ECE 2100 Circuit Analysis - Homepages at WMUhomepages.wmich.edu/~dlitynsk/ECE 2100 Lec PDF final/ECE...
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ECE 2100 Circuit Analysis Lesson 35 Chapter 8: Second Order Circuits Daniel M. Litynski, Ph.D.
Transcript of ECE 2100 Circuit Analysis - Homepages at WMUhomepages.wmich.edu/~dlitynsk/ECE 2100 Lec PDF final/ECE...
ECE 2100
Circuit Analysis
Lesson 35
Chapter 8: Second Order Circuits
Daniel M. Litynski, Ph.D.
ECE 2100
Circuit Analysis
Lesson 32-34
Chapter 7: First Order Circuits
(Natural response RC & RL circuits, Singularity
functions, Step response RC & RL circuits,
General Solution)
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ECE 2100 Circuit Analysis
Chapter 7
First-Order Circuits
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First-Order CircuitsChapter 7
7.1 Introduction
7.2 The Source-Free RC Circuit
7.3 The Source-Free RL Circuit
7.4 Unit-step Function
7.5 Step Response of an RC Circuit
7.6 Step Response of an RL Circuit
ECE 2100
Circuit Analysis
Lesson 35
Chapter 8: Second Order Circuits
Daniel M. Litynski, Ph.D.
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ECE 2100 Circuit Analysis
Chapter 8
Second-Order Circuits
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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Second-Order CircuitsChapter 8
8.1 Introduction
8.2 Finding Initial and Final Values
8.3 The source-free series RLC circuit
8.4 The source-free parallel RLC circuit
8.5 Step response of a series RLC circuit
8.6 Step response of a parallel RLC
8.7 General Second-Order Circuits
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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 second-order differential equation. It consists of resistorsand the equivalent of two energy storage elements.
RLC Series RLC Parallel RL T-config RC Pi-config
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8.2 Finding Initial and Final Values
Finding Initial
and Final Values of
v(0), i(0), dv(0)/dt, di(0)/dt, v(∞), i(∞)
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8.2 Finding Initial and Final Values
Key Points for Finding Initial and Final Values:
1. Polarity of v(t) and i(t) using passive sign convention (see Figs. 6.3 & 6.23)
2. Capacitor voltage is always continuous so v(0+) = v(0-) for switch at t = 0
3. Inductor current is always continuous so i(0+) = i(0-) for switch at t = 0
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8.2 Finding Initial and Final Values
Example 8.1: The switch in Fig. 8.2 has been closed for a long time. It is opened at t = 0. Find: (a) v(0+), i(0+), (b) dv(0+)/dt, di(0+)/dt, (c) v(∞), i(∞)
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8.2 Finding Initial and Final Values
Example 8.2: In the circuit of Fig. 8.5, calculate: (a) iL(0
+), vC(0+), vR(0
+), (b) diL(0
+)/dt, dvC(0+)/dt, dvR(0
+)/dt, (c) iL(∞), vC(∞), vR(∞),
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8.3 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?
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8.3 Source-Free SeriesRLC Circuits (2)
Method is illustrated
on pp. 320-324 of textbook (review)
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8.3 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 and
=> 02 2
02
2
idt
di
dt
id
LCand
L
R 1
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General 2nd order Form
where
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02 2
02
2
idt
di
dt
id
There are three possible solutions for the following 2nd order differential equation:
1. If > o, over-damped case
tstseAeAti 21
21)( 2
0
2
2,1swhere
2. If = o, critical damped casetetAAti )()( 12 2,1swhere
3. If < o, under-damped case
)sincos()( 21 tBtBeti dd
twhere
22
0d
8.3 Source-Free Series RLC Circuits (4)
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8.3 Source-Free Series RLC Circuits (5)
Example 1
If R = 10 Ω, L = 5 H, and C = 2 mF in Fig. 8.8, find α, ω0, s1 and s2.
What type of natural response will the circuit have?
• Practice Problem 8.3
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8.3 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.
• Practice Problem 8.4. Please refer to lecture or textbook for more detailed elaboration.
Answer: i(t) = e–2.5t[5cos1.6583t – 7.538sin1.6583t] A
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8.4 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
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LCRCv
dt
dv
dt
vd 1and
2
1 where0 2 0
2
02
2
There are three possible solutions for the following 2nd order differential equation:
1. If > o, over-damped case
tstseAeAtv 21 )( 21
2
0
2
2,1 swhere
2. If = o, critical damped casetetAAtv )( )( 12
2,1swhere
3. If < o, under-damped case
)sincos()( 21 tBtBetv dd
twhere
22
0d
8.4 Source-Free Parallel RLC Circuits (2)
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8.4 Source-Free Parallel RLC Circuits (3)
Example 3
Refer to the circuit shown below.
Find v(t) for t > 0.
• Practice Problem 8.6 modified. Please refer to lecture or textbook for more detailed elaboration.
Answer: v(t) = 66.67(e–10t – e–2.5t) V
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8.5 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 s
2
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.
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8.5 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.
tsts
t eAeAtv 21
21)( (over-damped)
t
t etAAtv )()( 21 (critically damped)
)sincos()( 21 tAtAetv dd
t
t (under-damped)
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8.5 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.
• Practice Problem 8.7. Please refer to lecture or textbook for more detailed 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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8.6 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 s12
2
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8.6 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(∞) = IsThe values of A1 and A2 are obtained from the initial conditions:
i(0) and di(0)/dt.
tsts
t eAeAti 21
21)( (over-damped)
t
t etAAti )()( 21 (critical damped)
)sincos()( 21 tAtAeti dd
t
t (under-damped)
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8.6 Step-Response Parallel RLC Circuits (3)
Example 5
Find i(t) and v(t) for t > 0 in the circuit shown
below:
• Practice Problem 8.8 modified. Please refer to lecture or textbook for more detail elaboration.
Answer: v(t) = Ldi/dt = 5x20sint = 100sint V
