AC Phasor 2KM1324 AC Phasor 2

35
Lecture 15 Lecture 15 Sinusoidal Signals, Complex N b Ph Numbers, Phasors ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

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Transcript of AC Phasor 2KM1324 AC Phasor 2

  • Lecture 15Lecture 15Sinusoidal Signals, Complex

    N b PhNumbers, Phasors

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Steady-State Sinusoidal Analysis1. Identify the frequency, angular frequency,

    peak value rms value and phase of apeak value, rms value, and phase of a sinusoidal signal.

    2. Solve steady-state ac circuits using phasors and complex impedancesphasors and complex impedances.

    3 Compute power for steady state ac3. Compute power for steady-state ac circuits.

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Steady-State Sinusoidal Analysisy y

    4. Find Thvenin and Norton equivalent circuits.

    5. Determine load impedances for maximum power transfermaximum power transfer.

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Sinusoidal Currents and Voltagesg)cos()( += tVtv m

    Vm is the peak value

    is the angular frequency in radians per q y psecond

    i h h l is the phase angle

    T is the period

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    T is the period

  • )cos()( += tVtv m

    )()( mHambley mixes units; t in radians, in degrees

    Frequency

    ==sec

    1 cyclesHzT

    f

    2 radians

    A f

    =secT

    Angular frequency

    f 2=

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    ( ) ( )o90cossin = zz

  • Root Mean Square (RMS) Valuesq ( )

    tVtV m )cos()( += The average value of tVtV m )cos()( + cos(t+) is zero. This is the squared qversion of the signal, and its mean value is .

    tVTR

    dtR

    tVT

    pdtT

    PT

    m

    Tm

    T

    Average0

    22

    0

    22

    0

    )(cos11)(cos11

    +=+==

    RVrms

    2000

    =

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    R

  • Root Mean Square (RMS) Valuesq ( )

    ( )dttvT

    VT

    2rms

    1 = ( )dttiTIT

    2rms

    1 =( )T 0rms ( )T 0rms VP

    2rms

    avg = RIP 2rmsavg =Ravg rmsavg

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • RMS Value of a Sinusoid

    V2rmsmVV =

    The rms value for a sinusoid is the peak l di id d b th t f t Thivalue divided by the square root of two. This

    is not true for other periodic waveforms such as square waves or triangular waves.

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • AC Power in the US

    60-Hz 115-V power is a RMS value for60-Hz 115-V power is a RMS value for the voltage. The peak voltage is:

    VVV RMSm 16321152 ===

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Power Delivered to a 50 Resistor

    )100cos(100)( ttv = 100

    50)71.71( 22 W

    RvP rmsavg ===

    100

    m

    m

    vv

    Vv

    ==

    50)100(cos100)()(

    50222 t

    Rtvtp

    R==

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    2RMSv = )100(cos200 2 t=

  • Phasor Definition

    ( ) ( )111 cos :function Time tVtv +=( ) ( )111111 :Phasor V =V

    rePhas rePhas

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Phasor Arithmetic66.85

    2310

    2110)60sin(10)60cos(1060101 jjjZ +=+=+== ooo

    54.354.3225

    225)45sin(5)45cos(54552 jjjZ +=+=+== ooo

    10550455601021 ZZ == ooo

    1524556010

    2

    1

    ZZ =

    = ooo

    2.1254.854.354.366.8521 jjjZZ +=+++=+

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    12.546.154.354.366.8521 jjjZZ +=+=

  • Eulers Formula

    += )sin()cos(j je

    =+=

    )sin()cos()sin()cos(

    j

    j

    je

    je

    =)I ()i (

    )Re()cos(j

    je

    =+==

    1)(sin)(cos

    )Im()sin(22j

    j

    e

    e

    = 1)()(

    je

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Eulers Formula

    )sin()cos( += j je1)sin()cos(

    )sin()cos(

    =+=+=

    j

    j

    je

    je

    11lim....028759045235367182818284.2 +==n

    e

    ....264389793238461415926535.3

    1lim....028759045235367182818284.2

    = +

    n n

    e

    1

    1

    ==

    jej

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Adding Sinusoids Using Phasorsg g

    Step 1: Determine the phasor for each termStep 1: Determine the phasor for each term.

    Step 2: Add the phasors using complex arithmetic.Step 2: Add the phasors using complex arithmetic.

    Step 3: Convert the sum to polar form.Step 3: Convert the sum to polar form.

    Step 4: Write the result as a time function.p

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Using Phasors to Add Sinusoids( ) ( )o45cos201 = ttv ( ) ( )( )

    o60sin102 += ttv ( )( )ooo

    3010

    9060cos10 += t( )o30cos10 = to4520V o45201 =V

    o3010VELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    30102 =V

  • Using Phasors to Add Sinusoidsg

    45201 = oV)45sin(20)45cos(20

    1

    j += oo

    14.1414.142220

    2220 jj ==

    30102 = oVj

    5668110310

    )30sin(10)30cos(10 += oo

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    jj 566.82110

    2310 ==

  • Using Phasors to Add Sinusoids

    566.814.1414.1421s

    +=+= VVV

    jj

    g

    14.1980.22 = jjj

    8.29)14.19()80.22( 22

    =+=Vs

    o01.408.2214.19

    8.2214.19)( 1 =

    == TanTan s

    o01.408.29s =V

    ( ) ( )ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    ( ) ( )o01.40cos8.29 = ttvs

