ultimo 15 supch08.pdf

CHAPTER 8 - SECOND-ORDER CIRCUITS

List of topics for this chapter :Finding Initial ValuesThe Source-Free Series RLC CircuitThe Source-Free Parallel RLC CircuitStep Response of a Series RLC CircuitStep Response of a Parallel RLC CircuitGeneral Second-Order CircuitsSecond-Order Op Amp Circuits

FINDING INITIAL VALUES

Problem 8.1 Given the circuit shown in Figure 8.1, which has existed for a long time, find)0(v 1C , )0(v 2C , )0(i 1L , and )0(i 2L .

Figure 8.1

When a circuit reaches steady state, an inductor looks like a short circuit and a capacitor lookslike an open circuit. So, use the following circuit to find the initial values.

15

20

+ vC2

10 +

vC1

5

10 10 A

10

iL1 iL2

L1 L2

5

15

20

+ vC2

10 5

10 10 A10

iL1 iL2

+

vC1

5

Use source transformations to simplify the circuit.

Now, it is evident that

11C v)0(v = 5v)0(i 21L =

322C vv)0(v = 10v)0(i 32L =

Use nodal analysis to find 1v , 2v , and 3v .

At node 1 : 05vv

778.744.44v 211

=

+

0)vv(778.7)44.44v(5 211 =+2.222v778.7v778.12 21 =

At node 2 : 05v

5vv 212

=+

0v2v- 21 =+

2v

v1

2 =

At node 3 : 010v

10v 33

=+

0v2 3 =0v3 = volts

Substitute the equation from node 2 into the equation for node 1.

2.2222v

778.7v778.12 11 =

2.222v889.8 1 =25v1 = volts

Then, 5.12225

2v

v1

2 === volts

+ vC27.778

5 10 44.44 V

10

iL1 iL2

+

vC1

+

v3v1 v25

Therefore,=)0(v 1C volts25 =)0(i 1L amps5.2=)0(v 2C volts5.12 =)0(i 2L amps0

Problem 8.2 Given the circuit shown in Figure 8.1, which has existed for a long time, find)0(iL and )0(vC .

Figure 8.1

=)0(iL amp1 =)0(vC volts5

THE SOURCE-FREE SERIES RLC CIRCUIT

Problem 8.3 Given the circuit in Figure 8.1, which has reached steady state before theswitch closes, find )t(i for all 0t > .

Figure 8.1

Use KVL to write a loop equation for 0t > .

+

10 V

10 10 H30

i(t) 1/20 Ft = 0

+

10

10

iL

L

+

10i 5

5

+

vC

20 V C

i

0dt)t(iC1

dt)t(di

L)t(Ri =++ Multiply by L1 and differentiate with respect to time.

0)t(iLC1

dt)t(id

dt)t(di

LR

2

2

=++

Rearranging the terms and inserting the values for R, L, and C,

0)t(i2dt)t(di

3dt)t(id

2

2

=++

Assume a solution of stAe .0Ae2sAe3Aes ststst2 =++

0Ae)2s3s( st2 =++

Thus,0)2s)(1s( =++

which gives real and unequal roots at 1-s1 = and 2-s2 = .

Hence,-2t

2-t

1 eAeA)t(i +=

At += 0t , the circuit is

So,21 AA0)0(i +== or 12 -AA =

Also,

10-dtdi(0)

10)0(vL ==+ volts or -1dt)0(di=

and 210

20

1 A2-AeA2e-Adt)0(di

==

So,11121 AA2-AA2-A1- =+==

Hence,

+0 V +10 V

i(0) = 0

+ vL

-1A1 = and 1A2 =

Therefore,=)t(i (((( )))) 0tampsee- -2t-t >>>>++++


Figure 8.1

At = 0t ,0)0(i)0(i == + amps and 10)0(v)0(v CC == + volts

For 0t > ,

0dt)t(i10/11

dt)t(di

10)t(i20 =++ Multiply by 101 , differentiate with respect to time, and rearrange the terms.

0)t(idt)t(di

2dt)t(id

2

2

=++

Again, using a solution of stAe ,0AesAe2Aes ststst2 =++

0Ae)1s2s( st2 =++

Thus,0)1s( 2 =+

which gives a real and repeated root at 1-s 2,1 = .

