ECE53A Introduction to Analog and Digital Circuits Lecture Notes Second-Order Analog Circuit and...
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ECE53A
Introduction to Analog and Digital Circuits
Lecture Notes
Second-Order Analog Circuit and System
ECE 53A Fall 2007
• Order of an analog circuit or system depends on the number of energy storing elements L’s and C’s.
• Simplest second-order circuits are ones containing two C’s plus resistors and their dual networks containing two L’s; e.g.
• Simplest RLC-circuit is the series RLC-circuit shown in Figure 2(a) and its dual, the parallel (or shunt) RLC network.
R1
C1
R2
C2RL R1 R2 RL
L1 L2
( )gv t
+
-( )gv t
+
-
( )gi t
+
-
R CL
V(t)R
C
L
( )gv t
+
-i(t)
Figure 1(a) Figure 1(b)
Figure 2(a)
Single loop RLC-circuitFigure 2(b)
Single node-pair RLC-circuit
ECE 53A Fall 2007Series-RLC Circuit
( ) ( ) (1)
1( ) ( ) (2)
c g
t
c c
diRi L v t v t
dt
v t i dC
Figure 1 Single-loop series RLC circuit
R
C
L
i(t)
iL(t)
+
-
Vc(t)( )gv t+
-
2
2
2
2
( ) ( ) ( )( )
( ) ( ) 1 1( ) ( ) (3)
g
g
di t d i t i tR L v tdt dt C
d i t R di ti t v t
dt L dt LC L
First of all, even before we write the network equations to solve for the single unknown which is the loop current i(t), we need to satisfy two initial conditions (I.C) Vc(0) and iL(0).
KVL (loop) equation:
Differentiate Eq.(1) using (2),
[Note iL(t)=i(t)]
[Note ic(t)=i(t)]
or
• Eq.(3) is in the form of a generic second-order (n=2) ORDINARY differential equation with constant coefficients,
• Where can be a current or voltage and is the input or forcing function. The circuit we are considering is linear and lumped containing two(2) energy storage circuit elements (In our case, the circuit contains a inductor and a capacitor).
• For the series RLC circuit, to solve for , we need two(2) I.C, vc(0) and iL(0). Let , substituting this into the homogeneous equation
• We obtain the quadratic equation
• This is known as the characteristic equation. For the second-order circuits, the generic characteristic equation can be written as
2
1 02
( ) ( )( ) ( ) (4)
d x t dx ta a x t f t
dt dt
ECE 53A Fall 2007
( )x t ( )f t
( ), 0i t t ( ) sti t ke
2
2
( ) ( ) 1( ) 0 (5)
d i t R di ti t
dt L dt LC
2 10 (6)
Rs s
L LC
2 2 0 (7)n ns s
• Where
• Comparing Eq.(6) and (7), we have
• If (No damping) and the characteristic eq. becomes
• Let be , then for ,
• This justifies the naming of as the un-damped natural or the un-damped resonant frequency of the circuit.
• The roots of the generic quadratic characteristic equation are
Damping factor
Undamped natural or undamped resonant frequency in rad/secn
ECE 53A Fall 2007
0, 0R
2 or 2 or (8)2n
R R LC C R CR
L L L L
2 2 2 1ns s
LC
2n n
1 1 or (9)
LC LC
s s j 0( 0)R 2
n n
1 1 or in rad/sec
LC LC
n
2 2 22
1,2
2 4 41 (10)
2n n n
ns
ECE 53A
• If can be shown that does not affect the behavior of the response of the circuit by using frequency, thus, the behavior of the circuit response depends solely on the damping factor . Without lose of generality, let us assume . Thus, Eq.(10) reduces to
• Case 1: If ,(roots are real and equal. Double order) Critical damping
• Case 2: If , (both roots are real and unequal) Un-damped
• Case 3: If , (roots are complex and complex conjugal) Over-damped
21,2 1 (11)s
n 1rad/sec
1 21, ,s s
21 21, , 1s s
21 21, , 1s s j
j
j
j
Case 1 Case 2 Case 3
Fall 2007
ECE 53A
• Standard method is used to solve for the complete solutions of the ordinary second-order differential equation as
• where
( ) ( ) ( ) (12)h px t x t x t
( ) solution of the homogeneous equation
( ) partial solution due to the forcing functionh
p
x t
x t
1 2s s1 2( ) e e (13)t t
hx t k k
Fall 2007
KCL (Nodal) Equation: Since , we have
ECE 53A
( ) ( )cv t v t
2
2
( ) ( )( ) ( )
1( ) ( )
( )( ) 1 ( ) 1 1( ) (14)
L g
t
L
g
v t dv ti t C i t
R dt
i t v dL
di td v t dv tv t
dt RC dt LC C dt
Parallel-RLC Circuit
Figure 2. Single node-pair parallel RLC circuit
or
Note this equation is identical to (3) if ( ) ( )
1
g g
v t i t
R GR
C L
v i
In fact the parallel circuit in Fig. 2 is the dual network for the series circuit in Fig. 1.
