KONTROL SİSTEMİ
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Transcript of KONTROL SİSTEMİ
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7/30/2019 KONTROL SSTEM
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26 September 2004 21:30 2-1 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Eyll 2004
www.altas.org
Blok
Diyagramlar
ve
aret AkGrafikleri
26 September 2004 21:30 2-2 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Block Diagram
BLOCK MANIPULATION RULES
Gain Block
Summation Junction
Pick-off Point
G(s)Ea(s) m (s)
+ _
R(s) E(s)
Y(s)
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Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Block DiagramBLOCK MANIPULATION RULES
RULE
#
PROCESS ORIGINAL BLOCK
DIAGRAM
EQUIVALENT BLOCK
DIAGRAM
1
2
3
COMBINING
SERIAL
BLOCK
COMBINING
PARALLEL
BLOCK
CLOSING A
FEEDBACK
LOOP
G1
G1
G2
G2
+_G1 G2
G1G2
G G
+_
1 GHH
+ +_
-+
+
X
X
X
Y
Y
Y X Y
X
X
Y
Y
26 September 2004 21:30 2-4 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
4
5
6
7
MOVING
A SUMMING
JUNCTION
AHEAD OF
A BLOCK
MOVING
A SUMMING
JUNCION
PAST A BLOCK
MOVING A
PICKOFF POINT
AHEAD OF
A BLOCK
MOVING A
PICKOFF POINT
PAST A BLOCK
G
GG
G
G
G
G
G
1/G
1/G
X
X
X
X
X
X
X X
Y
Y
Y
Y
Y
ZZ
Z
Z
XX
Y
Y
+-
+
+
+
++
++
_
_
_
Y
Y
Y
G
G
Block Diagram
BLOCK MANIPULATION RULES
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26 September 2004 21:30 2-5 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Block Diagram Reduction
An overall input/output transfer function can be
obtained from the block diagram by applying someblock diagram reduction rules.
Series rule:
Feedback rule:
G1(s) G2(s) G1(s) G2(s)
G(s)
H(s)
)(1
)(
sGH
sG
++ _
26 September 2004 21:30 2-6 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Feedback rule derivation:
G(s)
H(s)
+ _R(s) E(s) C(s)
)()(1
)(
)(
)(
)]()()()[()(
)()()()()()()(
sHsG
sG
sR
sC
sCsHsRsGsC
sEsGsCsCsHsRsE
+=
=
==
This is a rule
that is usedextensively.
then
and
Block Diagram Reduction
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26 September 2004 21:30 2-7 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Servomotor Block Diagram Reduction
Td(s)
Ea(s) m(s)+
+
+
--
Ki
Kb
aa LsR +
1
mm JsB +
1
s
1
Td(s)
Ea(s) ++
+
--
Kiaa LsR +
1
mm JsB +
1
s
1G1(s) G2(s) G3(s) G4(s)
H1(s)
With Td(s)=0 , first combine the inner forward path.
G1 G2 G3(s)
H1(s)
+ _G4(s)
Ea(s) m(s)
26 September 2004 21:30 2-8 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Next combine the feedback loop.
)(1
)(
1321
321
sHGGG
sGGG
+ G4(s)
m(s)Ea(s)
The final series combination is
)(1
)(
1321
4321
sHGGG
sGGGG
+
Ea(s)m(s)
Note: The defined values of the components may be substituted in
to get the final transfer function in terms of system parameters.
Servomotor Block Diagram Reduction
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26 September 2004 21:30 2-9 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Block Diagram
Block Diagram Reduction
An ExampleGiven a control system represented inthe block diagram shown. Determine
the relationship Y(s)/R(s).
R(s) +G1 G2 G3
H2
H1
Y(s)
_
+
+
_
+
(a)
26 September 2004 21:30 2-10 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Block Diagram
Block Diagram Reduction
An Example
Y(s)R(s)
H2
G1
G1 G2
H1
+ _
+
+G3
+
_
(b)
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Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Block Diagram
Block Diagram Reduction
An Example
Y(s)R(s)
H2
G1
G3+
_+
_
G1G2
1- G1G2H1
26 September 2004 21:30 2-12 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Block Diagram ReductionAn Example
+_R(s) Y(s)
(d)
G1G2G3
1- G1G2H1+G2G3H2-
R(s) Y(s)
(e)
G1G2G3
1- G1G2H1+G2G3H2+G1G2G3
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26 September 2004 21:30 2-13 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Signal Flow Graphs
The block diagram reduction method works well for
relatively simple block diagrams, but it gets very
confusing for more complicated models.
A signal flow graph represents the same information
as the block diagram, however it leads to a set ofrules that allow a systematic approach to finding the
overall input/output transfer function.