A repeated root gives the following solution,-t

2-t

1 teAeA)t(i +=

At 0t = ,1

02

01 Ae)0(AeA0)0(i =+== or 0A1 =

+

10 V

10 10 H20

i(t) 1/10 Ft = 0

Also,

-10dt)0(di

10)0(vL == or -1dt)0(di=

and 20

20

2 Ae)0(AeA0dt)0(di

=+=

Hence,A2 = 1

Therefore,=)t(i (((( )))) 0tampste- -t >>>>


Figure 8.1

Writing a loop equation for 0t > gives,

0dt)t(i20/11

dt)t(di

10)t(i20 =++ Multiply by 101 , differentiate with respect to time, and rearrange the terms.

0)t(i2dt)t(di

2dt)t(id

2

2

=++

Again, using a solution of stAe ,0Ae2sAe2Aes ststst2 =++

0Ae)2s2s( st2 =++

Thus,0)j1s)(j1s( =+++

which gives complex roots at j1-s 2,1 #= .

Hence, we have a solutiontj)(-1

2tj)-(-1

1 eAeA)t(i ++=

+

10 V

10 10 H20

i(t) 1/20 Ft = 0

At 0t = ,21

02

01 AAeAeA0)0(i +=+== or 12 -AA =

Also,0

20

1 eAj)-1(eA)j1-(-1dt)0(di

++==

Using 12 A-A = , we get)j)(-A-1(A)j-1(1- 11 ++=11 Aj-2Aj)1j-1(1- =+=

or2j1

A1 = and 2j1-

A2 =

Therefore,t)j-1(tj)(-1 ej2

1-e

2j1)t(i + +=

= j2ee

-e)t(i-jtjt

t-

=)t(i (((( )))) 0tamps)tsin(e- t- >>>>

THE SOURCE-FREE PARALLEL RLC CIRCUIT

Problem 8.6 Given the circuit in Figure 8.1, find )t(vC for all 0t > .

Figure 8.1

Carefully DEFINE the problem.Each component is labeled, indicating the value and polarity. The problem is clear.

20 V+

vC(t)

1/10 F 10 H+

10

iL

t = 0

PRESENT everything you know about the problem.The goal of the problem is to find )t(vC for all 0t > .

There is a switch which opens at 0t = . So, there are two circuits. The first circuit, when theswitch is closed, is used to find the initial values of the capacitor and inductor. Note thatthere is a dc source. At dc, a capacitor is an open circuit and an inductor is a short circuit.Thus, we have the following circuit.

Recall that the voltage of a capacitor cannot change instantaneously and the current throughan inductor cannot change instantaneously.

)0(v)0(v)0(v CCC + == and )0(i)0(i)0(i LLL + ==

The second circuit, after the switch opens, is used to find the final solution.

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.The first circuit was simplified by applying the characteristics of capacitors and inductorswith a dc source. Ohms law should provide the answer you need when the circuit consists ofa dc voltage source and a resistor.

For the second circuit, there is only one node or one loop. In this case, the use of KCL orKVL should provide the desired equation to find the solution to the problem. Because thecomponents are in parallel, the voltage across each component is the same. So, use KCL tofind the currents in terms of the voltage Cv .

ATTEMPT a problem solution.

Begin by finding the initial values of the capacitor and inductor.

10 H

iLiC

vC

1/10 F

20 V +

10

iL(0)+vC(0)

At 0t = ,

It is evident from the circuit that0)0(vC = volts.

Using Ohms law,21020)0(iL == amps

After the switch opens, the circuit becomes

Using KCL,0ii LC =+

0dt)0v(101

dt)0v(d

101

CC

=+

Multiply both sides of the equation by 10 and differentiate both sides with respect to time.

0vdtvd

C2C

2

=+

but stAe must be a solution.

