• Here, order refers to the number of capacitors and inductors
• First Order Circuits : The circuits which contain only one capacitor or only one inductor.
• First Order RC Circuits : Such circuits contain independent and dependent sources, resistors + one capacitor.
R
T i
C i
+ v
− v
T
+ v
C
−
C
• KVL : v
T
•
= R
T i
C
+ v
C i
C
= C dv
C dt
⇒ v
T
= R
T
C dv
C dt
+ v
C dv
C dt
+
1
R
T
C v
C
=
1
R
T
C
V
T
• Result is a first order ODE
+ v i v s
C
−
• v
C
= v s
, i
C
= C dv s dt i s
+ i v
−
C i
C
= i s
, v
C
= v
C
( t
0
)+
1
C
R t t
0 i s
( τ ) dτ
1

• First Order RL Circuits : Such circuits contain independent and dependent sources, resistors + one inductor.
i
+ v
− i
N
+
R
N
− v
L i
L
L
• KCL : i
N
•
= G
N v
L
+ i
L v
L
= L di
L dt
⇒ i
N
= G
N
L di
L dt
+ i
L di
L dt
+
1
G
N
L i
L
=
1
G
N
L i
N
G
N
=
R
1
N
• Result is a first order ODE v s
+ v i
−
L i s
+ i v
L
−
• v
L
= v s
, i
L
= i
L
( t
0
)+
1
L
R t t
0 v s
( τ ) dτ i
L
= i s
, v
L
= L di s dt
2

• Step Response of First Order RC / RL Circuits :
• Here we have DC sources, i.e.
V
T and i
N are constants
• dv
C dt
+
1
R
T
C v
C
=
1
R
T
C
V
T di dt
L +
1
G
N
L i
L
=
1
G
N
L i
N
• These two equations can be combined into a single equation :
• dx dt
+
1
τ x =
1
τ x
∞
• For RC circuits, x = v
C
, τ = R
T
C , x
∞
= V
T
• For RL circuits, x = i
L
, τ = G
N
L =
R
L
N
, x
∞
= i
N
• τ : Time Constant.
• Unit : [ τ ] = Ω sec
Ω
(for RC case) =
Ω sec
Ω
(for RL case) = sec
• Solution of the ODE : Since x
∞ is constant, dx
∞ dt
= 0
• d ( x − x
∞
) dt
+
1
τ
( x − x
∞
) = 0
• Define a new variable y ( t ) = x ( t ) − x
∞
⇒ dy dt
+
1
τ y = 0
• Solution is : y ( t ) = y ( t
0
) e − t − t
0
τ
•
•
• x ( t ) − x
∞
= ( x ( t
0
) − x
∞
) e − t − t
0
τ
For RC Case :
For RL Case : v
C
( t ) − V
T
= ( v
C
( t
0
) − V
T
) e
− t − t
0
RT C i
L
( t ) − i
N
= ( i
L
( t
0
) − i
N
) e
− t − t
0
GN L
3

• Why x
∞
? assume that τ > 0 and take limit as t → ∞
• lim t →∞ x ( t ) = x
∞
+ lim t →∞
( x ( t
0
) − x
∞
) e − t − t
0
τ
= x
∞
3
2.5
4
3.5
5
4.5
2
1.5
1
0 1 2 3 4 5 t (sec)
6 7 8 9 10
• Question : Practically, how long should we wait till we safely assume that x ( t ) ≈ x
∞
. This is called the steady state
• Practically, we have x ( t ) ≈ x
∞
, for t − t
0
> 5 τ
• x ( t ) − x
∞
≈ ( x ( t
0
) − x
∞
) e − 5 ≈ 0 .
006( x ( t
0
) − x
∞
)
• In the above figure, we have τ = 1 sec .
4