Basic definitions
26 September 2004 21:30 2-14 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Signal Flow Graphs
DEFINITION: - It is agraphicaltool for control systems
analysis and design
- It consists ofnodes and branches
- The relationship between the inputs(s) andoutput(s) are determined byMasons gain
formula
PROPERTIES OF FLOW GRAPHS:
Each branch is unilateral (one direction)
Each node trasnsmits the sum of all entering signals along each output branch
Aforward path is the path travelled by the signal in a forward directionA loop is formed when the signal travels and returns to its original source
Special nodes: Source node - no inputs
Sink node - no outputs
Basic definitions
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Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
The main steps are as follows:
a) construct the signal flow graph either from a
block diagram or from the basic physical
connection of system components (the transfer
functions of the components must be known).
b) Identify and calculate the various paths and
loops in the signal flow graph.
c) With the results from b), apply a formula,
Masons formula, to determine the overall
transfer function.
Signal Flow Graphs Basic definitions
26 September 2004 21:30 2-16 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Nodes, branches and transmission elements
Summation node Distribution node
node
x1 x2t12
nodebranch
x2 = t12x1
t12 x2x1
t12 G12(s)
x2
x1
t12t13
t14 x3
x4
t14t24
t34x2
x3
x4
x1
x4 = t14x1 + t24x2 + t34x3
x2 = t12x1
x3 = t13x1x4 = t14x1
Signal Flow Graphs Construction
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Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
ConstructionSignal Flow Graphs
1. A SINGLE BRANCH
V1,V2 are called nodes and T12 is called a branch
This single branch represents the equation V2 = T12 V1
Note: V1 = V2/T12 (each branch is unilateral)
2. SUM OF TWO BRANCHES
V3 = T13V1 + T23V2
V1 V2
T12
V1 V3
V2
T13
T23
26 September 2004 21:30 2-18 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Construction
3. PARALLEL BRANCHES
V2 = (T12a + T12b) V1
V2 = T12 V1
V1 = T21 V2
V1 V2
T12b
T12a
T12
V1
Signal Flow Graphs
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26 September 2004 21:30 2-19 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Construction
4. CASCADED BRANCHES
V3 = T12 + T23 V1
5. NODE ELIMINATION
V3 = T13V1 + T23V2 and V4 = T34V3 , then
V4 = (T34T13)V1 + (T34T23)V2
V1 T12 V2 V3T23
T13 T34V1 V3
T23
V2
V4 V1T34 T13
V4
T34 T23
Signal Flow Graphs
26 September 2004 21:30 2-20 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Write down and label the nodes from input to
output, representing all the important signals.
Draw in all the branches connecting the nodes and
write down their transmission functions.
Check for any additional nodes and branches
required in the feedback paths.
ConstructionSignal Flow Graphs
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26 September 2004 21:30 2-21 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Td(s)
Ea(s) ++
+
--
Ki
Kb
aa LsR +
1
mm JsB +
1
s
1
Td(s)
Ea(s)
+
--
Kiaa LsR +
1
mm JsB +
1
s
1
G1(s) G2(s) G3(s) G4(s)
H1(s)
x1 x2 x3 x4
x5
m(s)
x1 G1x2 G2
x3 x4G3 G41Ea(s)m(s)
-H1
Td(s)1
Example 1: Servomotor System
Signal Flow Graphs
26 September 2004 21:30 2-22 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Source node: only has outgoing branches.
Sink node: only has incoming branches. Path: continuous unidirectional succession of branches
(passes through no node more than once).
Forward path: a path from input to output.
Feedback path or loop: originates and terminates at thesame node.
Non-touching paths: paths with no common nodes.
Path gain or loop gain: product of branch gains ortransmission functions along the path.
Definitions
Masons Formula
Signal Flow Graphs
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Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
1. Identify all forward paths and write the path gainsMk.
2. Identify all loops and write the loop gains.3. Identify all non touching loop pairs and write down
the loop gain products.
4. Do the same for groups of 3, 4, non touching loops.
5. Calculate as defined.
6. Identify all loops not touching forward path k, and
repeat steps 2 -> 5 to calculate k .
7. Apply Masons formula to calculate the overall
transfer function.
A Systematic ApproachMasons Formula
Signal Flow Graphs
26 September 2004 21:30 2-26 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Forward paths:
M1 Eax1x2x3x4 m Gain = G1G2G3G4 Feedback loops:
L1
x1
x2
x3
x4
x1
Loop gain = - G1
G2
G3
H1
Non touching loop pairs: none
x1 G1 x2 G2 x3 x4G3 G41Ea(s) m(s)
-H1
Td(s)
)(
)(
sE
s
a
m
Masons FormulaExample 1
Servomotor System
Signal Flow Graphs
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26 September 2004 21:30 2-27 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
then, = 1 + G1G2G3H1Loops not touching forward path 1: none
then, 1 = 1
Apply Masons formula.
( )1321
432111
a
m
HGGG1GGGGM
)s(EsT
+=
==
Masons FormulaExample 1
Servomotor System ...continued
Signal Flow Graphs
26 September 2004 21:30 2-28 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Consider the transfer function from the disturbance
input, Td(s) to the output, m(s) , with (Ea = 0). The forward path is now
M1 Tdx3x4 m Gain = G3G4 The loops are not changed, so and 1 are unchanged.
