Secong Order Circuit

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Transcript Secong Order Circuit

EKT101
Electric Circuit
Theory
Chapter 5
First-Order and Second
Circuits
1
First-Order and Second Circuits
Chapter 5
5.1 Natural response of RL and RC Circuit
5.2 Force response of RL and RC Circuit
5.3 Solution of natural response and force
response in RL and RC Circuit
5.4 Natural and force response in series
RLC Circuit
5.5 Natural and force response in parallel
RLC Circuit
2
5.1 Natural response of RL and
RC circuit (1)
• A first-order circuit is characterized by a firstorder differential equation.
By KCL
iR  iC  0
Ohms law
v
dv
C
0
R
dt
Capacitor law
• Apply Kirchhoff’s laws to purely resistive circuit results in
algebraic equations.
• Apply the laws to RC and RL circuits produces differential
equations.
3
5.1 Natural response of RL and
RC circuit (2)
• The natural response of a circuit refers to the behavior
(in terms of voltages and currents) of the circuit itself,
with no external sources of excitation.
Time constant
RC
Decays more slowly
Decays faster
• The time constant  of a circuit is the time required for the
response to decay by a factor of 1/e or 36.8% of its initial value.
• v decays faster for small  and slower for large .
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5.1 Natural response of RL and
RC circuit (3)
DC source
disconnected
The key to working with a source-free RC circuit is
finding:
v (t )  V0 e  t / 
where
RC
1. The initial voltage v(0) = V0 across the
capacitor.
2. The time constant  = RC.
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5.1 Natural response of RL and
RC circuit (4)
Example 1
Refer to the circuit below, determine vC, vx, and
io for t ≥ 0.
Assume that vC(0) = 30 V.
• Please refer to lecture or textbook for more detail elaboration.
Answer: vC = 30e–0.25t V ; vx = 10e–0.25t ; io = –2.5e–0.25t A
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Solution 1
vC(0) = 30 V.
= vC = 30e–0.25t V
vx = 10e–0.25t
io = –2.5e–0.25t A
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5.1 Natural response of RL and
RC circuit (5)
Example 2
The switch in circuit below is opened at t = 0,
find v(t) for t ≥ 0.
• Please refer to lecture or textbook for more detail elaboration.
Answer: V(t) = 8e–2t V
8
Solution 2
1/6 F
V(t) = 8e–2t V
9
5.1 Natural response of RL and
RC circuit (6)
• A first-order RL circuit consists of a inductor
L (or its equivalent) and a resistor (or its
equivalent)
By KVL
vL  vR  0
di
L
 iR  0
dt
Inductors law
Ohms law
di
R
  dt
i
L
i (t )  I 0 e
Rt / L
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5.1 Natural response of RL and
RC circuit (7)
A general form representing a RL
i (t )  I 0 e
where
•
•
•
t / 
L

R
The time constant  of a circuit is the time required for the response
to decay by a factor of 1/e or 36.8% of its initial value.
i(t) decays faster for small  and slower for large .
The general form is very similar to a RC source-free circuit.
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5.1 Natural response of RL and
RC circuit(8)
Comparison between a RL and RC circuit
A RL source-free circuit
i (t )  I 0 e
t / 
where
L

R
A RC source-free circuit
v(t )  V0 e  t /
where
  RC
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5.1 Natural response of RL and
RC circuit(9)
The key to working with a source-free
circuit is finding:
i (t )  I 0 e
 t /
where
RL
L

R
1. The initial voltage i(0) = I0 through the
inductor.
2. The time constant  = L/R.
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5.1 Natural response of RL and
RC circuit(10)
Example 3
Find i and vx in the circuit.
Assume that i(0) = 5 A.
Answer: i(t) = 5e–53t A
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3
Solution 3
find rth
1
5
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5.1 Natural response of RL and
RC circuit(11)
Example 4
For the circuit, find i(t) for t > 0.
• Please refer to lecture or textbook for more detail elaboration.
Answer: i(t) = 2e–2t A
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17
Unit-Step Function (1)
• The unit step function u(t) is 0 for negative
values of t and 1 for positive values of t.
 0,
u (t )  
1,
t0
t0
 0,
u (t  to )  
1,
t  to
t  to
 0,
u(t  to )  
1,
t   to
t   to
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Unit-Step Function (2)
Represent an abrupt change for:
1. voltage source.
2. for current source:
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5.2 Force response of RL and
RC Circuit (1)
• The step response of a circuit is its behavior when the
excitation is the step function, which may be a voltage
or a current source.
• Initial condition:
v(0-) = v(0+) = V0
• Applying KCL,
dv v  Vs u (t )
c 
0
dt
R
or
v  Vs
dv

u (t )
dt
RC
• Where u(t) is the unit-step function
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5.3 Solution of natural and force
response in RL and RC Circuit
• Integrating both sides and considering the initial
conditions, the solution of the equation is:
t0
V0
v(t )  
t / 
V

