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