Ac Sinusoidal Ckt

8/10/2019 Ac Sinusoidal Ckt

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Chapter 5Steady-State Sinusoidal Analysis

1 . Identify the frequency, angular frequency, peak value,rms value, and phase of a sinusoidal signal.

2 . Solve steady-state ac circuits using phasors andcomplex impedances.

3 . Compute power for steady-state ac

4 . Find Th venin and Norton equivalent circuits.

5 . Determine load impedances for maximum powertransfer.


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5. Steady -State Sinus oid al An alysis

* In most circuits, the t r ans ien t respon se (i.e., thecomplimentary solution) decays rapidly to zero, the s teady- s ta te respon se (i.e., the forced response or the particular

solution) persists.* In th is c hapter, we learn eff ic ient method s fo r f indin g th e

s teady-s ta te respon ses for s inu soid al sou rces .


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5. Steady -State Sinus oid al An alysis

5 .1 Sinuso idal Currents and Vol tages

5.1.1 Phase and Phase A ng le

* Consider the sinusoidal voltage as shown,

f 2 T

2

T

1 f : ) s (or H zin frequencythe

s,i n period thei s T , 2 T : have we

cycle per by 2 increases angletheSince

degree) i n (usuallyanglephasethei s rad/s in frequencyangular thei s

voltage theof valuePeak thei s V

where t cos V ) t ( v

1 -

m

m


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5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage

5.1.1 Phase and Phase A ng le

* We usually use cos ine func t ion to model a sinusoidal signal.In case there is a sine function, we can use the following

conversion:

60 - i s (t) v of anglephasethethat saycan wethus and

60 t 200 cos 10

90 30 t 200 cos 10 t v

30 t 200 sin 10 t v

: example For

90) - cos(z sin(z)

x

x

x


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5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage 5.1.2 Roo t-Mean-Squ are (RMS) Values (or Effec tiv e Values )

R I P havewe,dt t i

T

1 I

: as current periodictheof valuerm s thedefinewei f Similarly,

R

V P havewe,dt t v

T

1 V

: as v(t) voltageperiodictheof value(rms) squre- m ean - root thedefiningBy

R / dt ) t ( v T 1 dt

R t v

T 1 dt t p

T 1

T E P

: i s period per poweaverageThe

dt t p E : i s period per delivered energythe

,R

t v t p bygiven i s resistance theto delivered power The

R.resistance ato T period with v(t) voltageperiodicaapplyingConsider

2 rms g av

T

0

2 rms

2 rms

g av

T

0

2 rms

2

T

0

2 T

0

2 T

0

T av

T

0 T

2


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5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage

5.1.3 RMS Value of a Sinu so id

m m

rms

m rms m m

rms

2 m

T

0

2 m

T

0

2 2 m

T

0

2 rms

m

I 0.7071 2

I I Similarly,

H z 60155.5V),(V 110V : power l residentia : Note V 0.7071 2

V V

2 sin 2

1 2 T 2 sin

2

1 T

T 2

V

dt 2 t 2 cos 1 T 2

V

dt t cos V T

1 (t)dt v

T

1 V

t

cos V t v

: by given voltagesinusoidal aConsider


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5. Steady -State Sinus oid al Analys is 5 .1 Sinu so idal Current and Vo ltage Exam ple 5.1 Pow er del ivered to a res is tance by a s inus oidal sou rce

W t 200 cos 100100

W t 100 cos 200 50 t

100 cos 100 R t v t P

W 100 50

) 71 .70 ( R

V P

70.71V 2 / V V

m s 201/f T ,H 50 /2 f

t p an d

P power averagethe,V : F i n d

resistance 50ato applied i s

V t) 100cos(100 v(t) a: Given

2

2 2 2

2 2 rms

g av

m rms

z

av g r ms


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5. Steady -State Sinus oid al Analys is 5.2 Phasors

5.2.1 Definitio n

* A phasor is a vector in complex number plane that represents

the magnitude and phase of a sinusoid.* In ac circuit analysis, voltages and currents are usually

represented as phasors.