Substituting,( )

0AedtAed

st2

st2

=+

0AeAes stst2 =+0Ae)1s( st2 =+

Thus,

10 H

iLiC

vC

1/10 F

20 V +

10

iL(0)+vC(0)

0)js)(js()1s( 2 =+=+which gives complex roots at js,s 21 =

So, we have a solutionjt-

2jt

1C eAeA)t(v +=

Now, to solve for 1A and 2A .0AA)0(v 21C =+= or 12 A-A =

Now,-jt

1jt

1C eAeA)t(v =

-2)0(-i)0(i LC == amps

)ejAejA(101

dt)0(dv

101)0(i 0101CC +==

or -210A2j 1

=

Hence,

10jj10-

j220-

A1 ===

Therefore,jt-jt

C e10je10j)t(v =[ ]jt-90j-jt90jC eeee10)t(v +=[ ])90t(j-)90t(jC ee10)t(v ++ += [ ])90tcos(210)t(vC +=

0tvolts)90tcos(20)t(vC >+=

EVALUATE the solution and check for accuracy.The circuit must satisfy the conservation of energy. In this case, the energy supplied fromone component is absorbed by the other component.

Recall, from Chapter 6 Capacitors and Inductors, that the energy stored in a capacitor is2vC

21

w =

and the energy stored in an inductor is2iL

21

w =

Thus,

20)20(101

21

Cv21

w 22 =

== joules

and

20)2)(10(21

Li21

w 22 =

== joules

These two answers match. This circuit satisfies conservation of energy.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=)t(vC (((( )))) 0tvolts90tcos20 >>>>++++

Problem 8.7 [8.23] In the circuit in Figure 8.1, calculate )t(io and )t(vo for 0t > .

Figure 8.1

At 0t = ,

24)30(828)0(vo =+= and 0)0(io =

t = 0, note this is amake before break

switch so theinductor current is

not interrupted.

+

2

8

1 H

+

vo(t)

1/4 F30 V

io(t)

+

8 +

vo(t)

1/4 F30 V

io(t)2 1 H

For 0t > , we have a source-free parallel RLC circuit.

41

)4/1)(8)(2(1

RC21

===

2)41)(1(1

LC1

o ===

Since is less than o , we have an underdamped response with a damping frequency of

9843.1)16/1(422od ===and the natural response is

t-d2d1o e))tsin(A)tcos(A()t(v +=

Use the initial values to find the unknowns.1o A24)0(v ==

Also,

0dt)0(dv

C)0(i oo ==

where t-d2dd1dt-

d2d1o

e))tcos(A)tsin(A-(e))tsin(A)tcos(A(-dtdv

+++=

So,

2d1o AA-0dt

)0(dv+==

Thus,

024.39843.1)24)(41(

AA 1d

2 ==

=

Therefore,=)t(vo [[[[ ]]]] 0tvoltse))tsin(024.3)tcos(24( 4t-dd >>>>++++

8

1 H

+

vo(t)

1/4 F

io(t)

Problem 8.8 Given the circuit in Figure 8.1, which has reached steady state before theswitch closes, find )t(v for all 0t > .

Figure 8.1

=)t(v (((( )))) 0tvolts90tcose40 -t >>>>++++

STEP RESPONSE OF A SERIES RLC CIRCUIT

Problem 8.9 [8.25] A branch voltage in a series RLC circuit is described by

24v8dtdv

4dtvd2

2

=++

If the initial conditions are dt)0(dv

0)0(v == , find )t(v .

Recall the general loop equation for a series RLC circuit.

SVviRdtdi

L =++

where dtdvCi = . Then,

LCV

LCv

dtdv

LR

dtvd S2

2

=++

The complete response of this circuit has a forced response and a natural response. To find theforced response, realize that 24V8 S = which means that 3VS = . This is the forced response.Now, to find the natural response, let 0VS = and

0v8dtdv

4dtvd2

2

=++

0v)8s4s( 2 =++

10 t = 0

10

10 H4 A 1/20 F

v(t)

Thus,08s4s2 =++

which gives complex roots at 2j2-2

32164-s 2,1 =

= .

So, we have the solution-2t

21S e))t2sin(A)t2cos(A(V)t(v ++=

Solve for the unknowns.11s A3AV0)0(v +=+== or 3-A1 =

2t-21

2t-21 e))t2cos(A2)t2sin(-2A(e))t2sin(A)t2cos(-2(Adt

dv+++=

21 A2A-20dt)0(dv

+== or -3AA 12 ==

Therefore,=)t(v volts]e))t2sin()t2(cos(33[ -2t++++

Problem 8.10 [8.31] Calculate )t(i for 0t > using the circuit in Figure 8.1.