• For RC Case : lim t →∞ v
C
( t ) = V
T
⇒ lim t →∞ i
C
( t ) = 0
• In the steady state, capacitor behaves like an open circuit .
• For RL Case : lim t →∞ i
L
( t ) = i
N
⇒ lim t →∞ v
L
( t ) = 0
• In the steady state, inductor behaves like a short circuit .
• Example 1 : Example 7.8. p. 300.
• Since the switch is open for a long time, we may assume that inductor reaches its steady state → short circuit !
• By shorting the inductor → i
L
(0) =
R
1
V
A
+ R
2
• This shows a way to set up the initial condition for inductors.
• For t > 0, the switch is closed. If we wait long enough, we may assume that inductor reaches its steady state → short circuit !
• i
L
( ∞ ) = x
∞
= i
N
=
V
A
R
1
• For t > 0, the switch is closed → τ = G
1
L =
L
R
1
• i
L
( t ) =
V
A
R
1
+(
R
1
V
A
+ R
2
−
V
A
R
1
) e
− t − t
0
G
1
L
=
V
A
R
1
+(
R
1
V
A
+ R
2
−
V
R
A
1
) e −
R
1( t − t
0)
L
• v
L
( t ) = L di
L
( t ) dt
= −
1
G
1
(
R
1
V
A
+ R
2
−
V
A
R
1
) e
− t − t
0
G
1
L
→ 0
5

• Example 2 : Example 7.9. p. 301.
• Since the switch is closed for a long time, we may assume that capacitor reaches its steady state → open circuit !
• By opening the capacitor → v
C
(0) =
V
R
1
A
R
+ R
1
2
• For t > 0, the switch is opened. If we wait long enough, we may assume that capacitor reaches its steady state → open circuit !
• v
C
( ∞ ) = x
∞
= V
T
= V
A
• For t > 0, the switch is opened → τ = R
2
C
• v
C
( t ) = V
A
+ (
R
V
A
1
R
+ R
1
2
− V
A
) e
− t − t
0
R
2
C
→ V
A
• i
C
( t ) = C dv
C dt
( t )
= −
R
1
2
(
V
A
R
1
R
+ R
1
2
− V
A
) e
− t − t
0
R
2
C
→ 0
• After finding v
C
( t ), i
C
( t ), v
L
( t ), i
L
( t ), how can we find the remaining voltages or currents ?
• By substitution . i.e. Replace the capacitor/inductor by a voltage and/or current source with the found solution.
6

+
− v i
C i v
+ v
− i
• Here v ( t ) = v
C
( t ), i ( t ) = i
C
( t ), which are already found.
+
− v i
C i v
• Here v ( t ) = v
L
( t ), i ( t ) = i
L
( t ), which are already found.
+ v
− i
7

• Example :
4 A 1
1
2
1 F
+ vC
− vT
R
T
1 F
+ v C
−
1
4 A
1
2
1
1 io
2
+ voC =vT
−
Req = R
T
• R
T
= 1 Ω, By current division ⇒ i
0
= 1
4
4 = 1 A ⇒ v oc
= v
T
= 2 V
• τ = R
T
C = 1 sec , v
C
( ∞ ) = v
T
= 2 V . Assume that v
C
(0) = 1 V .
• v
C
( t ) − v
C
( ∞ ) = ( v
C
(0) − v
C
( ∞ )) e t − t
0
τ
• ⇒ v
C
( t ) = 2 − e − t ⇒ i
C
( t ) = C dv
C dt
( t )
= e − t
• Suppose that we want to find out, say, i
1 and v
2
.
⇒ Use substitution.
+ v2
−
+ i1
1
4 A 1 2 vC
−
• Here, v
C
= 2 − e − t . Simple node analysis yields :
• i
1
( t ) = 3 − 0 .
5 e − t , v
2
( t ) = 1 + 0 .
5 e − t .
8

• Example :
2 A
2 H i
L
1
2
1
2 A
1
2
1 iN
1
2
1
Req=R
T iN R
T i
L
2 H
• R
T
= 1 Ω, By KCL ⇒ 2 = v
1
+ i
N
, i
N
= v
1
⇒ i
N
= 1 A .
• τ = G
T
L = 2 sec , i
L
( ∞ ) = i
N
= 1 A . Assume that i
L
(0) = 3 V .
• i
L
( t ) − i
L
( ∞ ) = ( i
L
(0) − i
L
( ∞ )) e t − t
0
τ
• ⇒ i
L
( t ) = 1 + e − t
2
⇒ v
L
( t ) = L di
L dt
( t )
= − 2 e − t
2
• Suppose that we want to find out, say, i
1 and i
2
.
⇒ Use substitution.
i
L i1
2 i2
2 A
1
1
• Here, i
L
= 1 + e − t
2
. Simple node analysis yields :
• i
1
( t ) = 1 − e − t
2
, i
2
( t ) = 1 + e − t
2
.
9

