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 10e
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
k1 + k 2 e
−
t
τ
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 also consist of a natural response and a forced response:
v(t) = vf(t)+vn(t)
The complete response must satisfy both the initial
conditions and the “final conditions” of the forced response.
Engr228 - Chapter 8, Nilsson 10e
2
Source-Free Parallel RLC Circuits
We will first study the natural response of second-order circuit
by studying a source-free parallel RLC circuit.
iC(t)
Parallel RLC Circuit
C
iR(t)
+
R v(t) L
-
iL(t)
iR + iL + iC = 0
v(t ) 1
dv (t )
+ ∫ v(t )dt + C
=0
R L
dt
d 2 v(t ) 1 dv (t ) 1
C
+
+ v(t ) = 0
dt 2
R dt
L
Second-order
Differential equation
Source-Free Parallel RLC Circuits
This second-order differential equation can be solved by
assuming the form of a solution.
st
Assume a solution is of the form: v(t ) = Ae
C
d 2v(t ) 1 dv (t ) 1
+
+ v(t ) = 0
dt 2
R dt
L
1
1
Ase st + Ae st = 0
R
L
1
1
Ae st (Cs 2 + s + ) = 0
R
L
1
1
Cs 2 + s + = 0
R
L
CAs 2 e st +
which means
• This is known as the characteristic equation.
Engr228 - Chapter 8, Nilsson 10e
3
Source-Free Parallel RLC Circuits
Cs 2 +
1
1
s+ =0
R
L
Using the quadratic formula, we get
2
s1 = −
1
1
⎛ 1 ⎞
+ ⎜
⎟ −
2 RC
⎝ 2 RC ⎠ LC
2
1
1
⎛ 1 ⎞
s2 = −
− ⎜
⎟ −
2 RC
⎝ 2 RC ⎠ LC
Both v(t ) = A1e s t and v(t ) = A2e s t are solutions to the equation.
1
2
st
Therefore, the complete solution is v(t ) = A1e 1 + A2e
s 2t
A1 and A2 are determined by the initial conditions of the circuit.
Source-Free Parallel RLC Circuits
2
1
1
⎛ 1 ⎞
s1, 2 = −
± ⎜
⎟ −
2 RC
⎝ 2 RC ⎠ LC
1
Define resonant frequency:
ω0 =
LC
1
Define damping factor:
α=
2 RC
Then:
s1, 2 = −α ± α 2 − ω02
We will now divide the circuit response into three cases
according to the sign of the term under the radical.
Engr228 - Chapter 8, Nilsson 10e
4
Second-Order Differential Equation Solution
2
1
1
⎛ 1 ⎞
s1, 2 = −
± ⎜
⎟ −
2 RC
⎝ 2 RC ⎠ LC
s1, 2 = −α ± α 2 − ω02
• Case 1 – overdamped response
α > ω0 (inside the square root is a positive value)
• Case 2 – critically damped response
α = ω0 (inside the square root is zero)
• Case 3 – underdamped response
α < ω0 (inside the square root is a negative value)
Circuit Responses
Engr228 - Chapter 8, Nilsson 10e
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Overdamped Case (α > ω0 )
Find v(t) in the circuit
at the right.
Given initial conditions:
vc (0) = 0, iL(0) = -10A
α=
1
= 3.5
2 RC
ω0 =
1
= 6
LC
α > ω0 therefore this is an overdamped case
s1, 2 = −α ± α 2 − ω02
s1 = -1, s2 = -6
Overdamped Case - continued
−t
The solution is in form of v(t ) = A1e + A2e
−6t
Use initial conditions to find A1 and A2
From vc (0) = 0 at t = 0:
v(0) = 0 = A1e0 + A2e0 = A1 + A2
From KCL taken at t = 0:
iR + iL + iC = 0
v ( 0)
dv (t )
+ (−10) + C
=0
R
dt t =0
0
1
+ (−10) +
− A1e −t − 6 A2 e −6t
R
42
(− A1 − 6 A2 ) = 420
(
Engr228 - Chapter 8, Nilsson 10e
)
t =0
=0
6
Overdamped Case - continued
Solving the two equations, we get A1 = 84 and A2 = -84
The solution is
v(t ) = 84e −t − 84e −6t = 84(e −t − e −6t )V
v(t)
v(t ) = 84(e −t − e −6t )
t
Example: Overdamped RLC Circuit
Find vC(t) for t > 0
vC(t) = 80e−50,000t − 20e−200,000t V for t > 0
Engr228 - Chapter 8, Nilsson 10e
7
Critically Damped Case (α = ω0 )
Find v(t) in the circuit
at the right.
