Electric Circuit Analysis
E927500
I. Course Topics: (3 Units)
AC Circuits
10. Sinusoidal Steady-State Analysis
11. AC Steady-State Power
12. Three-Phase Circuits
Lab I: Transformer Using PSpice (3 hrs.)
Signal and System in Circuit Analysis
14. Laplace Transform
15. Fourier Series & Fourier Transform
13. Frequency Response
02-25-2015
Schedule
Mon
Wed
(導談) 2/25
3/02
3/09
3/16
3/23
3/30
4/06 (Off)
4/13
4/15
4/20
4/21
CA
EXAM
Mid Term I (Ch. 10~11)
CA 1 (PSpice)
4/27
5/04
5/11
5/18
5/25
5/27
Mid Term II (Ch. 12,14,15,Lab I)
6/01
Lab II: Bode Plot Using Matlab (3 hrs.)
16. Filter Circuits
6/03
CA 2 (Matlab)
6/08
6/15
6/22
6/25
Final (Ch. 13,16,Lab II)
II. Textbook: R. C. Dorf and J. A. Svoboda, Introduction to Electric Circuit, 8th Ed., Wiley, 2011.
References: C. K. Alexander and M. O. Sadiku, Fundamentals of Electric Circuits, 5th Ed.,
McGraw Hill, 2013.
III. Grade:
Mid Term I & II: 60 (30 + 30), Final: 30; Total: (60+30) × 110 /100 = 99.
CA: 2 × 3 = 6; 小抄: 3 × 3 = 9; Overall: 99 + 6 + 9 = 114.
IV. Handout Server: 成大 Moodle 數位學習平台
V. Time & Place:
1. Lecture:
Place: 41102 at Engineering Science
Time: 9:10~12:00, Monday (除導談外)
2. Exam:
Place: 41102 at Engineering Science
Time: 19:10~22:00, Wednesday
3. Lab:
Place: 301 at 電算中心
Time: 19:10~22:00, Wednesday
Instructor: Shean-Jen Chen (陳顯禎)
Phone: (O) (06) 2757575 ext. 63351, (H) (06) 2368441, Mobile Phone: 0931178072
E-mail: sheanjen@mail.ncku.edu.tw
TA: 林承翰、黃馨儀; (06) 2385822; Adaptive Photonics Lab. at Eng. Sci. Bld. B1
Chapter10:SinusoidalSteadyͲStateAnalysis
10.1 Motivation
10.2 Sinusoids’Features
10.3 Phasors
10.4 Phasor RelationshipsforCircuitElements
10.5 ImpedanceandAdmittance
10.6 Kirchhoff’sLawsUsingPhasors
10.7NodeVoltage&MeshCurrentAnalysisUsingPhasors
10.8Superposition,SourceTransformation,Thevenin &
NortonEquivalentCircuits
10.9OpAmpACCircuits
10.10Applications
10.11Summary
10.1Motivation (1)
Howtodeterminevt andit?
vst=10V
DCїAC
Howcanweapplywhatwehavelearnedbeforetodetermine
it andvt?
10.1Motivation(2)
‡ The output of the circuit will consist of two parts:
9a transient part that dies out as time increases;
9a steadyͲstate part that persists.
Typically, the transient part dies out quickly, perhaps in a couple
of milliseconds.
‡ AC circuits are the subject of this chapter. In particular,
9 It's useful to associate a complex number with a sinusoid.
Doing so allows us to define phasors and impedances.
9 Using phasors and impedances, we obtain a new
representation of the linear circuit, called the “frequencyͲ
domain representation.”
9 We call analyze ac circuits in the frequency domain to
determine their steadyͲstate response.
10.2Sinusoids(1)
‡ Asinusoid isasignalthathastheformofthesineorcosine
function.
‡ Ageneralexpressionforthesinusoid,
vt Vm VLQZt I where
VP =theamplitude ofthesinusoid
Ȧ =theangularfrequencyinradians/s
Ȧt =theargumentofthesinusoid
I =thephase
SHULRGT
S
Z
10.2Sinusoids(2)
Aperiodicfunction isonethatsatisfiesvt = vt + nT,forallt
andforallintegersn.
f
Hz
T
Z
Sf
v leadsv1 byorv
I
1 lagsv by I
• Onlytwosinusoidalvalueswiththesamefrequencycanbe
comparedbytheiramplitudeandphasedifference.
• Ifphasedifferenceiszero,theyareinphase;ifphase
differenceisnotzero,theyareoutofphase.
