Download Second Order Circuits I

Circuit Theory
Chapter 8
Second-Order Circuits
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Second-Order Circuits
Chapter 8
8.1
8.3
8.5
8.2
8.4
Examples of 2nd order RLC circuit
The source-free parallel RLC circuit
Step response of a parallel RLC
The source-free series RLC circuit
Step response of a series RLC circuit
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8.1 Examples of Second
Order RLC circuits (1)
What is a 2nd order circuit?
A second-order circuit is characterized by a secondorder differential equation. It consists of resistors
and the equivalent of two energy storage elements.
RLC Series
RLC Parallel
RL T-config
RC Pi-config
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8.3 Source-Free Parallel
RLC Circuits (1)
0
Let
1
i (0)  I 0   v(t )dt
L
v(0) = V0
Apply KCL to the top node:
t
v 1
dv
  vdt  C  0
R L 
dt
Taking the derivative with
respect to t and dividing by C
The 2nd
order of
expression
d 2 v 1 dv 1


v0
2
RC dt LC
dt
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8.3 Source-Free Parallel
RLC Circuits (2)
There are three possible solutions for the
following 2nd order differential equation:
d 2v
dv
 2
 02v  0
2
dt
dt
where  
1
2RC
and 0 
1
LC
1. If  > o, over-damped case
v(t )  A1 e s1t  A2 e s2t where
s1, 2      2  0
2
2. If  = o, critical damped case
v(t )  ( A2  A1t ) e t
where
s1, 2   
3. If  < o, under-damped case
v(t )  e t ( B1 cos d t  B2 sin d t ) where
d  02   2
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8.3 Source-Free Parallel
RLC Circuits (3)
Example 3
Refer to the circuit shown below.
Find v(t) for t > 0.
• Please refer to lecture or textbook for more detail elaboration.
Answer:
v(t) = 66.67(e–10t – e–2.5t) V
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8.5 Step-Response Parallel
RLC Circuits (1)
• The step response
is obtained by the
sudden application
of a dc source.
The 2nd
order of
expression
d 2i 1 di
i
Is



2
dt RC dt LC LC
It has the same form as the equation for source-free parallel
RLC circuit.
• The same coefficients (important in determining the
frequency parameters).
• Different circuit variable in the equation.
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8.5 Step-Response Parallel
RLC Circuits (2)
The solution of the equation should have two components:
the transient response vt(t) & the steady-state response vss(t):
i(t )  it (t )  iss (t )
 The transient response it is the same as that for source-free case
it (t )  A1e s1t  A2 e s2t
(over-damped)
it (t )  ( A1  A2t )e t
(critical damped)
it (t )  e t ( A1 cos d t  A2 sin d t )
(under-damped)
 The steady-state response is the final value of i(t).
 iss(t) = i(∞) = Is
 The values of A1 and A2 are obtained from the initial conditions:
 i(0) and di(0)/dt.
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8.5 Step-Response Parallel
RLC Circuits (3)
Example 5
Find i(t) and v(t) for t > 0 in the circuit shown in
circuit shown below:
• Please refer to lecture or textbook for more detail elaboration.
Answer:
v(t) = Ldi/dt = 5x20sint = 100sint V
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8.2 Source-Free Series
RLC Circuits (1)
• The solution of the source-free
series RLC circuit is called as the
natural response of the circuit.
• The circuit is excited by the energy
initially stored in the capacitor and
inductor.
The 2nd
order of
expression
d 2 i R di
i


0
2
L dt LC
dt
How to derive and how to solve?
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8.2 Source-Free Series
RLC Circuits (3)
There are three possible solutions for the following
2nd order differential equation:
d 2 i R di
i


0
2
L dt LC
dt
=>
d 2i
di
2

2



0i 0
2
dt
dt
where

R
2L
and 0 
1
LC
General 2nd order Form
The types of solutions for i(t) depend
on the relative values of  and .
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8.2 Source-Free Series
RLC Circuits (4)
There are three possible solutions for the following
2nd order differential equation:
d 2i
di
2

2



0i 0
2
dt
dt
1. If  > o, over-damped case
i(t )  A1e s1t  A2e s2t
2
where s1, 2       0
2
2. If  = o, critical damped case
i(t )  ( A2  A1t )et
where
s1, 2   
3. If  < o, under-damped case
i (t )  e t ( B1 cos  d t  B2 sin  d t ) where d  02   2
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8.2 Source-Free Series
RLC Circuits (5)
Example 1
If R = 10 Ω, L = 5 H, and
C = 2 mF in 8.8, find α,
ω0, s1 and s2.
What type of natural
response will the circuit
have?
•
Please refer to lecture or textbook for more detail elaboration.
Answer: underdamped
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8.2 Source-Free Series
RLC Circuits (6)
Example 2
The circuit shown below
has reached steady state
at t = 0-.
If the make-before-break
switch moves to position b
at t = 0, calculate i(t) for
t > 0.
• Please refer to lecture or textbook for more detail elaboration.
Answer: i(t) = e–2.5t[5cos1.6583t – 7.538sin1.6583t] A
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8.4 Step-Response Series
RLC Circuits (1)
• The step response
is obtained by the
sudden application
of a dc source.
The 2nd
order of
expression
vs
d 2 v R dv v



2
L dt LC LC
dt
The above equation has the same form as the equation for
source-free series RLC circuit.
• The same coefficients (important in determining the
frequency parameters).
• Different circuit variable in the equation.
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8.4 Step-Response Series
RLC Circuits (2)
The solution of the equation should have two components:
the transient response vt(t) & the steady-state response vss(t):
v(t )  vt (t )  vss (t )
 The transient response vt is the same as that for source-free case
vt (t )  A1e s1t  A2 e s2t
(over-damped)
vt (t )  ( A1  A2t )e t
(critically damped)
vt (t )  e t ( A1 cos d t  A2 sin d t ) (under-damped)
 The steady-state response is the final value of v(t).
 vss(t) = v(∞)
 The values of A1 and A2 are obtained from the initial conditions:
 v(0) and dv(0)/dt.
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8.4 Step-Response Series
RLC Circuits (3)
Example 4
Having been in position for a long time, the
switch in the circuit below is moved to position b
at t = 0. Find v(t) and vR(t) for t > 0.
•
Please refer to lecture or textbook for more detail elaboration.
Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V
vR(t)= [2.31sin3.464t]e–2t V
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1)determine the qualitative form of the response vc(t) as
being either overdamped, underdamped, or critically damped.
Overdamped
Underdamped
Underdamped
Critically damped
2) Determine the initial value of Vr(t), and
dVr(t)/dt, and the final value of Vr(t)
http://www.rose-hulman.edu/CLEO/video/player.php?id=34&embed
Vr(0) = 6V;
dVr(0)/dt = 3V/s;
Vr(infinity) = 0V
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3) For both the inductor current and voltage, determine their
initial values, the initial values of their derivatives, and their
final values. The capacitor has 9 J of stored energy before the
switch closes.
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4)Plot the inductor current iL(t) and inductor voltage vL(t) for time t =
-0.1 seconds to t = 3 seconds. Confirm your results using a circuit
simulator.
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iL(t) = 2 + 14.5 sin(4.15t) e
-1.67t
, t>=0
http://www.rose-hulman.edu/CLEO/browse/?path=1/2/76/89/224/253
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