Figure 8.1Before 0t = , the capacitor acts like an open circuit while the inductor acts like a short circuit.

0)0(i = amps and 20)0(v = volts

For 0t > , the LC circuit is disconnected from the voltage source as shown below.

+

20 V5

1/16 F

+ v i

1/4 Ht = 0

1/16 F

+ v i

1/4 H

This is a lossless, source-free series RLC circuit.

0L2

R== 8)41()161(

1LC1

o =

+==

8jjj-j-s o22od ====

Since is equal to zero, we have an undamped response. Hence,)t8sin(A)t8cos(A)t(i 21 +=

where 1A0)0(i == .

So,)t8sin(A)t(i 2=

To solve for 2A , we know that

80-)20)(4-()0(vL1-)0(v

L1

dt)0(di

L ====

However,

)t8cos(A8dtdi

2=

and 2A8-80dt)0(di

==

which leads to -10A2 = .

Therefore,=)t(i 0tampssin(8t)10- >>>>

STEP RESPONSE OF A PARALLEL RLC CIRCUIT

Problem 8.11 Given the circuit in Figure 8.1, find )t(v for all 0t > .

Figure 8.1

20 4 A

t = 0

15 H 1/30 F20

v(t)

=)t(v (((( )))) 0tvoltse120e120 -2t-t >>>>


Figure 8.1

At = 0t ,0)0(vC = volts and 0)0(iL = amps.

At += 0t ,0vvv RLC === volts

So, the 4 amp current source must all go through the capacitor or4)0(iC =+ amps.

Hence,

4)0(idt)0(dv

C C == ++

or 1203014

C4

dt)0(dv

===

+

volts/sec

For 0t > , the circuit becomes

where the 20 ohm resistor in parallel with the 60 ohm resistor is equivalent to a 15 ohm resistor.

20 4 A

t = 0

15 H 1/30 F

v(t)

60

4 A 15 H 1/30 F

v

15

Using nodal analysis,

0dt)0v(d

301

dt)0v(151

150v

4- =

++

+ Multiplying by 30 and differentiating with respect to time yields

0v2dtdv

2dtvd2

2

=++

Substituting the solution of stAe produces0Ae2sAe2Aes ststst2 =++

0Ae)2s2s( st2 =++

Hence,0)j1s)(j1s(2s2s2 =+++=++


So, we have the solutiont)j(1-

2t)j(1-

1 eAeA)t(v + +=

At 0t = ,0AAeAeA)0(v 210201 =+=+= or 12 A-A =

Also,

[ ] [ ] 120Aj)-(1-Aj)(1-dt)0(dv

11 =+=

120A-2jAjAAjA- 11111 ==+j

1 e60j60j2-120

A === and -j12 e60-AA ==

Therefore,t)j(1-j-t)j(1-j ee60ee60)t(v + +=

{ }jjtj-jt-t eee60 + += )]90t(cos2[e60 -t =

=)t(v (((( )))) 0tvolts90tcose120 -t >>>>

The answer can also be written as=)t(v (((( )))) 0tvoltstsine120 -t >>>>


Figure 8.1At 0t = ,

0)0(iL = amps and 0)0(vC = volts

For complicated circuits that can be simplified using either a Thevenin or Norton equivalent, doso immediately!

Solving for ocV ,

Using nodal analysis,

080i80V

16080V ococ

=

+

0)i80V(2)80V( ococ =+0i16080V3 oc =

where 160

V80i oc

= .

Hence,

+=

160V80

16080V3 ococ

80 80 V

t = 0

30 H 1/60 F40 +

i

160

+

80i

v(t)

80 V +

i

160

+

Voc

80

+

80i

( )ococ V8080V3 +=

160V4 oc =40Voc = volts

Solving for scI ,

Using KCL,

800i80

160080

Isc

+

=

where 21

160080

i =

= amp

Hence,

121

21

i21

80i80

16080

Isc =+=+=+= amp

Finding a Norton equivalent circuit,1II scN == amp

and 40140

IV

RRsc

oc

Neq ==== ohms

Now, we can work with the following simplified circuit,

The two 40 ohm resistors in parallel convert to a 20 ohm resistor, so we can simplify the circuitfurther to become

80 80 V +

i

160

+

80iIsc

1 A 30 H 1/60 F40 40

v

Writing a single node equation results in

0dt)0v(d

601

dt)0v(301

200v

1- =

++

+ Multiplying by 60 and differentiating with respect to time leads to

0v2dtdv

3dtvd2

2

=++

Substituting the solution stAe , we get0Ae2sAe3Aes ststst2 =++

0Ae)2s3s( st2 =++Hence,

0)2s)(1s(2s3s2 =++=++which gives real and unequal roots at -1s1 = and -2s2 = .