• Response to a Pulse : This will depend on the relation between τ and T : v(t)
+
R vC A v(t)
C
− t
T
0 < t < T t > T
A
R
C
+ vC
R
C
+ vC
−
−
• If T À τ , v
C
( T ) ' v
C
( ∞ ) = A . Otherwise, v
C
( T ) = v
C
( ∞ ) = A vC(t) vC(t)
A
A
T
Time Constant < < T t
• Example : R = 1 k Ω, C = 1 µF , A = 10 V , T = 10 ms , v
C
(0) = 0 V .
• ⇒ τ = RC = 1 ms << T
• 0 < t < 10 ⇒ v
C
( t ) = 10 − 10 e − t ⇒ v
C
(10) ' 10 V
• t > 10 ⇒ v
C
( t ) = 10 e − ( t − 10)
• Example : R = 10 k Ω, C = 1 µF , A = 10 V , T = 10 ms , v
C
(0) = 0 V .
• ⇒ τ = RC = 10 ms = T
• 0 < t < 10 ⇒ v
C
( t ) = 10 − 10 e − 0 .
1 t ⇒ v
C
(10) = 10 − 10 e − 1 = 6 .
32
• t > 10 ⇒ v
C
( t ) = 6 .
32 e − 0 .
1( t − 10)
10
T
Time Constant comparable with T t

• Sinusoidal Response of First Order RC / RL Circuits :
• Here, the source term in the Th´evenin/Norton equivalent circuit is sinusoidal.
• dx dt
+
1
τ x =
1
τ x
∞ x
∞
= V
A cos ωt
• x ( t ) = x
N
( t ) + x
F
( t ) = Natural Response + Forced Response
• This is the same as SUPERPOSITION. Natural response is due to initial condition x (0) and is called ZERO INPUT RESPONSE. This is the solution of ODE when the source term is set to ZERO.
• Forced response is due to source term and is called ZERO STATE RE-
SPONSE. This is the solution of ODE when x (0) = 0.
•
• dx
N dt
+
1
τ x
N
= 0 dx
F dt
+
1
τ x
F
=
1
τ
V
A cos ωt x
N
( t ) = Ke − t
τ
• x
F
( t ) = V
F cos( ωt + φ ) = a cos ωt + b sin ωt
• a = V
F cos φ , b = − V
F sin φ
• ˙
F
( t ) = − aω sin ωt + bω cos ωt
• ( bω +
1
τ a ) cos ωt + ( − aω +
1
τ b ) sin ωt =
1
τ
V
A cos ωt
• ( bω +
1
τ a ) =
1
τ
V
A
( − aω +
1
τ b ) = 0
• Given τ, V
A and ω Find a and b ⇒ V
F and φ
11

• Example 7-12 , p. 307.
• Note that since the switch is open for a long time → v (0) = 0.
• For Th´evenin Equivalent circuit : R
T
= 4 k Ω k 4 k Ω = 2 k Ω
• v
T
= 0 .
5 v s
( t ) = 10 sin 1000 t → τ = 2 .
10 3 × 10 − 6 = 2 .
10 − 3 sec.
• x
N
( t ) = Ke − t
τ
= Ke − 500 t
• x
F
( t ) = V
F cos( ωt + φ ) = a cos ωt + b sin ωt
• ( bω +
1
τ a ) = 0 ( − aω +
1
τ b ) =
1
τ
V
A
ω = 1000 Hz.
• 1000 b + 500 a = 0 − 1000 a + 500 b = 5000 −→ a = − 4 , b = 2
• v ( t ) = Ke − 500 t − 4 cos 1000 t + 2 sin 1000 t
• v (0) = 0 = K − 4 −→ K = 4
• a = V
F cos φ = − 4 , b = − V
F sin φ = 2
• V
F
=
√ a 2 + b 2 = 4 .
47, tan φ = − 2 / − 4 −→ φ = − 153 deg.
• v ( t ) = 4 e − 500 t + 4 .
47 cos(1000 t − 153)
• Note that as t → ∞ , −→ v ( t ) → 4 .
47 cos(1000 t − 153)
• This is called Sinusoidal Steady State . We will find it directly in
Chp. 8 in a simple way → Phasor Analysis.
12

