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Automatic Control Systems
Lecture-5Block Diagram Representation of Control Systems
Introduction• A Block Diagram is a shorthand pictorial representation of
the cause-and-effect relationship of a system.
• The interior of the rectangle representing the block usuallycontains a description of or the name of the element, gain,or the symbol for the mathematical operation to beperformed on the input to yield the output.
• The arrows represent the direction of information or signal flow.
dt
dx y
Introduction• The operations of addition and subtraction have a special
representation.
• The block becomes a small circle, called a summing point, withthe appropriate plus or minus sign associated with the arrowsentering the circle.
• The output is the algebraic sum of the inputs.
• Any number of inputs may enter a summing point.
• Some books put a cross in the circle.
Introduction
• In order to have the same signal or variable be an inputto more than one block or summing point, a takeoff (orpickoff) point is used.
• This permits the signal to proceed unaltered alongseveral different paths to several destinations.
Example-1• Consider the following equations in which 𝑥1 , 𝑥2 , 𝑥3 , are
variables, and 𝑎1, 𝑎2 are general coefficients or mathematicaloperators.
522113 xaxax
Example-1
522113 xaxax
Example-2• Draw the Block Diagrams of the following equations.
11
2
22
13
11
12
32
11
bxdt
dx
dt
xdax
dtxbdt
dxax
)(
)(
Canonical Form of A Feedback Control System
Characteristic Equation
• The control ratio is the closed loop transfer function of the system.
• The denominator of closed loop transfer function determines thecharacteristic equation of the system.
• Which is usually determined as:
)()(
)(
)(
)(
sHsG
sG
sR
sC
1
01 )()( sHsG
Example-31. Open loop transfer function
2. Feed Forward Transfer function
3. control ratio
4. feedback ratio
5. error ratio
6. closed loop transfer function
7. characteristic equation
8. Open loop poles and zeros if 9. closed loop poles and zeros if K=10.
)()()(
)(sHsG
sE
sB
)()(
)(sG
sE
sC
)()(
)(
)(
)(
sHsG
sG
sR
sC
1
)()(
)()(
)(
)(
sHsG
sHsG
sR
sB
1
)()()(
)(
sHsGsR
sE
1
1
)()(
)(
)(
)(
sHsG
sG
sR
sC
1
01 )()( sHsG
)(sG
)(sH
Reduction techniques
2G1G21GG
1. Combining blocks in cascade
1G
2G21 GG
2. Combining blocks in parallel
3. Eliminating a feedback loop
G
HGH
G
1
G
1H
G
G
1
Example-4: Reduce the Block Diagram to Canonical Form.
Example-4: Continue.
Example-5• For the system represented by the following block diagram
determine:1. Open loop transfer function
2. Feed Forward Transfer function
3. control ratio
4. feedback ratio
5. error ratio
6. closed loop transfer function
7. characteristic equation
8. closed loop poles and zeros if K=10.
Example-5– First we will reduce the given block diagram to canonical form
1s
K
Example-5
1s
K
ss
Ks
K
GH
G
11
1
1
Example-5 (see example-3)1. Open loop transfer function
2. Feed Forward Transfer function
3. control ratio
4. feedback ratio
5. error ratio
6. closed loop transfer function
7. characteristic equation
8. closed loop poles and zeros if K=10.
)()()(
)(sHsG
sE
sB
)()(
)(sG
sE
sC
)()(
)(
)(
)(
sHsG
sG
sR
sC
1
)()(
)()(
)(
)(
sHsG
sHsG
sR
sB
1
)()()(
)(
sHsGsR
sE
1
1
)()(
)(
)(
)(
sHsG
sG
sR
sC
1
01 )()( sHsG
)(sG
)(sH
Example-6• For the system represented by the following block diagram
determine:1. Open loop transfer function
2. Feed Forward Transfer function
3. control ratio
4. feedback ratio
5. error ratio
6. closed loop transfer function
7. characteristic equation
8. closed loop poles and zeros if K=100.
Reduction techniques
4. Moving a summing point behind a block
G G
G
G G
G
1
5. Moving a summing point ahead a block
7. Moving a pickoff point ahead of a block
G G
G G
G
1
G
6. Moving a pickoff point behind a block
8. Swap with two neighboring summing points
A B AB
Example-7
R
_+
_
+1G 2G 3G
1H
2H
++
C
• Reduce the following block diagram to canonical form.
