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MODELING OF ELECTRICAL SYSTEMS
1st METHOD
Current Law
(Kirchoff)
Voltage Law
2nd METHOD
Current Law
Magnetic Energy, Electrical Energy, Virtual Work:
Application of Lagrange’s Equation
2nd method is used in this course.
FUNDAMENTAL ELEMENTS OF ELECTRICAL SYSTEMS
Passive Elements
L
x
Analogy Between Mechanical and Electrical Systems
.
..
x
Velocity
(m/s)
x
Acceleration
(m/s2)
Generalized Coordinates
q
Charge
(Coulomb)
+
R
C
Displacement
(m)
Active Elements
i
Current
(Amper)
di
dt
m
Mass
(kg)
k
c
Spring Stiffness
Damping Constant
(N/m)
(Ns/m)
f  mx
1
E1  mx 2
2
f  kx
1
E 2  kx2
2
L
1/C
Inductance
C:Capacitance
(Farad)
(Henry)
di
dt
Generalized Charges
VL
Lagrange’s Equation
1
E1  Lq 2
2
Op-Amp
f
Force
(N)
f  cx
W   cx x
R
Resistance
(Ohm)
W  f x
V
Voltage
(Volt)
1
V  Ri
q
C
W  Vq
1 2 W   Rq q
E2 
q
2C
V
x
Analogy Between Mechanical and Electrical Systems
.
..
Displacement
(m)
x
Velocity
(m/s)
x
Acceleration
(m/s2)
m
Mass
(kg)
k
c
Spring Constant
Damping Constant
(N/m)
(Ns/m)
f
Force
(N)
θ
.
θ
..
θ
IG
Kr
Cr
Angular
Dispalement
Angular
Velocity
Angular
Acceleration
Mass
moment of
inertia
Rotational spring
constant
Rotational damping
constant
(Nm/rad)
Nm/(rad/s)
M
Moment
(Nm)
V
Voltage
(Volt)
δW
(rad)
(rad/s)
(rad/s2)
(kg-m2)
q
Charge
(Coulomb)
i
Current
(Amper)
L
1/C
Inductance
(Henry)
C:Capacitance
(Farad)
R
Resistance
(Ohm)
E2
E1
E1
E2
δW
1 2
q
2C
W   Rq q
1
E1  Lq 2
2
E2 
W  Vq
Current through R1= q  q 2
Modeling of Electrical Systems:
V1: Input q and q2 : Generalized charges
1 2
1 2
E1  Lq
E2 
q2
2
2C
W  V1q  R 1 (q  q 2 )(q  q 2 )  R 2q 2 q 2
Example 3.1:
L
+
V1
C
q 2
q
-
R1
R2
 ( V1  R 1q  R 1q 2 )q  ( R 1q  R 1q 2  R 2q 2 )q 2
Qq
Qq2
d   (E1  E 2 )   (E1  E 2 )
 Qq

dt 
q

q


  V1  R 1q  R 1q 2
Lq
d   (E1  E 2 )   (E1  E 2 )

 Qq 2


dt 
q 2
q 2


1
q 2  R 1q  ( R 1  R 2 )q 2
C
 R 1   q  0 0   q   V1 
   0 1      

R 1  R 2  q 2  
q
0
 C   2   
L=3.4 mH, C=286 µF, R1=3.2 Ω, R2=4.5 Ω
   R 1
 L 0  q

 0 0  q



  2   R1
Ls 2  R 1s
 R 1s
 R 1s
1 0
( R 1  R 2 )s 
C
D(s)=0.02618s3+26.288s2+11188.81s=0
Eigenvalues: 0, -502.06±418.70i (ξ=0.768)
Example 3.2 C1
q 4
C2
R1
q 1
V1
q 3
q 3
q 2
q f
No current flow through Op-Amp
q 1  q 2  q 3  q 4  q 4  q 1  q 2  q 3
q 4  q 3  q f
R3
E1  0 E2 
+
R2
 q f  q 1  q 2  q 3  q 3 q 1  q 2
1
1 2
(q 1  q 2  q 3 ) 2 
q3
2C1
2C2
W  V1q1  V2 (q1  q 2 )  R 1q 1q1
V2
 R 2q 2q 2  R 3q 3q 3
Input : V1 Generalized charges: q1, q2, q3
1
(q1  q 2  q 3 )  V1  V2  R 1q 1
C1


1
(q1  q 2  q 3 )  V2  R 2q 2
C1
1
1
(q1  q 2  q 3 )  q 3   R 3q 3
For Op-Amp : V+=V-=0
C1
C2
 1
1
1 


 C

C
C
1
1   q1 
 R 1 0  R 3   q 1   1
 V1 
1
1
   
0 R
 q     1

R
q 2    0 
2
3  2 

 C1 C1
C1     


0
R 3  q 3   1
 0
1
1
1  q 3   0 



 C1 C1 C1 C 2 
 R 3q 3  V2
R1
0

 0
0
R2
0
 1
 C
 R 3   q 1   1
1
 
R 3  q 2    

 C1


R 3  q 3   1

 C1
R 1s 
1
C1
1
C1
1

C1


1
C1
R 2s 
1
C1
1
C1
1
C1
1
C1
1
C1

1 
C1   q   V 
 1
1
1



 q    0 
2
C1     
1
1  q 3   0 


C1 C 2 

1
C1
1
R 3s 
0
C1
1
1
R 3s 

C1 C 2
 R 3s 
R1=15.9 kΩ, R2=837 Ω, R3=318 kΩ, C1=C2=0.005 µF
Eigenvalues: 0, -628.93±12561.76i (ξ=0.05)
GAIN CIRCUIT
R2
q
q 
R1
V1
+
q
V1  R 1q  0
0  R 2q  V2
V2
V1
R1
 R2
V1
 V2
R1
V2  
V2  
R2
V1
R1
R1
V1
q
R2
R2
V1
R1
R
q
R
+
V2'
+
V2
R

V2   2 V1
R1
V2  
 R
R 
R R
V2     2 V1   2 V1
R
R  R1  R1
INTEGRAL CIRCUIT
C
q
R
q
V1
+
V2
V2  
1
V1dt

RC
Rq  V1
V1  Rq  0
0
1
q  V2
C
q 
1
V1
R
q   q dt 
V2  
1
V1dt

R
1
V1dt
RC 
Negative sign can be eliminated bu adding an extra gain circuit with gain=1
q 3
R
R
Vc  (V1  V2  V3 )
R
V3
R
V2
V1
q 1
q
V1  Rq 1  0  q 1 
+
Vc
V V V 
 R  1  2  3   Vc
R R R
V2
 Vc  V1  V2  V3 
R
R
q 1
V
V2
, q 3  3
R
R
q  q 1  q 2  q 3
0  Rq  Vc
q 2 
R
V1
V1
R
Vc  V1  V2
R
+
R
q
+
V1
Vc
q 2
+
V2
0  q R  Vc
0  q R  Vc
q  q 1  q 2
R
V1  Rq 1  0
R
V
q 1   1
R

V2  q 2 R  0
q 2 
V2
R
 V V 
Vc  R   1  2 
 R R
Vc  V1  V2
Vc
R2
V2  K P V1  K I  V1dt  K D
R1
+
R
dV1
dt
R2
1
K D  R 4C4
KP 
KI 
R1
R 3C 3
C3
R
R3
+
R
+
R4
V1
C4
+
R
V2
PID Control Circuit
PID circuit is used frequently in Control Systems.