)90 ( I and I writecanwe

t sin I t iand t cos I t i for Similarly,

)90 ( V

therefore ,90 t cosV t v toit convert canwe

t sinV t v formtheof is sinusoid the If V :as phasor thedefinewe

t cosV t v formtheof voltagel sinusoidaa For

222111

222111

222

222

222

111

111

I I

V

V


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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.1 Definitio n

* Eulers Identity :

* In phasor application:

) (Ae Re ) t Acos(

) t jAsin( ) t Acos( Ae

sin jA cos AAe

sin j cos e l exponentia complex

) t j(

) t j(

j

j

1 1 1 1 1 1

j

jwt

j ) t j(

V V as presented i s ) t cos( V (t) v

voltage sinusoidal theexam ple,F or

) t cos( A AAform phasor i ts sim plyor ) Re(Ae as presented i s )

t

cos( A

result,aAs .Ae term thedeletingby

Ae form theto simplified i s Ae n ,applicatio circuit In


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5. Steady -State Sinus oid al Analys is 5 .2 Phasor s 5.2.4 Phaso r vs . Sinu so ids

* The phasor is simply a snapshot of arotating vector at t=0 .

m

mm

mt j

m

t j

m

m

V

:as defined is v(t) for phasor The

V cosV v(0) 0,t at

and )t ( V eV eV Ret v

:vector rotating a of part real the just s It'

t cosV t v

: shownas voltage al sinusoid the Consider

V

)(Ae Re ) t Acos(

) t jAsin( ) t Acos( Ae

sin jA cos A Ae

j sin cose l exponentiacomplex

) t j(

) t j(

j

j


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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.2 Phaso r Sum m ation

90 .9 t cos 54 .14

e 54 .14 Re t v

e e 54 .14 Re t v

e 54 .14 90 .9 54 .14 .5 2 j 33 .14 5 j 50 .2 j 33 .4 10 90 5 30 5 0 10

: number complextheConsider

e 90 5 30 5 0 10 Re t v

: form) polar in (or Identitys Euler' In e e 5 e 5 10 Re

e 5 e 5 e 10 Re t v

e 5 Re e 5 Re e 10 Re

90 t cos 5 30 t cos 5 t cos 10 t v form cosineto functions theal l rewritefirst We

) 90 t cos( 5 ) 60 t sin( 5t) cos( 10v(t)

voltages threeof summation theConsider

) 90 .9 t ( j

t j 90 .9 j

90 .9 j

t j

t

j 90 j 30 j

) 90 t ( j ) 30 t ( j t j

) 90 t ( j ) 30 t ( j t j


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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.2 Phaso r Sum m ation

Now w e use Phasor nota t ion to s impl i fy ou r ca lcu lat ion :

Note: In us ing ph asors to add s inuso ids , a ll o f the t erms mu s thave the same f requency .

90.9t cos54.14t v

9.9014.54

90.954.14

5.2 j33.14

5 j50.2 j33.410

905305010

90t cos530t cos5t cos10t v

90t cos560t sin5t cos10t v

321

V

V V V V


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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs Exam ple 5.2 Us ing phaso r s t o Add S inuso ids

7 .39t cos97 .29t v7 .3997 .29

14.19 j06 .23

5 j660.814.14 j14.1430104520

3010 and 4520 :are phasors The

(t)v(t)v(t)v Find

60t sin10t v

45t cos20t v that Suppose

s

21 s

21

21 s

2

1

V V V

V V


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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.3 Fun dam ental Phaso r Operations

/2V : Root Square

-V

1

V

11 : Reciprocal

-V V

: DivisionV V :tion Multiplica

then ,V ,V If

111

11111

212

1

2

1

212121

222111

V

V

V

V V V

V V


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5. Steady -State Sinus oid al Analys is 5.2 Phaso rs 5.2.5 Phase Relat ions hip s

.60bylags or 60byleads

:that may sayweand 204 and 403 :as d represente be canThey

20t cos4t v and 40t cos3t v

:voltagestheConsider

1221

21

21

V V V V

V V


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5. Steady -State Sinus oid al Analys is 5 .3 Com plexImpedances

5.3.1 Impedance

* Impedance means com plex res i s t ance .

* The impedance concept is equivalent to stating that capacitors andinductors act as f requency -dependent res is to rs .

* By using impedances, we can solve sinusoidal steady-state circuitwith relatively ease compared to the methods of Chapter 4.

* Except for the fact that we use complex arithmetic, s inus o ida l s t eady- s ta te analys is i s the same as the analys is of res is t ive c i rcu i ts .


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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es 5.3.2 Indu ct anc e

form. phasor in law sOhm' is This

Z :have We

90 L L j Z :as inductance the of impedance the Define

L j )90( I 90 L :as voltage Rewrite90byvoltage the lags current the : Note

V I L and )90( I : forms phasor their In

t cos I Ldt

t di Lt v

:inductor an through f low t sin I t i current a Consider

L L L

L

Lm L

mm Lm L

m L

L

m L

I V

I V

V I


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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es 5.3.3 Capac itanc e

C C C

C

C m m C

m m C m C

m m C

C

m C

I Z V : have we

90 - C

1

C j

1 Z : as ecapacitanc of im pedancetheDefine

I 90 - C

1 90 I 90 - C

1 V V as voltageRewrite

90 byvoltageleads current the: Note

90 I 90 CV I , V V : forms phasor their In

) 90 t cos( CV ) t sin( CV - dt

dv C (t) i

capacitor aacross ) t cos( V (t) v voltageaConsider

d d l l l d


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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es

* EL I the IC E m an.* Impedance that are pure imaginary are called reactance .

