-11.@
-o4#a4d1Pat$ ?1(s)?(s)? @(s)?K(s)? (s)? (s)? (s)?* (s)? (s)?8(s)
-o4#a4d1Pat$ *a!" 76.39 k
s
3
INDU:IUAL LOOP *AIN) *AIN) ? A1?K(s) 1(s) ?1>s+(-15)
−150
s
/&uation-11.K GGGGGGGGGGGGGGGGGGGGGGGGGGGG. /&uation-11.K
? A ? ? (s) (s) (s) 1>s+(-1.@)
−1.32
?
s
11. GGGGGGGGGGGGGGGGGGGGGGGGGGGG /&uation 11.
? A@ ?(s)+?@(s)+ ?K(s)+?(s)+? (s)+? (s)+?* (s)+? (s)+?8(s)+ @(s)
k+15+1>s+5.8+1>s+1>s+5.+-15>N
−150∗0.8 k ∗0.2∗10
3
s π I"d!@!dua I"d!@!dua Loop
/&uation 11. 11. GGGGGGGGGGGGGGGGGGG. /&uation
I"d!@!dua I"d!@!dua Loop *a!" *a!"
−150
?K(s) 1(s)
−1.32
? (s) (s) (s) ?
37 | P a g e
s s
FET/BSEE/F-13 2016
−150∗0.8 k ∗0.2∗10
(s)?* (s)? (s)?8(s) @(s) ?(s)?@(s)?K(s)?(s)? (s)?
3
s π
T0O NON1TOUCHIN* LOOP) From the the given loops we can see that only only two non-touching non-touching loops eOists ?A1 and ? and ? A.
No"1touc$!"% oops ? A1? A1 + ? A
No"1touc$!"% oop %a!" %a!"
−150
s
∗−1.32 s
198
s²
!hree non-touching loops does not eOists 6ow we have to find values of of ∆ ∧∆ k
∆ =1−¿ #um of loop gains Jsum of two non-touching loop gains ∆ =1−¿ 9? (s) J ? A(s) J ? A@ : J 9? ? A : GGGGGGGGGGGGG. /&uation /&uation 11.* 11.* A1 A1 ∆ =1 +
150
s
+
1.32
s
+
76.29 k
198
s³
s²
+
∆ =¿ s@J11.@s J1H8sJ*.Hk>sU GGGGGGGGGGGGGGG..G. /&uation-11.8 GGGGGGGGGGGGGGG..G. /&uation-11.8 we know that !here eOists one forward-path so k1
∆
1
∆ −loo p gains terms touching touching the forward-path forward-path GGGGGGGGG... /&uation /&uation 11.H 11.H ∆
1
1
MASONFs MASONFs RULE) RULE) 2asons rule formula formula
( s ) ?(s) R (s )
%ccording to formula
38 | P a g e
∑ T k∆ k k
∆
GGGGGGGGGGGGG. /&uation /&uation 11.15 11.15
FET/BSEE/F-13 2016
T 1 1∗∆ 1 ?(s) ∆ 76.39 k
∗1
?(s)
s³ s ³ + 151.32 s ² + 198 s + 76.39 k s³ 76.39 k ∗1
s ³ + 151.32 s ² + 198 s +76.39 k
76.39 k
/&uation 11.1 11.11 1 s ³ + 151.32 s ² + 198 s + 76.39 k GGGGGGGGGGGGG. /&uation
+++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ ++++++++++++++++++++2% +++++++2%!A%$4=/+++ !A%$4=/+++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + 2FIA/D
M407#%/D
39 | P a g e
FET/BSEE/F-13 2016
?0% (step response)D
40 | P a g e
FET/BSEE/F-13 2016
+++++++++++++++++++++++++++++++++++++++++++++%0!=++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++ ++++++++ $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 D 6ow we have, replace the power amplifier with the transfer function function of unity and the we have to evaluate peak time, percentage percent age overshoot, and settling settling time for k 6ow consider the block block diagram of antenna antenna a"imuth position position control control system and replace replace k1555
.LOC/ DIA*RAM) DIA*RAM)
(Fig-H) Me have
km ? p(s) k+1+ s ( s + am) + kg 0.8
? p(s) k+1+ s (s + 1.32) + 5. .16 k
s ( s + 1.32)
$y using feedback formula formula we obtain .16 k
?f(s) s ( s + 1.32 ) + 0.16∗3.18 k .16 k
?f(s) 41 | P a g e
s ( s + 1.32 ) + 0.508 k
FET/BSEE/F-13 2016
.508 k
?(s)
/&uation 1.1 s ( s + 1.32 ) +0.508 k GGGGGGGGGGGGG. /&uation
%s k then .508∗5
?(s)
s ( s + 1.32 ) +.508∗5 2.54
?(s)
s ( s + 1.32 ) +2.54 2.54
?(s)
s ² + 1.32 s + 2.54 GGGGGGGGGGGGGGGGGGGG. /&uation-1.
!his is a second order e&uation so Me have to use the formulaD
T ( s )=
!n s ² + 2 "!n+ !n ² GGGGGGGGGGGGGGGGGGGG. /&uation-1.@
6ow compare the denominator of e&uation Ha with @5a we have 6%!C0%A 6%!C0%A 0/#46#/D
Pn<.K Pn 1.HK =%2I6? 0%!I4D 0%!I4D
QPn 1.@ 1.32
Q
2∗1.594
Q5.K1K
PEA/ TIME) TIME) eak time is the time time given for the maOimum maOimum peak of the system or time time for first overshoot overshoot in the system. system. !o calculate the peak time standard formula is given
42 | P a g e
FET/BSEE/F-13 2016
π !#
!p
secGGGGGGGGGGGGGGGGGGGG.. /&uation-1.K secGGGGGGGGGGGGGGGGGGGG
Mhere
!#
1
!n
√ 1− " ² ²
sec
π
!p
sec
1.594 √ 1− 0.414²
!p .1Ksec
PERCENTA*E PERCENTA*E O:ERSHOOT O:ERSHOOT)) −π"
R4# e
√ 1− " ² ²
∗100
GGGGGGGGGGGGGGGG.. /&uation-1.
−π (0.414 )
e
e
√ 1−( 0.414) ²
−3.14∗0.414 √ 1−(0.414) ²
∗100
+155
R4#5.@H*+155
R4#@.H*R
SETTLIN* TIME) TIME) !ime taken by the system to reach H8R of steady-state response is called settling time.
!sK $ sec 4
!s !s "!n secGGGGGGGGGGGGGGGGGGGG. /&uation-1.
