Chapter 8
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 8 Objectives
•
Be able to determine the natural and the step response of parallel RLC circuits.
•
Be able to determine the natural and the step response of series RLC circuits.
Engr228 - Chapter 8, Nilsson 9E 1

RL and RC Circuit Review
• Transient, natural, or homogeneous response
– Fades over time.
– Resists change.
• Forced, steady-state, particular response
– Follows the input.
– Independent of time passed.
•
Assume that the total response is of the form k
1
+ k
2 e
−
τ t
The RLC Circuit
• RLC circuits contain both an inductor and a capacitor.
• These circuits have a wide range of applications, including oscillators and frequency filters.
• They can also model automobile suspension systems, temperature controllers, airplane responses, etc.
• The response of RLC circuits with DC sources and switches will consist of the natural response and the forced response: v(t) = v f
(t)+v n
(t)
The complete response must satisfy both the initial conditions and the “final conditions” of the forced response.
Engr228 - Chapter 8, Nilsson 9E 2

Source-Free RLC Circuits
•
We will first study the natural response of second-order circuit by studying a source-free parallel RLC circuit.
Parallel RLC Circuit i
R
+ i
L
+ i
C
= 0 v ( t )
+
R
1
L
v ( t ) dt + C dv ( t ) dt
= 0
C d
2 v ( t ) dt
2
1
+
R dv ( t )
+ dt
1
L v ( t ) = 0
Second-order
Differential equation
Source-Free RLC Circuits
•
This second-order differential equation can be solved by assuming the form of a solution. If the form of the solution is correct, then we can determine the unknown constants.
Assume a solution is of the form: v ( t ) = Ae st which means
C d
2 v ( t )
+ dt
2
1
R dv ( t ) dt
1
+
L v ( t ) = 0
CAs
2 e st
+
1
R
Ase st
+
1
L
Ae st
Ae st
( Cs
2
+
1
R s +
1
L
) = 0
= 0
Cs
2
+
1
R s +
1
L
= 0
•
This is known as the characteristic equation.
Engr228 - Chapter 8, Nilsson 9E 3

Source-Free RLC Circuits
Cs
2
1
+
R s +
1
L
= 0
Using the quadratic formula, we get s
1
1
= −
2 RC
+


1
2 RC


2
1
−
LC s
2
1
= −
2 RC
−



1
2 RC
 2


1
−
LC
Both v ( t ) = A
1 e s
1 t and v ( t ) = A
2 e s
2 t are solutions to the equation.
Therefore, the complete solution is v ( t ) = A
1 e s
1 t
+ A
2 e s
2 t
A
1 and A
2 are determined by the initial conditions of the circuit.
Source-Free RLC Circuits s
1 , 2
=
1
−
2 RC
±
Define resonant frequency:
Define damping factor:


1
2 RC


2
−
1
LC
1
ω
0
=
LC
α
1
=
2 RC
Then: s
1 , 2
= − α ± α 2
− ω
0
2
We will now divide the circuit response into three cases according to the sign of the term under the radical.
Engr228 - Chapter 8, Nilsson 9E 4

Second-Order Differential Equation Solution s
1 , 2
= −
1
2 RC
±


1
2 RC


2
−
1
LC s
1 , 2
= − α ± α
2
− ω
0
2
•
Case 1 – overdamped response
α > ω
0
(inside the square root is a positive value)
•
Case 2 – critically damped response
α = ω
0
(inside the square root is zero)
•
Case 3 – underdamped response
α < ω
0
(inside the square root is a negative value)
Circuit Responses
Engr228 - Chapter 8, Nilsson 9E 5

Mechanical Analogue
A pendulum is an example of an underdamped second-order mechanical system.
displacement(t) t
Overdamped Case ( α > ω
0
)
Find v(t) in the circuit at the right.
Given initial conditions: v c
(0) = 0, i
L
(0) = -10A
α
1
=
2 RC
= 3 .
5
ω
0
=
1
LC
= 6
α > ω
0 therefore this is an overdamped case s
1 , 2
= − α ± α 2
− ω
0
2 s
1
= -1, s
2
= -6
Engr228 - Chapter 8, Nilsson 9E 6