iC(t)
1/42f
Given initial conditions:
vc (0) = 0, iL(0) = -10A
α=
1
= ω0 =
2 RC
+
v(t)
-
iR(t)
iL(t)
8.573Ω 7H
1
= 2.45
LC
Critically damped when α = ω0
s1 = s2 = -2.45
The complete solution in this case is of the form
v(t ) = A1test + A2e st
Critically Damped Case - continued
Use initial conditions to find A1 and A2
From vc (0) = 0 at t = 0:
v(0) = 0 = A1 (0)e0 + A2e0 = A2
−2.45t
Therefore A2 = 0 and the solution is reduced to v(t ) = A1te
Find A1 from KCL at t = 0:
iR + iL + iC = 0
v(0)
dv (t )
+ (−10) + C
=0
R
dt t =0
0
1
+ (−10) +
A1t (−2.45)e − 2.45t + A1e − 2.45t
R
42
1
− 10 + ( A1 ) = 0
42
(
Engr228 - Chapter 8, Nilsson 10e
)
t =0
=0
8
Critically Damped Case - continued
Solving the equation, A1 = 420
The solution is
v(t ) = 420te−2.45tV
v(t)
v(t ) = 420te−2.45t
t
Critically Damped Example
Find R1 such that the circuit is critically damped for t > 0 and R2
so that v(0)=2V.
Answer: R1 = 31.63 kΩ, R2=0.4Ω
Engr228 - Chapter 8, Nilsson 10e
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Underdamped Case (α < ω0 )
s1, 2 = −α ± α 2 − ω02
For the underdamped case, the term inside the bracket will be
negative and s will be a complex number.
Define
ωd = ω02 − α 2
s1, 2 = −α ± jωd
Then
v(t ) = A1e( −α + jωd )t + A2e( −α − jωd )t
v(t ) = e −αt ( A1e jωd t + A2e − jωd t )
Underdamped Case - continued
v(t ) = e −αt ( A1e jωd t + A2e − jωd t )
Using Euler’s Identity
e jθ = cosθ + j sin θ
v(t ) = e −αt ( A1 cos ωd t + jA1 sin ωd t + A2 cos ωd t − jA2 sin ωd t )
v(t ) = e −αt (( A1 + A2 ) cos ωd t + j ( A1 − A2 ) sin ωd t
v(t ) = e −αt ( B1 cos ωd t + B2 sin ωd t )
v(t ) = e −αt ( B1 cos ωd t + B2 sin ωd t )
Engr228 - Chapter 8, Nilsson 10e
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Mechanical Analogue
A pendulum is an example of an underdamped second-order
mechanical system.
displacement(t)
t
Underdamped Case - Example
iC(t)
Find v(t) in the circuit
at the right.
+
v(t)
-
1/42f
Given initial conditions:
vc (0) = 0, iL(0) = -10A
α=
1
=2
2 RC
ω0 =
iR(t)
10.5Ω
iL(t)
7H
1
= 6
LC
α < ω0 therefore, this is an underdamped case
ωd = ω02 − α 2 = 2
v(t) is of the form
Engr228 - Chapter 8, Nilsson 10e
v(t ) = e −2t ( B1 cos 2t + B2 sin 2t )
11
Underdamped Case - continued
Use initial conditions to find B1 and B2
From vc (0) = 0 at t = 0:
v(0) = e0 ( B1 cos 0 + B2 sin 0) = B1
Therefore B1 = 0 and the solution is reduced to
v(t ) = e −2t ( B2 sin 2t )
Find B2 from KCL at t = 0:
iR + iL + iC = 0
v(0)
dv (t )
+ (−10) + C
=0
R
dt t =0
0
1
+ (−10) +
2 B2 e − 2t cos 2t − 2 B2 e − 2t sin 2t
R
42
1
− 10 + ( 2 B2 ) = 0
42
(
)
t =0
=0
Underdamped Case - continued
Solving: B2 = 210 2 = 297
The solution is: v(t ) = 297 e −2t sin 2tV
v(t)
v(t ) = 297 e −2t sin 2tV
t
Engr228 - Chapter 8, Nilsson 10e
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Underdamped Example
Find iL for t > 0
iL = e−1.2t (2.027 cos 4.75t + 2.561 sin 4.75t) A
Summary of Transient Responses