43 | P a g e
FET/BSEE/F-13 2016
4 0.414∗1.594
!s !s
sec
!s.51 !s.51 sec +++++++++++++++++++++++++++++++++++++++++++++%0!/++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 D D E=erive the system of part d to a closed loop step respons responsee of the system
CLOSEDLOOP STEP RESPNSE O- THE SYSTEM) From e&uation-@5 e&uation- @5
( s ) 2.54 = ?(s) R ( s ) s ² + 1.71 s + 2.54 GGGGGGGGGGGGGGGGGGGG. /&uation-1.*
R ( s )=
1
s
( s )∗2.54 ¿ R (s) s ² + 1.32 s + 2.54 1
(s)
¿
s
∗2.54
s ² + 1.32 s + 2.54
¿
2.54
(s)
¿ (s) s ³ + 1.32 s ² + 2.54 s
s ( s ² + 1.32 s + 2.54 )
2.54
GGGGGGGGGGGGGGGGGGGG. /&uation-1.8
$y using partial partial fraction methodD methodD
% &s + ¿ + (s) s ( s ² + 1.32 s + 2.54 ) Me can get the value of % by putting value s5S
44 | P a g e
GGGGGGGGGGGGGGGGGG. /&uation-1.H
FET/BSEE/F-13 2016
2.54 s
%T s5 s ( s ² + 1.32 s + 2.54 ) (s is canceled from numerator and denominator) 2.54
%T s5
( 0 ) ² + 1.71 (0 )+ 2.54 2.54
%
2.54
%1 6ow we have to calculate the the value of $ and For this we can find find their values by comparing comparing coefficient method method (s) % (
s ² + 1.32 s + 2.54 ¿ J $s+s J +sGGGGGGGGG.GGGG. /&uation-1.15
(s) % (
s ² + 1.32 s + 2.54 ¿ J $s< J s
Mhere %1 #o (s) s
51 J $,
s ² + 1.32 s + 2.54 J $s< J s $-1
51.@ J ,
-1.@
1 s + 1.32 ¿ − (s) s ( s ² + 1.32 s + 2.54 )
1
(s)
¿ − s
s + 1.32 ( s ² + 1.32 s +2.54 )
$y completing s&uare s&uare method method
s+sJ (1.@>)
45 | P a g e
GGGGGGGGGGG.GGG../&uation-1.11
FET/BSEE/F-13 2016
s
(s) ¿ 1 −
( s + 0.66 )+.66
( 1.37 ) 1.37
(s +0.66 ) ² +(1.37 ) ²
s
( s + 0.855)+ 1.37
(0.66 )
(s) ¿ 1 −
1 ( s + 0.66 ) +( 1.37 )0.48 ¿ (s) s − (s + 0.66 ) ² +(1.37 ) ²
s
1.37
( s +0.66 ) ² +(1.37 ) ²
GGGGGGGGGGGGGGGGG. /&uation-1.1@
1 s + 0.66 ( 1.37 ) 0.48 ¿ − (s) s ( s + 0.66 ) ² +( 1.37 ) ² J ( s +0.66 ) ² +( 1.37 ) ²
GGGGGGGGGGGGG. /&uation-1.1K
6ow by taking Inverse laplas !ransform !ransform we get c (t) 1- e
− 0.66
/&uation-1.1 (cos1.@*t J5.K8sin1.@*t) GGGGGGGGGGGG.... /&uation-1.1
+++++++++++++++++++++++++++++++++++++++++++++%0!F++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + $B C#I6? 46FI?C0%!I46 46FI?C0%!I46 D EFinding the value of k that gives the specific R4# 6ow it is given that R4# is 1R and the power amplifier is unity unity so we can find the value value of k ower amplifier unity unity
46 | P a g e
FET/BSEE/F-13 2016
.508 k
?(s)
s ( s + 1.32 ) + 0.508 k
OS ?
√ 1−" ² ²
−π"
e
∗100
−π"
e
OS( ?
√ 1−" ² ²
$y taking natural log (ln) 5n both sides we have ln
'() 100
−ln
√
=ln e
'() 100
−ln J?
− π"
=
π"
√ 1−"
100
'() 100
$y solving e&uation e&uation 1@ we have
J?
47 | P a g e
√
2
'()
π ² + ln ²
−ln
√ 1− " ² ²
15 100
π ² + ln ²
15 100
GGGGGGGGGGG../&uation-1@
FET/BSEE/F-13 2016
1.90
J?
3.66
Q5.1H
T ( s )=
!n s ² + 2 "!n + !n ²
6ow by comparing ?(s) with !(s) we have =amping ratio formulaD
QPn 1.@ 1.32
Pn
2∗0.52
Pn1.*
%nd natural response response
!n is e&ual to formula
Pn<5.58k From here we can calculate the value of kD
(1.*)<5.58k
(1.27 ) ² k
0.508
k @. &&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&CHA &&&&&&CHAPTRT'&&&&&& PTRT'&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&& EFI6= !/ 0%6?/ 4F0/%2AIFI 4F0/%2AIFI/0 /0 ?%I6 0/VCI/= !4 7// A4#/= A44 #!%$A/
ANNTENA CONTROL) STA.ILIT STA.ILITY Y DESI*NE IN TERMS O- *AIN) .Y USIN* USIN* CON-I*URAT CON-I*URATION ION () %s we know that stable stable system have their close close loop poles in the the left half of s plane plane as loop gain is changed changed location of poles also changes changes .roper gain settings are are essential for for the stability of closed closed loops systems. systems. 04$A/2 #!%! #!%!/2/6!D /2/6!D
48 | P a g e
FET/BSEE/F-13 2016
Find the range of of pre amplifier amplifier gain re&uired re&uired to to keep the closed loop system system stable 6.63 k
? (s)
3
s + 101.7 s ² + 171 s + 6.63 k
04C! !%$A/D !%$A/D
# @ # # 1 #5
1
1*1
151.*1
.@7
1*1-5.57
5
.@7
5
17392.4 − 6.63 K 101.71
5
5.571*1 171
7
0.065
7@.@
%s there is no sign change in the first column if 5W7W@ system will b stable for the value of k eOist in this range.
+++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++++++++%AA/6?/@+ ++++%AA/6?/@++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++ +++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ C#I6? 46FI?C0%!I46 D EFI6= !/ 0%6?/ 4F0/%2AIFI 4F0/%2AIFI/0 /0 ?%I6 0/VCI/= !4 7// A4#/= A44 #!%$A/ 76.4 k
? (n)
04C! !%$A/D !%$A/D
49 | P a g e
3
s + 151.32 s ² + 198 s + 76.4 k
FET/BSEE/F-13 2016
# @ #
1
1H8
11.@
*.K7
# 1 #5
1H8-5.57
5
*.K7
5
29898 −76.4 K
5
151.32
198−0.505 K =0
198
7
0.505
7@H.58
%s there is no sign change in the first column if 5W7W@H.58 system will b stable for the value of k eOist in this range. +++++++++++++++++++++++++++++++++++++++%!/0K++++++++++++++++++++++++++++++++++++++++ E!/ 4/6 A44 0/#46#/ #B#!/2
ANNTENA ANNTENA CONTROL) CONTROL) OPEN1LOOP OPEN1LOOP RESPONSE RESPONSE SYA SYATEM) TEM) C#I6? 46FI?C0%!I46 1D 6ow we have to find the open-loop response of the antenna a"imuth position control control system 04$A/2D Find the open loop response system and the open loop angular velocity. ++++++++++++++++++++++++++++++++++++%0!%+++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + E4/6 A44 0/#46#/ #B#!/2 1
!o ( s ) =
50 | P a g e
s
∗20.83
( s + 100 ) (s + 1.71)
FET/BSEE/F-13 2016
$y using partial fraction method
% & + + Po(s) s s + 100 s + 1.71 GGGGGGGGGG../&uation [email protected]
$y taking laplas inverse inverse Po (t) %J$e-155t Je-1.*1t GGGGGGGGGG../&uation 1@.
0/=C!I46D
s ( s + 1.71 ) ( s +100 ) ¿ * ( n )=
100∗2.083∗0.1∗s
¿
R+s,
C+s, 20.83
++++++++++++++++++++++++++++++++++++%0!$+++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + C#I6? 46FI?C0%!I46 1D E=%2I6? 0%!I4 %6= 6%!C0%A 0/#46#/ 4F 4/6 A44 #B#!/2 4pen loop transfer function is given as
51 | P a g e
FET/BSEE/F-13 2016
* ( s )=
20.83
s ² + 101.71 s + 171
NATURAL RESPONSE) Pn<1*1 171 Pn √ 171
Pn[email protected]*
DAMPIN* RAT RATIO) IO)
QPn151.*
" =
101.71 2∗13.07
Q@.8H %s we know that if QX1 so the system has over damped response
++++++++++++++++++++++++++++++++++++%0!++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + C#I6? 46FI?C0%!I46 1D E=/0I3/ !/ %6?CA%0 3/A4I!B 0/#46#/ !4 #!/ I6C! Me obtained the transfer function from above
* ( s )=
20.83
s ² + 101.71 s + 171
Mhere as we know that ( s ) ? (s) R ( s )
#o ( s ) 20.83 = ? (s) R ( s ) s ² + 101.71 s + 171 52 | P a g e
FET/BSEE/F-13 2016
%nd input is given which is a step response response 1
0(s) s #o 1
!o ( s ) =
s
∗20.83
( s + 100 ) (s + 1.71)
!o ( s )=
20.83
s ( s ² + 101.71 s + 171 ) GGGGGGGGGG../&uation 1@.@
$y using partial partial fraction method
% & + + Po(s) s s + 100 s + 1.71 Me can calculate the value of % when s5 20.83 s
%T s5
s ( s ² + 101.71 s + 171 ) 20.83
%T s5
( 0 ) ² +101.71 ( 0 )+ 171
20.83
%
171
%5.1
6ow we have to calculate $ when s-155 20.83 ( s + 100)
$T s-155
s ( s + 100 )( s + 1.71) 20.83
$T s-155
53 | P a g e
s ( s + 1.71)
FET/BSEE/F-13 2016
20.83
$
−100 (−100 + 1.71 )
$-5.551 For put s-1.*1 20.83 ( s + 171)
T s-1.*1
s ( s + 100 )( s + 1.71) 20.83
T s-1.*1
s ( s + 100 ) 20.83
−1.71 (−1.71+ 100)
-5.1K
0.122
Po(s)
s
−
0.0021
s + 100
+
0.124
s + 1.71 GGGGGGGGGGG.GGG../&uation
[email protected]
!aking !aking inverse laplas transformD Po (t) 5.1-5.551 e
−100t
−0.124 e−
1.71 t
GGGGGGGGGG../&uation 1@.
