Microsoft PowerPoint - Electrical and Electromechanical Systems.ppt - Compatibility Mode-compressed

System Dynamics Electrical and
Electromechanical Systems
Indrawanto
FTMD-ITB
Sem 2 2022-2023
1
ELECTRICAL ELEMENTS
Voltage and current are the primary variables used to describe a circuit’s behavior.
Current is the flow of electrons. It is the time rate of change of electrons passing
through a defined area, such as the cross section of a wire. Because electrons are
negatively charged, the positive direction of current flow is opposite to that of the
electron flow. The mathematical description of the relation between the number of
electrons (called charge Q) and current i is
The unit of charge is the coulomb (C), and the unit of current is the ampere (A), which
is one coulomb per second.
2
1
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the resistor is
given by Ohm’s law i-e
v R (t )  iR (t )R
• The Laplace transform of the above equation is
VR ( s )  I R ( s )R
3
Parallel and Series Circuits
4
2
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the Capacitor
is given as:
vc ( t ) 
1
 ic (t )dt
C
• The Laplace transform of the above equation (assuming there is no
charge stored in the capacitor) is
Vc ( s ) 
1
Ic (s)
Cs
5
Basic Elements of Electrical Systems
• The time domain expression relating voltage and current for the inductor
is given as:
v L (t )  L
di L (t )
dt
• The Laplace transform of the above equation (assuming there is no
energy stored in inductor) is
VL ( s )  LsI L ( s )
6
3
V-I and I-V relations
Component
Symbol
V-I Relation
I-V Relation
Resistor
v R (t )  iR (t )R
iR (t ) 
Capacitor
vc ( t ) 
Inductor
v L (t )  L
v R (t )
R
dvc (t )
1
 ic (t )dt ic (t )  C
C
dt
di L (t )
dt
iL (t ) 
1
 v L (t )dt
L
7
7
Power and Energy
The energy E stored in a capacitor is
1
 dv 
E   Pdt   i v dt    C v dt  C  v dv  C v 2
2
 dt 
The energy E stored in an inductor is
1
 di 
E   Pdt   i v dt   i L  dt  L  i di  Li 2
2
 dt 
8
4
Voltage-current and energy
relations for circuit elements
Tabel 6.1.2
v iR
P  R i2 
Capacitance:
t
Q
1
v   i dt  0
C0
C
1
E  Cv 2
2
Inductance:
vL
Resistance:
di
dt
E
v2
R
1 2
Li
2
9
Example#1
• The two-port network shown in the following figure has vi(t) as
the input voltage and vo(t) as the output voltage. Find the transfer
function Vo(s)/Vi(s) of the network.
vi( t)
v i ( t )  i( t ) R 
vo (t ) 
i(t)
C
vo(t)
1
 i ( t ) dt
C
1
 i ( t ) dt
C
10
10
5
Example#1
1
1
vo (t ) 
 i ( t ) dt
 i ( t ) dt
C
C
• Taking Laplace transform of both equations, considering initial
conditions to zero.
v i ( t )  i( t ) R 
Vi ( s )  I ( s )R 
Vo (s) 
1
I(s)
Cs
1
I(s)
Cs
• Re-arrange both equations as:
V i ( s )  I ( s )( R 
CsV o ( s )  I ( s )
1
)
Cs
11
11
Example#1
1
)
Cs
• Substitute I(s) in equation on left
V i ( s )  I ( s )( R 
CsV o ( s )  I ( s )
V i ( s )  CsV o ( s )( R 
Vo ( s )

Vi ( s )
1
)
Cs
1
Cs ( R 
1
)
Cs
Vo ( s )
1

Vi ( s ) 1  RCs
12
12
6
Example#1
Vo ( s )
1

Vi ( s ) 1  RCs
• The system has one pole at
1  RCs  0
 s
1
RC
13
13
Example#2
• Design an Electrical system that would place a pole at -3.
Vo ( s )
1