10.2Sinusoids(3)
Example:
VLQ S t q calculateitsamplitude,
Givenasinusoid,,
phase,angularfrequency,period,andfrequency.
Solution:
Amplitude=5,phase=–60o,angularfrequency=4 ʋ rad/s,
period=0.5s,frequency=2Hz.
Example:
i FRVt o Findthephaseanglebetweenand
i VLQ t o ,doesi1 leadorlagi2?
Solution:
SinceVLQȦtR FRVȦt
i1 leadsi2 by155o.
10.3Phasor (1)
‡ Aphasor isacomplexnumberthatrepresentstheamplitude
andphase ofasinusoid.
i t 5H^I P e jI e jZt ` Ÿ
I P e jI
,
I P ‘I
whereI iscalledaphasor.
‡ Phasors maybeusedwhenthecircuitislinear,thesteadyͲstate
responseissought,andallindependentsourcesaresinusoidal
andhavethesamefrequency.
‡ Itcanberepresentedinoneofthefollowing:
a. Rectangular
z x jy r FRV I j VLQ I z r ‘I
b. Polar
r
x y
jI
z re
c. Exponential
where
y
I
WDQ x
10.3Phasor (2)
Mathematicoperationofcomplexnumber:
1.
Addition
z z x x j y y 2.
Subtraction
z z
x x j y y 3.
Multiplication
z z 4.
Division
z
z
5.
Reciprocal
z
6.
Squareroot
7.
Complexconjugate
z
8.
Euler’sidentity
e r jI
z
rr ‘ I I
r
‘I I
r
‘ I
r
r ‘I x jy
r ‘I
re jI
FRV I r j VLQ I
10.3Phasor (3)
ĺ
10.3Phasor (4)
‡ Transform a sinusoid from the time domain to the phasor
domain:
v t V m FRV Z t I 9 Vm ‘ I
(time domain)
(phasor domain)
• Amplitude and phase difference are two principal concerns in
the study of voltage and current sinusoids.
• Phasor will be defined from the cosine function in all our
proceeding study. If a voltage or current expression is in the form
of a sine, it will be changed to a cosine by subtracting from the
phase.
10.3Phasor (5)
v t dv
dt
³ vdt
9
V ‘I
Laplacetransform
L > f t @
F s
jZ 9
9
jZ
³
f
f t e st dt
Noinitialcond.
&sets = jȦ
Thedifferencesbetweenvt and9:
‡ vt isinstantaneousortimeͲdomain representation
9 isthefrequencyorphasorͲdomainrepresentation.
‡ vt istimedependent,9 isnot.
‡ vt isalwaysrealwithnocomplexterm,9 isgenerallycomplex.
Note:Phasor analysisappliesonlywhenfrequencyisconstant;when
itisappliedtotwoormoresinusoidsignalsonlyiftheyhave
thesamefrequency.
10.3Phasor (6)
Example:Usephasorapproach,determinethecurrentit ina
circuitdescribedbytheintegroͲdifferentialequation.
i ³ idt Solution:
di
dt
FRV t q
a
a
I
q I
q
10.3Phasor (7)
‡ Wecanderivethedifferentialequationsforthefollowing
circuitinordertosolveforvot inphasedomain9R.
d vo dv
v
dt dt
VLQ t o However,thederivationmaysometimesbeverytedious.
‡ Insteadoffirstderivingthedifferentialequationandthen
transformingitintophasor tosolvefor9R wecantransformall
theRLC componentsintophasor first,thenapplytheKCL laws
andothertheoremstosetupaphasor equationinvolving9R
directly.
10.4Phasor RelationshipsforCircuitElements(1)
Resistor:
Inductor:
V leadsI by90Ʊ
Capacitor:
I leadsV by90Ʊ 10.4Phasor RelationshipsforCircuitElements(2)
Example:Ifvoltagevt= FRVt – Risappliedtoa ʅF
capacitor,calculatethecurrent,it,throughthecapacitor.
Answer: iW FRVt RP$
10.5ImpedanceandAdmittance(1)
• Theimpedance= ofacircuitistheratioofthe
phasor voltage9 to
9
=
R jX
thephasor current,,
,
measuredinohmsȍ.
whereR 5H=istheresistanceandX ,P=isthereactance.
PositiveX isforL (orlagging)andnegativeX isforC (orleading).
• Theadmittance< isthereciprocalof
impedance,
,
<
G jB
Unit:siemens (6)
= 9
10.5ImpedanceandAdmittance(2)
‡ After we know how to convert RLC components from time to
phasor domain, we can transform a time domain circuit into a
phasor/frequency domain circuit.