++++++++++++++++++++++++++++++++++++%0!/+++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ C#I6? 46FI?C0%!I46 1D Find the open loop angular velocity using a step response as input in 2%!A% $ E2%!A%$ 4==/ F40 !0%6#F/0 FC6!I46 04?0%2D
54 | P a g e
FET/BSEE/F-13 2016
!0%6#F/0 FC6!I46D
?0%D
+++++++++++++++++++++++++++++++%AA/6?/K+++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ +
55 | P a g e
FET/BSEE/F-13 2016
C#I6? 46FI?C0%!I46 D
%0!-(%) E4/6 A44 0/#46#/ #B#!/2 1
!o ( s ) =
s
∗24
( s + 150 ) (s + 1.32)
$y using partial fraction method
% & + + Po(s) s s + 150 s + 1.32
$y taking laplas inverse inverse GGGGGGGGGGGGGGGGGGG../&uation-1K Po (t) %J$e-15t Je-1.@t GGGGGGGGGGGGGGGGGGG../&uation-1K
0/=C!I46D
s ( s + 1.32 ) ( s + 150 ) ¿ * ( n )=
150∗0.8∗0.2∗s
¿
R+s,
C+s, 24
56 | P a g e
FET/BSEE/F-13 2016
++++++++++++++++++++++++++++++++++++%0!$+++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + C#I6? 46FI?C0%!I46 D E=%2I6? 0%!I4 %6= 6%!C0%A 0/#46#/ 4F 4/6 A44 #B#!/2 4pen loop transfer function is given as
* ( s )=
24
s ² + 151.32 s + 198
NATURAL RESPONSE) Pn<1H8 198 Pn √ 198
Pn1K.5*
DAMPIN* RAT RATIO) IO)
QPn11.@
" =
151.32 2∗14.07
Q.@8 %s we know that if QX1 so the system has over damped response
++++++++++++++++++++++++++++++++++++%0!++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + C#I6? 46FI?C0%!I46 D E=/0I3/ !/ %6?CA%0 3/A4I!B 0/#46#/ !4 #!/ I6C! Me obtained the transfer function from above
* ( s )=
24
s ² + 151.32 s + 198
57 | P a g e
FET/BSEE/F-13 2016
Mhere as we know that ( s ) ? (s) R ( s )
#o ( s ) 24 = ? (s) R ( s ) s ² + 151.32 s + 198 %nd input is given which is a step response response 1
0(s) s #o 1
!o ( s ) =
s
∗24
( s + 150 ) (s + 1.32)
!o ( s )=
24
s (s ² + 151.32 s + 198 ) GGGGGGGGGG../&uation 1K.1
$y using partial partial fraction method
% & + + Po(s) s s + 150 s + 1.32 GGGGGGGGGG../&uation 1K. Me can calculate the value of % when s5 24 s
s ( s ² + 151.32 s + 198 )
%T s5
20.83
%T s5
( 0 ) ² + 151.32 ( 0 )+ 198
24
% 198
58 | P a g e
FET/BSEE/F-13 2016
%5.11
6ow we have to calculate $ when s-15 24 ( s + 150 )
$T s-15
s ( s + 150 )( s + 1.32) 24
$T s-15
s ( s + 1.32) 24
$
−150 (−150 + 1.32 )
$-5.5515* For put s-1.@ 24 ( s + 1.32)
T s-1.@
s ( s + 150 )( s + 1.32) 24
T s-1.@
s ( s + 150 ) 24
−1.32 (−1.32+ 150)
-5.1
0.1212
Po(s)
s
−
0.00107
s + 150
+
0.1222
s + 1.32 GGGGGGGGGGGGG..G../&uation 1K.@
!aking !aking inverse laplas transformD −150 t
Po (t) 5.11-5.5515* e
−0.1222 e−
1.32 t
GGGGGGGGGG../&uation 1K.K
+++++++++++++++++++++++++++++++++++%0!/++++++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ C#I6? 46FI?C0%!I46 D
59 | P a g e
FET/BSEE/F-13 2016
Find the open loop angular velocity using a step response as input in 2%!A%$. E2%!A%$ 4=/ 4F 4/6-A44 %6?CA%0 3/A4I!B
04?0%2D
!0%6#F/0 FC6!I46D
?0%D
60 | P a g e
FET/BSEE/F-13 2016
++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++%!/0 ++++++%!/0*+++++++++++ *++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++ ++++++ E#!/%=B #!% #!%!/ !/ /0040 %6!/66% 46!04AD #!/%=B#!/%=B-#!% #!%!/ !/ /0040 M.0.! M.0.! ?%I6D ?%I6D C#I6? 46FI?C0%!I46 1D 04$A/2D EFind the steady state error in terms of gain 7 for different input responses. #teady state response of unit step input #teady state response of ramp input #teady state response response of parabolic input input ++++++++++++++++++++++++++++++++++++++++++%0!%+++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + E#!/%=B #!%!/ #!%!/ 0/#46#/
61 | P a g e
FET/BSEE/F-13 2016
In the following block block diagram we shift shift input potentiometer potentiometer to right side side of summing 'unction 'unction so we have
0(s)
(s) 6.63
Me obtain a feed back system 4pen loop control system obtained from the block diagram isD 6.63 k
?p (s)
2
s ( s + 101.7 s + 171) GGGGGGGGGGGGGGG
/&uation1K. /&uation1K. oncluded from the closed loop feedback control system block diagram. %ccording %ccording to the transfer transfer function we obtained from the the block diagram we can see see that the power of s is 1 so we have to calculate the steady state error for type 1 for all of the cases mentioned in the problem. problem.