• Second Order Circuits : The circuits which contain total of two capacitor- inductor combination.
→ 2 capacitors, 2 inductors, or 1 capacitor+
1 inductor.
• Before considering the most general case, we will first consider two important cases : series and parallel RLC circuits.
• Series RLC circuits i C C
+ v
C
−
L i
L
+ v
L
−
R
T
+ v
R v
T
−
C
+ v
C
−
+ v
L
−
• KVL : v
T
= v
R
+ v
C
+ v
L
KCL : i
R
= i
C
= i
L
• From the first order case, we know that v
C and i
L are state variables
(i.e. variables of ODE)
• i
C
= C dv
C dt
= i
L
−→ dv
C dt
= i
L
/C (1)
• v
T
= R
T i
R
+ v
C
+ v
L
= Ri
L
+ v
C
+ L di
L dt
• di
L dt
= −
1
L v
C
−
R
L
T i
L
+
1
L v
T
(2)
• (1) and (2) are called state equations . They are first order coupled
ODE’s. They can be written in matrix form as :
13

• d dt
 v
C i
L

=

0 1 /C
− 1 /L − R
T
/L
  v
C i
L

+

0
1 /L

V
T
• General form : −→ ˙ = Ax + bu
(*)
• x :(vector) state variable, u : input, A : matrix, b : vector.
• Alternative form : Scalar second order equation. Differentiate (1), use
(1) and (2) to eliminate i
L
:
• ¨
C
+
R
T
L v ˙
C
+
1
LC v
C
=
1
LC v
T
(**)
• So we have to either solve (*) or (**). Since they are equivalent, they give the same result. Usually we are given v
C
(0) and i
L
(0).
• To solve (**), we need ˙
C
(0). This comes from (1) : v ˙
C
(0) = i
L
(0) /C
• Before finding the solution of (**), let us consider the other case :
14

• Parallel RLC circuits i
C
C i
L
L i
N i
R
R i
C
C i
L
L
• KCL : i
N
= i
R
+ i
C
+ i
L
, KVL : v
R
= v
C
= v
L
,
• From the first order case, we know that v
C and i
L are state variables
(i.e. variables of ODE)
• v
L
= L di dt
L = v
C
−→ i
˙
L
= v
C
/L (1)
• i
N
= Gv
R
+ C v ˙
C
+ i
L
= Gv
C
+ C v ˙
C
+ i
L
• v ˙
C
= − G/Cv
C
− 1 /Ci
L
+ 1 /Ci
N
(2)
• (1) and (2) are called state equations and can be written also as :
• d dt
 v
C i
L

=

− G/C − 1 /C
1 /L 0
  v
C i
L

+

1 /C
0
 i
N
(*)
• Alternative form : Scalar second order equation. Differentiate (1), use
(1) and (2) to eliminate v
C
:
15

• i
L
+
G
C i
˙
L
+
1
LC i
L
=
1
LC i
N
(**)
• So we have to either solve (*) or (**). Since they are equivalent, they give the same result. Usually we are given v
C
(0) and i
L
(0).
• To solve (**), we need ˙ i
L
(0). This comes from (1) : i
˙
L
(0) = v
C
(0) /L
• Solution = zero input response + zero state response.
• Zero Input Response ( v
T
= 0, i
N
= 0)
• ¨ + a x + bx = 0
• Series RLC : x = v
C
, a =
R
T
L
, b =
1
LC
• Parallel RLC : x = i
L
, a =
G
C
, b =
1
LC
• x (0) and ˙ (0) are given.
• Assume solution of the form x ( t ) = e st ⇒ ˙ ( t ) = se st ⇒ ¨ ( t ) = s 2 e st
• ⇒ ( s 2 + as + b ) e st = 0 −→ ( s 2 + as + b ) = 0
• This is called the characteristic polynomial of the circuit, here s could be real of complex.
16