5.3.4 Resis tanc e

R R

R

R :havewe

R Z asresistanceof impedancethe Define

I V

5 S d S Si id lA l i 5 3C l I d


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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es



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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Qu iz - Ex erc is es 5.6, 5.7, 5.8



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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Qu iz - Ex erc is es 5.6, 5.7, 5.8

5 St d St t Si id lA l i 5 3C l I d


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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Ad di t ional Example: represen t the c i rcu i t s how n in the

f requency do main us ing im pedances and p hasors .

5 St d St t Si id lA l i 5 3C l I d


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5. Steady -State Sinus oid al Analys is 5.3 Com plex Im pedanc es Ad di t ional Example: represen t the c i rcu i t s how n in the

f requency do main us ing im pedances and p hasors .

5 Steady StateSinus oid alAnalys is 5 4Circui tA nalysis


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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is

5 .4 Circui t A nalys is

* Im pedance c i rcui t analysis is the same as res is t ive c i rcui t analysis, we can directly apply KCL , KVL, nodal analys is ,

m esh analys i s ,

* The above phasor approach can only apply for steady st atewi th s inu so ids o f the same f requency .

0- 0(t)i-(t)i(t)i

:equations KCL

0 0t vt vt v

:equation KVL

321321

321321

I I I

V V V



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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Exam ple 5.3 S teady-State AC A nalys is of a Ser ies Circu i t* Find the steady-state current, the phasor voltage across each

element, and construct a phasor diagram.

15t 500cos707 .0t i

15707 .0454.141

30100 Z

454.141100 j10050 j150 j100 Z Z R Z

50 jC

1 j Z ,150 j L j Z ,30100

s

C Leq

C L s

V I

V

1054.3515707 .09050

C

1 j

751.106 15707 .090150 L j

157 .7015707 .0100 R

C

L

R

I V

I V

I V


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Exam ple 5.4 Ser ies /Paral le l Com bin at ion of Com plex Impedanc es

* Find the voltage across the capacitor, the phasor currentthrough each element, and construct a phasor diagram

901.09010018010

100 j18010

Z V

1801.0100

18010 RV

1351414.050 j509010

50 j50100 j9010

Z Z

t 1000cos10180t 1000cos10t v

1801050 j50100 j

4571.709010

Z Z Z

50 j50 Z

4571.704501414.0

01 )100 j( 11001

1 Z 1 R1

1 Z

100 jC

1 j Z ,100 j L j Z ,90-10

C

C C

C R

RC L

s

C

RC L

RC sC

RC

C RC

C L s

I

I

V I

V V

V



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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Exam ple 5.5 Steady -State AC No de-Voltage An alysis* Find the voltage at node 1 using nodal analysis

7 .29t 100cos1.16 t v or 7 .291.16 : for Solve

5.11.0 j2.0 j2 j2.0 j2.0 j1.0

05.15 j10 j

90-2 j5--

10

2node and node1 at KCL Write

11

1

21

21

122

211

V

V

V V

V V

V V V

V V V



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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Exerc is e 5.11 Steady-Sta te AC Mesh-Current An alys is* Solve for the mesh currents

5 S d S Si id lA l i 5 4Ci i A l i


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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Ad di t iona l Examp le :

)A51.87 - (3t Bcos i(t) when

Land B e min Deter

5 S d S Si id lA l i 5 4Ci i A l i


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5 St d St t Si id lA l i 5 4Ci i tA l i


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Ad di t iona l Examp le :

5 St d St t Si id lA l i 5 4Ci i tA l i


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5. Steady -State Sinus oid al Analys is 5 .4 Circui t A nalys is Ad di t iona l Examp le : Comm erc ia l Ai r l iner Door B r idge Ci rcu i t



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Ad di t iona l Examp le :



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y y yQuiz Exerc is e 5.10* Find the phasor voltage and phasor current at each element

5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t


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y y5.5 Power in A C Circui t

5.5.1 Voltage, Current an d Im pedanc e

)- where , I - I I

then ,V If :(Note

Z V

I where

I

Z

0V

Z

iscurrent phasor The

(X/R)tan , X R Z where

jX R Z Z impedancethe

0V or t)cos( V v(t) network,the In

ivimvm

vm

mm

mm

1-22

mm

V

V I

V



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y y5.5.2 Voltage, Curren t and Po w er for a Resist iv e Load