62 | P a g e
FET/BSEE/F-13 2016
“STEADY1STATE “STEADY1STATE ERROR -OR STEP RESPONSE”
e ( + ) =, ( + )−- ( + ) is
!he steady-state error for the given system For step response response u (t),
sR ( s ) e ( ∞ ) e step( ∞ ) 1 + lim * ( s ) GGGGGGGGGG../&uation 1 s.0
e ( + ) =lim s* ( s )= lim s 0
s0
e ( +)
sR ( s ) 1+* ( s)
1 1 + kp
GGGGGGGGGGGGGGGGGGGGGGG/&uation-1.1
1
Mhere 0(s)
(unit step response)
s
-OR TYPE () ' 1
s 1
s* ( s ) =lim { s 0
1+
6.63 k
1
s ( ¿ ¿ 2 + 101.71 s 1 + 171) e ( + )= lim ¿ s0
1
e ( + ) =lim s* ( s )=({ s 0
1+
6.63 k 0
1
so
e ( ∞ )
e ( ∞ ) 5
63 | P a g e
+
,
kp∞
})
}
FET/BSEE/F-13 2016
“STEADY1STATE “STEADY1STATE ERROR -OR RAMP INPUT” For ramp input tu (t),
s R ( s) e ( ∞ ) eramp( ∞ ) lim * ( s ) GGGGGGGGGGGGG.G../&uation 1. s 0
0(s)1>s 1
e ( ∞ ) eramp( ∞ )
lim
s * (s )
s 0
the steady-state error is 1
e ( ∞ )
lim
where
kv
kv
s* ( s )
s 0
s lim
eramp( ∞ )
s 0
s
6.63 k
s ( ¿ ¿ 2 + 101.71 s1 + 171) 1
¿ -OR TYPE () ?( 6.63 k 2
1
(( 0) + 101 .71(0 ) + 171 ) ¿ ¿ eramp( ∞ ) ¿ 1
¿ 6.63 k 171
eramp( ∞ )
¿ ¿ ¿ 1
¿ 64 | P a g e
FET/BSEE/F-13 2016
1
k
k
eramp( ∞ )
,
25.79
kv 25.79
25.79
eramp( ∞ )
k
“STEADY1STATE “STEADY1STATE ERROR -OR PARA.OLIC PARA.OLIC INPUT” For ramp input t u (t),
s R ( s) e ( ∞ ) e parabolic( ∞ ) lim * ( s ) GGGGGGGGGG../&uation 1.@ s 0
0(s)1>s@ 1
e ( ∞ ) e parabolic( ∞ )
lim
2
s * ( s)
s 0
the steady-state error is 1
e ( ∞ )
where
ka
ka
s s ( ¿ ¿ 2 + 101.71 s1 + 171 ) 6.63 k
e parabolic( ∞ )
¿ lim s 0
1
¿ -OR TYPE () ?(
65 | P a g e
2
s ¿
lim s 0
2
s * ( s)
FET/BSEE/F-13 2016
0
¿ 0
¿ 6.63 k
e parabolic( ∞ ) ( ¿ 2 + 101.71 ( ¿¿ 1 + 171 )¿ ) ¿¿ ¿ ( 0 )¿ 1
¿ 1
e parabolic( ∞ )
0
1
e parabolic ( ∞ )
,
0
ka5
e parabolic ( ∞ ) ∞ ++++++++++++++++++++++++++++++++++++++++++%0!$+++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + EFind the value of k ,to yield 15R steady-st steady-state ate error in response %s we can see that according according to part part a we can obtain steady-state steady-state error error only with the ramp input input so we have to calculate calculate the value of gain(k) from the given e&uation because this is the only input that yields the non"ero finite valve or we can say a non-"ero error. error. 25.79
eramp( ∞ )
k
eramp( ∞ ) 15R 25.79
15R
k
∗100
25.79
66 | P a g e
5.1
k
FET/BSEE/F-13 2016
25.79
k
0.1
k*.H
and we can see that in chapter according to e&uation ()the range of the system is 5WkW@.H and we can see that k is *.H hence it is the range of the system so the given system is stable. +++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ ++++++++++++++++++%AA/6?/ +++++%AA/6?/+++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++ +++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ +
E#!/%=B #!%!/ #!%!/ 0/#46#/ C#I6? 46FI?C0% 46FI?C0%!I46 !I46 D From e&uation e&uation we have 76.4 k
?(s)
s ( s + 150 ) ( s + 1.32 )
oncluded from the closed loop feedback control system block diagram
“STEADY1STATE “STEADY1STATE ERROR -OR STEP RESPONSE” !he steady-state error for the given system
e ( + ) =, ( + )−- ( + ) is
For step response response u (t),
sR ( s ) e ( ∞ ) e step( ∞ ) 1 + lim * ( s ) GGGGGGGGGG../&uation 1.K s.0
e ( + ) =lim s* ( s )= lim s 0
e ( +)
67 | P a g e
s0
sR ( s ) 1+* ( s)
1 1 + kp
GGGGGGGGGGGGGGGGGGGGGGG/&uation-1.
FET/BSEE/F-13 2016
1
Mhere 0(s)
s
(unit step response)
-OR TYPE () ' 1
s 1
s* ( s ) =lim { s 0
76.4 k
1+
}
s ( ¿ ¿ 2 + 151.32 s1 + 198 s ) e ( + )= lim ¿ s0
1
e ( + ) =lim s* ( s )= lim { s 0
s0
1+
76.4 k
}
0
1
so
e ( ∞ )
+
kp∞
,
e ( ∞ ) 5
“STEADY1STATE “STEADY1STATE ERROR -OR RAMP INPUT” For ramp input tu (t),
s R ( s) e ( ∞ ) eramp( ∞ ) lim * ( s ) GGGGGGGGGGGGGGGGGGGGGGG/&uation-1. s 0
0(s)1>s 1
e ( ∞ ) eramp( ∞ )
lim
s * (s )
s 0
2
whose laplas transform is
6.63 k / s
the steady-state error is
1
e ( ∞ )
68 | P a g e
kv
lim
where
kv
s 0
s* ( s )
FET/BSEE/F-13 2016
s lim
eramp( ∞ )
s
s 0
76.4 k
1
s ( ¿ ¿ 2 + 151.32 s 1 + 198) 1
¿ -OR TYPE () ?( 76.4 k 2
(( 0 ) + 151.32( 0 )+198 ) ¿ ¿ eramp( ∞ ) ¿ 1
¿ 1
eramp( ∞ )
76.4 k
(198 )
k
,
kv
(2.59 )
1
eramp( ∞ )
k
( 2.59) 2.59
eramp( ∞ )
k
“STEADY1STATE “STEADY1STATE ERROR -OR PARA.OLIC INPUT” For ramp input t u (t),
s R ( s) e ( ∞ ) e parabolic( ∞ ) lim * ( s ) GGGGGGGGGGGGGGGGG/&uation-1.* s 0
0(s)1>s@ 1
e ( ∞ ) e parabolic( ∞ )
69 | P a g e
lim s 0
2
s * ( s)
FET/BSEE/F-13 2016
2
whose laplas transform is
6.63 k / s
the steady-state error is
1
e ( ∞ )
where
ka
ka
lim s 0
s lim
e parabolic( ∞ )
76.4 k
2
s
s ( ¿ ¿ 2 + 151.32 s 1 + 198 )
s 0
1
¿
-OR TYPE () ?(
s
(¿ ¿ 2 + 151.71 s + 198 ) 1
¿
76.4 k
¿
1¿
e parabolic( ∞ )
lim
¿
s
s 0
1
¿ 0
¿ 0
¿ 76.4 k
e parabolic( ∞ ) ( ¿ 2 + 151.32 ( ¿ ¿ 1 + 198 ( 0 )) ¿ ) ¿¿ ¿ ( 0 )¿ 1
¿ 1
e parabolic( ∞ )
70 | P a g e
0
2
s * ( s)
FET/BSEE/F-13 2016
1
e parabolic ( ∞ )
,
0
ka5
e parabolic ( ∞ ) ∞ ++++++++++++++++++++++++++++++++++++++++++%0!$+++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + EFind the value of k ,to yield 5R steady-st steady-state ate error in response %s we can see that according according to part part a we can obtain steady-state steady-state error error only with the ramp input input so we have to calculate calculate the value of gain(k) from the given e&uation because this is the only input that yields the non"ero finite valve or we can say a non-"ero error. error. 2.59
eramp( ∞ )
k
GGGGGGGGGGG/&uation-1
eramp( ∞ ) 5R 2.59
5R
k
∗100
2.59
5.
k 2.59
k
0.2
k1.H
and we can see that in chapter according to e&uation ()the range of the system is 5WkW@H. 6.63
and we can see that
s ( s + 100 ) ( s + 1.71 )
k is 1.H hence it is the range of the system so the given system is stable.