• ( s 2 + as + b ) = 0 −→ roots are s
1 and s
2
• Roots can be found as : s
1 , 2
= ±
− a ±
√
2 a 2 − 4 b
• Case 1 : Distinct Roots, s
1
= s
2
• x ( t ) = K
1 e s
1 t + K
2 e s
2 t
• Here K
1 and K
2 are constants (real or complex) to be found from the initial conditions.
• x (0) = K
1
+ K
2
, ˙ (0) = s
1
K
1
+ s
2
K
2
• K
1
= s
2 x (0) − ˙ (0) s
2
− s
1
K
2
= s
1 x (0) − ˙ (0) s
1
− s
2
• These formulas are valid even if s
1 and s
2 are complex.
• What if the roots are complex ?
• s = s
1 s
2
= − α + β ⇒ K
1
K
2
= r/ 2 e − θ
• x ( t ) = K
1 e s
1 t + K
2 e s
2 t = K
1 e st K
1 e = 2 < ( K
1 e st )
• x ( t ) = 2 < (( r/ 2) e − αt +  ( βt − θ ) ) = re − αt cos( βt − θ )
• Need to find r and θ by using initial conditions.
17

• x ( t ) = re − αt cos( βt − θ )
• x (0) = r cos θ −→ ˙ (0) = − αr cos θ + βr sin θ
• r cos θ = x (0) −→ r sin θ =
˙ (0)+ αx (0)
β
• Alternatively we could find the solution as :
• x ( t ) = C
1 e − αt cos βt + C
2 e − αt sin βt
• x (0) = C
1
, ˙ (0) = − αC
1
+ βC
2
−→ C
2
=
˙ (0)+ αx (0)
β
• Case 2 : Repeated Roots, s
1
= s
2
= s
• x ( t ) = K
1 e st + K
2 te st
• x (0) = K
1
, ˙ (0) = sK
1
+ K
2
−→ K
2
= ˙ (0) − sx (0)
18

• Stability of solutions
• s
1
= s
2
−→ x
N
( t ) = K
1 e s
1 t + K
2 e s
2 t
• s
1
= s
2
= s −→ x
N
( t ) = K
1 e st + K
2 te st
• If < ( s i
) < 0 −→ x
N
( t ) → 0 as t → ∞ −→ Stable Case .
• In this case, if the independent sources are bounded −→ solutions are bounded.
• If < ( s i
) > 0 , or s
1
= s
2
≥ 0 −→ x
N
( t ) → ∞ as t → ∞ −→ Unstable
Case .
• In this case, solutions are unbounded.
• For practical reasons, we want stable systems .
• From geometrical point of view, the roots of characteristic polynomial are in the open left half of the complex plane in the stable case.
• We will do further classification for the stable case, as explained in the next example.
• Example 7-15 , p. 318.
• Consider a series RLC circuit with no independent sources. (Hence we find zero input response, i.e. only the natural response). Let C = 0 .
25 µF ,
L = 1 H , v
C
(0) = 15 V , i
L
(0) = 0 A . Find the solution for : (a) R = 8 .
5 k Ω,
(b) R = 4 k Ω, (c) : R = 1 k Ω.
19

• ¨
C
+
R
T
L v ˙
C
+
1
LC v
C
=
1
LC v
T
= 0 (**)
• Characteristic polynomial :
• ⇒ ( s 2 + as + b ) = 0 −→ a =
• Roots can be found as : s
1 , 2
=
R
T
L
, b =
1
LC
− a ±
√
2 a 2 − 4 b
• Case 1 : a =
R
L
T = 8500, b =
1
LC
= 4 .
10 6 ⇒ s
1
= − 500, s
2
= − 8000.
• Roots are negative ⇒ Stable case.
• v
C
( t ) = K
1 e − 500 t + K
2 e − 8000 t
• v
C
(0) = K
1
+ K
2
= 15.
• v ˙
C
(0) = i
L
(0) /C = 0 ⇒ − 500 K
1
− 8000 K
2
= 0
• ⇒ K
1
= 16 , K
2
= − 1 ⇒ v
C
( t ) = 16 e − 500 t − e − 8000 t V
• C dv
C dt
= i
L
= − 2 .
10 − 3 e − 500 t + 2 .
10 − 3 e − 8000 t A
• This is called Overdamped Case (i.e. two distinct real-negative roots)
Overdamped Case
6
5
4
3
2
1
0
0 0.1
0.2
0.3
0.4
0.5
t (sec.)
0.6
0.7
0.8
0.9
1
20

