Download V 1 Generalized charges

Modeling of Electrical Circuits
The electrical circuit, which will be found the mathematical model, is
shown in Example 3.1.
In the electrical circuit, V1 is the input, q and q2 are the generalized
charges.
The currents (qdot and q2dot) are shown on the lines. The current
passing through the resistor R1 is not independent due to the first
kirchoff’s law.
Example 3.1:
L
+
V1
-
Current in the resistor R1 = q  q 2
C
q
R1
q 2
R2
V1: Input
q ve q2 : Gen. charges
Example 3.1 (Continue) :
C
L
+
V1
-
q
V1: Input
Current in the resistor R1 = q  q 2
q 2
R1
q ve q2 : Gen. charges
R2
The magnetic energy and electrical energy equations can be stated as
follows.
1
1 2
E1  Lq 2
E2 
q2
2
2C
Then, the virtual work expression which is related to the input and
resistors is written.
W  V1q  R 1 (q  q 2 )(q  q 2 )  R 2q 2 q 2
General voltages are found by taking parentheses of q and q2. The
term in the first parenthesis is the general voltage for the charge q. The
term in the second parenthesis is the general voltage for the charge q2.
 ( V1  R 1q  R 1q 2 )q  ( R 1q  R 1q 2  R 2q 2 )q 2
Qq
Qq2
Example 3.1 (Continue) :
L
+
V1
-
C
q
V1: Input
Current in the resistor R1 = q  q 2
q 2
R1
q ve q2 : Gen. charges
R2
1
E1  Lq 2
2
E2 
1 2
q2
2C
W  ( V1  R 1q  R 1q 2 )q  ( R 1q  R 1q 2  R 2q 2 )q 2
When the Lagrange equation is applied to the charges to q and q2,
respectively, the equations of motion are obtained.
d   (E1  E 2 )   (E1  E 2 )

 Qq


dt 
q
q

d   (E1  E 2 )   (E1  E 2 )

 Qq 2


dt 
q 2
q 2

   R 1
 L 0  q

 0 0  q



  2   R1

  V1  R 1q  R 1q 2
Lq

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   
Example 3.1 (Continue) :
L
+
V1
-
C
q
V1: Input
Current in the resistor R1 = q  q 2
q 2
R1
  V1  R 1q  R 1q 2
Lq
q ve q2 : Gen. charges
R2
L=3.4 mH, C=286 µF, R1=3.2 Ω, R2=4.5 Ω
1
q 2  R 1q  ( R 1  R 2 )q 2
C
This is a set of differential equations, which are similar to mathematical
models of mechanical systems considered in previous lecture. These
equations can be organised in matrix form.
   R 1
 R 1   q  0 0   q   V1 
 L 0  q

   0 1      

 0 0  q

  2    R 1 R 1  R 2  q 2   C  q 2   0 
Eigenvalue equation can be found substituting the numerical values into
the matrix equation as follows. We can get the results by manual solution
or Matlab.
D(s)=0.02618s3+26.288s2+11188.81s=0
2
Ls  R 1s
 R 1s
1  0 Eigenvalues: 0, -502.06±418.70i (ξ=0.768)
 R 1s
( R 1  R 2 )s 
C
Example 3.2:
Another electrical circuit with an op-amp element, which will be found the
mathematical model, is shown in Example 3.2. Op-amps are active
elements that works with an external energy.
C1
q 4
q 3
C2
R1
q 1
V1
q 3
q 2
R2
q f
R3
Input : V1
Generalized charges: q1, q2, q3
+
V2
In the electrical circuit, V1 is the input, q1, q2 and q3 are the generalized
charges.
The first kirchoff’s law is applied to the circuit in Example 3.2.
q 1  q 2  q 3  q 4  q 4  q 1  q 2  q 3
q 4  q 3  q f
 q f  q 1  q 2  q 3  q 3  q 1  q 2
Example 3.2 (Contiune):
C1
q 4
q 3
R1
q 1
V1
q 3
q 2
R2
Generalized charges: q1, q2, q3
q 4  q 1  q 2  q 3
q f
R3
C2
Input : V1
q f  q 1  q 2
+
V2
The magnetic energy and electrical energy equations are written as
follows.
E1  0
1
1 2
2
E2 
(q1  q 2  q 3 ) 
q3
2C1
2C2
Then, the virtual work expression is written.
W  V1q1  V2 (q1  q 2 )  R 1q 1q1  R 2q 2q 2  R 3q 3q 3
Example 3.2 (Contiune):
C1
q 4
q 3
C2
R1
q 1
V1
q 3
q 2
R2
Input : V1
q 4  q 1  q 2  q 3
q f
R3
Generalized charges: q1, q2, q3
+
V2
E1  0
q f  q 1  q 2
1
1 2
2
E2 
(q1  q 2  q 3 ) 
q3
2C1
2C2
W  V1q1  V2 (q1  q 2 )  R 1q 1q1  R 2q 2q 2  R 3q 3q 3
The Lagrange equation is applied to the charges to q1, q2 and q3. Then,
the equations of motion are obtained for the electrical system.
d   ( E1  E 2 )   ( E1  E 2 )