(1) Current is in phase with voltage.(2) Energy flows continuously from

source to load.

rms rms m m

m m

2

m m

m m

m

I V 2

I V P power averageThe

t) cos2 2

1

2

1 ( I V

t cos I V t i t v t p

t cos I t) cos( R

V t i

t cos V t v 0 0,R Z

resistive,pureis load theIf



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y y5.5.3 Voltage, Curren t and Pow er for an Ind uc tive Lo ad

(1) Current lags the voltage by 90 degree ( ELI )(2) Half of the pow er is pos i t ive , energy is delivered to the inductance

and stored in the magnetic field; the other hal f of the pow er isnegat ive , the inductance returns energy to the source.

(3) The average po w er is zero , and we say react ive po wer flows backand forth in-between the source and the load.

0 P power averageThe

t sin2 I V t 2 sin 2

I V

t

sin t

cos I V

t i t v t p

t sin I

90 t cos I t i t cos V t v

90 ,90 LZ

,inductance purei s load theIf

rms rms m m

m m

m

m

m



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y y5.5.4 Voltage, Curren t and Po w er for a Capacit ive Lo ad

(1) Current leads the voltage by 90 degree ( IC E )(2) The average pow er is zero: react ive po wer flows back and forth in-

between the source and the load.(3) Reactive power is negative (positive) for a capacitance

(inductance ).

(4) Reactive power in inductance and in capacitance cancel eachother.

0 P power averageThe

t 2 sin I V t 2 sin 2

I V

t sin t cos I V

t i t v t p

t sin I

90 t cos I t i

t cos V t v

-90 ,90 -

C

1 Z

e,capacitanc pureis load theIf

rms rms m m

m m

m

m

m



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5.5.5 Pow er Calculat ion for a General (RLC) Lo ad

A ctiv e (Real) load d ue to R

React ive load d ue to L, C

remains.term second theonlyreactive, pure :90

remains;term1st theonlyresistive, pure:0

t 2 sin sin2 I V t 2cos1cos

2 I V

t sint cos sin I V t coscos I V

t cost cos I V t p

t cos I t i

t cosV t v

9090- , Z jX R Z load, RLC general For

mmmm

mm2

mm

mm

m

m



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t 2 sin sin2 I V

t 2cos1cos2

I V t p

t sint cos sin I V

t coscos I V p(t)t cost cos I V t p

t cos I t i

t cosV t v

mm

mm

mm

2mm

mm

m

m

) t ( 2 sin sin2 I V

) t ( 2cos1 cos2 I V t p

t sin t cos sin I V

t cos cos I V t p

t cos t cos I V t p ) - t cos( I

- , t cos I t i

t cosV t v

vmm

vmm

vvmm

v2

mm

vvmm

vm

ivim

vm

cos I V P have we

,2 / I I and 2 / V V using ,cos2 I V

P

:is P power (real)average the zero,of values average have ))t sin(2( and ))t cos(2( involving terms the Since

rmsrms

mrmsmrmsmm

vv



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Av erage Power :

Pow er Factor :

Power factor is often stated as percentage, e.g.,90% lagging (i.e., current lags voltage, inductive load)60% leading (i.e., current leads voltage, capacitive load)

Reactiv e Pow er: The last term in power formula is the power flowing back and forth

between the source and the energy-storage elements. Reactivepower is its peak power.Ap paren t Power :

Note: 5kW load is d i fferent f rom 5kVA load .

t 2 sin sin 2

I V t 2 co s 1 cos

2

I V t p m m m m

W cos I V cos 2

I V P rms rms

m m

angle power thecalled i s , cos PF i v

Reactive) Amperes(Volt VAR sinrms I rmsV sin2m I mV Q

2 rms rms 2 2

rms rms 2 2

rms rms 2 2

rms rms

I V sin I V cos I V Q P

Am pere) - (Volt VA I V S



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y y5.5.6 Im pedanc e tr iangle and Pow er Triang le

* The impedance triangle:

* The Power triangle:Apparent power, average (real) power, and reactive power form a

triangle.