???????????????????????? ??????????? ?????????????????????????? ??????????????????@A%B'06C?? ?????@A%B'06C?????????????? ????????????????????????? ?????????????????????????? ?????????????????? ????? E044! A4C# !/6IVC/#
71 | P a g e
FET/BSEE/F-13 2016
%6!/66% 46!04A #B#!/2D!0%6I/6! #B#!/2D!0%6I/6! =/#I?6 M.0.! M.0.! ?%I6D 04$A/2 #!%! #!%!/2/6!D /2/6!D EFind the preamplifier gain for R overshoot ++++++++++++++++++++++++++++++++++++++++++%0!%+++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++++++++++++ +++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++++++ +++++++++++++++++++++++++ +++++++++++++++++++++++++ ++++++++++++++++++++++++++ ++++++++++++++++++++++ +++++++++ + C#I6? 46FI?C0%!I46 1D
6.63 k
%ccording %ccording to following block block diagram we have obtained obtained from from open loop system we we have a transfer function function From e&uation e&uation 1.1 6.63 k
?(s)
s ( s + 100 ) ( s + 1.71 )
6ow we have to sketch root locus form the the given transfer function
STEP) (+a,) (+a,) 5!"d t$e t$e poese4os poese4os a"d "o o5 64a"c$es 64a"c$es 54om t$e %!@e" %!@e" ope" oop oop t4a"s5e4 t4a"s5e4 5u"ct!o" 5u"ct!o" Poes) %ccording to e&uation 1.1 !here are three poles in the given transfer function
#5S
#-155S
#-1.*1S
Ze4os) !here are no "eros in the given transfer function %s we have no "eros the given transfer function function so all of the poles will terminate at infinity infinity ( ∞ )
NO O- .RANCHES) .RANCHES) 72 | P a g e
FET/BSEE/F-13 2016
6o of branches6p-"
@-5 6 @
STEP) (+6,) (+6,) 5!"d t$e t$e sta4t!"% a"d e"d!"% e"d!"% po!"ts o5 poes a"d e4oes e4oes STARTIN* POINT) !he starting point of poles are calculated as
#5S
#-155S
#-1.*1S
%nd the terminating point the poles are infinity infinity as "eros "eros are not present present so
∞;
∞;
∞;
STEP) 8) D4a# a 4oot 4oot ocus7 #e can draw root locus at the point where we have od d number of poles. %t positive aOis as we ha ve no pole there so we donLt have any root locus on positive aOis. 6ow we will check root locus on the negative aOis we have @ poles one pole p ole on origin as odd no of pole so root locus is forms on negative O-aOis afterwards the origin till the pole at s-1.*1.!hen after this point we h ave no root locus till s-155 because we have two n o. of poles there and root locus eOist there where odd no. of poles eOists .6ow after s-155 we hav e three no. of poles so
73 | P a g e
FET/BSEE/F-13 2016
their root locus eOists till the infinity.
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
STEP) =) -!"d t$e A"%e A"%ess o5 asymptotes asymptotes !o find the angle of asymptotes we have the formula
( 2 / + 1 ) π ¿ a no. o0 0ini 0inite te pole poless −no . o0 0inite 0inite 1e,os 1e,os /&uation-1.1
6o. of finite poles p@ 6o. of finite "eros"5 "eros"5 %nd &5,J1,J 6ow from e&uation no. 1.1 we can calculate angle of asymptotes
-o4 ? 1 ¿
74 | P a g e
(2 (0 )+ 1) 180 p − 1
GGGGGGGGGGGG
FET/BSEE/F-13 2016
(2 (0 )+ 1) 180 ¿ 1 3− 0 1 ( 180 )
1
3
15°
-o4 ?( ¿
(2 (1 )+1 ) 180 p− 1
(2 + 1) 180 ¿ 3 −0 3 ( 180 )
3
185°
-o4 ?8 ¿
¿
(2 (2 )+ 1) 180 p− 1
(4 +1) 180 3− 0 5 ( 180)
3
@55°
75 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw -jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
76 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
77 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
78 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
79 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw -jw
#!/D KD entriode
a
0inite pole pole −∑ 0inite 1e,os ∑ 0inite
no. o0 0ini 0inite te pole poless −no . o0 0inite 0inite 1e,os 1e,os GGGGGGGGGGGG.G/&uation-1.
( 0 + (−1.71 ) +(−100 ))−(0 ) a 3− 0 −1.71−100 a
3
−101.71 a
3
a -@@.H5 !he a intersects the real real aOis at -@@.H5 at an angle.
STEP) <) -!"d t$e 64ea9a#a 64ea9a#ayy po!"t7 6.63 k
1J?(s) (s) 1J s ( s + 100 ) ( s + 1.71 ) GGGGGGGG.G/&uatio-1.@
80 | P a g e
FET/BSEE/F-13 2016
s ( s + 100 ) ( s + 1.71 ) + 6.63 k s ( s + 100 )( s + 1.71)
1J?(s)(s)5
#o
s ( s + 100 ) ( s + 1.71 ) + 6.63 k 5 s ( s + 100 )( s + 1.71) s ( s + 100 ) ( s + 1.71 ) + 6.63 k
5
s@ D#$#!#s D##sD2!23 <$ GGGGGGGGGGGG.G/&uation-1.K
− s ³ −101.71 s ² −171 s k
6.63
GGGGGGGGGGGG.G/&uation-1.
6ow we have to take derivative derivative of k w.r. w.r.tt s 3
2
2 k −2 ( 0.15 s ) 2 (15.3 s ) 2 ( 25.7 s ) = − − 2s 2s 2s 2s
2 k =−3 ( 0.15 ) s2− 2 ( 15.3 ) s −25.8 2s 2 k =−0.45 s2 −30.6 s −25.8 GGGGGGGGGGG/&uation-1. 2s 2
−0.45 s −30.6 s −25.8 =0 0.45 s
2
+ 30.6 s + 25.8=0
GGGGGGGGGGG/&uation.1.*
#o by using &uadratic e&uation we have value of s
s=
81 | P a g e
−30.6 3 √ (− (−30.6 ) ²− 4 ( 0.45 ) (25.8 ) 2 ( 0.45 )
FET/BSEE/F-13 2016
s=
−30.6 3 √ 936.4 936.4 −4 ( 11.61)
s=
−30.6 3 √ 936.4 936.4 −46.4
s=
−30.6 3 √ 890
0.9
0.9
0.9
s has two different different values
s=
s=
−30.6 + √ 890 0.9
−30.6 + 29.8 0.9
s-5.8
S
s=
S
s=
S
−30.6− √ 890 890 0.9
−30.6−29.8 0.9
s-*.1
6ow we have to check the validity validity of each point in order order to do so we have to put the value of of both of s in e&uation-1. e&uation-1. k
−0.15 s ³ −15.3 s ²− 25.8 s
k
−0.15 (−0.852) ³ −15.3 (−0.852) ² −25.8 (−0.852 )
for s-5.8 we have
k15.H for s-*.1 k
−0.15 (−67.16) ³ −15.3 (−67.16 ) ²−25.8 (−67.16)
k-18H8.H* so break away point lies on root locus so the valid point is at k -5.8
STEP) ) -!"d t$e po!"t po!"t o5 !"te4sect!o" o" t$e aB!s aB!s 6ow we have to calculate the the point of intersection intersection on 'P aOis in order order to do so we have have to solve the e&uation e&uation using routh routh table.
82 | P a g e
FET/BSEE/F-13 2016
# @
1
#
151.*1
1*1 .@k
17392.41 −6.63 k
# 1
5
101.71
# 5
.@k
5
#o at k@.@ the system is marginally stable. %(s) 151.*1s< J.@kGGGGGGGGGGGGGGGGGGe&uation-1 J.@kGGGGGGGGGGGGGGGGGGe&uation-1. . utting value of of k in e&uation
%(s) 151.*1s< J.@ (@.@)
%(s) <151.*1s< J1*@H.K1 %(s) 5 151.*1s< J1*@H.K15 #J1@'
S #-1@'
-IND *AIN 9 8< O:ERSHOOT D 6ow we have to find the gain gain for R overshoot, overshoot, for this we have to calculate calculate Q !o calculate Q we have form formula from e&uation 1@ −π"
OS( ?
e
√ 1−" ² ²
$y taking natural log (ln) 5n both sides we have ln
'() 100
−ln
83 | P a g e
−π"
=ln e
'() 100
=
√ 1−" ² ²
π"
√ 1−"
2
FET/BSEE/F-13 2016
−ln J?
√ √
100
π ² + ln ²
−ln J?
'() '()
GGGGGGGGGGG../&uation-1@
100
25 100
π ² + ln ²
25 100
1.38
J?