Overdamped Case - continued
The solution is in form of v ( t ) = A
1 e
− t
+ A
2 e
− 6 t
Use initial conditions to find A
1 and A
2
From v c
(0) = 0 at t = 0: v ( 0 ) = 0 = A
1 e
0
+ A
2 e
0
= A
1
+ A
2
From KCL taken at t = 0: i
R
+ i
L
+ i
C
= 0 v ( 0 )
+
R
( − 10 ) + C dv ( t ) dt t = 0
= 0
0
R
(
−
+
A
1
( − 10 )
− 6 A
2
+
1 (
−
42
)
= 420
A
1 e
− t
− 6 A
2 e
− 6 t
) t = 0
= 0
Overdamped Case - continued
Solving the two equations, we get A
1
The solution is
= 84 and A
2
= -84 v ( t ) = 84 e
− t
− 84 e
− 6 t
= 84 ( e
− t
− e
− 6 t
) V v(t) v ( t ) = 84 ( e
− t
− e
− 6 t
) t
Engr228 - Chapter 8, Nilsson 9E 7

Example: Overdamped RLC Circuit
Find v
C
(t) for t > 0 v
C
(t) = 80e −50,000t − 20e −200,000t V for t > 0
Critically Damped Case ( α = ω
0
)
Find v(t) in the circuit at the right.
Given initial conditions: v c
(0) = 0, i
L
(0) = -10A
α =
1
2 RC
= ω
0
=
1
LC
= 2 .
45
Critically damped when α = ω
0 s
1
= s
2
= -2.45
The complete solution in this case is of the form v ( t ) = A
1 te st
+ A
2 e st
Engr228 - Chapter 8, Nilsson 9E 8

Critically Damped Case - continued
Use initial conditions to find A
1 and A
2
From v c
(0) = 0 at t = 0: v ( 0 ) = 0 = A
1
( 0 ) e
0
+ A
2 e
0
= A
2
Therefore A
2
= 0 and the solution is reduced to v ( t ) = A
1 te
− 2 .
45 t
Find A
1 from KCL at t = 0: i
R
+ i
L
+ i
C
= 0 v ( 0 )
+ ( − 10 ) + C
R dv ( t ) dt t = 0
= 0
0
R
+ ( − 10 ) +
1
42
(
A
1 t ( − 2 .
45 ) e
− 2 .
45 t
− 10 +
1
42
( A
1
) = 0
+ A
1 e
− 2 .
45 t
) t = 0
= 0
Critically Damped Case - continued
Solving the equation, A
1
The solution is
= 420 v ( t ) = 420 te
− 2 .
45 t
V v(t) v ( t ) = 420 te
− 2 .
45 t t
Engr228 - Chapter 8, Nilsson 9E 9

Critically Damped Example
Find R
1 such that the circuit is critically damped for so that v(0)=2V.
t > 0 and R
2
Answer: R
1
= 31.63 k Ω , R
2
=0.4
Ω
Underdamped Case ( α < ω
0
) s
1 , 2
= − α ± α 2
− ω
0
2
For the underdamped case, the term inside the bracket will be negative and s will be a complex number.
Define ω d
= ω
0
2
− α 2
Then s
1 , 2
= − α ± j ω d v ( t ) = A
1 e
( − α + j ω d
) t
+ A
2 e
( − α − j ω d
) t v ( t ) = e
− α t ( A
1 e j ω d t
+ A
2 e
− j ω d t
)
Engr228 - Chapter 8, Nilsson 9E 10