‡ Hence, we can apply the KCL laws and other theorems to directly
set up phasor equations involving our target variables for solving.
Example:Determinevt andit.
vs
FRV t 9
Answers: it = FRVt – R $;vt = FRVt +
R 9
10.6Kirchhoff’sLawsUsingPhasors (1)
‡ Both KVL and KCL are hold in the phasor domain or more
commonly called frequency domain.
‡ Moreover, the variables to be handled are phasors, which are
complex numbers.
‡ All the mathematical operations involved are now in complex
domain.
same
frequency
complex
numbers
10.6Kirchhoff’sLawsUsingPhasors (2)
10.6Kirchhoff’sLawsUsingPhasors (3)
• ThefollowingprinciplesusedforDCcircuitanalysisallapplyto
ACcircuit.
‡ Forexample:
a. voltagedivision
b. currentdivision
c. circuitreduction
d. impedanceequivalence
e. YͲȴtransformation
10.6Kirchhoff’sLawsUsingPhasors (4)
Example:FindtheinputimpedanceatȦ =50rad/s.
Example:DeterminetheinputimpedanceatȦ =10rad/s.
Answer: =LQ =± j
10.7NodeVoltage&MeshCurrentAnalysis
‡ 3StepstoAnalyzeACCircuits:
1. Transform thecircuittothephasor orfrequencydomain.
2. Solve theproblemusingcircuittechniques(nodalanalysis,
meshanalysis,superposition,etc.).
3. Transform theresultingphasor tothetimedomain.
Phasor
Laplace xform
Fourier xform
TimetoFreq
Phasor
Inv. Laplace xform
Fourier xform
Solvevariables
inFreq
FreqtoTime
‡ SinusoidalSteadyͲStateAnalysis:
FrequencydomainanalysisofACcircuitviaphasors ismuch
easierthananalysisofthecircuitinthetimedomain.
10.7aNodeVoltageAnalysis
ThebasicofNodalAnalysisisKCL.
Example: Usingnodalanalysis,findv andv inthefigure.
10.7bMeshCurrentAnalysis
ThebasicofMeshAnalysisisKVL.
Example: Find,o inthefollowingfigureusingmeshanalysis.
10.8aSuperpositionTheorem
Whenacircuithassourcesoperatingatdifferentfrequencies,
• Theseparatephasor circuitforeachfrequencymustbesolved
independently, and
• ThetotalresponseisthesumoftimeͲdomainresponsesofall
theindividual phasor circuits.
Example:Calculatevo inthecircuitusingthesuperpositiontheorem.
5.3SuperpositionTheorem(1)
Ͳ Superposition statesthatthevoltageacross(orcurrentthrough)
anelementinalinearcircuitisthealgebraicsum ofthevoltage
across(orcurrentsthrough)thatelementduetoEACH
independentsourceactingalone.
Ͳ Theprincipleofsuperpositionhelpsustoanalyzealinearcircuit
withmorethanoneindependentsourcebycalculatingthe
contributionofeachindependentsourceseparately.
Ͳ StepstoApplySuperpositionPrinciple:
1. Turnoffallindep.sourcesexceptonesource.Findtheoutput(v ori)due
tothatactivesourceusingtechniquesinChapters2&3.
2. Repeat Step1foreachoftheotherindep.sources.
3. FindTotal contributionbyaddingallcontributionsfromindep.sources.
Note:InStep1,thisimpliesthatwereplaceeveryvoltagesourceby9 (ora
shortcircuit),andeverycurrentsourceby$ (oranopencircuit).
28
Dependentsourcesareleftintactbecausetheyarecontrolledbyothers.
10.8bSourceTransformation(1)
Example: Find,o usingtheconceptofsourcetransformation.
10.8bSourceTransformation(2)
5.2SourceTransformation(1)
Ͳ LikeseriesͲparallelcombinationandwyeͲdeltatransformation,
sourcetransformationisanothertoolforsimplifyingcircuits.
Ͳ Anequivalentcircuitisonewhosev-i characteristicsareidentical
withtheoriginalcircuit.
Ͳ Asourcetransformationistheprocessofreplacingavoltage
sourcevs inserieswitharesistorR byacurrentsourceis in
parallelwitharesistorR,andviceversa.
‡ Transformationofindependentsources
+
+
Ͳ
‡ Transformationofdependentsources
+
Ͳ
+
Ͳ
Ͳ
9 Thearrowofthecurrent
sourceisdirectedtoward
thepositiveterminalofthe
voltagesource.