• Case 2 : a =
R
L
T = 4000, b =
1
LC
= 4 .
10 6 ⇒ s
1
= s
2
= − 2000.
• v
C
( t ) = K
1 e − 2000 t + K
2 te − 2000 t
• v
C
(0) = K
1
= 15,
• v ˙
C
(0) = i
L
(0) /C = 0 ⇒ − 2000 K
1
+ K
2
= 0 ⇒ K
2
= 30000
• v
C
( t ) = (15 + 30000 t ) e − 2000 t V
• C dv
C dt
= i
L
= 15 te − 2000 t A
• This is called Critically Damped Case (i.e. two repeated real-negative roots)
Critically Damped Case
7
6
5
4
3
2
1
0
0 0.1
0.2
0.3
0.4
0.5
t (sec.)
0.6
0.7
0.8
0.9
1
• Case 3 : a =
R
L
T = 1000, b =
1
LC
= 4 .
10 6 ⇒ s
1 , 2
= − 500 ±  500
√
15.
• α = 500 , β = 500
√
15
• v
C
•
( t ) = Ke − 500 t
= K
1 e − 500 t cos(500 cos(500
√
√
15 t + φ )
15 t ) + K
2 e − 500 t sin(500
√
15 t )
21

• v
C
(0) = 15 = K
1
• v ˙
C
(0) = i
L
(0) /C = 0 ⇒ − 500 K
1
+ 500
√
15 K
2
= 0 ⇒ K
2
=
√
15
• v
C
( t ) = 15 e − 500 t cos(500
√
15 t ) +
√
15 e − 500 t sin(500
√
15 t )
• K cos φ = K
1
= 15 , K sin φ = − K
2
= −
√
15
• K =
√
15 2 + 15 = 15 .
5 , tan φ = −
√
15 ⇒ φ = − 75 .
52 deg.= − 1 .
32 rad.
• v
C
( t ) = 15 .
5 e − 500 t
• = 15 .
5 e − 500 t cos(500
√
15 t − 1 .
32) (True notation) cos(500
√
15 t − 75 .
52) (False but acceptable)
• This is called Underdamped Case (i.e. two complex conjugate roots with negative real part)
Underdamped Case
1
0
−1
−2
−3
−4
−5
0
3
2
0.1
0.2
0.3
0.4
0.5
t (sec.)
0.6
0.7
0.8
0.9
1
• Here s = − α + β ; α determines the decay, and β determines the frequency.
22

Im s2 s1
Overdamped
Re
Im s1=s2=s
Critically Damped
Re b
−a
Underdamped
• Step Response
• In this case, the standard ODE becomes :
• ¨ ( t ) + a x ( t ) + bx ( t ) = A
• x ( t ) = x
N
( t ) + x
F
• x
N
( t ) is the natural response and can be found as before.
• x
F
= B −→ ¨
F x
F
= 0 ⇒ x
F
= A/b
• Example 7-19, p. 328 . Series RLC circuit with V
T
= 10 V , C =
0 .
5 µF , L = 2 H , R = 1 k Ω, v
C
(0) = 0, i
L
(0) = 0.
• ¨
C
+
R
T
L v ˙
C
+
1
LC v
C
=
1
LC v
T
• ¨
C
( t ) + 500 ˙
C
( t ) + 10 6 v
C
( t ) = 10 7
(**)
23