Underdamped Case - continued v ( t ) = e
− α t
( A
1 e j ω d t
+ A
2 e
− j ω d t
)
Using Euler’s Identity e j θ
= cos θ + j sin θ v ( t ) = e
− α t
( A
1 cos ω d t + jA
1 sin ω d t + A
2 cos ω d t − jA
2 sin ω d t ) v ( t ) = e
− α t
(( A
1
+ A
2
) cos ω d t + j ( A
1
− A
2
) sin ω d t v ( t ) = e
− α t
( B
1 cos ω d t + B
2 sin ω d t ) v ( t ) = e
− α t
( B
1 cos ω d t + B
2 sin ω d t )
Underdamped Case - continued
Find v(t) in the circuit at the right.
Given initial conditions: v c
(0) = 0, i
L
(0) = -10A
α =
1
2 RC
= 2 ω
0
=
1
LC
= 6
α < ω
0 therefore, this is an underdamped case
ω d
= ω
0
2
− α 2
= 2 v(t) is of the form v ( t ) = e
− 2 t
( B
1 cos 2 t + B
2 sin 2 t )
Engr228 - Chapter 8, Nilsson 9E 11

Underdamped Case - continued
Use initial conditions to find B
1 and B
2
From v c
(0) = 0 at t = 0: v ( 0 ) = e
0
( B
1 cos 0 + B
2 sin 0 ) = B
1
Therefore B
1
= 0 and the solution is reduced to v ( t ) = e
− 2 t
( B
2 sin 2 t )
Find B
2 from KCL at t = 0: i
R
+ i
L
+ i
C
= 0 v ( 0 )
+ ( − 10 ) + C dv ( t )
R dt t = 0
= 0
0
R
+ ( − 10 ) +
1
42
(
− 10 +
1
42
(
2 B
2 e
− 2 t cos
2 B
2
) = 0
2 t − 2 B
2 e
− 2 t sin 2 t
) t = 0
= 0
Underdamped Case - continued
Solving: B
2
= 210 2 = 297
The solution is: v ( t ) = 297 e
− 2 t sin 2 tV v(t) v ( t ) = 297 e
− 2 t sin 2 tV t
Engr228 - Chapter 8, Nilsson 9E 12

Find i
L for t > 0
Underdamped Example i
L
= e −1.2t (2.027 cos 4.75t + 2.561 sin 4.75t) A
Summary of Transient Responses
Engr228 - Chapter 8, Nilsson 9E 13

Source –Free Series RLC Circuit v
R
+ v
L
+ v
C
= 0 i ( t ) R + L di ( t ) dt
1
+
C
i ( t ) dt = 0
L d
2 i ( t ) dt
2
+ R di ( t ) dt
1
+
C i ( t ) = 0
Series and Parallel RLC Circuits
Parallel RLC
C d
2 v ( t ) dt
2
1
+
R dv ( t )
+ dt
1
L v ( t ) = 0 v ( t ) = A
1 e s
1 t
+ A
2 e s
2 t
Series RLC
L d
2 i ( t ) dt
2
+ R di ( t ) dt
1
+
C i ( t ) = 0 i ( t ) = A
1 e s
1 t
+ A
2 e s
2 t s
1 , 2
=
1
−
2 RC
±


1
2 RC


2
1
−
LC s
1 , 2
=
R
−
2 L
±


R
2 L


2
−
1
LC s
1 , 2
= − α ± α
2
− ω
0
2 s
1 , 2
= − α ± α 2
− ω
0
2
α
1
=
2 RC
ω
0
=
1
LC
R
α =
2 L
ω
0
=
1
LC
Engr228 - Chapter 8, Nilsson 9E 14

