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THE TRANSFER FUNCTION
OBJECTIVES
General Objective : To understand and interpret the concept of transfer function.
Specific Objectives : At the end of the unit you will be able to :
State the definition of transfer function.
Identify the transfer function from a simple electric circuit.
Solve the transfer function from block diagram of open-loop control system and closed-loop control system.
Reduce a block diagram of multiple subsystem to a single block representing the transfer function from input to output.
Rewrite the reduction method of block diagram.
UNIT6
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6.0 EXPLANATION OF THE TRANSFER FUNCTION
In this unit we discuss how to find a mathematical model, called a Transfer Function, for linear, time-invariant electrical, mechanical and electromechanical systems. The transfer functions is define as G(s) = C(s)/R(s), or the ratio of the Laplace transform of the output to the Laplace Transform of the input. This relationship is algebraic and also adapts itself to modeling interconnected subsystems.
We realize that the physical world consists of more systems than what we have illustrated in this unit. For example, we could apply transfer function modeling to electrical systems. Of course, we must assume these systems to be linear, or make linear approximations, in order to use this modeling technique.
6.1 DEFINITION OF TRANSFER FUNCTION
INPUTINPUT
The “transfer function” of a system is :The ratio of Laplace Transform of the output variable to the Laplace Transform of the input variable with all the initial conditions zero.
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Whenever a system is perturbed by some input signal all, of the dependent variables in the system vary as a result. The “transfer function” of a system is the ratio of Laplace Transform of the output variable to the Laplace Transform of the input variable with all the initial conditions zero. When a physical system is analysed, a mathematical model is developed by writing differential equations with the help of various physical laws governing the system.
The steps involved in obtaining the transfer function are as follows :(i) Write the differential equations governing the system.(ii) Laplace Transform the equations i.e, to replace the terms
involving d/dt by s and ∫ dt by 1/s .(iii) Obtain the ratio of Transformed output to input variables.
6.2 DERIVATION OF TRANSFER FUNCTION FROM THE EASIER ELECTRIC CIRCUIT
There are assumptions made while deriving Transfer Function of electrical systems.
(i) When a device is a part of a larger system, it should not load the source which provides it input signal, i.e., ideally the source should have zero (or low) internal impedance.
(ii) The output of the device should not be loaded by the component that receives its output signal. The systems is approximated by linear lumped parameters model with suitable assumptions
The loading effects occur with electrical, mechanical, electromechanical and fluid devices and should be taken into account while deriving transferfunctions.
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Let us now implement the procedure of the deriving the transfer functions through various examples to follow :
Example 6.1 : Obtain the transfer function of the circuit shown in Figure 6.1
Figure 6.1 : An RLC electrical circuit(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
Assume the current response i (t) due to a change in applied voltage v (t) is required.
Application of Kirchhoff’s voltage law around the circuit gives
(6.2.1)
(6.2.2)
Laplace Transforming equations (6.2.1) and (6.2.2) we have,
(6.2.3)
R L
C vo(t)vi(t)
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(6.2.4)
Hence we have from equations (6.2.3) and (6.2.4),
Example 6.2 : Derive the transfer function of the circuit shown in Figure 6.2
Figure 6.2 : An RC electrical circuit (Source : Katsuhiko Ogata (1990), Modern Control Engineering)
Writing the differential equations with the help of Kirchhoff’s voltage law
(6.2.5)
(6.2.6)
Laplace Transforming equations (6.2.5) and (6.2.6) we have,
(6.2.7)
R
C vo(t)vi(t)
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(6.2.8)
Hence we have from equations (6.2.7) and (6.2.8),
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Activity 6A
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!
6.1 Define the transfer function.
6.2 For the network given, what is the transfer function?
C
vo(t)vi(t)
R
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Feedback To Activity 6A
6.1 The “transfer function” of a system is the ratio of Laplace Transform of the output variable to the Laplace Transform of the input variable with all the initial conditions zero.
6.2 Writing the differential equations with the help of Kirchhoff’s voltage law
(6.2.9)
(6.2.10)
Laplace Transforming equations (6.2.9) and (6.2.10) we have,
(6.2.11)
(6.2.12)
Hence we have from equations (6.2.11) and (6.2.12),
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6.3 DERIVATION OF TRANSFER FUNCTION FROM BLOCK DIAGRAM SYSTEMS
We can make a simple transfer function from a complete diagram whether in open-loop control system or closed-loop control system.