• For x
N
, the characteristic polynomial is : s 2 + 500 s + 10 6 = 0
• Roots : s
1 , 2
= − 250 ±  968 −→ Underdamped case.
• v
CN
( t ) = K
1 e − 250 t cos 968 t + K
2 e − 250 t sin 968 t = Ke − 250 t cos(968 t + φ )
• V
CF
= 10 7 / 10 6 = 10
• v
C
( t ) = 10 + K
1 e − 250 t cos 968 t + K
2 e − 250 t sin 968 t
• v
C
(0) = 10 + K
1
= 0 −→ K
1
= − 10
• v ˙
C
(0) = i
L
(0) /C = 0 ⇒ − 250 K
1
+ 968 K
2
= 0 −→ K
2
= − 2 .
58
• v
C
( t ) = 10 − 10 e − 250 t cos 968 t − 2 .
58 e − 250 t sin 968 t
• K cos φ = K
1
= − 10 , K sin φ = − K
2
= 2 .
58
• K =
√
100 + 2 .
58 2 = 10 .
12, φ = 104 .
46 deg.
• The characteristic polynomial can also be written as :
• s 2 + 2 ξω
0 s + ω 2
0
= 0
• ξ : damping ratio, ω
0
: undamped natural frequency.
• s
1 , 2
= ω
0
( − ξ ±
√
ξ 2 − 1)
• ξ > 1 −→ Overdamped Case.
• ξ = 1 −→ Critically damped Case.
• 0 < ξ < 1 −→ Underdamped case.
• ξ = 0 −→ Lossless Case. In this case, the natural response is a pure sinusoid .
24

• General Second Order Circuits
• Note that we have v
C and i
L as variables, and by element relations we have i
C
= C v ˙
C and v
L
= L
˙ i
L
.
• Hence the general strategy is as follows :
• Step 1 : Use any method we have seen (node, mesh, combined constraints etc.) to obtain i
C and v
L in terms of v
C
, i
L and independent sources only. (This is a process of writing equations, and eliminating undesired variables ⇒ Linear Algebra)
• At the end of step 1 , we obtain the following equations :
• i
C
= c
11 v
C
+ c
12 i
L
+ d
1 u
1
• v
L
= c
21 v
C
+ c
22 i
L
+ d
2 u
2
• Here, coefficients c ij
, d i depend on circuit parameters, u i depend on independent sources.
• Step 2 : Use i
C
= C v ˙
C and v
L
= L i
˙
L
:
• C v ˙
C
= c
11 v
C
+ c
12 i
L
+ d
1 u
1
• L i
˙
L
= c
21 v
C
+ c
22 i
L
+ d
2 u
2
• These equations can be written either in component form :
• v ˙
C
= a
11 v
C
+ a
12 i
L
+ b
1 u
1
(*)
•
˙ i
L
= a
21 v
C
+ a
22 i
L
+ b
2 u
2
(**)
• Or in matrix form :
25

• d dt
 v
C i
L

=
 a
11 a
21 a
12 a
22
  v
C i
L

+
 b
1 u
1 b
2 u
2

• General form : −→ ˙ = Ax + bu
(*)
• x :(vector) state variable, u : input, A : matrix, b : vector or matrix.
• Alternative form : Scalar second order equation. Differentiate ( ∗ ) or
( ∗∗ ), use these equations to eliminate i
L or v
C
:
• Case 1 : a
12
= 0
• ¨
C
= a
11 v ˙
C
+ a
12 i
˙
L
+ b
1 u
1
= a
11 v ˙
C
+ a
12
( a
21 v
C
+ a
22 i
L
+ b
2 u
2
) + b
1 u
1
• = a
11 v ˙
C
+ a
12 a
21 v
C
+ a
22 a
12 i
L
+ a
12 b
2 u
2
+ b
1 u
1
• = a
11 v ˙
C
+ a
12 a
21 v
C
+ a
22
( ˙
C
− a
11 v
C
− b
1 u
1
) + a
12 b
2 u
2
+ b
1
˙
1
• = ( a
11
+ a
22 C
+ ( a
12 a
21
− a
11 a
22
) v
C
− a
22 b
1 u
1
+ a
12 b
2 u
2
+ b
1 u
1
• ¨
C
− ( a
11
+ a
22
) ˙
C
+ ( a
11 a
22
− a
12 a
21
) v
C
= − a
22 b
1 u
1
+ a
12 b
2 u
2
+ b
1
˙
1
• Case 2 : a
21
= 0. Change v
C with i
L and indexes 1 ←→ 2, we obtain :
• i
L
− ( a
22
+ a
11
)˙ i
L
+ ( a
22 a
11
− a
21 a
12
) i
L
= − a
11 b
2 u
2
+ a
21 b
1 u
1
+ b
2 u
2
• Let us see the relation with A :
• ( a
11
+ a
22
) = Trace( A ) = T
• ( a
11 a
22
− a
12 a
21
) = det A = D
• ¨ − T ˙ + Dx = u s
• v
C
Case : u s
= − a
22 b
1 u
1
+ a
12 b
2 u
2
+ b
1 u
1
• i
L
Case : u s
= − a
11 b
2 u
2
+ a
21 b
1 u
1
+ b
2
˙
2
26