9 Thesourcetransformation
isnotpossiblewhenR =0
forvoltagesource andR =
’ forcurrentsource. 31
5.2SourceTransformation(2)
AvoltagesourcevV connectedinserieswitharesistorRV anda
currentsourceiV is connectedinparallelwitharesistorRS are
equivalentcircuitsprovidedthat
Rp
32
Rs vs
Rs is
10.8cThevenin &NortonEquivalentCircuits
TheveninEquivalent
NortonEquivalent
Example:FindtheThevenin equivalentatterminals a–b.
5.4Thevenin’s Theorem(1)
ItstatesthatalineartwoͲterminalcircuit
(Fig.a)canbereplacedbyanequivalent
circuit(Fig.b)consistingofavoltage
sourceV7K inserieswitharesistorR7K,
where
9 V7K istheopenͲcircuitvoltageatthe
terminals.
9 R7K istheinputorequivalent
resistanceattheterminalswhenthe
independentsourcesareturnedoff.
34
5.4Thevenin’s Theorem(2)
TofindR7K :
9 Case1:Ifthenetworkhasnodependentsources,weturnoffall
indep.Source.R7K istheinputresistanceofthenetworklooking
btwterminalsa &b.
9 Case2:Ifthenetworkhasdepend.Sources.Depend.sources
arenottobeturnedoffbecausetheyarecontrolledbycircuit
variables.(a)Applyvo ata &b anddeterminetheresultingio.
ThenR7K voio.Alternatively,(b)insertio ata &b and
determinevo.AgainR7K voio.
(a)
(b)
35
5.5Norton’sTheorem(1)
ItstatesthatalineartwoͲterminalcircuit(Fig.a)canbereplaced
byanequivalentcircuit(Fig.b)consistingofacurrentsourceIN in
parallelwitharesistorRN,
(a)
(b)
where
9 IN istheshortͲcircuitcurrentthrough theterminals.
9 RN istheinputorequivalentresistanceattheterminalswhen
theindepen.sourcesareturnedoff.
36
10.9OpAmpACCircuits(1)
Thekeytoanalyzingopampcircuitsistokeeptwoimportant
propertiesofanidealopampinmind:
‡ Nocurrententerseitherofitsinputterminals.
‡ Thevoltageacrossitsinputterminalsiszero.
Example: ComputetheclosedͲloopgainandphaseshift.Assume
thatR R NȍC ȝ)C ȝ)andȦ UDGV.
,QYHUWLQJ$PSOLILHU
6.2IdealOpAmp
Anidealopamphasthefollowingcharacteristics:
1. InfiniteopenͲloopgain,A §’
2. Infiniteinputresistance,Ri §’
3. Zerooutputresistance,Ro §
Example: Determinethevalueofio.
ZKHQvs
38
9
6.4ConfigurationsofOpamp(1)
‡ InvertingAmplifier:reversesthepolarity oftheinputsignal
whileamplifyingit.
Negativefeedbackbtwthe
invertinginput(vi &output(vo)
vi isconnectedtothe
invertinginputviaR
Tofindtherelationshipbtwvi &vo:
%\.&/DWQRGH
v v
i i Ÿ i R
' v
v
v vo
Rf
IRUDQLGHDORSDPS
VLQFHWKHQRQLQYHUWLQJWHUPLQDOLVJURXQGHG
? vo
Rf
R
vi
&ORVHGORRSYROWDJHJDLQLV
Equivalentcircuit
Noninvertinginputisgrounded
Av
vo vi
R f R
Example: Findvo &i inR ifvi 9,R Nȍ,&Rf Nȍ.
39
6.4ConfigurationsofOpamp(3)
‡ NoninvertingAmplifier:designedtoproducepositivevoltagegain.
Tofindtherelationshipbtwvi & vo:
%\.&/DWLQYHUWLQJWHUPLQDO
i
' v
v
i Ÿ
vi Ÿ vo
9ROWDJHJDLQAv
vi connectedtononinvertinginputterminal
Rf
VKRUWFLUFXLWRU
R
f RSHQFLUFXLW
Ÿ vo
v
R
v vo
Rf
Rf ·
§
¨ ¸ vi
R
©
¹
vo vi R f R
‡ Toisolatetwocascadedstages:
vi DVYROWDJHIROORZHU
Hasaveryhighinputimpedanceand
40
thuseliminateinterstageloading
10.9OpAmpACCircuits(2)
Oscillator:acircuitproducesanacwaveformasoutputwhen
poweredbydcinput.