6.3.1 Open-loop Control System
A system without feedback is called an open-loop system.
Figure 6.3 : Block diagram of open-loop system(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
For open loop system,
INPUTINPUT
G(s)
R(s) C(s)
Input Output
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C(s) = G(s) R(s)
6.3.2 Closed-loop control systemAll automatic control systems are of the closed-loop type of control system. This is necessitated by the introduction of feedback for comparing the reference input R(s), with the controlled output C(s).
Figure 6.4 : Block diagram of closed-loop system(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
Let us assume that,
R(s) = input (reference) or controlling variableC(s) = output or controlled variableB(s) = feedback signalG(s) = C(s) / E(s) = forward path transfer functionE(s) = Actuating signal
B(s)
R(s) C(s)
E(s)
G(s)
H(s)
_
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H(s) = feedback transfer function
= G(s) H(s) = (open) loop transfer function
M(s) = = closed-loop transfer function (control ratio)
From Figure 6.4, we have
C(s) = G(s) E(s) (6.3.1)
E(s) = R(s) – B(s)= R(s) – H(s) C(s) (6.3.2)
Eliminating E(s) from equation (6.3.1) and (6.3.2), we have
C(s) = G(s) R(s) – G(s) H(s)C(s)
= M(s) = (6.3.3)
Hence, the system shown in Figure 6.4 can be reduced to single block shown in Figure 6.5.
Figure 6.5 : Reduce form of Figure 6.4
)()(1
)(
sHsG
sG
R(s) C(s)
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Equation (6.3.3) is valid for negative feedback system. Hence, for a positive feedback system we have
= M(s) =
In general, for a positive/negative feedback systems, the control ratio is given by
= M(s) = as the case may be.
6.3.3 Block Diagram Reduction Rules
Rule (1) : Combining blocks in cascade
Figure 6.6 (a) : Blocks in cascade
G1 G2
R1R1G1
R1G1G2G1G2
G1 R1G1G2
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Rule (2) : Combining blocks in Parallel
Figure 6.6 (b) : Blocks in Parallel
Rule (3) : Moving a pick-off point after a block
Figure 6.6 (c) : Moving a pick-off point after a block
G1
G2
RG1
R
+RG1RG2
G1G2
RG1RG2R
RG2
G
RG
R
R
G
R RG
G
1R
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Rule (4) : Moving a take-off point ahead of a block
Figure 6.6 (d) : Moving a take-off point ahead of a block
Rule (5) : Moving a summing point after a block
Figure 6.6 (e) : Moving a summing point after a block
R
G
RG
RG
G
G
RG
RG
R
R1
R2
R1R2
G
G[R1R2]
G
R1+
R1G+
G
G[R1R2]
R2G
R2
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Rule (6) : Moving a summing point ahead of a block
Figure 6.6 (f) : Moving a summing point ahead of a block
Rule (7) : Eliminating a feedback loop
R2
R1R2/G
G
R1GR2+G
R1 R1G+
R1GR2
R2
1/G
R1
G
H
R
+C
R
GH
G
1
C
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Figure 6.6 (g) : Eliminating a feedback loop
Figure 6.6 (a,b,c,d,e,f,g) : Block diagram reduction rules(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
6.3.4 A Block Diagram Reduction
The block diagram of a multiple-loop feedback control system isshown in Figure 6.6. Use block diagram reduction to simplify this to a single block relating C(s) to R(s). Note that, for clarity, the dependency upon s has been omitted from the transfer functions within the blocks.
G1(s) G2(s) G3(s)
C(s)
H1(s)
H2(s)
H3(s)
R(s) +
+_ _
++
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Figure 6.7 Block Diagram
(a)
+
G1(s) G3(s)G2(s)
H3(s)
H2(s)
H1(s)
R(s) C(s)
- -
G1(s) G2 (s)G3 (s)
H1(s)H2(s)H3(s)
R(s) C(s)
-
+
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(b)
(c)
Figure 6.8 (a,b,c) : Steps to reduce the block diagram(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
Steps to reduce the block diagram :(Figure6.8(a)) collapse summing junctions;(Figure6.8(b)) form equivalent cascaded system in the forward
path and equivalent parallel system in the feedback path;
(Figure6.8(c)) form equivalent feedback system and multiply by cascaded G1(s)
Finally, the feedback system is reduced and multiplied by G1(s) to yield the equivalent transfer function shown in Figure 6.7 (c ).
