Chapter 2
Material We Will Cover Next
• We need to write differential equations representing the system or
subsystem.
• Then write the Laplace transform of the system.
• Then we will write the overall input-output relationship of the
interconnected components. That is the Transfer Function T(s), also
called G(s).
• Section 2.2 in the text includes tables, in particular Table 2.2, p. 44, of
various variables (elements) used in modeling physical systems and
differential equations describing ideal elements of physical systems.
• It is expected that the student is already familiar with these concepts.
In general the student will have access to this information for exams if
needed.
• We will investigate making linear approximations of non-linear
elements.
• We will briefly mention the Laplace Transform
• Get our first exposure to poles and zeros and to the Characteristic
Equation
• Discuss Partial Fraction Expansion
• Introduce Final and Initial Value Theorems
1
Control Systems are Often More
Complex Than Single Loops
• One forward path
• Four transfer functions in the forward path G1, G2, G3, G4.
• Three Feedback Loops.
• Three Summing Junctions
2
Block Diagram = Flow Graph
Besides Block diagrams, control systems
are often represented using Flow Graphs
Block Diagram
Flow Graph
H3
R(s)
1
G1
G2
-H1
1
G3
G4
1
C(s)
-H2
Note that two of the H transfer functions in the Flow Graph have
negative signs. This is necessary since the summing nodes do
not have any signs associated with them as in the block diagram.
3
Solve For The Transfer Function
R(S)
+
E
C(s)
G1
_
F
H
Solution:
C(s) = G1E
F = C(s)H
E = R(s)-F
E = R(s)- C(s)H
C(s) = G1[R(s)- C(s)H]
C(s) +C(s)HG1 = G1(R(s)
C(s)[1 + HG1] = G1(R(s)
(1)
(2)
(3)
(4)
(5)
C(s)/R(s)=G1/(1+G1H)
4
Solve For The Transfer Function
R(S)
+
E
G1
A
G2
B
+
+
C(s)
_
F
Solution:
C(s) = A + B
A = G1E
B = G2E
F = HB
E = R(s)-F
H
(1)
(2)
(3)
(4)
(5)
Substitute (2) and (3) into (1);
Substitute (5) into (6);
Substitute (3) into (4);
Substitute (5) into (8);
Solve (9) for F;
Substitute (10) into (7);
C(s) = G1E  G 2E  (G1  G 2 )E
C(s) = (G1  G 2 )[R(s)  F]
F = HG 2E
F = HG 2[R(s)  F]
HG 2R(s)
F =
1  G 2H
HG 2R(s)
C(s) = (G1 +G 2 )[R(s) 
]
1  G 2H
1  G 2H  HG 2
C(s) = (G1  G 2 )
R(s)
1  G 2H
C(s) (G1  G 2 )
T(s) =

R(s)
1  G 2H
(6)
(7)
(8)
(9)
(10)
(11)
5
Attempt to Solve For The Transfer Function
D
A
B
E
F
J
M
C
K
6
Overall Transfer Function
Mason’s Gain Rule
The overall Transfer function can be obtained from the formula:
T ( s) 
M

k
k
(1)
• Mk is the transmittance of each forward path between a source and a sink node
(2)
•   1  L1  L2   L3 
• ∑L1 is the sum of the transmittances of each closed path.
• ∑L2 is the sum of the product of the transmittances of all possible
combinations of two non-touching loops.
• ∑L3 is the sum of the product of the transmittances of all possible
combinations of three non-touching loops.
• ∆k is the cofactor of Mk. It is the determinant of the remaining sub-graph
when the forward path which produces Mk is removed. Thus it does not
include any loops which touch the forward path in question. It is equal to
unity when the forward path touches all the loops in the graph of when the
graph contains no loops.
7
Applying Mason’s Gain Rule
H3
1
R(s)
G1
G2
M

T ( s) 
 M 1  G1G2G3G4

L
L

1  1
k
2
 G1G2G3G4 H1 H 2
L L
M

T(s) 