Series RLC Circuit Solution
Define ω
0
=
1
LC
Then if:
α > ω
0
(overdamped):
R
α =
2 L s
1
= − α + α 2 − ω
0
2 s
2
= − α − α 2
− ω 2
0
α = ω
0
(critically damped):
α < ω
0
(underdamped): i ( t ) = A
1 e s
1 t
+ A
2 e s
2 t i ( t ) = e
− α t (
A
1 t + A
2
) i ( t ) = e − α t ( B
1 cos ( ω d t ) + B
2 sin ( ω d t ) )
Example: Initial Conditions
Find the labeled voltages and currents at t = 0 and t = 0 + .
Answer: i i i
R
(0 − ) = −5 A v
R
(0 − ) = −150 V
L
C
(0
(0 −
− ) = 5 A
) = 0 A v
L
(0 − ) = 0 V v
C
(0 − ) = 150 V i i i
R
(0 + ) = −1 A v
R
(0 + ) = −30 V
L
C
(0
(0 +
+ ) = 5 A
) = 4 A v
L
(0 + ) = 120 V v
C
(0 + ) = 150 V
Engr228 - Chapter 8, Nilsson 9E 15

Example: Initial Slopes
Find the first derivatives of the labeled voltages and currents at t = 0 + .
Answer: di
R
/dt(0 + ) = −40 A/s dv
R
/dt(0 + ) = -1200 V/s di
L
/dt(0 + ) = 40 A/s dv
L
/dt(0 + ) = -1092 V/s di
C
/dt(0 + ) = -40 A/s dv
C
/dt(0 + ) = 108 V/s
The Lossless LC Circuit
• The resistor in the RLC circuit serves to dissipate initial stored energy.
• When this resistor becomes 0 in the series RLC or infinite in the parallel RLC, the circuit will oscillate.
•
For t > 0, v(t) =2 sin 3t if i(0) = -1/6 A and v(0) = 0 V
Engr228 - Chapter 8, Nilsson 9E 16

Summary: Solving RLC Circuits
1. Identify the series or parallel RLC circuit.
2. Find α and ω
0
.
3. Determine whether the circuit is overdamped, critically damped, or underdamped.
4. Assume a solution (natural response + forced response).
A
1 e s
1 t
+ A
2 e s
2 t
+ V f
A
1 te st
+ A
2 e st
+ V f e
− α t ( B
1 cos ω d t + B
2 sin ω d t ) + V f
Overdamped
Critically damped
Underdamped
5. Find A , B , and V f using initial and final conditions.
Find V c
(t)
Forced Response Example t = 0 sec
4A
S1 i
L
(t) 3H
30
+
Vc
-
1/27f
α
R
=
2 L
= 5 ω
0
=
1
LC
= 3 s
1 , 2
= − α ± α
2
− ω
0
2
= − 1 , − 9
This is an overdamped case, so the solution is of the form
A
1 e
− t
+ A
2 e
− 9 t
+ V f
5A
Engr228 - Chapter 8, Nilsson 9E 17

Forced Response Example - continued v
C
( t ) = A
1 e
− t
+ A
2 e
− 9 t
+ V f
The initial conditions are v
C
(0) = 150 V and i
L
(0) = -5 A
As time goes to infinity, v
C
( ∞ ) = 150 V and i
L
( ∞ ) = -9 A
Using v
C
(0) = 150 V , we get 150 = A
1
+ A
2
+ V f
Using v
C
( ∞ ) = 150 V , we get 150 = 0 + 0 + V f
Therefore, V f
= 150 and A
1
+A
2
= 0
Forced Response Example - continued
Also, initial conditions determine that i
C
(0) = 4A i
C
( t ) = C dv ( t ) dt
=
1
27
( − A
1 e
− t
− 9 A
2 e
− 9 t
) i
C
( 0 ) = 4 =
1
27
( − A
1 e
0
− 9 A
2 e
0 )
108 = − A
1
− 9 A
2
A
1
= 13.5, A
2
= -13.5
Therefore, v
C
( t ) = 13 .
5 e
− t
− 13 .
5 e
− 9 t
+ 150 V
Engr228 - Chapter 8, Nilsson 9E 18

Chapter 8 Summary
•
Showed how to determine the natural and the step response of parallel RLC circuits.
• Showed how to determine the natural and the step response of series RLC circuits.
Engr228 - Chapter 8, Nilsson 9E 19