(
V

V
)
e
0
s
 s
Final value
at t -> ∞
Complete Response =
=
Natural response
(stored energy)
V0
e–t/τ
Initial value
at t = 0
+
+
t 0
Source-free
Response
Forced Response
(independent source)
Vs
(1–e–t/τ)
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5.2 Force response of RC
Circuit(3)
Three steps to find out the step response
of an RC circuit:
1. The initial capacitor voltage v(0).
2. The final capacitor voltage v() — DC
voltage across C.
3. The time constant .
v (t )  v ()  [v (0)  v ()] e
t /
Note: The above method is a short-cut method. You may also
determine the solution by setting up the circuit formula directly
using KCL, KVL , ohms law, capacitor and inductor VI laws.
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5.2 Force response of RC
Circuit(4)
Example 5
Find v(t) for t > 0 in the circuit in below.
Assume the switch has been open for a long
time and is closed at t = 0.
Calculate v(t) at t = 0.5.
• Please refer to lecture or textbook for more detail elaboration.
Answer:
v(t )  15e 2t  5 and v(0.5) = 0.5182V
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5.2 Force response of RL
Circuit(1)
• The step response of a circuit is its behavior when the
excitation is the step function, which may be a voltage or
a current source.
•
Initial current
i(0-) = i(0+) = Io
•
Final inductor current
i(∞) = Vs/R
•
Time constant  = L/R

t
Vs
Vs 
i (t )   ( I o  )e u (t )
R
R
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5.2 Force response of RL
Circuit(2)
Three steps to find out the step response
of an RL circuit:
1. The initial inductor current i(0) at t = 0+.
2. The final inductor current i().
3. The time constant .
i (t )  i ()  [i (0)  i ()] e
t /
Note: The above method is a short-cut method. You may also
determine the solution by setting up the circuit formula directly
using KCL, KVL , ohms law, capacitor and inductor VI laws.
25
5.2 Force response of RL
Circuit(3)
Example 6
The switch in the circuit shown below has been
closed for a long time. It opens at t = 0.
Find i(t) for t > 0.
• Please refer to lecture or textbook for more detail elaboration.
Answer:
i(t )  2  e 10t
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Example 6 (solution)
Apply source
transformation
30V
i ( 0)  3 A
30
2A
i(t )  2  e 10t
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Second order circuit
5.4
NATURAL AND FORCE
RESPONSE
IN SERIES RLC CIRCUIT
28
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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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 2i R di
i


0
2
dt
L dt LC
How to derive and how to solve?
30
Source-Free Series
RLC Circuits (2)
Method will be
illustrated
during the lecture
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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
a

w
0i 0
2
dt
dt
where
a
R
2L
and w0 
1
LC
General 2nd order Form
The types of solutions for i(t) depend
on the relative values of a and w.
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Source-Free Series
RLC Circuits (4)
There are three possible solutions for the following
2nd order differential equation:
d 2i
di
2

2
a

w
0i 0
2
dt
dt
1. If a > wo, over-damped case
i(t )  A1e s1t  A2e s2t
2
where s1, 2   a  a  w0
2
2. If a = wo, critical damped case
i(t )  ( A2  A1t )eat
where
s1, 2   a
3. If a < wo, under-damped case
i (t )  e at ( B1 cos w d t  B2 sin w d t ) where wd  w02  a 2
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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
34
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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t>0
under-damped case
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5.4 NATURAL AND FORCE
RESPONSE IN PARALLEL
RLC CIRCUIT
38
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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Source-Free Parallel
RLC Circuits (2)
There are three possible solutions for the following
2nd order differential equation:
d 2v
dv
 2a
 w02v  0
2
dt
dt
where a 
1
2RC
and w0 
1
LC
1. If a > wo, over-damped case
v(t )  A1 e s1t  A2 e s2t where
s1, 2   a  a 2  w0
2
2. If a = wo, critical damped case
v(t )  ( A2  A1t ) e at
where
s1, 2   a
3. If a < wo, under-damped case
v(t )  e at ( B1 cos wd t  B2 sin wd t ) where
wd  w02  a 2
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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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42
:
v(t) = 66.67(e–10t – e–2.5t) V
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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.
44
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 at
(critically damped)
vt (t )  e at ( A1 cos wd t  A2 sin wd 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.
45
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
46
47
48
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.
49
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 at
(critical damped)
it (t )  e at ( A1 cos wd t  A2 sin wd 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.
50
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
51
v(t) = Ldi/dt = 5x20sint = 100sint V
52