  • Exercise 5.4

    )90cos(10)cos(10

    )sin(10)cos(10)(1o+=

    +=tt

    tttv

    [ ][ ]00110

    )90sin()90cos()0sin()0cos(109010010

    ++=+++=

    +=

    jjj

    V

    [ ][ ]110

    00110=

    ++j

    j

    45)1(11)(

    14.142101010

    1

    22

    o=====+=

    TanTan V

    4514.141

    o=V

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    )45cos(14.14)(1o= ttv

  • Exercise 5.4

    6053010

    )60cos(5)30cos(10

    )30sin(5)30cos(10)(

    oo

    oo

    oo

    +=++=

    +++=tt

    ttti

    V[ ] [ ]

    )23

    215

    21

    2310

    )60sin()60cos(5)30sin()30cos(106053010

    +

    +=

    +++=+=

    jj

    jjV

    67.02.1133.45.2566.8

    )2222

    +=++=

    jjj

    jj

    67.067.0

    2.11)67.0()2.11(

    1

    22

    o=+=V

    42.32.11

    42.32.11

    67.02.11

    67.0)( 1

    o

    o

    ==

    == TanTan

    V

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    )42.3cos(2.11)( o+= tti

  • Exercise 5.4

    6015020)60cos(15)cos(20

    )60cos(15)90sin(20)(2o

    oo

    +=

    ++=tt

    ttti

    V[ ] [ ][ ] 31

    )60sin()60cos(15)0sin()0cos(206015020

    +++=

    +=jj

    V

    [ ]135.27135.720

    23

    21150120

    =+=

    ++=

    jj

    jj

    1313

    4.30)13()5.27( 22

    =+=V

    3.255.27

    135.27

    13)( 1 =

    == TanTan

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    )3.25cos(4.30)(2o= tti

  • Phase RelationshipsSinusoids can be visualized as the real-visualized as the realaxis projection of vectors rotating in thevectors rotating in the complex plane. The phasor for a sinusoidphasor for a sinusoid is a snapshot of the corresponding rotatingcorresponding rotating vector at t = 0.

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Phase RelationshipsTo determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise. Then when standing at aThen when standing at a fixed point, if V1 arrives first followed by V2 after a rotation of we say that Vrotation of , we say that V1leads V2 by . Alternatively, we could say that V2 lags V1by . (Usually, we take as the smaller angle between the two phasors.)

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    wo p so s.)

  • Phase RelationshipsTo determine phase relationships betweenrelationships between sinusoids from their plots versus time, find the shortest time interval tpbetween positive peaks of th t f Ththe two waveforms. Then, the phase angle is = (tp/T ) 360 If the peak of v1(t) 360 . If the peak of v1(t)occurs first, we say that v1(t)leads v2(t) or that v2(t) lags

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    2( ) 2( ) gv1(t).

  • Complex Impedances-Inductorp p

    mL titi += )sin()( m

    LL

    mL

    tLidt

    tdiLtv

    titi

    +== )cos()()()s ()(

    i

    dt

    I = 90LLmL

    mL

    LLjZLjLLi

    iIIV

    I

    ===

    =

    90)90(

    90

    L LLjZ == 90

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

    LLL Z IV =

  • Complex Impedances InductorComplex Impedances-Inductor

    LL Lj IV = o90== LLjZL jL

    Z IV LLL Z IV =

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Complex Impedances-Capacitor

    Z IV

    p p p

    CCC Z IV =o90111 ===

    CCjCjZC CCjC

    RIV RR RIV =

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Impedances-Resistorp

    RR RIV =

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Exercise 5.5

    o

    o

    )30(10)(

    )30cos(10)(1 =tt

    ttv

    oo

    o

    )45cos(10)45sin(10)(

    )30cos(10)(

    3

    2

    =+=+=

    tttv

    ttv

    o30101 =V V1 lags V2 by 60 Find the phasor for each voltage and state phase relationship between them

    o

    o

    4510

    30102

    1

    +=

    V

    V1 2

    V2 leads V3 by 75 V l d V b 15 45103 =V V1 leads V3 by 15

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Exercise 5.6

    250)200cos(100

    ==L

    HLtv Find the impedance of the inductance, the phasor current and the phasor

    voltage. Draw the phasor diagram.

    905050)250)(200(0100

    25.0

    =

    =L

    jHjLjZ

    HLV

    voltage. Draw the phasor diagram.

    90290500100

    905050)25.0)(200(

    ===

    ====L

    L

    L

    Z

    jHjLjZVI

    9050LL Z

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Exercise 5.7

    100)200cos(100

    ==c

    FCtv

    Find the impedance of the capacitor, the phasor current and the phasor voltage. Draw the phasor diagram.

    1110100

    100

    =

    =C

    j

    FCV

    voltage. Draw the phasor diagram.

    9020100

    9050102)10100)(200(

    11126

    =====

    C

    C xj

    xj

    Cj

    CjZ

    V

    90290500100 +=

    ==C

    CC Z

    VI

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

  • Exercise 5.8

    50)200cos(100)(

    ==R

    Rttv Find the impedance of the resistor, the

    phasor current and the phasor voltage. Draw the phasor diagram.

    0500100

    50

    =

    =R

    RZ

    RV

    Draw the phasor diagram.

    020100050

    =====

    CR

    R RZVI 02

    050RR Z

    I

    ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.