Vi ( s ) 1  RCs
vi( t)
• System has one pole at
s
i(t)
C
v2(t)
1
RC
• Therefore,

1
 3
RC
if
R  1 M
and
C  333 pF
14
14
7
Example#3
• Find the transfer function G(S) of the following
two port network.
L
vi(t)
C
i(t)
vo(t)
15
15
Example#3
• Simplify network by replacing multiple components with
their equivalent transform impedance.
L
Z
Vi(s)
I(s)
C
Vo(s)
16
16
8
Transform Impedance
(Resistor)
iR(t)
IR(S)
+
+
Transformation
vR(t)
VR(S)
ZR = R
-
-
Impedance, a generalization of the electrical resistance concept. This
concept enables us to derive circuit models more easily, especially for
more complex circuits, and is especially useful for obtaining models of
circuits containing operational amplifiers.
17
17
Transform Impedance
(Inductor)
iL(t)
IL(S)
+
vL(t)
-
ZL=LS
+
VL(S)
LiL(0)
-
18
18
9
Transform Impedance
(Capacitor)
ic(t)
Ic(S)
+
vc(t)
ZC(S)=1/CS
-
+
Vc(S)
-
19
19
Equivalent Transform Impedance
(Series)
• Consider following arrangement, find out equivalent
transform impedance.
ZT  Z R  Z L  Z C
Z T  R  Ls 
1
Cs
20
20
10
Equivalent Transform Impedance
(Parallel)
1
1
1
1



ZT
Z R Z L ZC
1
1
1
1
 

1
ZT
R Ls
Cs
21
21
Equivalent Transform
Impedance
• Find out equivalent transform impedance of
following arrangement.
L2
L2
R1
R2
22
22
11
Back to Example#3
L
Z
Vi(s)
Vo(s)
C
I(s)
1
1
1


Z ZR ZL
1 1
1
 
Z R Ls
Z
RLs
1  RLs
23
23
Example#3
Z
RLs
1  RLs
L
Z
Vi(s)
C
I(s)
V i ( s )  I ( s )Z 
1
I(s)
Cs
Vo(s)
Vo (s) 
1
I(s)
Cs
24
24
12
Circuit examples
25
Loop Currents
v1  R1i1  R2i2  0
v2  R2i2  R3i3  0
v1  R1i A  R2i A  R2iB  0
v2  R3iB  R2iB  R2i A  0
26
13
Capacitance and Inductance in
Circuits
From Kirchhoff’s voltage law
1
i  v s  v1 
R
For the capacitor
v1 
t
Q
1
i dt  0

C0
C
The capacitance and inductance
circuit can be modeled as
dv1 1
 i
dt
C
RC
dv1
1
v s  v1 

dt
RC
dv1
 v1  v s
dt
27
Pulse Response of a Series RC
Circuit
The capacitance and inductance
circuit can be modeled as
dv
RC 1  v1  V
dt
Solution of the model



v1 t   v1 0e t / RC  V 1  e t / RC  V 1  e t / RC

When the switch is moved back to point
B at time t = D
RC
dv1
 v1  0
dt
Solution of the model


v1 t   v1 D e  t  D  / RC  V 1  e  D / RC e  t  D  / RC
28
14
Series and Parallel
Impedances
Z s   Z1 s   Z 2 s 
1
1
1


Z s  Z1 s  Z 2 s 
29
State-variable Models of
Circuits
From Kirchhoff’s voltage law
di
vs  Ri i  L  vi  0
dt
Solve this for di/dt
1
di 1
R
 v s  vi  i
dt L
L
L
1
v1   i dt
C
dv1 1
 i
dt
C
The state-variable model is
 di   R
 dt   L
 dv    1
 i 
 dt   C
1
  i   1 
L
     L vs
0  v1   0 