2rmsrms22rmsrms22rmsrms22rmsrms

I V sin I V cos I V Q P

Ampere)-(Volt VA I V S



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y y5.5.6 Im pedanc e tr iangle and Pow er Triang le

reactance theacross voltagethei s V whereX

V Q

resistance theacrtoss voltagethei s V whereR

V

P : Also

load) capacitive 0,X load; inductive0,(X

X I sin 2

I V Q power reactivetheSimilarly,

R I R 2

I ) Z I V : (Note

Z

R

2

I V cos

2

I V P power averageThe

Z

X sin ,Z

R cos ,X j R Z Z

Xrms

2 Xrms

Rrms

2 Rrms

2 rms

m m

2 rms

2 m

m m

m m m m

R

V

R

1

2

V

R

1

Z

R

2

V

Z

R

2

V

Z

R

2

I V P

2 Rrms

2 Rm

2

22m

2

2mmm

t 2 sin sin2 I V

t 2cos1 cos2 I V

t p mmmm



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y yExam ple 5.6 AC Pow er Calcula t ion

VAR5.045 sin1.0071.7

sin I V Q power reactiveThe

W 5.045cos1.0071.7

cos I V P : power The

A1.02

1414.0

2 I

V 071.7 2

10

2V

45 )135( 90

sourcethe fromtaken power reactiveand power the Find (1)

rmsrms , s

rmsrms , s

rms

srms , s

iv

I

V

5. Steady -State Sinus oid al Analys is 5 .5 Power in A C Circui t E l 5 6 ACP C l l i


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Exam ple 5.6 AC Pow er Calcula t ion

.resistancethebyabsorbed is sourcethebydelivedrd power theof All

0 P ,0 P

0isinductanceand ecapacitancthebyabsorbed power thecourse,Of

W 5.010021.0 R

2 R I P

:resistancethetodelivered power (real)The(4)

)QQQ: Note( VAR5.0 )100( 2

1.0 X I Q

:capacitor thetodelivered power reactiveThe(3)

VAR0.1 )100( )1.0( X I Q

:inductor thetodelivered power reactiveThe(2)

C L

22

R2rms , R R

C L

2

C 2

rms ,C C

2 L

2rms L

I


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5. Steady -Sta te Sinu so idal A nalys is 5 .5 Power in A C Circui t


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Ad dit ional Example:



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Exam ple 5.7 Using Pow er Tr ianglesFind power, reactive power, and power factor for the source

and the phasor currents as shown.

We first find the power and reactive power for each load, thensum over to obtain the power and reactive power for thesource.



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Exam ple 5.7 Using Pow er Tr iangles

kVAR559.3101.5660.8QQQ

kW 1055 P P P

: sourcethebydelivered power reactiveand power The

kVAR101.5Q

57 .45tan5000tan P Q

57 .45 )7 .0arccos( B,load For

kVAR660.8500010

P I V Q

kW 5 )5.0( 10cos I V P

5.0cos:havewe A,load For

B A

B A

B

B B B

B

224

2 A

2 Armsrms A

4 A Armsrms A

A



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Exam ple 5.7 Using Pow er Tr iangles

59.4915 ,59.49 )59.19( 30

A15 I 2 I

A61.10 /V I V I :current effectiveThekV 12 / V :voltage sourceeffectiveThe

kVA61.10Q P I V S : power apparent The

leading.94.21%or 9421.0 cos : factor power The

59.19Q/P tan :angle power The

kVAR559.3QQQ ,kW 10 P P P

:by sourcedelivered power reactiveand power The

ivi

rmsm

rmsrmsrmsrms

rms

22rmsrms

1

B A B A

I I

I

V

v

i


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60

Exam ple 5.8 Pow er-Facto r Correct ionA 50kW load operates from a 60-Hz 10kV-rm s line with a power

factor of 60% lagging. Compute the capacitance that mustbe placed in parallel with the load to achieve a 90% laggingpower factor.

F 126 .1

2356 377

1

X

1C

0.377 60 2

2356 4245010

QV

X

kVAR45.42QQQ

kVAR22.24 tan P Q84.25 )9.0( cos

kVAR67 .66 tan P Q

13.53 )6 .0( cos

C

24

C

2

rmsC

LnewC

newnew

1new

L

1 L

66.67kVAQ L

50kW P

22.42kVAQ new

0.6 PF 50kW



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61

Qu iz - Exerc ise 5.12 Pow er in A C Ci rcu i t s.

5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton


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5.6 Thevenin and Norton Equivalent Circui ts

* A two terminal circuit composed of sinusoidal sources (of thesame frequency), resistances, capacitances, andinductances can be simplified to Thevenin or Nortonequivalent c i rcu i t .

5.6.1 Thevenin Equ ivalent Circu its

* The Thevenin impedance can also beobtained by zeroing sources.