Example-7
R
_+
_
+1G 2G 3G
1H
1
2
G
H
++
C
Example-7
R
_+
_
+21GG 3G
1H
1
2
G
H
++
C
Example-7
R
_+
_
+21GG 3G
1H
1
2
G
H
++
C
R
_+
_
+121
21
1 HGG
GG
3G
1
2
G
H
C
Example-7
R
_+
_
+121
321
1 HGG
GGG
1
2
G
H
C
Example-7
R
_+232121
321
1 HGGHGG
GGG
C
Example-7
Example 8
Find the transfer function of the following block diagram
2G 3G1G
4G
1H
2H
)(sY)(sR
1. Moving pickoff point A ahead of block2G
2. Eliminate loop I & simplify
324 GGG B
1G
2H
)(sY4G
2G
1H
AB3G
2G
)(sR
I
Solution:
3. Moving pickoff point B behind block324 GGG
1GB
)(sR
21GH 2H
)(sY
)/(1 324 GGG
II
1GB
)(sRC
324 GGG
2H
)(sY
21GH
4G
2G A3G 324 GGG
4. Eliminate loop III
)(sR
)(1
)(
3242121
3241
GGGHHGG
GGGG
)(sY
)()(
)(
)(
)(
32413242121
3241
1 GGGGGGGHHGG
GGGG
sR
sY
)(sR
1GC
324
12
GGG
HG
)(sY324 GGG
2H
C
)(1 3242
324
GGGH
GGG
2G4G1G
4H
2H
3H
)(sY)(sR
3G
1H
Example 9
Find the transfer function of the following block diagrams
Solution:
2G4G1G
4H
)(sY
3G
1H
2H
)(sR
A B
3H4
1
G
4
1
G
I1. Moving pickoff point A behind block
4G
4
3
G
H
4
2
G
H
2. Eliminate loop I and Simplify
II
III
443
432
1 HGG
GGG
1G
)(sY
1H
B
4
2
G
H
)(sR
4
3
G
H
II
332443
432
1 HGGHGG
GGG
III
4
142
G
HGH
Not feedbackfeedback
)(sR )(sY
4
142
G
HGH
332443
4321
1 HGGHGG
GGGG
3. Eliminate loop II & IIII
143212321443332
4321
1 HGGGGHGGGHGGHGG
GGGG
sR
sY
)(
)(
Example-10: Reduce the Block Diagram.
Example-10: Continue.
Example-11: Simplify the block diagram then obtain the close-loop transfer function C(S)/R(S). (from Ogata: Page-47)
Example-11: Continue.
Superposition of Multiple Inputs
Example-12: Multiple Input System. Determine the output C due to inputs R and U using the Superposition Method.
Example-12: Continue.
Example-12: Continue.
Example-13: Multiple-Input System. Determine the output C due to inputs R, U1 and U2 using the Superposition Method.
Example-13: Continue.
Example-13: Continue.
Example-14: Multi-Input Multi-Output System. Determine C1 and C2 due to R1 and R2.
Example-14: Continue.
Example-14: Continue.
When R1 = 0,
When R2 = 0,
Block Diagram of Armature Controlled D.C Motor
Va
iaT
Ra La
J
c
eb
(s)IK(s)cJs
(s)V(s)K(s)IRsL
am
abaaa
Block Diagram of Armature Controlled D.C Motor
(s)V(s)K(s)IRsL abaaa
Block Diagram of Armature Controlled D.C Motor
(s)IK(s)cJs ama
Block Diagram of Armature Controlled D.C Motor
amplifier Motor Gearing Valve
Actuator
Water
container
Processcontroller
Float
measurement (Sensor)
Error
Feedback signal
resistance comparator
Desired
water level
Input
Actual
water level
Output
Fig. 1.8
For the Fig.1.1, The water level control system:
Figure 1.1
The DC-Motor control system
Desi red
rot at e speed nRegul at or Tr i gger Rect i f i er DC
mot or
Techomet er
Act uat or
Processcont rol l er
measurement ( Sensor )
comparat or
Act ual
rot at e speed n
Error
Feedback si gnal
Ref erence
i nput ur
Out put n
Fi g. 1. 9
auk
ua
uf
e
Fundamental structure of control systems
1) Open loop control systems
Cont rol l er Act uat or Process
Di st urbance
( Noi se)
I nput r ( t )
Ref erence
desi red out put
Out put c( t )
( act ual out put )
Cont rol
si gnal
Act uat i ng
si gnal
uk
uact
Fi g. 1. 10
Features: Only there is a forward action from the input to the output.
2) Closed loop (feedback) control systems
Cont rol l er Act uat or Process
Di st urbance
( Noi se)
I nput r ( t )
Ref erence
desi red out put
Out put c( t )
( act ual out put )
Cont rol
si gnal
Act uat i ng
si gnal
uk
uact
Fi g. 1. 11
measurementFeedback si gnal b( t )
+-
( +)
e( t ) =
r ( t ) - b( t )
Features:
1) measuring the output (controlled variable) . 2) Feedback.
not only there is a forward action , also a backward action between the output and the input (measuring the output and comparing it with the input).
Chapter 2 Mathematical models of systems2.1 Introduction
Controller Actuator Process
Disturbance
Input r(t)
desired outputtemperature
Output T(t)
actual
outputtemperature
Control
signal
Actuating
signal
uk uac
Fig. 2.1
temperaturemeasurement
Feedback signal b(t)
+
-(－)
e(t)=
r(t)-b(t)
1) Easy to discuss the full possible types of the control systems—in terms of the system’s “mathematical characteristics”.
2) The basis — analyzing or designing the control systems.
For example, we design a temperature Control system :
The key — designing the controller → how produce uk.
2.1.1 Why？
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