So, we have the solution-2t

2-t

1 eAeA)t(v +=

At 0t = ,0)0(v = volts and 0)0(iL = amps.

Since 0)0(v = volts,0vR = volts and 0iR = amps.

Hence, the entire 1 amp of current flows through the capacitor. So, we have

1)0(iC = amp and dt)0(dv

C)0(iC =

Thus,

60)1(6011)0(iC

1dt

)0(dvC === volts/sec

but 60A2A-eA2eA-dt)0(dv

210

20

1 ===

1 A 30 H 1/60 F

v

20

Also,0AAeAeA)0(v 210201 =+=+=

implies that 12 -AA = .

So,60A2A- 21 =

or 60AA2A- 111 ==+

Hence,60A1 = and -60A2 = .

Therefore,=)t(v (((( )))) 0tvoltse60e60 -2t-t >>>>

The power of using a Norton equivalent circuit is clearly demonstrated by this problem.

GENERAL SECOND-ORDER CIRCUITS

Problem 8.14 Given the circuit in Figure 8.1, find )t(vC for all 0t > .

Figure 8.1

This is a second-order circuit with a forced response and a natural response.

Solve for initial conditions.At 0t = ,

)0(v)0(v CC + = = 0 and 13030)0(i)0(i LL === + amp.

Hence,

1dt

)0(dvC)0(i CC ==+

+ amp

+

30 V

30 10 H

1/20 F+

vc(t)

t = 0

or 20)1(2011)0(i

C1

dt)0(dv

CC

===+

+

volts/sec

Solving for final values,0)(iC = amps and 30)(vC = volts

which is also the forced response,30v

fC = volts.

Solving for the natural response,

The loop equation is

0vdtdi

LRi C =++

where dtdv

Ci C= .

So, we now have,

0vdtvd

LCdtdv

RC C2C

2C

=++

Substituting for the values of R, L, and C

0vdtvd

2010

dtdv

2030

C2C

2C

=++

Simplifying and rearranging the terms,

0v2dtdv

3dtvd

CC

2C

2

=++

Substituting a solution stC Aev = yields0Ae2sAe3Aes ststst2 =++

0Ae)2s3s( st2 =++

Thus,0)2s)(1s(2s3s2 =++=++

which gives real and unequal roots at 1-s1 = and -2s2 = .

30 10 H

1/20 F+

vc

i

Hence, the natural response is-2t-t

C BeAev n +=

and the complete response is-2t-t

CCC BeAe30vvv nf ++=+=

Now, we need to solve for the unknowns, A and B.0BeAe30)0(v 00C =++=

-30BA =+ or 30-AB =

Also,

202B-Ae-2B)(e)A-(0dt)0(dv 00C

==++=

Now,20-B2A =+

which leads to20-30)-A)(2(A =+

Hence,-2060A- = or -40A =

Then,1030-(-40)B ==

Therefore,=)t(vC (((( )))) 0tvoltse10e4030 -2t-t >>>>++++

Problem 8.15 Given the circuit shown in Figure 8.1, find )t(iL for all 0t > .

Figure 8.1

Carefully DEFINE the problem.Each component is labeled, indicating the value and polarity. The problem is clear.

5 A 15 15 H 1/30 F

iL(t)

t = 0

PRESENT everything you know about the problem.The goal of the problem is to find )t(iL for all 0t > .

There is a switch in the circuit. So, we will need to look at two different circuits. The firstcircuit, when the switch is open, is used to find the initial values of the capacitor andinductor. Note that there is a dc source. At dc, a capacitor is an open circuit and an inductoris a short circuit. Thus, we have the following circuit.

Recall that the voltage of a capacitor cannot change instantaneously and the current throughan inductor cannot change instantaneously.