5.6.2 Nort on Equ ivalent Circu its

sc

t

sc

oct

oct

Z I V

I V

V V

scn

sc

t

sc

oct I

Z

I I

V

I

V

5. Steady -State Sinus oid al Analys is 5 .6 Thevenin and No rton


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63

Exam ple 5.9 Thevenin and Norton Eq uivalents

V 90100

4571.7045414.1

Z I

A45414.1 9011

A01100

0100100V

terminalsat circuit t Apply shor

50 j50

4571.704501414.0

1 )100 j( 11001

1 Z

sources zeroing by Z f ind We

t sct

s R sc

s R

t

t

V

I I I

I



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65


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5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton E l 5 10 M i P T f


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67

Exam ple 5.10 Maxim um Pow er Trans fer

W 25 )50( 2

1 R I P

90150 j5050 j50

90100 Z Z

50 j50 Z Z

when power output max.havewill we(a)

50 j50 Z Since

2

load 2

arms

load t

t a

*t load

t

V I

W 71.2071.70

2

7653.0 R I P

50.67 765.050.2266 .130

90100

71.7050 j50

90100

Z Z

71.7050 j50 Z R

resistive,betohasload the If (b)

2

load 2

brms

load t

t b

t load

V I

5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton Q i E i 5 14 dE i 5 15


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68

Quiz Exerc is e 5.14 and Ex ercis e 5.15

5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton Q i E i 5 14


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Quiz Exerc is e 5.14

5. Steady -State Sinu so idal An alysis 5 .6 Thevenin and No rton Q i E i 5 15


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Quiz Exerc is e 5.15


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5. Steady -State Sinu so idal An alysis SUMMARY


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5. Steady -State Sinu so idal An alysis SUMMARY


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Chapter 6


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Chapter 6Frequency Response, Bode Plots,

and Resonance1. State the fundamental concepts of Fourier analysis.2 . Determine the output of a filter for a given input

consisting of sinusoidal components using the

filter

s transfer function.3 . Use circuit analysis to determine the transferfunctions of simple circuits.

4 . Draw first-order lowpass or highpass filter circuitsand sketch their transfer functions.

5.

Understand decibels, logarithmic frequency scales,and Bode plots.6 . Calculate parameters for series and parallel

resonant circuits.


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6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns 6.1.1 Fourier A nalys is


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77

y

* All real -wo r ld s igna ls a re sums of s inu so ida l com ponents having various frequencies, amplitudes, and phases.

* The square wave is a special example:

* Most of the real-world signals areconfined to finite range of frequency.

* It is important to learn how the circuits

respond to components havingdifferent frequencies.

.. .t ) s in(5 5 4A

t ) s in(3 3 4A

t ) s in (

4A(t ) v 0 0 0 sq

T 2 where0

6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns 6.1.2 Filt ers


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78

* Filters process the sinusoidal components of an input signaldifferently depending of the frequency of each component.

Often, the goal of the f i l ter i s to re ta in the com pon ents incer ta in f requency ranges and reject com pon ents in o therf requency ranges .

6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns 6.1.3 Fil ters and Transfer Fun ctio ns


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79

* Since the impedances of inductances and capacitanceschange with frequency, RLC c i rcu i t s p rov ide one w ay torealize electri cal fil ters .

* The t ransfer fun ct ion of a two-port filter is defined as:

90- f 2

190-C 1 Z ,90 fL290 L Z C L

phaset h eis- H(f)

magni tudetheisH(f) w h e r e

H(f) H(f)) f H

in o u t

in

o u t

VV

V

V)(

6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns Exam ple 6.1 Us ing Transfe r Func t ion to F ind Outpu t


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80

For the transfer functions shown, find the output signal,given the input: )40t 2000cos( 2 )t ( vin

in

out 303 )1000( H

Hz 1000 f is signal input theof f requencyThe

V

V

402inV 706 402303* )1000( H inout V V )702000cos(6)( t t vout


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6. Frequ ency Respo nse 6 .1 Fou rier An alysis , Fi l ters , Trans fer Fun ctio ns Exam ple 6.2 Mul t i -inpu t com po nents , Superpos i t ion Pr inc iple


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82

)70t 4000cos( )t 2000( sco2 )t ( vin

)10t 4000cos( 2 )30t 2000cos( 6 )t ( vout

6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Ideal Filters


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83

6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters 6.2 First Ord er Low -Pass Fil ters


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84

A low -pass f i l ter is designed to p ass low-f requencycom ponents and re jec t h igh- f requency c om ponents . Inother words, for low frequencies, the output magnitude isnearly the same as the input; while for high frequencies, theoutput magnitude is much less than the input.