v s
R1
+ v
C1
−
R2
C1
+ v
C2
−
C2
• KCL at C
1
: i
C 1
+ G
1
( v
C 1
− v s
) + G
2
( v
C 1
− v
C 2
) = 0
• ⇒ C
1 v ˙
C 1
= − ( G
1
+ G
2
) v
C 1
+ G
2 v
C 2
+ G
1 v s
• KCL at C
2
: i
C 2
+ G
2
( v
C 2
− v
C 1
) = 0
• ⇒ C
2 v ˙
C 2
= G
2 v
C 1
− G
2 v
C 2
• ⇒ a
11
= − ( G
1
+ G
2
) /C
1
, a
12
= G
2
/C
1
, a
21
= G
2
/C
2
, a
22
= − G
2
/C
2
, b
1
= G
1
/C
1
, u
1
= v s
, b
2
= 0, u
2
= 0
• ( a
11
+ a
22
) = Trace( A ) = T = − (( G
1
+ G
2
) /C
1
+ G
2
/C
2
)
• ( a
11 a
22
− a
12 a
21
) = det A = D = G
1
G
2
/C
1
C
2
• By using Case 2 : u s
= − a
11 b
2 u
2
+ a
21 b
1 u
1
+ b
2 u
2
= ( G
1
G
2
/C
1
C
2
) v s
• ¨
C 2
+ (( G
1
+ G
2
) /C
1
+ G
2
/C
2 v
C 2
+ ( G
1
G
2
/C
1
C
2
) v
C 2
= ( G
1
G
2
/C
1
C
2
) v s
• Given initial conditions v
C 1
(0) and v
C 2 the second state equation given above : v
C 2
(0) from
• ⇒ C
2 v ˙
C 2
(0) = G
2 v
C 1
(0) − G
2 v
C 2
(0)
• Given the parameters G i
, R i and the source term v s
, we can solve this
ODE to find v
C 2
.
• Then we can find ˙
C 2 by differentiation. By using the second equation, we can find v
C 1
...etc...
27

• Example on p. 334 , Fig. 7-14
+ v
C1 v s
R1
+ v
1
−
R2
+ v
2
−
C1
C2
−
+
−
−
+ v o
• KCL for Node 1 : i
C 1
+ G
1
( v
1
− v s
) + G
2
( v
1
− v
2
) = 0
• KCL for Node 2 : i
C 2
+ G
2
( v
2
− v
1
) = 0
• Op-amp eqn : v
2
= v
+
= v
−
= v o
−→ v
C 1
= v
1
− v o
= v
1
− v
2
, v
C 2
= v
2
• C
1 v ˙
1
= C
1 v ˙
1
− C
1 v ˙
2
= − ( G
1
+ G
2
) v
1
+ G
2 v
2
+ G
2 v s
• C
2 v ˙
2
= G
2 v
1
− G
2 v
2
• v ˙
2
= ( G
2
/C
2
) v
1
− ( G
2
/C
2
) v
2
• C
1 v ˙
1
= C
1
( G
2
/C
2
) v
1
− C
1
( G
2
/C
2
) v
2
− ( G
1
+ G
2
) v
1
+ G
2 v
2
+ G
2 v s
• v ˙
1
= ( G
2
/C
2
− (( G
1
+ G
2
) C
1
) ) v
1
+ ( ( G
2
/C
1
) − ( G
2
/C
2
)) v
2
+ ( G
2
/C
1
) v s
• applying Case 2, we obtain :
• ¨
2
+ (( G
1
+ G
2
) /C
1 2
+ ( G
1
G
2
/C
1
C
2
) v
2
= ( G
1
G
2
/C
1
C
2
) v s
• As before, given the parameters, and the initial conditions, we can solve this ODE. Then, we can find v
1
. Then we can find all of the remaining voltages and currents ...etc...
28