Applying Masons formula
( )
1321
4311
1)( HGGG
GGM
sT
sT
d
m
+=
==
Note: The denominator has not changed.
Masons FormulaExample 1
Servomotor System ...continued
Signal Flow Graphs
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26 September 2004 21:30 2-29 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Forward Paths:
M1 RR x
3x
4C Gain = G
6G
4G
5
M2 RR x1x2x3x4 C Gain = G1G2G3G4G5
C(s)G1x1 G2
x2 x3G3 G41R(s)
-H1
x4 G5
R(s)
-H2
G6
Masons FormulaExample 2
26 September 2004 21:30 2-30 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Feedback loops:
L1 x1x2x1 Loop gain =G2H1L2 x3x4x3 Loop gain =G4H2
Non touching loop pairs:
L1L2 Loop gain = G2G4H1H2
then = 1 (G2H1G4H2) + (G2G4H1H2)= 1 + G2H1 + G4H2 + G2G4H1H2
Masons FormulaExample 2
... continued
Signal Flow Graphs
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26 September 2004 21:30 2-33 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Masons Formula Example 3Low frequency ac model of a single transistor circuit ... continued
CIRCUIT EQUATIONS:
SIGNAL FLOW GRAPH:
Vs +1 Vb +1 Vb`e Vout
-Rs -R b
Is
1/R1
-GmRL
(1) Vb = Vs - Rs Is
( 2) Vb`e = Vb -R b Is
(3) Is = Vb`e / Ri
(4) Vout = -Gm Vb`e RL
Signal Flow Graphs
26 September 2004 21:30 2-34 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Masons Formula Example 3
Low frequency ac model of a single transistor circuit ... continued
FP1 = Vs + Vb + Vb`e + Vout
= (1) (1) (-Gm RL)But ,
FP2 Vs + Vb + Vb`e + Is + Vb`e +Vout (This contains a loop)
LOOP #1 = Vb + Vb`e + Is + Vb
= (1) (1/R1) (-Rs) = - Rs/Ri
LOOP #2 = Vb`e + Is + Vb`e
= (1/Ri) (-Rb`) = -Rb`/Ri
LOOP #3 = (1) (-Rb`) (-Rs)
Signal Flow Graphs
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26 September 2004 21:30 2-35 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Masons Formula Example 3
Low frequency ac model of a single transistor circuit ... continued
T1 = Vout/Vs =(1) (1) (-Gm RL) [ 1 - 0]
1 - (-Rs/Ri -Rb`/Ri)
FP1
=-GmRLRi
Ri + Rs + Rb`
T2 = Vb`e/Vs =(1) (1) [1 - 0 ]
1 + Rs/Ri + Rb`/Ri
}
FP2
=Ri
Ri + Rs + Rb`
Signal Flow Graphs
26 September 2004 21:30 2-36 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Masons Formula Example 4
CVin A V1 B
F E
V4
V2 V3 DVout
Vout / Vin =
ABCD
1 - CEF
Vout / Vin =FP
1 - LP
Signal Flow Graphs
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Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Masons Formula Example 5
Vin A V1 B
G
C
D
V2 E
F
Vout
FP1 = ACE , FP2 = ABDE
LOOP #1 = BG , LOOP #2 = EF
T = Vout/Vin = ACE [ 1 - 0 ] + ABDE [ 1 - 0 ]1 - ( BG + EF ) + ( BGEF )
=ACE + ABDE
1 - ( BG + EF ) + ( BGEF )
Signal Flow Graphs
26 September 2004 21:30 2-38 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Masons Formula Example 6
Vin A B
C
D
I
E F
H
G Vout
.F.P.1 = ACFG , LOOP #1 = DI
F.P.2 = ABDEFG , LOOP #2 = FH
T = Vout/Vin =ACFG [ 1 - DI ] + ABDEFG [ 1 - 0 ]
1 - ( DI + FH ) + ( DIFH )
= ACFG - ACFGDI + ABDEFG
1 - DI - FH + DIFH
Signal Flow Graphs
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26 September 2004 21:30 2-39 i.h Alta - 2003
Department of Electrical and Electronics Engineering - Electrical and Control Area
ELKE 405 Automatic Control Systems
Masons Formula Example 7
Vin +1
H
B
CA D E
G
F
+1 Vout.
FP1 = (1) (B) (1) = B LOOP #1 = CH
FP2 = (1) (A) (C) (D) (E) (F) (1) = ACDEF LOOP #2 = EG
T = Vout/Vin =B [ 1 - ( CH + EG ) + ( CHEG ) ] + ACDEF [ 1 - 0 ]
1 - ( CH + EG ) + ( CHEG )
T = Vout/Vin =B - BCH + BEG + BCHEG + ACDEF
1 - CH - EG + CHEG
Signal Flow Graphs