Transcript Lecture7.ppt

EE 529 Circuits and
Systems Analysis
Lecture 7
Mustafa Kemal Uyguroğlu
Analysis of Circuit Containing OP-AMP
Show that
 R R 
v0
  R1 1  3  3 
is
 R1 R2 
R1
R2
B
R3
A
C
is
+
v0
D
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_
Analysis of Circuit Containing OP-AMP
Equivalent circuit
R1
R2
B
R1
R3
R2
B
R3
A
A
C
is
C
+
v0
D
_
+
is

o
D
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v0
_
Analysis of Circuit Containing OP-AMP
Circuit Graph
R1
R2
B
a
R3
A
(R1)
b
(R3)
C
(R2)
+
is

o
D
(is)
(o)
()
v0
_
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d
c
Analysis of Circuit Containing OP-AMP
Fundamental Cut-set Equations
a
(R1)
b
(R3)
i1  is  io  is
c
i3  i2  io  is  i2  is
(R2)
(is)
(o)
()
i  is  i0  i2  i2  i2
d
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Analysis of Circuit Containing OP-AMP
By using the terminal equations, the fundamental cut-set equations can
be written as:
i1  is  io  is
v1
 is
R1
i3  i2  io  is  i2  is
v3
v2
 is 
R3
R2
i  is  i0  i2  is  i2
v2
i  is 
R2
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Analysis of Circuit Containing OP-AMP
Fundamental Circuit Equations
v0  v1  v3  v  0
a
(R1)
b
(R3)
c
 v1  v3  v
v2  v3  v
(R2)
(is)
(o)
()
d
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Analysis of Circuit Containing OP-AMP
Subst. of fundamental circuit equations into fundamental cut-set
equations yields:
v1
 is
R1
v3  v
 is  v3  is R1  v
R1
v3
v
1
 is  2  is   v3  v 
R3
R2
R2
 1
v
1 

v

i


 3 s
R2
 R3 R2 
 1
v
1 


i
R

v

i




 s 1 
s
R2
 R3 R2 
 1
1 1
v
1 
1 
 is  is R1     is R1    
R3
 R3 R2 
 R1 R3 R2 
R
R 
v
  R1  3  1  3 
is
R2 
 R1
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Mathematical Models of Energy Storage Elements
 Capacitor
A
a
i(t)
v(t)
C
B
dv  t 
i t   C
dt
t
1
v(t )   i  t  dt v(0)
C0
b
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Mathematical Models of Energy Storage Elements
 Inductor
A
a
i(t)
v(t)
L
B
di  t 
v t   L
dt
t
1
i  t    v  t  dt i  0 
L0
b
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Analysis of Circuits Containing Energy Storage Elements
 RLC Circuit
● Branch Voltages Method
(L)
a
A
L
B
(C)
(S)
Vs
b
R
C
o
O
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(R)
Analysis of Circuits Containing Energy Storage Elements
 RLC Circuit
● Branch Voltages Method
♦ Fundamental Cut-Set Equations
(L)
a
(C)
(S)
o
iC  iL  iR
b
(R)
• Terminal Equations
dvC
iC  C
dt
t
1
iL   vL dt iL  0 
L0
iR  GvR
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Analysis of Circuits Containing Energy Storage Elements
 Subst. of terminal equations into
fundamental cut-set equation yields:
ic  iL  iR
t
dvc 1
C
  vL dt  iL  0   GvR ............(1)
dt L 0
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Analysis of Circuits Containing Energy Storage Elements
 RLC Circuit
● Branch Voltages Method
♦ Fundamental Circuit Equations
(L)
a
(C)
(S)
vL  VS  vC
b
(R)
vR  vC
o
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Analysis of Circuits Containing Energy Storage Elements
 Subst. of fundamental circuit equations into
Eq.(1) yields:
t
dvc 1
C
  VS  vC  dt  iL  0   GvC
dt L 0
or
t
t
dvc 1
1
C
  vC dt  GvC   VS dt  iL  0 
dt L 0
L0
Integro-Differential Equation
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Analysis of Circuits Containing Energy Storage Elements
 Taking the derivate wrt t and dividing by C
results in
d 2vc G dvC
1
1