‡ Inorderforasinewaveoscillatortosustainoscillations,itmust
meettheBarkhausen criteria:
1. Theoverallgainoftheoscillatormustbeunityorgreater.
Thuslossesmustbecompensated
fo
1.
forbyanamplifyingdevice.
S RC
2. Theoverallphaseshift(from
theoutputandbacktothe
input)mustbezero.
‡ Threecommontypesofsinewave
oscillatorsarephaseͲshift,twinT,
andWeinͲbridge oscillators.
2.
10.9OpAmpACCircuits(3)
CapacitanceMultiplier:tocreatealargecapacitance.
Example:
10.10Applications(1)
• SeriesRC shiftcircuits:
Ͳ LeadingoutputͲ Laggingoutput
10.10Applications(2)
Fig. (a) shows two sinusoidal voltages, one labeled as input and the
other labeled as output. We want to design a circuit that will
transform the input sinusoid into the output sinusoid.
Fig. (b) shows a candidate circuit. We must first determine
whether this circuit can do the job. Then, if it can, we will design
the circuit, that is, specify the required values of R1, R2, and C.
The input sinusoid is v t VLQS t FRVS t q 9
and the corresponding phasor is 9 e j q
The output sinusoid is
v t VLQS t q
‘ q 9
FRVS t q 9
and the corresponding phasor is 9 e j q 9
j q
The ratio of these phasors is 9 e
e jq
j q
9 e
The magnitude of this ratio, called the gain, G, of the circuit used
to transform the input sinusoid into the output sinusoid is G 9 9
The angle of this ratio is called the phase shift, ș, of the required
circuit: T ‘ 9 q
9
Therefore, we need a circuit that has a gain of 2 and a phase shift
of 120Ʊ.
State the Goal: Determine whether it is possible to design the
circuit shown in Fig. (b) to have a gain of 2 and a phase shift of
120Ʊ. If it is possible, specify the appropriate values of R R C.
Generate a Plan: Analyze the circuit shown in Fig. (b) to determine
the ratio of the output phasor to the input phasor, 99.
Determine whether this circuit can have a gain of 2 and a phase
shift of 120Ʊ. If so, determine the required values of R R C.
Act on the Plan: The impedance = corresponds to the resistor R1
in Fig. (b), and impedance = corresponds to the parallel
combination of resistor R and capacitor C. That is,
=
? 9
9
R = =
=
R jZC R jZC
R jZCR R
R
jZCR
R R
jZCR
The phase shift: T
‘
9
9
‘
R R
jZCR
q WDQ ZCR
What values of phase shift are possible? Notice that Ȧ, C, and R
are all positive, which means that
q d WDQ ZCR d q
Therefore, the circuit shown in Fig. (b) can be used to obtain phase
shifts between 90Ʊ and 180Ʊ. Hence, we can use this circuit to
produce a phase shift of 120Ʊ.
The gain: G
To find R
R
9
9
R R
jZCR
R R
R R
Z C R
WDQ q T WDQq T Ÿ R
ZC
R G
WDQ q T These equations can be used to design the circuit. For Ȧ
UDGV, C ȝ), G , and ș Ʊ we calculate
R
: DQG
R
N:
10.12Summary(1)
‡ With the pervasive use of ac electric power in the home and
industry, it is important for engineers to analyze circuits with
sinusoidal independent sources.
‡ The steadyͲstate response of a linear circuit to a sinusoidal input
is itself a sinusoid having the same frequency as the input signal.
‡ Circuits that contain inductors and capacitors are represented by
differential equations. When the input to the circuit is sinusoidal,
the phasors and impedances can be used to represent the
circuit in the frequency domain. In the frequency domain, the
circuit is represented by algebraic equations.
‡ The steadyͲstate response of a linear circuit with a sinusoidal
input is obtained as follows:
1. Transform the circuit into the frequency domain, using
phasors and impedances.
10.12Summary(2)
2. Represent the frequencyͲdomain circuit by algebraic
equation, for example, mesh or node equations.
3. Solve the algebraic equations to obtain the response of the
circuit.
4. Transform the response into the time domain, using phasors.
‡ A circuit contains several sinusoidal sources, two cases:
9 When all of the sinusoidal sources have the same frequency,
the response will be a sinusoid with that frequency, and the
problem can be solved in the same way that it would be if
there was only one source.
9 When the sinusoidal sources have different frequencies,
superposition is used to break the timeͲdomain circuit up into
several circuits, each with sinusoidal inputs all at the same
frequency. Each of the separate circuits is analyzed separately
and the responses are summed in the time domain.