九十七學年度第二學期 電路學（二）授課綱要

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九十七學年度第二學期 電路學（二）授課綱要

課本： Fundamentals of Electric Circuits, 3rd edition. by Charles K. Alexander and Matthew N. O. Sadiku

計分方式：

平時： 40 % （出席： 20% ，小考、演習課： 20% ）期中考： 30 %期末考： 30 %

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內容：•Second-Order Circuits

•Sinusoids and Phasors

•Sinusoidal Steady-State Analysis

期中考•AC Power Analysis

•Three-Phase Circuits

•Magnetically Coupled Circuits

•Frequency Response 

期末考

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Chapter 8Chapter 8

Second-Order CircuitsSecond-Order Circuits

電路學電路學 (( 二二 ))

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Second-Order CircuitsSecond-Order CircuitsCChapter 8hapter 8

8.1Examples of 2nd order RCL circuit8.2 Finding Initial and Final Values8.3The source-free series RLC circuit8.4The source-free parallel RLC circuit8.5Step response of a series RLC circuit8.6Step response of a parallel RLC8.7 General Second-Order Circuits

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8.1 Examples of Second 8.1 Examples of Second Order RLC circuits (1)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

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8.2 Finding Initial and Final 8.2 Finding Initial and Final Values (1)Values (1)

v(0), i(0), dv(0)/dt, di(0)/dt, i(∞), v(∞)

The voltage of a capacitor is always continued

(0 ) (0 )v v

The current of a inductor is always continued

(0 ) (0 )i i

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8.2 Finding Initial and Final 8.2 Finding Initial and Final Values (2)Values (2)

Example 1 The switch has been closed for a long time. Find (a) v(0+), i(0+), (b) dv(0+)/dt, di(0+)/dt, (c) i(∞), v(∞)

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8.2 Finding Initial and Final 8.2 Finding Initial and Final Values (3)Values (3)

Example 2 In the circuit, 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 8.3 Source-Free Series RLC Circuits (1)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 Series 8.3 Source-Free Series RLC Circuits (2)RLC Circuits (2)

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

02 202

2

idt

di

dt

id LC

andL

R 1

2 0

General 2nd order Form

where

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02 202

2

idt

di

dt

id

There are three possible solutions for the following 2nd order differential equation:

1. If > o, over-damped casetsts eAeAti 21

21)( 20

22,1 swhere

2. If = o, critical damped casetetAAti )()( 12 2,1swhere

3. If < o, under-damped case

)sincos()( 21 tBtBeti ddt

where 220 d

8.3 Source-Free Series 8.3 Source-Free Series RLC Circuits (3)RLC Circuits (3)

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8.3 Source-Free Series 8.3 Source-Free Series RLC Circuits (4)RLC Circuits (4)

Example 3 If R = 10 Ω, L = 5 H, and C

= 2 mF in the circuit, find α, ω0, s1 and s2.

What type of natural response will the circuit have?

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8.3 Source-Free Series 8.3 Source-Free Series RLC Circuits (5)RLC Circuits (5)

Example 4 (p. 324) Find i(t) in the circuit. Assume the circuit has reached steady state at t = 0-.

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8.3 Source-Free Series 8.3 Source-Free Series RLC Circuits (6)RLC Circuits (6)

Example 5 (p.326) The circuit shown below has reached steady state at t = 0-.

If the make-before-break switch moves to position b att = 0, calculate i(t) for t > 0.

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8.4 Source-Free Parallel 8.4 Source-Free Parallel RLC Circuits (1)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

202

2

There are three possible solutions for the following 2nd order differential equation:

1. If > o, over-damped casetsts eAeAtv 21 )( 21 2

02

2,1 s where

2. If = o, critical damped casetetAAtv )( )( 12

2,1swhere

3. If < o, under-damped case

)sincos()( 21 tBtBetv ddt

where 220 d

8.4 Source-Free Parallel 8.4 Source-Free Parallel RLC Circuits (2)RLC Circuits (2)

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8.4 Source-Free Parallel 8.4 Source-Free Parallel RLC Circuits (3)RLC Circuits (3)

Example 6 (p.328)

In the parallel circuit, find v(t) for t > 0, assuming v(0) = 5 V, i(0) = 0, L = 1 H, and C = 10 mF. Consider these cases: R = 1.932 , R = 5 , and R = 6.25 .

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8.4 Source-Free Parallel 8.4 Source-Free Parallel RLC Circuits (4)RLC Circuits (4)

Example 7 (p.330)

Find v(t) for t > 0 in the RLC circuit.