30
15
Isolated Amplifier
The input impedance Zi is very large
The output impedance Zo is very low
31
Operational Amplifiers
The voltage gain G of an operational amplifier is typically very large
(greater than 105)
Op amps are widely used in instruments and control systems for
multiplying, integrating and differentiating signals.
There are two inputs port, inverting input (-) and non-inverting input (+)
32
16
General Op-Amp Input-Output
Relation
33
Op-amp Circuits
34
17
General Op-Amp Input-Output
Relation (Continued)
35
General Op-Amp InputOutput Relation (Continued)
36
18
Electric Motors
•
Electric motors is electromechanical
systems consists of an electrical
subsystem and a mechanical
subsystem with mass and possibly
elasticity and damping
Relation between the force f to
the current i:
f  BLi
When the direction of the field,
the conductor, and its velocity are
mutually perpendicular, then:
vb  BLv
When there is no power
loss
vbi  fv  Bl i v
mv  f  BLi
37
Electric Vehicle
38
19
DC Motor
39
DC Motor
• Type of electric motors:
– Direct current
– Alternating current
• Basic elements:
– Stator
– Rotor
– Armature
– Commutator
• Stator:
– Permanent magnet
– electromagnet
40
20
Armature-controlled DC
Motor
Torque on the armature is
T  nBLia r  nBLr ia  K T ia
Kt is motor’s torque constant
Angular velocity is
vb  nBLv  nBLr   K b
where Kb = nBLr is the motor’s back
emf constant
Kirchhoff’s voltage law gives
v a  R a i a  La
dia
 K b  0
dt
From Newton’s law applied to the inertia I,
I
d
 T  c  TL  K T ia  c  TL
dt
41
Motor Block Diagram
I a s  
1
Va s   K b s 
La s  R a
 s  
1
K T I a s   TL s 
Is  c
42
21
Motor Transfer Functions
For the output I(s)
I a s 
Is  c

Va s  La Is 2  Ra I  cLa s  cRa  K b K T
I a s 
Kb

TL s  La Is 2  Ra I  cLa s  cRa  K b K T
For the output Ω(s)
KT
s 

Va s  La Is 2  Ra I  cLa s  cRa  K b K T
La s  R a
s 

TL s  La Is 2  Ra I  cLa s  cRa  K b K T
The characteristic equations
La Is 2  Ra I  cLa s  cRa  K b K T  0
43
Motor Dynamic Response
44
22
State-Variable of the Motor
Model
dia
1
va  Ra ia  K b 

dt
La
d 1
 K T ia  c  TL 
dt
I
Letting x1 = ia and x2 = w, the state equation become
 R
dx1
1
v a  Ra x1  K b x 2    a

dt
La
 La
dx 2 1
K
 K T x1  cx 2  TL    T
dt
I
 I

K b   x1   1


La   x 2   La
 v 
0  a 
 TL 
c  x  
1  v 
   1   0    a 
I   x2  
L  TL 
45
State-Variable of the Motor
Model (Continued)
Define u as
v 
u   a
TL 
The vector-matrix form is
x  Ax  Bu
K 
 Ra
 b
 L
La
A a

c
 KT
 
I 
 I
1

0 

B   La

 0  1
I 

46
23
Field Controlled DC Motor
T  nB i f Lia r  nLria B i f   K T i f
v f  Rf i f  Lf
I
di f
dt
d
 T  c  TL  K T i f  c  TL
dt
47
The characteristic roots are all real  no oscillation output for a step input
48
24
Motor Transfer Functions
(Armature Control)
For the output I(s)
I a s 
Is  c

Va s  La Is 2  Ra I  cLa s  cRa  K b K T
I a s 
Kb

TL s  La Is 2  Ra I  cLa s  cRa  K b K T
For the output Ω(s)
KT
s 

Va s  La Is 2  Ra I  cLa s  cRa  K b K T
La s  R a
s 

TL s  La Is 2  Ra I  cLa s  cRa  K b K T
The characteristic equations
La Is 2  Ra I  cLa s  cRa  K b K T  0
49
Steady-State Motor
Response (Armature Control)
ia 
cVa  K bTL
cRa  K b K T

K T Va  RaTL
cRa  K b K T
50
25
SENSORS AND
ELECTROACOUSTIC DEVICES
• A TACHOMETER
There is no applied voltage va, thus at steady state condition
The voltage across the resistor vt which is equal to IaRa becomes
51
SENSORS AND
ELECTROACOUSTIC DEVICES
• Accelerometer
Using Newton’s law gives
Substituting y for x - z
The transfer function between input z and out y is
52
26