vC 
VS
2
dt
C dt LC
LC
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Analysis of Circuits Containing Energy Storage Elements
 RLC Circuit
● Chord Currents method
(L)
a
A
L
B
(C)
(S)
Vs
b
R
C
o
O
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(R)
Analysis of Circuits Containing Energy Storage Elements
 RLC Circuit
● Chord Currents Method
♦ Fundamental Circuit Equations
(L)
a
(C)
(S)
vL  VS  vC
b
vR  vC
(R)
• Terminal Equations
o
diL
dt
vR  RiR
vL  L
t
1
vC   iC dt vC  0 
C0
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Analysis of Circuits Containing Energy Storage Elements
 Subst. of terminal equations into fundamental
circuit equations yield:
vL  VS  vC
t
diL
1
L
 VS   iC dt vC  0  ...................(1)
dt
C0
vR  vC
t
1
iR R   iC dt vC  0  ..................(2)
C0
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Analysis of Circuits Containing Energy Storage Elements
♦ Fundamental Cut-Set Equations
(L)
a
(C)
(S)
iC  iL  iR
b
(R)
o
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Analysis of Circuits Containing Energy Storage Elements
 Subst. of fundamental cut-set equation into
Eqns. (1) and (2) yields:
t
diL
1
L
 VS    iL  iR  dt vC  0  ...................(3)
dt
C0
t
1
iR R    iL  iR  dt vC  0  ..................(4)
C0
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Analysis of Circuits Containing Energy Storage Elements
 Taking the derivate wrt t results in
d 2iL d
1
1
L 2  VS  iL  iR
dt
dt
C
C
or
d 2iL
d
iR  LC 2  iL  C VS ..................(5)
dt
dt
diR 1
1
R
 iL  iR ..................(6)
dt C
C
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Analysis of Circuits Containing Energy Storage Elements
 Subst. Eq. (5) into (6) gives:
d 2iL
d 2iL
d
d  1
1
d 
R  LC 2  iL  C VS   iL   LC 2  iL  C VS 
dt 
dt
dt  C
C
dt
dt 
d 3iL
diL
d 2 iL 1
d2
1
d
RLC 3  R
 RC 2 VS  iL  L 2  iL  VS
dt
dt
dt
C
dt
C
dt
or
d 3iL
d 2iL
diL
d2
d
RLC 3  L 2  R
 RC 2 VS  VS
dt
dt
dt
dt
dt
Taking the integral wrt t and dividing by RLC results in
d 2iL
1 diL
1
1 d
1


iL 
VS 
VS
2
dt
RC dt LC
L dt
RLC
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Solutions of n-th order DE
 Consider an n-th order DE
dn
d n 1
d n 2
y  t   a1 n 1 y  t   a2 n 2 y  t  
n
dt
dt
dt
...............(1)
 an 1
d
y  t   an y  t   u  t 
dt
Let
x1  t   y  t 
d
d
y  t   x1  t 
dt
dt
d2
d
x3  t   2 y  t   x2  t 
dt
dt
x2  t  
d n 1
d
xn  t   n 1 y  t   xn 1  t 
dt
dt
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Solutions of n-th order DE
 Therefore Eq.(1) can be written as
d
xn  t   a1 xn  t   a2 xn 1  t  
dt
 an 1 x2  t   an x1  t   u  t 
Companion matrix
or in matrix form
 x1  t    0

 
x
t


 2
  0
d 

 
dt 
x
t


 n 1   0
 x  t     an
 n
 
1
1
 an 1
 a2
  x1  t    0 
  

x
t


 0
 2
    u t 

  