 Q q1


dt 
q 1
q 1


1
(q1  q 2  q 3 )  V1  V2  R 1q 1
C1
1
d   (E1  E 2 )   (E1  E 2 )

 Qq 2  
(q1  q 2  q 3 )  V2  R 2q 2


dt 
q 2
q 2
C1

d   ( E1  E 2 )   ( E1  E 2 )

 Qq 3


dt  q 3
q 3

 
1
1
(q1  q 2  q 3 )  q 3   R 3q 3
C1
C2
Example 3.2 (Contiune):
C1
q 4
q 3
C2
R1
q 1
V1
q 3
q 2
Input : V1
q f
R3
1
(q1  q 2  q 3 )  V1  V2  R 1q 1
C1
+
R2
Also, in op-amp: V+=V-=0
Generalized charges: q1, q2, q3
V2
1
 (q1  q 2  q 3 )  V2  R 2q 2
C1

1
1
(q1  q 2  q 3 )  q 3   R 3q 3
C1
C2
 R 3q 3  V2
These equations can be organised in matrix form as they are linear
differential equations.
 1
1
1 


 C

C
C
1
1
1

R
0

R
q
 1
  q1   V1 
3  1  
1
1
   
0 R
 q     1

R
q 2    0 
2
3  2 



C
C1
C1
q   0 
0
R 3  q 3   11
 0

1
1
1  3  



C
C
C
C

1
1
1
2
Example 3.2 (Contiune):
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
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 

Using numerical values given for Example 3.2, the eigenvalues of the
circuit with an op-amp can be found as follows.
R1=15.9 kΩ, R2=837 Ω, R3=318 kΩ, C1=C2=0.005 µF
R 1s 
1
C1
1
C1
1

C1


1
C1
R 2s 
1
C1
1
C1
1
C1
1
R 3s 
0
C1
1
1
R 3s 

C1 C 2
 R 3s 
Eigenvalues: 0, -628.93±12561.76i (ξ=0.05)
Circuits with op-amps are widely used instruments and control systems
for multiplying, integrating, and differentiating signals.
C
R2
R
R1
+
V1
+
V1
V2
V2
V2  
R2
V1
R1
V2  
R
C
+
V1
V2
V2   RC
dV1
dt
1
V1dt

RC
These are examples of op-amp circuits such as a inverting-summing
amplifier and a difference amplifier.
R
R
R
V3
Vc  (V1  V2  V3 )
R
V2
+
V1
Vc
R
R
R
+
V1
R
V2
Vc  V1  V2
R
+
V1
Vc
+
V2
Vc
PID control circuit:
R2
V2  K P V1  K I  V1dt  K D
R1
+
R
KP 
R2
1
K D  R 4C4
KI 
R1
R 3C 3
C3
R
R3
+
R
+
R4
V1
C4
+
R
dV1
dt
V2