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8.4 Source-Free Parallel 8.4 Source-Free Parallel RLC Circuits (5)RLC Circuits (5)

Example 8 (p.331)

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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1 1 H

1 FWhat type of response is exhibited by the circuit ?

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1 F1 H1

What type of response is exhibited by the circuit ?

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8.5 Step-Response Series 8.5 Step-Response Series RLC Circuits (1)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.

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8.5 Step-Response Series 8.5 Step-Response Series RLC Circuits (2)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()( tAtAetv ddt

t 21

(under-damped)

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8.5 Step-Response Series 8.5 Step-Response Series RLC Circuits (3)RLC Circuits (3)

Example 9 (p.333)

For the circuit, find v(t) for t > 0. Consider these cases: R = 5 , R = 4 , and R = 1 .

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8.5 Step-Response Series 8.5 Step-Response Series RLC Circuits (4)RLC Circuits (4)

Example 10 (p.336)

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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8.6 Step-Response Parallel 8.6 Step-Response Parallel RLC Circuits (1)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 s 12

2

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8.6 Step-Response Parallel 8.6 Step-Response Parallel RLC Circuits (2)RLC Circuits (2)

The solution of the equation should have two components:the transient response it(t) & the steady-state response iss(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()( tAtAeti ddt

t 21

(under-damped)

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8.6 Step-Response Parallel 8.6 Step-Response Parallel RLC Circuits (3)RLC Circuits (3)

Example 11 (p.337)

In the circuit, find i(t) and iR(t) for t > 0.

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8.6 Step-Response Parallel 8.6 Step-Response Parallel RLC Circuits (4)RLC Circuits (4)

Example 12 (p.339)

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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8.7 General Second-Order 8.7 General Second-Order Circuits (1)Circuits (1)

Given a 2nd-order circuit, we determine its step response x(t) by following steps:

1. Determine the initial conditions x(0) and dx(0)/dt.

2. Turn off the independent sources and find the form of the transient response xt(t).

3. Obtain the steady-state response xss(t) = x(∞).

4. The total response x(t) = xt(t) + xss(t)Determine the constants associated with xt(t) by imposing the initial conditions.

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Driving Source y(t) Steady-State Response xss(t)

1. Constant 2. Me bt 3. Msin(t + )

a constant Ne bt A sin t + B cos t

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8.7 General Second-Order 8.7 General Second-Order Circuits (2)Circuits (2)

Example 13 (p.339)

Find complete response v and then i in the circuit.

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8.7 General Second-Order 8.7 General Second-Order Circuits (3)Circuits (3)

Example 13 (p.341)

Find vo(t) for t > 0 in the circuit.

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8.8 Second-Order Op Amp 8.8 Second-Order Op Amp Circuits (1)Circuits (1)

Because inductors are bulky and heavy, we only consider RC second-order op amp circuit.

Example 14 (p.344)

In the op amp circuit, find vo(t) for t > 0 when vs = 10u(t) mV. Let R1 = R2 = 10 k, C1 = 20 μF, and C2= 100 μF

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8.8 Duality (1)8.8 Duality (1)Two circuits are said to be duals of one another if they are described by the same charactering equations with dual quantities interchanged.

Construct the dual circuit of a given circuit

• Place a node at the center of each mesh. Place the reference node outside the given circuit.

• Draw lines between these nodes such that each line crosses an element. Replace the element by its dual.

• Determine the polarity of voltage sources and direction of current sources. A voltage source produce clockwise mesh current has a dual current source entering to the corresponded node. A current source is in the same direction as the mesh current has a dual voltage source which positive terminal connected to the corresponded node.

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8.8 Duality (2)8.8 Duality (2)

Example 15 (p.351)

Construct the dual of the circuit.

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8.8 Duality (3)8.8 Duality (3)

Example 16 (p.351)

Obtain the dual of the circuit.

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8.9 Applications (1)8.9 Applications (1)

Example 17 (p.353)

Assuming that the switch in the figure is closed prior to t = 0-, find the inductor voltage vL(t) for t > 0.

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8.9 Applications (2)8.9 Applications (2)

Example 18 (p.353)

The output of a D/A converter is shown in the figure (a). If the RLC circuit in figure (b) is used as smoothing circuit, determine the output voltage vo(t).

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3, 4, 5, 10, 12, 14, 15, 19, 25, 29, 33, 37, 41, 43, 46, 49, 53, 55, 58, 62, 63, 65, 67, 75, 76, 78