)0(v)0(v)0(v CCC + == and )0(i)0(i)0(i LLL + ==

The second circuit, after the switch opens is used to find the final solution.

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.This is a second-order circuit with a forced response and a natural response. The forcedresponse is the response due to the current source at steady state. The natural response is theresponse of the parallel RLC circuit without the current source.

For simple resistive circuits such as the one used to find the initial values and final values, wewill use observation and Ohms law. For the parallel RLC circuit, use nodal analysis. Therewill be only one equation with this technique versus three equations with mesh analysis.

ATTEMPT a problem solution.

Solve for initial conditions using the circuit below.

5 A 15 iL(0) +

vC(0)

15 15 H 1/30 F

iL

vL

5 A

5 A 15 iL(0) +

vC(0)

At 0t = ,75)15)(5()0(v)0(v CC === + volts

0)0(i)0(i LL == +75)0(v)0(v CL == ++ volts

Thus,

51575

L)0(v

dt)0(di LL

===

++

amp/sec

Solving for final values, the forced response,

5i)(ifLL== amps

Now, solve for the natural response.

Writing the node equation,

0dt)0v(d

301

i15

0v LL

L=

++

where dtdi

15v LL =

So,

0dtid

3015

idtdi

1515

2L

2

LL

=++

Simplifying and rearranging the terms,

0i2dtdi

2dtid

LL

2L

2

=++

15 15 H 1/30 F

iL

vL

5 A 15 iL()+

vC()

Substituting the solution stL Aei = yields0Ae2sAe2Aes ststst2 =++

0Ae)2s2s( st2 =++

Thus,0)j1s)(j1s()2s2s( 2 =+++=++


Hence,( ) ( ) tj1-tj1-

L BeAei n+ +=

This means that the complete response is( ) ( ) tj1-tj1-

LLL BeAe5iii nf+ ++=+=

At 0t = ,0BA5BeAe5)0(i 00L =++=++=

-5BA =+ or 5-AB =

Also,

5B)j1(A)j1(0dt)0(diL

=+=

5jBBAjA- =+55)(-Aj5)(-AjAA- =+

5j552jA- =+

25-

2j-5j55

A =+

=

Then,

25-

525

5-AB ===

Therefore,( ) ( ) tj1-tj1-

L e)25-(e)25-(5i + ++=

+

=

2ee

e55)t(ijt-jt

t-L

[ ] 0tamps)tcos(e55)t(i -tL >= EVALUATE the solution and check for accuracy.

Check to see if the answer satisfies the initial conditions.

055)0cos(e55)0(i -0L === amps

This matches the initial condition that0)0(i)0(i LL == + amps

)tcos(e5)tsin(e50dt)t(di

t-t-L ++=

550)0cos(e5)0sin(e5dt)0(di 0-0-L

=+=+= amps/sec

This matches the initial condition that

51575

L)0(v

dt)0(di LL

===

++

amp/sec

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=)t(iL (((( ))))[[[[ ]]]] 0tampstcose55 -t >>>>

SECOND-ORDER OP AMP CIRCUITS

Problem 8.16 Given the circuit shown in Figure 8.1, find ov in terms of iv .

Figure 8.1

Clearly,

1

i1 R

vi =

+vi

+

10 F

100 k

i1 i2

+

10 F

+vo

100 k

v2

and

== dtvCR 1-dtRvC1-v i111i12where 1)10)(10(CR -5511 == .Hence, = dtv-v i2Also,

2

22 R

vi =

and

== dtvCR 1-dtRvC1-v 222222owhere 1)10)(10(CR -5522 == .

So, = dtv-v 2oTherefore,

=ov 2i dtvCircuits like these have many applications, especially in analog computers.

SearchHelpEWB Help PageWe want your feedbacke-Text Main MenuTextbook Table of ContentsProblem Solving WorkbookWeb LinksTextbook WebsiteOLC Student Center WebsiteMcGraw-Hill Website

PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsFinding Initial ValuesThe Source-Free Series RLC CircuitThe Source-Free Parallel RLC CircuitStep Response of a Series RLC CircuitStep Response of a Parallel RLC CircuitGeneral Second-Order CircuitsSecond-Order Op Amp Circuits

Chapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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