1 ekt101 electric circuit theory chapter 5 first-order and second circuits
TRANSCRIPT
1
EKT101Electric Circuit
TheoryChapter 5
First-Order and Second Circuits
2
First-Order and Second CircuitsChapter 5
5.1Natural response of RL and RC Circuit5.2Force response of RL and RC Circuit5.3Solution of natural response and force
response in RL and RC Circuit5.4 Natural and force response in series
RLC Circuit5.5Natural and force response in parallel
RLC Circuit
3
5.1 Natural response of RL and RC circuit (1)
• A first-order circuit is characterized by a first-order differential equation.
• Apply Kirchhoff’s laws to purely resistive circuit results in algebraic equations.
• Apply the laws to RC and RL circuits produces differential equations.
Ohms law Capacitor law
0 dt
dvC
R
v0 CR iiBy KCL
4
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.
• 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 t and slower for large t.
CRTime constantDecays more slowly
Decays faster
5
5.1 Natural response of RL and RC circuit (3)
The key to working with a source-free RC circuit is finding:
1. The initial voltage v(0) = V0 across the capacitor.
2. The time constant = RC.
/0)( teVtv CRwhere
DC source disconnected
6
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
7
Solution 1
vC(0) = 30 V.
= vC = 30e–0.25t V
vx = 10e–0.25t
io = –2.5e–0.25t A
8
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
9
Solution 2
1/6 F
V(t) = 8e–2t V
10
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)
0 RL vvBy KVL
0 iRdt
diL
Inductors law Ohms law
dtL
R
i
di LtReIti /
0 )(
11
5.1 Natural response of RL and RC circuit (7)
• 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 t and slower for large t.• The general form is very similar to a RC source-free circuit.
/0)( teIti
R
L
A general form representing a RL
where
12
5.1 Natural response of RL and RC circuit(8)
/0)( teIti
R
L
A RL source-free circuit
where /0)( teVtv RC
A RC source-free circuit
where
Comparison between a RL and RC circuit
13
5.1 Natural response of RL and RC circuit(9)
The key to working with a source-free RL circuit is finding:
1. The initial voltage i(0) = I0 through the inductor.
2. The time constant = L/R.
/0)( teIti
R
Lwhere
14
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
Solution 3find rth
15
1
5
3
?
? ?
16
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
17
t=0, SC
i(0) = current division ??
18
Unit-Step Function (1)• The unit step function u(t) is 0 for negative
values of t and 1 for positive values of t.
0,1
0,0)(
t
ttu
o
oo tt
ttttu
,1
,0)(
o
oo tt
ttttu
,1
,0)(
19
Unit-Step Function (2)
1. voltage source.
2. for current source:
Represent an abrupt change for:
20
• Initial condition: v(0-) = v(0+) = V0
• Applying KCL,
or
• Where u(t) is the unit-step function
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.
0)(
R
tuVv
dt
dvc s
)(tuRC
Vv
dt
dv s
21
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:
0)(
0)(
/0
0
teVVV
tVtv
tss
Final value at t -> ∞
Initial value at t = 0
Source-free Response
Complete Response = Natural response + Forced Response (stored energy) (independent source)
= V0e–t/τ + Vs(1–e–t/τ)
22
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 .
/ )]( )0( [ )( )( tevvvtv 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.
23
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.
5.2 Force response of RC Circuit(4)
• Please refer to lecture or textbook for more detail elaboration.
Answer: and v(0.5) = 0.5182V515)( 2 tetv
Solution Ex 5• 1 -The initial capacitor voltage v(0).
• 2- final capacitor voltage v() – Use KCL at nod to get v(∞)
• 3- time constant . – Use Rth =RC
24
25
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 currenti(0-) = i(0+) = Io
• Final inductor current i(∞) = Vs/R
• Time constant t = L/R
)()()( tueR
VI
R
Vti
ts
os
26
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 .
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.
/ )]( )0( [ )( )( teiiiti
27
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.
5.2 Force response of RL Circuit(3)
• Please refer to lecture or textbook for more detail elaboration.
Answer: teti 102)(
Example 6 (solution)
28
30V
Apply source transformation
Ai 3)0(
30
2A
teti 102)(
5.4 NATURAL AND FORCE RESPONSE IN SERIES RLC CIRCUIT
Second order circuit
29
30
Examples of Second Order RLC circuits (1)
What is a 2nd order circuit?
A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.
RLC Series RLC Parallel RL T-config RC Pi-config
31
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.
02
2
LC
i
dt
di
L
R
dt
idThe 2nd order of expression
How to derive and how to solve?
32
Source-Free Series RLC Circuits (2)
• At t=0,
• So,
• Eliminate integral, differentiate to t, rearrange
33
Source-Free Series RLC Circuits (3)
There are three possible solutions for the following 2nd order differential equation:
02
2
LC
i
dt
di
L
R
dt
id
The types of solutions for i(t) depend on the relative values of a and .w
=> 02 202
2
idt
di
dt
id LC
andL
R 1
2 0
General 2nd order Form
where
34
02 202
2
idt
di
dt
id
There are three possible solutions for the following 2nd order differential equation:
1. If a > wo, over-damped casetsts eAeAti 21
21)( 20
22,1 swhere
2. If a = wo, critical damped casetetAAti )()( 12 2,1swhere
3. If a < wo, under-damped case
)sincos()( 21 tBtBeti ddt
where 220 d
Source-Free Series RLC Circuits (4)
35
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
36
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
t>0
under-damped case
38
5.4 NATURAL AND FORCE RESPONSE IN PARALLEL RLC CIRCUIT
39
40
Source-Free Parallel RLC Circuits (1)
The 2nd order of expression
011
2
2
vLCdt
dv
RCdt
vd
0
0 )(1
)0( dttvL
IiLet
v(0) = V0
Apply KCL to the top node:
t
dt
dvCvdt
LR
v0
1
Taking the derivative with respect to t and dividing by C
41
LCRCv
dt
dv
dt
vd 1and
2
1 where0 2 0
202
2
There are three possible solutions for the following 2nd order differential equation:
1. If a > wo, over-damped casetsts eAeAtv 21 )( 21 2
02
2,1 swhere
2. If a = wo, critical damped casetetAAtv )( )( 12
2,1swhere
3. If a < wo, under-damped case
)sincos()( 21 tBtBetv ddt
where 220 d
Source-Free Parallel RLC Circuits (2)
42
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
43
44
: v(t) = 66.67(e–10t – e–2.5t) V
45
Step-Response Series RLC Circuits (1)
• The step response is obtained by the sudden application of a dc source.
The 2nd order of expression LC
v
LC
v
dt
dv
L
R
dt
vd s2
2
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.
46
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):
)()()( tvtvtv sst
The transient response vt is the same as that for source-free case
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.
tstst eAeAtv 21
21)( (over-damped)t
t etAAtv )()( 21 (critically damped)
)sincos()( 21 tAtAetv ddt
t (under-damped)
47
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
48
49
50
Step-Response Parallel RLC Circuits (1)
• The step response is obtained by the sudden application of a dc source.
The 2nd order of expression
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.
LC
I
LC
i
dt
di
RCdt
id s1
2
2
51
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):
)()()( tititi sst
The transient response it is the same as that for source-free case
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.
tstst eAeAti 21
21)( (over-damped)t
t etAAti )()( 21 (critical damped)
)sincos()( 21 tAtAeti ddt
t (under-damped)
52
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
53
v(t) = Ldi/dt = 5x20sint = 100sint V