π ² +( 1.38) ² √ π 1.38
J?
π ² + 1.921 √ π
J?7>>
6ow we have to find angle angle theta
cos-1 (Q) cos-1 (5.K5K) . 6ow we have to measure measure theta in real real negative aOis aOis in order to do so
∅ 180°- ∅ 180°-66.2° ∅
[email protected]
∅ [email protected] so that it cuts the root locus. #o the point of 6ow we have to sketch a line from origin at an angle intersection of the line and the root locus gives the point .
-5.8@@Z1.88' 6ow we have to apply ma%"!tude co"d!t!o"G as we have following e&uation from above 6.63 k
G(s) H(s) = s ( s + 100 ) ( s + 1.71 ) 84 | P a g e
FET/BSEE/F-13 2016
As s=p G(s) H(s) = 6.63 k
−0.833 + 1.88 4 (−0.833 +1.88 4 + 100 ) (−0.833 +1.88 4 +1.71 ) 6.63 k
G(s) H(s) = −0.833 + 1.88 4 ( 99.17 + 1.88 4 ) (−0.877 + 1.88 4 ) 6.63 k
G(s) H(s) = (−426.90 + 0.707 4 ) In magnitude condition ?(s) (s) 1 426.90
k
6.63
kK.K 1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw -jw
85 | P a g e
FET/BSEE/F-13 2016
B%6'-(+) ?????????????????????????????????????????E%'8%+C@O>0?????????????????????????????????????????? ???????????????????????? ??????????? ?????????????????????????? ?????????????????????????? ????????????????????????? ????????????????????????? ?????????????????????????? ??????????????????????? ?????????? B6O96%E I* E%'8%+5
@OE%*> I*>O5
86 | P a g e
FET/BSEE/F-13 2016
96%BA5
87 | P a g e
FET/BSEE/F-13 2016
????????????????????????????????????? ???????????????????????? ???????????????@A%880*90C ??@A%880*90C????????????? ????????????????????????? ????????????????????????? ????????????????????? ???????? ???????????????????????? ??????????? ?????????????????????????? ?????????????????????????? ????????????????????????? ????????????????????????? ?????????????????????????? ???????????????????? ??????? %ccording %ccording to following block block diagram we have obtained obtained from from open loop system we we have a transfer function function From e&uation e&uation 1.1 76.4 k
?(s)
s ( s + 150 ) ( s + 1.32 )
6ow we have to sketch root locus form the given transfer function
STEP) (+a,) (+a,) 5!"d t$e t$e poese4os poese4os a"d "o o5 64a"c$es 64a"c$es 54om t$e %!@e" %!@e" ope" oop oop t4a"s5e4 t4a"s5e4 5u"ct!o" 5u"ct!o" Poes) %ccording to e&uation !here are three three poles in the given transfer function
#5S
#-15S
#-1.@S
Ze4os) !here are no "eros in the given transfer function
88 | P a g e
FET/BSEE/F-13 2016
%s we have no "eros the given transfer function function so all of the poles will terminate at infinity infinity ( ∞ )
NO O- .RANCHES) .RANCHES) 6o of branches6p-"
@-5 6 @
STEP) (+6,) (+6,) 5!"d t$e t$e sta4t!"% a"d e"d!"% e"d!"% po!"ts o5 poes a"d e4oes e4oes STARTIN* POINT) !he starting point of poles are calculated as
#5S
#-15S
#-1.@S
%nd the terminating point the poles are infinity infinity as "eros "eros are not present present so
∞;
∞;
∞;
STEP) 8) D4a# a 4oot 4oot ocus7 #e can draw root locus at the point where we have od d number of poles. %t positive aOis as we ha ve no pole there so we donLt have any root locus on positive aOis. 6ow we will check root locus on the negative aOis we have @ poles one pole p ole on origin as odd no of pole so root locus is forms on negative O-aOis afterwards the origin till the pole at s-1.@.!hen after this point we h ave no root locus till s-15 because we have two n o. of poles there and root locus eOist there where odd no. of poles eOists .6ow after s-15 we hav e three no. of poles so
89 | P a g e
FET/BSEE/F-13 2016
their root locus eOists till the infinity.
1 10
ϭ
0 -160
-140
-120
-100
-80
-60
-40
-20
0
20
- -10 -1
-jw
STEP) =) -!"d t$e A"%e A"%ess o5 asymptotes asymptotes !o find the angle of asymptotes we have the formula
( 2 / + 1 ) π ¿ a no. o0 0ini 0inite te pole poless −no . o0 0inite 0inite 1e,os 1e,os /&uation-1*.1
6o. of finite poles p@ 6o. of finite "eros"5 "eros"5 %nd &5, J1, J 6ow from e&uation no. 1.1 we can calculate angle of asymptotes
-o4 ? 15°
-o4 ?( 185° 90 | P a g e
GGGGGGGGGGGG
FET/BSEE/F-13 2016
-o4 ?8 @55°
1
10
ϭ
0 -160
-140
-120
-100
-80
-60
-40
-20
0 -
-10
-1
-jw -jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
91 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
92 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
93 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
94 | P a g e
FET/BSEE/F-13 2016
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw
1 10
ϭ
0 -120
-100
-80
-60
-40
-20
0 - -10 -1
-jw -jw
95 | P a g e
FET/BSEE/F-13 2016
#!/D KD entriode
a
0inite pole pole −∑ 0inite 1e,os ∑ 0inite
no. o0 0ini 0inite te pole poless −no . o0 0inite 0inite 1e,os 1e,os GGGGGGGGGGGG.G/&uation-1*.
( 0 + (−1.32 ) +(−150 ))−(0 ) a 3− 0 −1.32−150 a
3
−151.32 a
3
a -5.KK !he a intersects the real aOis at -5.KK at an angle.
STEP) <) -!"d t$e 64ea9a#a 64ea9a#ayy po!"t7 76.4 k
1J?(s) (s) 1J s ( s + 150 ) ( s + 1.32 ) GGGGGGGG.G/&uatio-1*.@
s ( s + 150 ) ( s + 1.32 ) + 76.4 k s ( s + 150 ) ( s + 1.32)
1J?(s) (s) 5 #o
s ( s + 150 ) ( s + 1.32 ) + 76.4 k 5 s ( s + 150 ) ( s + 1.32) s ( s + 150 ) ( s + 1.32 ) + 76.4 k
96 | P a g e
5
FET/BSEE/F-13 2016
s@ D#1#!3s D#sD2!4 <$ GGGGGGGGGGGG.G/&uation-1*.K
− s ³ −151.32 s ² −198 s k
76.4
GGGGGGGGGGGG.G/&uation-1*.
6ow we have to take derivative derivative of k w.r. w.r.tt s 3
2
2 k −2 ( 0.013 s ) 2 ( 1.98 s ) 2 ( 2.59 s ) = − − 2s 2s 2s 2s
2 k =−3 ( 0.013 ) s2−2 ( 1.98 ) s −2.59 2s 2 k =−0.04 s 2−3.96 s− 2.59 GGGGGGGGGGG/&uation-1*. 2s 2
−0.04 s −3.96 s−2.59 =0 2
0.04 s
+3.96 s +2.59= 0
GGGGGGGGGGG/&uation.1*.*
#o by using &uadratic e&uation we have value of s
s=
s=
−3.96 3 √ 15.68 15.68− 4 ( 0.1036 )
s=
−3.96 3 √ 15.68 15.68− 0.4144
s=
−3.96 3 √ 15.27 15.27
s has two different different values
97 | P a g e
−3.96 3 √ (− (−3.96 ) ²− 4 ( 0.04 ) ( 2.59 ) 2 ( 0.04 )
0.08
0.08
0.08
FET/BSEE/F-13 2016
s=
s=
−3.96 + √ 15.27
S
0.08
−3.96 + 3.91
s=
S
0.08
s-5.