Fall 2007
R CL iL(t)+
-
Vc(t)( )gi t
( )gi t
1( )i t
Write KCL (Nodal) Equation by inspection, we have
ECE 53A
2
11 2
( ) ( )( ) ( )( ) (15)cg cg
v t v tv t v t dvC i t
R dt R
Parallel-RLC Circuit
Figure 3. Second-order passive RC-circuits
Under-damped values of R’s and C’s
Fall 2007
2 ( )v t
v 2 ( ) 0i t +
-
R1
C1
R2
C21( )v t+
-
1
22
c
c
v v
v v
2( )cv t
Li
ECE 53A Fall 2007
R1
RL
L1
C1
C2
1( )cv t
+
-
+ -1i
N=4
By inspection, nodal (or KVL) equations are given by
ECE 53A
1v 1 1 1 21
1 3
2 2 2 12
2 3
0 (1)
0 (2)
dv v v vCdt R R
dv v v vCdt R R
Second-Order RC-Network
(or its dual: RL-Network)
Node
Eq.(3) is in the form of the nodal admittance matrix and can be obtained from the circuit in Figure 1.
Fall 2007
Figure 1. Second-order RC-network
R1 R2
R3
C1 C22( )cv t
1( )cv t
+
-
+
-
2v1v
Node 1v
11 3 3 1
22
3 2 3
1 1 1
( ) 0 (3)
( )1 1 1 0
C sR R R v t
v tC s
R R R
To treat Equation (1) and (2) (or(3)) simultaneously, we first use (1) to determine in terms of , and then substitute the result into (2) to obtain a second-order differential equation in . We have
Note that Eq.(5) is of the generic form
ECE 53A
2v
312 3 1 1
1
21 1
21 1 2 2 3 1 3 2
11 2 1 2 2 3 1 2 2 3 1 2
1 (4)
1 1 1 1
1 1 10 (5)
Rdvv R C v
dt R
d v dv
dt RC R C R C R C dt
vR R C C R R C C R R C C
Fall 2007
1v
2
1 112
1 1 2 2 3 1 3 2
2
1 2 1 2 1 3 1 2 2 3 1 2
2 0 (6)
1 1 1 12 (7)
1 1 1 (8)
n n
n
n
d v dvv
dt dt
RC R C R C R C
R R C C R R C C R R C C
1v
Also, note that Eq.(1) and (2) are already in the form of state-variable formation. Thus, rewriting (1) and (2) as
ECE 53A
1 12
1 3 1 3 1
2 21
3 2 2 3 2
1 1 1 (11)
1 1 1 (12)
dv vv
dt R R C R C
dv vv
dt R C R R C
Therefore
Fall 2007
Since1 21 1 2 2
1 1 3 1 3 1
3 1 2 2 3 2
and , ,
1 1 1
where (13)1 1 1
c cx v v x v v x Ax
RC R C R CA
R C R C R C
1 1 2 2 3 1 3 2
12
1 2 1 2 1 3 1 2 2 3 1 2
1 1 1 1 1Damping Factor (9)
2
Undamped Resonant Frequency in rad/sec
1 1 1 (10)
n
RC R C R C R C
R R C C R R C C R R C C
j
For the special case of RC-network, the nodal equations are indeed the state equations using state variables .