2
1 1

C(s)
k
There is no M 2 or  2 or more.
 G1G2 H1  G2G3 H 3  G3G4 H 2
1
1

1
   1-
G4
-H2
-H1

G3
1
 1 G1G2 H1  G2G3 H 3  G3G4 H 2  G1G2G3G4 H1 H 2
G1G2G3G4
1  G1G2 H1  G2G3 H 3  G3G4 H 2  G1G2G3G4 H1 H 2
8
Applying Mason’s Gain Rule
G9
R(s)
G1
G2
G3
-H2
G5
G4
G6
-H4
-H1
G8
G7
C(s)
-H5
-H3
M 1  G 1G 2 G 3 G 4 G 5 G 6 G 7 G 8
1  1
M 2  G 1G 9 G 8
 2  1  [  G 4 H 1  G 6 H 4  G 3 G 4 G 5 G 6 H 3  G 4 G 5 G 6 H 2 ]  G 4 H 1G 6 H 4
= 1  G 4 H 1  G 6 H 4  G 3 G 4 G 5 G 6 H 3  G 4 G 5 G 6 H 2  G 4 H 1G 6 H 4
  1  [ G 4 H 1  G 6 H 4  G 3G 4 G 5 G 6 H 3  G 4 G 5 G 6 H 2  G 8 H 5 ]
 G 4 H 1G 6 H 4  G 4 H 1G 8 H 5  G 6 H 4 G 8 H 5  G 8 H 5 G 4 G 5 G 6 H 2  G 8 H 5 G 3 G 4 G 5 G 6 H 3
+ G 4 H 1G 6 H 4 G 8 H 5
  1  G 4 H 1  G 6 H 4  G 3G 4 G 5 G 6 H 3  G 4 G 5 G 6 H 2  G 8 H 5
+ G 4 H 1G 6 H 4  G 4 H 1G 8 H 5  G 6 H 4 G 8 H 5  G 8 H 5 G 4 G 5 G 6 H 2  G 8 H 5 G 3 G 4 G 5 G 6 H 3
+ G 4 H 1G 6 H 4 G 8 H 5
T ( s) 
C ( s ) M 1  1  M 2  2 G1G 2G 3G 4 G 5 G 6 G 7G 8  G1G 9 G 8 [1  G 4 H 1  G 6 H 4  G 3G 4 G 5 G 6 H 3  G 4 G 5 G 6 H 2  G 4 H 1G 6 H 4 ]


R( s )


9
Using Mason’s Gain Rule
Solve for the Transfer Function – C/R
1
R
A
G1
1
1
B
G2
G3
C
G4
1
-1
-1
-1
10
Applying Mason’s Gain Rule
1
R
A
G1
1
1
B
G2
G3
C
G4
1
-1
-1
-1
M 1  G 1G 2 G 3 G 4
1  1
M 2  G1
2  1
M 3  G 3G 4
 3  1  G1
M 4  1
 4  1  G1
M 5   G 3G 4 G1
5  1
  1  (  G 1  G 2  G 1G 2 G 3 G 4 )  G 1G 2  1  G 1  G 2  G 1G 2 G 3 G 4  G 1G 2
a)
C G1G 2G 3G 4  G1  G 3G 4 (1  G1 )  (1)(1  G1 )  G1G 3G 4

M
1  (  G 1  G 2  G 1G 2 G 3 G 4 )  G 1G 2

G 3 G 4  G 1G 2 G 3 G 4  1
1  G 1  G 2  G 1G 2 G 3 G 4  G 1G 2
11
Derive Transfer Function for Previous Control System Example
R(s)
+
Ka
-as  1
G1( s) 
4
Manipulated G 2( s) 
s 2  4s  2
Variable
G1(s)  Ka
Error
G2(s) 
Kb
s  4s  6
2
H(s)  0.5
Feedback
C(s)
H = 0.5
ApplyingMason's Rule
4
4K a
C(s)
G1( s)G 2( s)
s  4s  2
T (s) 


 2
4
R(s) 1  G1( s)G 2( s) H 1  K 
 0.5 ( s  4s  2)  2 K a
a
2
s  4s  2
Suppose we select K a  3 , then,
Ka 
T (s) 
2
C(s)
12
12
 2

R(s) s  4s  8 ( s  2  j 2)(s  2  j 2)
Note: The denominator, set equal to zero, is called the
Characteristic Equation
12
Spring Mass System
Input = f(t)
b
Process
K
Block Diagram
f  Ma
f (t )  My(t )  by(t )  Ky(t )
Differential Equation
F ( s) = Y ( s) Ms 2  Y ( s)bs  Y ( s) K  Y ( s)( Ms 2 + bs + K )
M
f
Output = y(t)
y
T ( s) 
Block Diagram
F(s)
Y ( s)
F ( s)
=
1
Ms  bs  K
Transfer Function
2
1
Ms2  bs  K
Y(s)
M
F(s)
General Quadratic Form
1/ M
s2  b s  K
F(s)
A
2
s2  2 s  
Natural Frequency
n 
K
M
Damping Ratio
2 n 
2