s=
S
−3.96− √ 15.27 15.27 0.08
−3.96−3.12 0.08
s-H8.K
6ow we have to check the validity validity of each point in order order to do so we have to put the value of of both of s in e&uation-1. e&uation-1. k
−0.013 s ³ −1.98 s ²− 2.59 s
k
−0.013 (−0.625) ³ −1.98 (−0.625 ) ²−2.59 (−0.625 )
for s-5. we have
k5.8K8 for s-H8.K k
−0.013 (−98.4 ) ³−1.98 (−98.4 ) ²−2.59 (−98.4 )
k-@5.8 so break away point lies on root locus so the valid point is at k -5 !s 8 sec
STEP) ) -!"d t$e po!"t po!"t o5 !"te4sect!o" o" t$e aB!s aB!s 6ow we have to calculate the the point of intersection intersection on 'P aOis in order order to do so we have have to solve the e&uation e&uation using routh routh table. # @
1
#
11.@
# 1
1H8
29961 −76.4 k
*.Kk 5
151.32
# 5
98 | P a g e
*.Kk
5
FET/BSEE/F-13 2016
#o at k@H.58 the system is marginally stable. %(s) 11.@s< J*.KkGGGGGGGGGGGGGGGGGGe&uation-1 J*.KkGGGGGGGGGGGGGGGGGGe&uation-1. . utting value of of k in e&uation
%(s) 11.@s< J*.K (@H.58)
%(s) <11.@s< JHHK.H1 %(s) 5 11.@s< JHHK.H15 #J1K.1'
S #-1K.1'
-IND *AIN 9 8< O:ERSHOOT D 6ow we have to find the gain gain for R overshoot, overshoot, for this we have to calculate calculate Q !o calculate Q we have form formula from e&uation 1@ −π"
OS( ?
e
√ 1−" ² ²
$y taking natural log (ln) 5n both sides we have ln
'() 100
−ln
√
100
99 | P a g e
√
=
π"
√ 1−"
2
'() 100
π ² + ln ²
−ln J?
=ln e
'()
−ln J?
− π" √ 1− " ² ²
'() 100
25 100
π ² + ln ²
25 100
GGGGGGGGGGG../&uation-1@
FET/BSEE/F-13 2016
1.38
J?
π ² +( 1.38) ² √ π 1.38
J?
π ² + 1.921 √ π
J?7>>
6ow we have to find angle angle theta
cos-1 (Q) cos-1 (5.K5K) . 6ow we have to measure measure theta in real real negative aOis aOis in order to do so
∅ 180°- ∅ 180°-66.2° ∅
[email protected]
∅ [email protected] so that it cuts the root locus. #o the point of 6ow we have to sketch a line from origin at an angle intersection of the line and the root locus gives the point .
-5.Z.H' 6ow we have to apply ma%"!tude co"d!t!o"G as we have following e&uation from above 76.4 k
G(s) H(s) = s ( s + 150 ) ( s + 1.32 ) As s=p 76.4 k
G(s) H(s) = −0.5 + 6.9 4 (−0.5 + 6.9 4 + 150 ) (−0.5 + 6.9 4 + 1.32) In magnitude condition ?(s) (s) 1 71294.23
k
100 | P a g e
76.4
FET/BSEE/F-13 2016
<33!#
????????????????????????????????????????????B%6'C+??????????????????????????????????????????????? ???????????????????????? ??????????? ?????????????????????????? ?????????????????????????? ????????????????????????? ????????????????????????? ?????????????????????????? ?????????????????????? ????????? E%'8%+ E%' 8%+ @O>0 @ O>0 B6O96%E5
@OE%*> I*>O5
101 | P a g e
FET/BSEE/F-13 2016
96%BA5
???????????????????????? ??????????? ?????????????????????????? ???????????????????@A%B'06C#$ ??????@A%B'06C#$???????????? ????????????????????????? ?????????????????????????? ?????????????????? ????? %*'0**% @O*'6O85 '%+I8I' >0I9* %*> '6%*I0*' B06FO6E B06FO6E%*@0 %*@0
INPUT POTENTIOMETER & OUTPUT POTENTIOMTER: SUBSYSTEM 1 & SUBSYSTEM 5:
V 10 10 = = =0.318 nπ ( 10 ) π 10 ( 3.147 ) PRE AMPLIFIER: SUBSYSTEM 2
Vp ( s ) = k Ve ( s ) PPWER AMPLIFIER: SUBSYSTEM 3:
Ea ( s ) = 100 Vp ( s ) s + 100 102 | P a g e
FET/BSEE/F-13 2016
MOTOR AND LOAD: SUBSYSTEM 4: 4: INERTIA:
9ear ratio =
N 1 1 kg = N 2 2 Jm
< $!$3g-m;
:ISCOUS DAMPER) =m5.5 6-m-s>rad
SPRIN*) SPRIN*)
7m.58@@ 6-m- >rad MOTOR AND LOAD POLE) POLE)
am1.*581.*1 θo ( s ) 0.2083 = Ea ( s ) s ( s + 1.71 ) +++++++++++++++++++++++++++++++++++++++%0!%++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + For using fre&uency response response techni&uesS Firstly transfer function function from above chap is
¿
?(s) (s)
6.63 k
s ( s + 100 ) ( s + 1.71 ) 6.63 k
s
(
s 100
+1
) ( 100
s 1.71
0.038 k
103 | P a g e
s
(
s 100
+1
)(
s 1.71
)
+1
)
+1
1.71
FET/BSEE/F-13 2016
6ow we have to substitute 'P instead of s in above e&uation 0.038 k
? ('P) ('P)
4!
(
4! 100
+1
)(
4! 1.71
+1
)
6ow we have to put the value of k1 in above e&uation 0.038
4!
(
4! 100
+1
)(
4! 1.71
+1
)
6ow we have to find magnitude
4! T ? ('P)T 5 log 15 15 5log 15 15 (5.5@8)- 5log 15 15 T('P)T -5log 15 15 (1J 6ow find the magnitude fre&uency response response
46#!%6!D 5.5@8k where k1 hence 5.5@8k 6ow Me have to find magnitude
5log 15 15 (Tk T) 5log 15 15 T5.5@8T TkTdb -8.@
A4! %! 40I?I6D - 5log 15 15 (T'PT) -5 db> decade passing through "ero db at w1 #I2A/ 4A/>[/04#D
4! a. when Pa
104 | P a g e
-5log 15 1J 15 ( 1J
1.71
)
1.71
4! )- 5log 15 15 (
100
+ 1 )
FET/BSEE/F-13 2016
-5log 15 15
√
-5log 15 15
√ 1+ 1
-5log 15 15
√ 2
1+(
1.71 1.71
)²
[email protected] b. when PWWa
-5log 15 15
√
-5log 15 15
√ 1+ .1 ²
1+(
0.71 1.71
)²
-5.5@K . when PXXa
-5log 15 15
√
-5log 15 15
√ 1+ 10 ²
1+(
17.1 1.71
)²
-5.5K when
4!