1   xn 1  t    0 
 a1   xn  t   1 


or in more compact form
d
x  t   Ax  t   bu  t 
dt
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Solutions of n-th order DE
d
x  t   Ax  t   bu  t 
dt
The right hand side of the equation may contain known ui(t)
functions and their different order of derivatives.
d
d
x  t   Ax  t   Bu  t   B1 u  t  
dt
dt
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dm
 Bm m u  t 
dt
State Variables Method
 So far, we explain, how a circuit containing energy
storage elements are analysed using Branch
Voltages method and Chord Currents Method. The
circuit equations are Integro-differential equations.
It means circuit equations contains integral and
derivative terms together.
 Now, another method called state variables method
will be introduced. In this method, the circuit
equations called circuit state equations do not
contain integral terms. The unknown quantities in
state equations are called state variables and as it
will be seen the state variables are capacitors
voltages and inductors currents
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State Variables Method
 Selection of tree is an important step in order to obtain the
state equations of a RLC circuit.
 The folllowing steps are used in selection of the tree.
● The edges in the system graph corresponding to the voltage sources
must be placed in the tree.
● The edges corresponding to as many capacitors as possible must
next be placed in the tree.
● If the tree is not complete, the edges corresponding to the resistors
must be chosen and as many resistors as possible must be included.
● If still the tree is not completed, then, the edges corresponding to
the inductors will be chosen until the tree is completed
● All the edges corresponding to the current sources must be placed
in the co-tree.
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State Variables Method
 After the selection of proper tree, the state
variables are branch capacitor voltages and chord
inductor currents. The terminal equations of
branch capacitors and chord inductors are:
d
CbC v bC  t   i bC  t 
dt
d
LcL i cL  t   v cL  t 
dt
where CbC and LcL are non-singular and diagonal
matrices.
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State Variables Method
 In order to obtain state equations ibC and vcL must
be written in terms of state variables and source
functions or/and derivatives of the source
functions.
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State Variables Method
EXAMPLE
 Consider the following RLC circuit.
A
Vs
L
D
R3
C
(L)
a
d
(C)
f
F
R2
(S)
B
(R3)
(R2)
b
In this circuit, there are one branch capacitor and one chord inductor.
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State Variables Method
EXAMPLE
 State variables are vc(t) and iL(t). The terminal equations
for states variables are
(L)
a
d
(C)
f
d
vC  t   iC  t 
dt
d
L iL  t   vL  t 
dt
C
(S)
(R3)
(R2)
b
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State Variables Method
EXAMPLE
 Now, let’s assume that the branch (C) is a voltage source
and chord (L) is a current source in the system graph.
(L)
a
(S)
d
(R3)
(C)
f
(R2)
Therefore, we have only one
unknown branch voltage v3,
and only one chord current i2..
• In order to find v3, branch
voltages method is used.
i3  iL  i2
G3v3  iL  G2v2
b
G3v3  iL  G2  vC  v3 
 G3  G2  v3  iL  G2vC
v3 
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G2
1
iL 
vC
G2  G3
G2  G3
State Variables Method
EXAMPLE
• In order to find i2, chord currents method is used.
(L)
a
d
(C)
f
v2  vC  v3
R2i2  vC  R3i3
(S)
R2i2  vC  R3  iL  i2 
(R3)
(R2)
 R2  R3  i2  vC  R3iL
i2 
R3
1
iL 
vC
R2  R3
R2  R3
b
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State Variables Method
EXAMPLE
• Now, let’s look at the terminal equations of the states
variables.
(L)
a
(S)
d
(C)
(R3)
f
(R2)
d
C vC  t   iC  t 
dt
d
L iL  t   vL  t 
dt
b
From the system graph, we can write that
iC  i2
and
vL  VS  v3
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State Variables Method
EXAMPLE
• Therefore, the circuit state equations are
C
0

1



R2  R3
0  d vC  t  



L  dt  iL  t   
G2
 G  G
3
 2
R3

R2  R3  vC  t   0 

    VS
1   iL  t   1 

G2  G3 
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