K
M

b
M
b
M
b
2 KM
M
Y(s)
n
n
Y(s)
2
n
System Constant
 n2 A 
A
1
M n
2
1
M

1
M (K  M )

1
K
13
More on the Characteristic Equation
Block Diagram
F(s)
1
Ms2  bs  K
General Quadratic Form
Y(s)
M
F(s)
A
2
s2  2 s  
M
2
n
s 2  2 n s   n 2  0
Roots of the C.E. are: s1 , s2   n   n  2  1   n  j n
If   1 Roots = s1 , s2   n
Are the same
If
If
 1
 1
Y(s)
Y(s)
n
n
Characteristic Equation (C.E.)
1/ M
s2  b s  K
F(s)
1  2
Roots are real but different
Roots are complex conjugates
Also many text books refer to   1   2
14
Characteristic Equation
•
The denominator polynomial of the transfer function when set equal to zero is
called the characteristic equation C.E..
s 2  4s  8  (s  2  j 2)(s  2  j 2)  0
•
•
•
•
•
•
The roots of the C.E. are the poles of the system.
The roots of the transfer function’s numerator are the zeros of the system.
Poles and zeros are critical frequencies
At the poles, the transfer function becomes infinite
At the zeros the transfer function becomes zero.
The complex frequency s-plane plot of the poles and zeros graphically
portrays the character of the natural transient response of the system.
T ( s) 
4( s  5)
( s  3  j 4)(s  3  j 4)
jω
x
4
4
-6 -5
-4 -3 -2 -1
x
-4
15
Root Location vs Stability of the 2nd order C.E.
jω
• Any time the roots are equal and real, the system is
critically damped, no overshoot.
Settling Time ~ Ts
1
Double
Root
0.9
0.8
0.7
Amplitude
0.6
X
 
0.5
0.4
0.3
0.2
0.1
0
jω
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time (sec)
• Any time the roots are unequal and real, the
system is overdamped, no overshoot.
Settling Time ~ Ts
1
0.9
0.8
X
0.7
0.6
Amplitude
X
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
Time (sec)
0.5
0.6
0.7
16
Root Location vs Stability of the 2nd order C.E.
Continued
jω
X
• Anytime the roots are complex conjugates,
the system experiences damped oscillation
Settling Time 2 ~ Ts
1.4
1.2
Amplitude
1
X
0.8
0.6
0.4
0.2
0
jω
X
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time (sec)
• If the poles lie on the imaginary axis the
system is marginally stable. It oscillates.
Settling Time ~ Ts
2
1.8
1.6
1.4
X
Amplitude
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
Time (sec)
0.5
0.6
0.7
17
Root Location vs Stability of the 2nd order C.E.
Continued
jω
• Anytime the roots are in the RHP the
system is unstable and diverges
X
Unit Step Response
80
60
40
Amplitude
20
X
0
-20
-40
-60
-80
-100
0
5
10
15
20
25
Time (sec)
jω
Unit Step Response #2
9
8
7
X
6
Amplitude
X
5
4
3
2
1
0
0
0.1
0.2
0.3
0.4
Time (sec)
0.5
0.6
0.7
0.8
18
Deriving a Transfer Function for a Control System Component
Consider a DC Torque Motor
Ra
ea
J, b
La
Ia
em
ө, ω
J = Motor Inertia, could also include load inertia
b = Friction
Ra = Motor winding resistance
La = Motor inductance
em = Back emf
KT = Motor Torque Sensitivity – A motor parameter
Km = Relates back emf to shaft speed, em = Kmω
T orqueDeveloped: Tm ( s)  KT I a ( s)
Back emf : em  K B; Assusme   constant hen
t e m  K m or Em ( s)  K m s ( s)
(1)
(2)
19
Deriving a Transfer Function for a DC Torque Motor
Continued
Ra
ea
J, b
La
Ia
em
ө, ω
KVL around loop:
E ( s )  Em ( s )
ea  I a R a  LI a e m or Ea ( s )  I a ( s )[ Ra  Ls ] Em ( s ) Solving for Ia ( s )  a
Ra  Ls
(3)
Motor and Load Inertia
Tm  J  b
or Js 2 (s)  Tm (s)  bs (s) solvingfor  (s) 
Tm (s)
Js 2  bs
(4)
Equations (1), (2), (3), and (4) are called Equilibrium Equations
20
Constructing Block Diagram for a DC Torque Motor
From Equilibrium Equations
Tm ( s )  K T I a ( s )
Ia(s)
(1)
Ia(s)
T (s)
 ( s )  2m
Js  bs
(4)
E m ( s )  K m s ( s )
Ia(s)
(2)
KT
KT
KT
Em(s)
E ( s )  Em ( S )
I a (s)  a
(3)
Ra  Ls
Ea(s) +
Ea - Em
-
1
Ra  Ls
Ia(s)
KT
Em(s)
Tm(s)
Tm(s)
1
Js 2  bs
Tm(s)
1
Js 2  bs
 (s)
 (s)
Kms
Tm(s)
1
Js 2  bs
 (s)
Kms
21
Constructing DC Motor Transfer Function from Block Diagram Using
Mason’s Rule
Ea(s) +
Ea - Em
-
1
Ra  Ls
Ia(s)
KT
Em(s)
M1 
Tm(s)
 (s)
Kms
1
1
KT
 KT  2

Ra  Ls
Js  Bs ( Ra  Ls)(Js 2  Bs)
  1
1
Js 2  bs
1  1
KT
 Kms
( Ra  Ls)(Js 2  Bs)
KT
1
( Ra  Ls)(Js 2  Bs)
Kt
 ( s) M 11



2
KT
Ea ( s )

s
[
JL
s
 ( JRa  bLa ) s  Ra b  K t K m ]
a
1
 Kms
2
( Ra  Ls)(Js  Bs)
 ( s)
Ea ( s)

Kt
s[ JLa s  ( JRa  bLa ) s  Rab  K t K m ]
2
22