-5log 15 1J 15 ( 1J
100
)
√
100
-5log 15 15
-5log 15 15
√ 1+ 1
a. when Pa
105 | P a g e
1+(
100
)²
FET/BSEE/F-13 2016
-5log 15 15
√ 2
[email protected] b. when PWWa
-5log 15 15
√
-5log 15 15
√ 1+ .1 ²
1+(
10 100
)²
-5.5K . when PXXa
-5log 15 15
√
1+(
1000 100
)²
-5.5K 6ow summing up the magnitude response (fre&uency) (fre&uency) as follows
? ('P) -5 -5.5K-5.5KJ5-8-8.58 [email protected]8[email protected] -5.5K-5.5K-5-8-8.58 [email protected]8-111.51 -5.5K-5-5-8-18.5K -K5-85-85-8-8 -5-155-155-8-88 -85-15-15-8-@K8
#C22/= C F0/VC/6B A4!D
106 | P a g e
w 5 .1 1 15 155 1555 15,555 155,555 1555,555
FET/BSEE/F-13 2016
6ow for phase fre&uency response response 46#!%6!D If constX5
phase angle 5 °
If constW5
phase angle 185°
If const is negative
phase angle-185°
A4! %! 40I?I6D !he plot at origin is
-H5°
#I2A/ 4A/#>[/04#
∅=
FET/BSEE/F-13 2016
Whee <
∅ 0°0°0°-0°=-0° 0°0°(-#"°)-0°=1"° 0°0°-0°-0°=-180° 0°-#"°-0°-180°=-1"° 0°-0°-0°-2!0°=-#"0° 0=60°-180°-0°=-0°
#ummed up %#/ A4!D
RAN*E -OR STA.IL STA.ILITY ITY O- /) +A, onsider the response at [email protected] rad>sec %t this point magnitude plot is -8.K1
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%nd phase is -180°
h+s hen =26.6 3,, 4a+se the the s5stem t& 'e ma3na,,5 sta',e sta',e Hen4e ane &7 3,, 'e 0<<26.6 0<<26.6 (B) PERCENTAGE OERS!OOT W!EN GAIN IS 3":
If we set gain as <3$ e first mae a nd order approG &assume that nd order transient response e.uation relating to percentage oershoot, damping ratio phase margin are true for for this system If <3$ magnitude cure of oer plot plot is $log3$ <!14db <!14db 'herefore the adHusted magnitude cure goes through "ero db at !<# %t this fre.uency, fre.uency, the phase angle is -#$! degree yielding a phase margin of 1!# degree (C) SETTELING TIME:
'o calculate calculate settling time, we mae a nd order response %pproG! the open loop magnitude response is - db hen the normali"ed magnitude response of our plotted graph is -32!14d+ 'hus we hae to estimate bandwidth +and-width<#! radJsec & according to 's formula 's becomes 's<4!1 sec (D) PEAK TIME:
Ksing calculated alue alue of band width from (c) we can estimate pea time time to be Bea time< 'p<!1 sec (E) RISE TIME:
For rise time ,we can use calculated band-width from ( c)
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and
L<$!2
e can find Mn
Mn<#!1
using normali"ed rise time and the Mn we can find 'r<#!14J#!1<#!#sec ?????????????????????????????????????????????E%'8%+C@O>0?????????????????????????????????????? ???????????????????????? ??????????? ?????????????????????????? ?????????????????????????? ????????????????????????? ????????????????????????? ?????????????????????????? ??????????????????????? ?????????? @O*FI9K6%'IO* #5 B6O96%E5
@OE%*> I*>O5
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96%BA5
????????????????????????????????????? ???????????????????????? ??????????????@A%880*90C? ?@A%880*90C?????????????? ????????????????????????? ????????????????????????? ????????????????????? ???????? ???????????????????????? ??????????? ?????????????????????????? ?????????????????????????? ????????????????????????? ????????????????????????? ?????????????????????????? ??????????????????????? ??????????
C#I6? 46FI?C0%!I46 D +++++++++++++++++++++++++++++++++++++++%0!%++++++++++++++++++++++++++++++++++++++++++++ +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ + For using fre&uency response response techni&uesS Firstly transfer function function from above chap is
¿
?(s) (s)
76.4 k
s ( s + 150 ) ( s + 1.32 ) 76.4 k
s
(
s 150
+1
) ( 150
s 1.32
0.38 k
s
(
s 150
+1
)(
s 1.32
6ow we have to substitute 'P instead of s in above e&uation
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)
+1
)
+1
1.32
FET/BSEE/F-13 2016
0.38 k
? ('P) ('P)
4!
(
4! 150
+1
)(
4! 1.32
+1
)
6ow we have to put the value of k1 in above e&uation 0.38
4!
(
4! 150
+1
)(
4! 1.32
+1
)
6ow we have to find magnitude
4! T ? ('P)T 5 log 15 15 5log 15 15 (5.@8)- 5log 15 15 T('P)T -5log 15 15 (1J 6ow find the magnitude fre&uency response response
46#!%6!D 5.5@8k where k1 hence 5.@8k 6ow Me have to find magnitude
5log 15 15 (Tk T) 5log 15 15 T5.@8T TkTdb -8.K5K
A4! %! 40I?I6D - 5log 15 15 (T'PT) -5 db> decade passing through "ero db at w1 #I2A/ 4A/>[/04#D
4! a. when Pa
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-5log 15 1J 15 ( 1J
1.32
)
1.32
4! )- 5log 15 15 (
150
+1 )
FET/BSEE/F-13 2016
-5log 15 15
√
-5log 15 15
√ 1+ 1
-5log 15 15
√ 2
1+(
1.32 1.32
)²
[email protected] b. when PWWa
-5log 15 15
√
-5log 15 15
√ 1+ .1 ²
1+(
0.132 1.32
-5.5@K . when PXXa
-5log 15 15
√
-5log 15 15
√ 1+ 10 ²
1+(
13.2 1.32
-5.5K when
4!
-5log 15 1J 15 ( 1J
150
%. when Pa -5log 15 15
√ 1+ 1
-5log 15 15 [email protected]
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√ 2
)
)²
)²
FET/BSEE/F-13 2016
$. when PWWa -5log 15 15
√ 1+ .1 ²
-5.5K . when PXXa
-5.5K 6ow summing up the magnitude response (fre&uency) (fre&uency) as follows
? ('P) -5 -5.5K-5.5KJ5-8-8.58 [email protected]8[email protected] -5.5K-5.5K-5-8-8.58 [email protected]8-111.51 -5.5K-5-5-8-18.5K -K5-85-85-8-8 -5-155-155-8-88 -85-15-15-8-@K8
#C22/= C F0/VC/6B A4!D
6ow for phase fre&uency response response 46#!%6!D
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w 5 .1 1 15 155 1555 15,555 155,555 1555,555
FET/BSEE/F-13 2016
If constX5
phase angle 5 °
If constW5
phase angle 185°
If const is negative
phase angle-185°
A4! %! 40I?I6D !he plot at origin is
-H5°
#I2A/ 4A/#>[/04#
∅=
∅=-#"°
Whee $$a
∅=-8#.28°
%& (') Whee <
∅ 0°0°0°-0°=-0° 0°0°(-#"°)-0°=1"° 0°0°-0°-0°=-180°
FET/BSEE/F-13 2016
100 1000 10000
0°-#"°-0°-180°=-1"° 0°-0°-0°-2!0°=-#"0° 0=60°-180°-0°=-0°
#ummed up %#/ A4!D
RAN*E -OR STA.IL STA.ILITY ITY O- /) +A, onsider the response at [email protected] rad>sec %t this point magnitude plot is -8.K1 %nd phase is -180°
h+s hen =2.08 3,, 4a+se the the s5stem t& 'e ma3na,,5 sta',e sta',e Hen4e ane &7 3,, 'e 0<<2.08 0<<2.08 (B) PERCENTAGE OERS!OOT W!EN GAIN IS 3":
If we set gain as <3$ e first mae a nd order approG &assume that nd order transient response e.uation relating to percentage oershoot, damping ratio phase margin are true for for this system If <3$ magnitude cure of oer plot plot is $log3$ <!14db <!14db 'herefore the adHusted magnitude cure goes through "ero db at !<#
117 | P a g e
FET/BSEE/F-13 2016
%t this fre.uency, fre.uency, the phase angle is -#$! degree yielding a phase margin of 1!# degree (C) SETTELING TIME:
'o calculate calculate settling time, we mae a nd order response %pproG! the open loop magnitude response is - db hen the normali"ed magnitude response of our plotted graph is -32!14d+ 'hus we hae to estimate bandwidth +and-width<#! radJsec & according to 's formula 's becomes 's< sec (D) PEAK TIME:
Ksing calculated alue alue of band width from (c) we can estimate pea time time to be Bea time< 'p<!1 sec (E) RISE TIME:
For rise time ,we can use calculated band-width from ( c) and
L<$!2
e can find Mn
Mn<#!1
using normali"ed rise time and the Mn we can find 'r<#!14J#!1<#!#sec ?????????????????????????????????????????????E%'8%+C@O>0?????????????????????????????????????? ???????????????????????? ??????????? ?????????????????????????? ?????????????????????????? ????????????????????????? ????????????????????????? ?????????????????????????? ??????????????????????? ?????????? @O*FI9K6%'IO* #5 B6O96%E
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COMAND WINDOW:
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GA!":
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