6.3.5 Block Diagram of Two Input System
)()()()()(1
)()()(
32132
123
sHsHsHsGsG
sGsGsG
R(s) C(s)
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In the present of more than one input to a system, the system may be a single output system called a multiple-input-single-output (MISO) system or a multiple output system called a multiple-input-multiple-output (MIMO) system. The output of the system are obtained by applying the ‘law of homogeneity’ or ‘Principle of Superposition’.
Let us consider a two-input linear system as shown in Figure 6.9.
Figure 6.9 : Block diagram of a two-input system(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
The response to the input R(s) is obtained by letting the disturbance signal D(s) = 0. The corresponding block diagram is shown in Figure 6.9, which gives
CR(s) = acting alone with D(s) = 0
D(s)
_
)(1 sG )(2 sG
H(s)
R(s) C(s)
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CR(s) = R(s) (6.3.4)
(a)
(b)
)(1 sG )(2 sG
H(s)
R(s) CR(s)
-
H(s)
R(s) CR(s)
-
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(c)
Figure 6.10(a,b,c) : Block Diagram Reduction with R(s) alone.(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
Similarly the response to the disturbance signal D(s) is obtained by assuming R(s) = 0. The block diagram for this case is shown in Figure 6.11, which gives
CD(s) = acting alone with R(s) = 0
CD(s) = D(s) (6.3.5)
The actual response of the system when both R(s) and D(s) are acting is obtained by adding the two individual responses CR(s) and CD(s).
)()()(1
)()(
21
21
sHsGsG
sGsG
R(s)
CR(s)
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(a)
(b)
)(2 sG
G1(s) H(s)
D(s) CD(s)
-
H(s)
D(s) CD(s)
-
G1(s)
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(c)
Figure 6.11 (a,b,c) : Block Diagram Reduction with D(s) alone.(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
This example shows how superposition may be used to handle system with more than one input :
Example 6.3: Determine the output Y(s) in the system shown below.
3sK
s
2
S + 1
R(s)
D(s)
Y(s)
-
+
+ +
)()()(1
)(
21
2
sHsGsG
sG
D(s)
CD(s)
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Figure 6.12 : Block diagram of a two input system in the Laplace domain(Source : Katsuhiko Ogata (1990), Modern Control Engineering)
Solution :1. Setting D(s) = 0, gives the transfer function between Y(s)
and R(s) as :
2. Setting R(s) = 0, gives the transfer function between Y(s) and D(s) as :
Since a Laplace transfer function is a linear operator, the principle of superposition is used to generate the overall output as the sum of the two input contributions:
or
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)1(2)3(
)()3(2)(2)(
sKss
sDssKRsY
Activity 6B
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!
6.3 The block diagram of a certain system is shown below. Determine the transfer function Y(s)/U(s).
6.4 The block diagram of a certain system is shown below. Determine the transfer function C(s)/R(s).
G1(s) G2(s)U(s)
-
Y(s)
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Feedback To Activity 6B
6.3 =
6.4 =
G(s)R(s)
-
C(s)
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KEY FACTS
1. The transfer functions is define as G(s) = C(s)/R(s), or the ratio of the Laplace transform of the output to the Laplace Transform of the input.
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SELF-ASSESSMENT
You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment given on the next page. If you face any problems, discuss it with your lecturer. Good luck.
Q6-1 State the steps to obtain the transfer function.
Q6-2(a) The transfer function E0(s)/E1(s) of the RC-network shown is
given by:
Eo(t)Ei(t)
R
C
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(b)
Q6-3Simplify the block diagram in the figure below and obtain the closed loop in transfer function C(s)/R(s).
Eo(t)Ei(t)
R
L
2G
1G
G4
R(s) C(s)
G3
-
+
+
+
-
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Feedback To Self-Assessment
Have you tried the questions????? If “YES”, check your answers now.
Q6-1 The steps involved in obtaining the transfer function are as follows :(i) To write the differential equations governing the system.(ii) To Laplace Transform the equations i.e, to replace the terms
involving d/dt by s and ∫ dt by 1/s .(iii) To obtain the ratio of Transformed output to input variables.
Q6-2 (a)
(b)
Q6-3
R(s) C(s)
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G1 + G2
G3 – G4
-
R(s)C(s)