Engr228 - Chapter 8, Nilsson 10e
13
Source - Free Series RLC Circuit
i(t)
+ vR(t) - R
+
vC(t)
vL(t) L
+
-
C
vR + vL + vC = 0
di (t ) 1
+ ∫ i (t )dt = 0
dt
C
d 2i (t )
di (t ) 1
L
+
R
+ i (t ) = 0
dt 2
dt
C
i (t ) R + L
Series and Parallel RLC Circuits
Parallel RLC
Series RLC
2
d v(t ) 1 dv (t ) 1
C
+
+ v(t ) = 0
dt 2
R dt
L
v(t ) = A1e s1t + A2e s2t
1
1
⎛ 1 ⎞
=−
± ⎜
⎟ −
2 RC
⎝ 2 RC ⎠ LC
s1, 2 = −α ± α 2 − ω02
α=
1
2 RC
Engr228 - Chapter 8, Nilsson 10e
ω0 =
L
d i(t )
di (t ) 1
+R
+ i(t ) = 0
2
dt
dt
C
i(t ) = A1e s1t + A2e s2t
2
s1, 2
2
1
LC
2
R
1
⎛ R⎞
s1, 2 = −
± ⎜ ⎟ −
2L
⎝ 2 L ⎠ LC
s1, 2 = −α ± α 2 − ω02
α=
R
2L
ω0 =
1
LC
14
Series RLC Circuit Solution
ω0 = 1
Define
LC
Then if:
α > ω0 (overdamped):
€
€
α = ω0 (critically damped):
α < ω0 (underdamped):
s1 = −α + α 2 − ω 02
R
α=
2L
s2 = −α − α 2 − ω 02
€i (t )
= A1e s1t + A2e s2t
i(t ) = e −αt ( A1t + A2 )
i (t ) = e −αt (B1 cos(ωd t ) + B2 sin (ωd t ))
Textbook Problem 8.49 Nilsson 10th
The circuit contains no initial energy. Find vo(t) for t ≥ 0
vo(t) = 16 - 16e−400tcos300t − 21.33e−400tsin300t V
Engr228 - Chapter 8, Nilsson 10e
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Example: Initial Conditions
Find the labeled voltages and currents at t = 0- and t = 0+ .
Answer:
iR(0−) = −5 A
iL (0−) = 5 A
iC (0−) = 0 A
vR(0−) = −150 V
vL (0−) = 0 V
vC (0−) = 150 V
iR(0+) = −1 A
iL (0+) = 5 A
iC (0+) = 4 A
vR(0+) = −30 V
vL (0+) = 120 V
vC (0+) = 150 V
Summary: Solving RLC Circuits
1. Identify the series or parallel RLC circuit;
2. Find α and ω0;
3. Determine whether the circuit is overdamped, critically
damped, or underdamped;
4. Assume a solution (natural response + forced response);
A1e s1t + A2e s2t + V f
A1test + A2e st + V f
e−αt ( B1 cos ωd t + B2 sin ωd t ) + V f
Overdamped
Critically damped
Underdamped
5. Find A, B, and Vf using initial and final conditions.
Engr228 - Chapter 8, Nilsson 10e
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Forced Response Example
t = 0 sec
Find Vc (t)
30
S1
4A
iL(t)
3H
+
Vc
1/27f
5A
-
α=
R
=5
2L
ω0 =
1
=3
LC
s1, 2 = −α ± α 2 − ω02 = −1,−9
This is an overdamped case, so the solution is of the form
A1e −t + A2e −9t + V f
Forced Response Example - continued
vC (t ) = A1e−t + A2e−9t + V f
The initial conditions are vC(0) = 150 V and iL(0) = -5 A
As time goes to infinity, vC(∞) = 150 V and iL(∞) = -9 A
Using vC(0) = 150 V , we get
150 = A1 + A2 + V f
Using vC(∞) = 150 V , we get 150 = 0 + 0 + V f
Therefore, Vf = 150 and A1+A2 = 0
Engr228 - Chapter 8, Nilsson 10e
17
Forced Response Example - continued
Also, initial conditions determine that iC(0) = 4A
dv (t ) 1
=
(− A1e −t − 9 A2 e −9t )
dt
27
1
iC (0) = 4 =
(− A1e 0 − 9 A2 e 0 )
27
108 = − A1 − 9 A2
iC (t ) = C
A1 = 13.5, A2 = -13.5
Therefore,
vC (t ) = 13.5e −t − 13.5e −9t + 150 V
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 10e
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