6.2.1 Transfer Fun ctio n

fC j21 R

have we ,V phasor ahaving l sinusoidaais signal input the shown,as

f ilter pass-loworder - f irst theConsider

in

in

V I

RC f j21 fC j21 R fC j21 inin V V I V

fC j21

out

fRC 2 j11 H(f)

in

out

V V

frequency power" -half " the

frequency,break" " the RC 21

f defineWe B

) f f j( 11

H(f) B

6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters 6.2.2 Magn itud e and Phase Plots of th e Trans fer Fun ctio n


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85

)(11

)( B f f j

f H 2)(1

1)(

B f f f H )arctan()(

B f f

f H

Power Half V P since ,V 2

1V ,2

1 H(f ) , f f As

90 H(f )also

rejected,components f requency-high 0 H(f ) , f f As

0 H(f )also passed,components f requency-low 1 ) f ( H 0, f As

2rmsrmsinrmsout B

B

6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Exam ple 6.3 Calcula t ion of RC Low -pass Outpu t


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86

1000 f 0,5

100, f 0,5 10, f 0,5

)t 2000cos( 5 )t 200cos( 5 )t 20cos( 5 )t ( v

33in

22in11in

in

V

V V

) f f ( j1

1 ) f ( H

B

Hz RC

f B 10010*10*)21000(*21

21

6

71.59950.0 )10( H

457071.0 )100( H 29.840995.0 )1000( H

71.5975.4 )10( H 1in1out V V

)71.5t 20cos( 975.4 )t ( v 1out

45535.3 )100( H 2in2out V V )45t 200cos( 535.3 )t ( v 2out

29.844975.0 )1000( H 3in3out V V

)29.84t 2000cos( 4975.0 )t ( v 3out

6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Exam ple 6.3 Calcula t ion of RC Low -pass Outpu t


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87

1000 f 0,5 100, f 0,5 10, f 0,5

)t 2000cos( 5 )t 200cos( 5 )t 20cos( 5 )t ( v

33in

22in11in

in

V V V

(t)v )29.84t 2000cos( 4975.0 (t)v )45t 200cos( 535.3 (t)v )71.5t 20cos( 975.4 )t ( v

3out

2out

1out out

6. Frequ ency Respo nse 6 .2 First-Order Lo w -Pass Fil ters Quiz Exercis e 6.4: A no ther First -Order Lo w -Pass Fil ter


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88

This i s a lso a low-pass f i l ter

L R/2 f where

) j(f /f 1

1 H(f )

is functiontransfer thethat Show

B

Bin

out

V

V


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6. Frequ ency Respo nse 6.3 Decibels and th e Casc ade Con nectio n 6.3.2 Casc ade two -Port Netw ork s


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90

2in

2out

1in

1out

1out

2out

1in

1out

1in

2out

in

out

) f ( H V V

V

V

V

V

V

V

V

V

V

V

)()()( 21 f H f H f H

dB2dB1dB ) f ( H ) f ( H ) f ( H

6. Frequenc y Respo ns e 6 .4 Bod e Plots 6.4 Bod e Plots


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91

2)(1

1)(

B f f f H

6. Frequ ency Respo nse 6 .4 Bo de Plots 6.4 Bo de Plots

ffb k


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92

2)(1

1)(

B f f f H

2

B

dB

) f f ( 1

1log 20 ) f ( H

BdB B

B

f f

log 20 ) f ( H f f For

dB0 H(f) f f For

B f f

f H arctan)(

90 H(f ) , f 10 f For

0 H(f ) /10, f f For

B

B

B f frequencybreak

6. Frequ ency Respo nse 6 .5 First -Order Hig h-Pass Fil ters 6.5 First-Order Hig h-Pass Fil ters


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93

6.5.1 Transfer Fu nc tion

)(1)(

)( B

B

in

out

f f j f f j

V V

f H

RC f B 2

1

2 B B

f f 1

f f ) f H(

B f f

f H arctan90)(

6. Frequ ency Respo nse 6 .5 First -Order Hig h-Pass Fil ters 6.5.2 Bo de Plo ts

2


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94

21)(

B

B

f f

f f f H

21log10log20)(

B BdB f

f f f

f H

0 f H , f 10 f For

90 f H /10, f f For

0 f H , f f For

f f log 20 ) f ( H , f f For

B

B

dB B

BdB B

6. Frequ ency Respo nse 6 .5 First -Order Hig h-Pass Fil ters Exerc is e 6.13 An oth er First -Order High-Pass Fil ter


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95

L R/2 f where

) f f ( j1 ) f f ( j

V V ) f ( H

:iscircuit theof

functiontransfer thethat Show

B

B

B

in

out

6. Frequ ency Respo nse 6.5 First -Order Filters First-Order Lo w -Pass Fil ters


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96

First-Order High -Pass Fil ters


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6. Frequ ency Respo nse 6 .6 Series Reson ances 6.6.1 Reson ant Circ uits (Secon d Ord er)