The roots of the characteristic equation are constrained to be only the negative -axis as shown below
ECE 53A
1 21 1 2( ) t tv t Ae A e
Where
Fall 2007
2
k ckx v
1
1 2 1 2
1 1 1 1
0, 0 and the constant and are
determined by the I.C. (0) (0) and (0) (0).c c
A A
x v x v
Can use Loop (KVL) analysis (3 loops) or use Nodal (KCL) analysis (2 node-pairs)
Better get use state-variable analysis
ECE 53A
1 2 3 4
Tx x x x x
High-Order (n=4) RLC-Network
Let Set up equations is straight forward – use a mixture of KCL equations (tree branch) and KVL equations (Cuts).
Solve by computing state and output equations.
1 1
2 2
3 1
4 2
c
c
L
L
x v
x v
x i
x i
to obtain 4 coupled 1st-rder differential equations.
Fall 2007
gv
2( )cv t
Li
Rg
RL
L1
C1
C2
1( )cv t
+
-
+ -1i
L2(1) (2)
+
-
N=4 (2L’s and 2C’s)
Tree: Connect all nodes – No closed path
Tree branches:
Loops:
Use a selected mixture of KCL and KVL equations
ECE 53A
1 1 2, , and g gv C C R
To obtain 4 coupled 1st-order differential equations which we can put into a compact matrix from I.C. 1 2 1 2(0) (0) (0) (0) (0)c c L Lx v v i i
1 2, and LL L R
Question: what kind of network is this?
Look at behavior of element (L’s and C’s) at
Fall 2007
First obtain directed graph:
4 KCL equations
3 KVL equations
No integration only 1st-order derivatives.
0 (DC) and s s
Rg L1
L2
RLC1
C2
Vg1
Let the capacitor C be charged to a voltage V0 and at t=0 let the switch be closed. The value of the resistance R will determine whether the system is
(a) over-damped (b) critically damped, (c) under-damped
ECE 53A
( ) ( )Li t i t
Fall 2007
Figure E-1. RC-network used to illustrate the finding the particular solution from the general
solution
Example E-1
R
Li(t)CV0 Vc
K
+
-
iL
Figure E2-2. Network response for the three cases: (a) over-damped (b) critically damped and © under-damped
In Eq.(1-3), K1 and K2 are arbitrary constant of integration.
Eq.(3) can be rewritten equivalently as
ECE 53A
2
2 21 2
1 1
2
( ) sin 1
tan
ntni t Ke t
K K K
K
K
General solutions in terms of
Fall 2007
2 2
2
1 11 1
1 2
11
11 or
2
( ) (1)
11 or
2
( ) (2)
11 or
2
( )
n nn
n
nn
t tt
t
j tt
Q
i t e K e K e
Q
i t K K e
Q
i t e K e
21
1 (3)nj tK e
and n
ECE 53A
Cascaded inverters with parasitic wiring inductance and gate capacitance
Fall 2007Two Cascaded Inverters
RLRL
RL RL
Vin1
Vout2
Vin2
Vout1
Vin1
Vout2
Vin2
Vout1
L1
Cgs
ECE 53A Fall 2007Circuit model of the cascaded inverters when the input at Vin1 is low
RLRL
Vin1
Vout2Vin2Vout1
L1
Cgs2RON RON
Series-RLC circuit: Ron L1 Cgs2
ECE 53A Fall 2007DUALITY and DUAL NETWORKS
Vs
Dual
L CR
Dual Quantities: 1 and
and
R G RL C
Ri RvLoop current and node-pair voltage
Loop and node pair
Short circuit and open circuit
Example 1: KCL and KVL
Example 2:
Dual
Network:
+
-i(t)
is LR C
V(t)
L
C
R
Vs
+
-
is
L
R C
1
33
1
2
2
Loop current ( ) node pair ( )i t v t
ECE 53A State-variable method
( ) ( ) ( )
( ) ( ) ( )
x t Ax t Bu t
y t Cx t Du t
1 2 1 2
2 2 constant matrix, known as state matrix
( ) state vector; and are state variables ( ) and/or ( )
( ) Input vector
( ) output vector
T
c L
A
x t x x x x v t i t
u t
y t
( ) ( ) ( )
( ) ( ) ( )
x t ax t bu t
y t cx t du t
I. First-order circuits and systems
For LTI systems, a,b,c,d are constants.
II. Second-order circuits and systems
For continuous-time LTI systems, A,B,C and D are constant matrices.
For LTI analog lumped circuits containing R, L, C and transformers.
I.C needed to solve the state equations: (0)x