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98

C f 21

L f 2

f requency" resonant " thecalled is f

f requencytheresistive, purelyis(f ) Z When

00

0

s

LC 2

1 f 0

)CR f 1/2( L/R f 2Q :resistancetheto frequencyresonant theat

inductancetheof reactancetheof ratiotheasdefined is factor qualityThe

00 s

f f

f f

jQ1 R ) f ( Z :asrewrittenbecanimpedancethe Now 00

s s

fC 2

1 j R fL2 j ) f ( Z

is sourcetheby seenimpedanceThe

s

6. Frequ ency Respo nse 6 .6 Series Reson ances 6.6.1 Reson ant Circ uits (Secon d Ord er)


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99

LC 2

1 f frequencyresonant With 0

)CR f 1/2( L/R f 2Q factor qualityand

00 s

f f

f f

jQ1 R 00

s fC 21

j R fL2 j ) f ( Z

is sourcetheby seenimpedanceThe

s

6. Frequ ency Respo nse 6 .6 Series Reson ances 6.6.2 Series Reson ant Circ uits as B and-Pass Fil ter


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100

f f

f f

jQ R f Z s s0

0

1)(

-

RV

) f f f f ( jQ1 R

f Z 00 s s

s

s V V I

) f f f f ( jQ1 R

00 s

s R

V I V

) f f f f ( jQ11

00 s s

R

V

V

rejected others pass, f naer freq.

Filter Pass- Band

) f f

- f f

( Q1 f H

0

-1/2

20

0

2 s

s

R

V

V

6. Frequenc y Respo ns e 6 .6 Series Reson ances 6.6.2 Series Reson ant Circu its as B and -Pass Fil ter


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101

s0 L H /Q f f f B

:asdefined iswidth- Band

L H f and f :asdenoted

21/ f H at occurs

: f requency power -half The

2 B

- f f ,2 B

f f 0 L0 H

-1/2

20

0

2 s

s

R ) f f

- f f

( Q1 f H V

V

6. Frequ ency Respo nse 6 .6 Series Reson ances Exam ple 6.5 Ser ies Resonant Circu i t


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102

Hz 1000 LC 2

1 f : frequency Resonant 0

10 R

L f 2Q: Factor Quality 0 s

Hz 10010

1000Q f

B:width- Band s

0

950Hz 2

B- f f , Hz 1050

2 B

f f 0 L0 H

6. Frequ ency Respo nse 6 .6 Series Reson ances Exam ple 6.5 Ser ies Resonant Circu i t


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103

j1000-C f -J/2 Z

1000 j L f 2 j Z

1000Hz f f requencyresonant At

0c

0 L

0

100 Z Z R Z c L s

001.0100

01 Z s

sV I

01001.0100 R R I V

9010001.01000 j Z L L I V

9010001.01000 j Z C C I V

6. Frequenc y Respo ns e 6.7 Parallel Reso nan ces


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104

6.7 Paral lel Reso nan ce

LC 2

1 f frequencyresonant With 0

CR) f (2 L) f R/(2Q factor qualityand 00 p

6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8 Ideal and Secon d-Order Fil ters

6 8 1IdealFilt ers


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105

6.8.1 Ideal Filt ers

6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8.1 Ideal Filt ers


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106

6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8.2 Secon d-Order L ow -Pass Fil ter


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107

) f f f f ( jQ1 ) f f ( jQ

V V

) f ( H 00 s

0 s

in

out

LC 21 f 0

R L f 2

Q 0 s

CR f 21

0

1Qchoose

passban d theinconstant elyapproximat beto gainthe

want we f ilter,adesign In

s

6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters 6.8.2 Seco nd -Order High -Pass Fil ter


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108


109/114


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6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters Exam ple 6.7 Fi l ter DesignD ig d d filt ith L 50 H th t


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111

Design a second-order filter with L=50mH that passescomponents higher in frequency than 1kHz , rejectscomponents lower than 1kHz .

We need a high-pass filter.To obtain a approximatelyconstant transfer functionIn the pass-band, we choose

LC 2

1 f since

1kHz f select and 1Q

0

0 s

F 507 .0 L f )2(

1C havewe

202

1.314Q

L f 2 R and

s

0

6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters * The Popu lar Sallen-Key Fil ters


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112

6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters


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113

* High er-ord er Fi l ters u s ing Cascade of 2 n d -ord er Fil ters

6. Frequ ency Respo nse 6.8 Ideal and Sec on d-Ord er Filters

* High er-ord er Fi l ters u s ing Cascade of 2 n d -ord er Fil ters


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