Lecture 15Lecture 15 Sinusoidal Signals, Complex Nb ... 15.pdf · Lecture 15Lecture 15 Sinusoidal...

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.


Steady-State Sinusoidal Analysisy y

4. Find Thévenin and Norton equivalent circuits.

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


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

Vm is the peak value

ω is the angular frequency in radians per q y psecond

θ i h h lθ is the phase angle

T is the period


T is the period

)cos()( θω += tVtv m

⎤⎡

)()( m

Hambley mixes units; ωt in radians, θ in degrees

Frequency ⎥⎦⎤

⎢⎣⎡ ==

sec1 cyclesHzT

f

⎤⎡2 radiansπ

⎦⎣

A f ⎥⎦⎤

⎢⎣⎡=

secTωAngular frequency

fπω 2=


( ) ( )o90cossin −= zz

Root Mean Square (RMS) Valuesq ( )

tVtV m )cos()( += ϑω The average value of tVtV m )cos()( +ϑωcos(ωt+ϕ) is zero.

This is the squared qversion of the signal, and its mean value is ½.

tVTR

dtR

tVT

pdtT

PT

m

Tm

T

Average0

22

0

22

0

)(cos11)(cos11⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

+== ∫∫∫ ϑω

ϑω

RVrms

2000

=

⎦⎣


R

Root Mean Square (RMS) Valuesq ( )

( )dttvT

VT

2rms

1∫= ( )dtti

TI

T2

rms1

∫=( )T 0

rms ∫ ( )T 0

rms ∫

VP2

rmsavg = RIP 2

rmsavg =Ravg rmsavg


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.


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 ===


Power Delivered to a 50Ω Resistor

)100cos(100)( ttv = π100

50)71.71( 22

WR

vP rms

avg ===

100

m

m

vv

Vv

=

=

50)100(cos100)()(

50222 t

Rtvtp

Rπ

==


2RMSv =

)100(cos200 2 tπ=

Phasor Definition

( ) ( )111 cos :function Time θtωVtv +=( ) ( )111

111 :Phasor θV ∠=V

rePhas rePhas


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

∠=∠∠

= oo

o

2.1254.854.354.366.8521 jjjZZ +=+++=+


12.546.154.354.366.8521 jjjZZ +=−−+=−

Euler’s Formula

θθθ += )sin()cos(j je

θθ

θθθ −=

+=− )sin()cos(

)sin()cos(j

j

je

je

θ

θθ

θ=

)I ()i (

)Re()cos(j

je

θθ

θθ

θ

=+=

=

1)(sin)(cos

)Im()sin(22j

j

e

e

θθ ∠= 1

)()(je


Euler’s Formula

)sin()cos( +=θ θθj je

1)sin()cos(

)sin()cos(

−=+=

+=π ππ

θθj

j

je

je

11lim....028759045235367182818284.2 ⎟⎞

⎜⎛ +==

n

e

....264389793238461415926535.3

1lim....028759045235367182818284.2

=

⎟⎠

⎜⎝

+∞→

πn n

e

1

1

−=

−=πje

j


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


Using Phasors to Add Sinusoids( ) ( )o45cos201 −= ttv ω

( ) ( )( )

o60sin102 += ttv ω

( )( )o

oo

3010

9060cos10 −+= tω

( )o30cos10 −= tω

o4520∠V o45201 −∠=Vo3010∠V


30102 −∠=V

Using Phasors to Add Sinusoidsg

45201 −∠= oV

)45sin(20)45cos(201

j −+−= oo

14.1414.142220

2220 jj −=−=

30102 −∠= oV

j

5668110310

)30sin(10)30cos(10 −+−= oo


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 θθ

o01.408.29s −∠=V

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

( ) ( )o01.40cos8.29 −= ttvs ω

Exercise 5.4

)90cos(10)cos(10

)sin(10)cos(10)(1o−+=

+=

tt

tttv

ωω

ωω

[ ][ ]00110

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

−++=−+−++=

−∠+∠=

jjj

V

[ ][ ]110

00110−=

++j

j

45)1(11)(

14.142101010

1

22

o−=−=−

=

==+=

−TanTan θθ

V

4514.141

o−∠=V


)45cos(14.14)(1o−= ttv ω

Exercise 5.4

6053010

)60cos(5)30cos(10

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

oo

oo

oo

−∠+∠=

−++=

+++=

tt

ttti

ωω

ωω

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 θθ

V


)42.3cos(2.11)( o+= tti ω

Exercise 5.4

6015020)60cos(15)cos(20

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

oo

∠∠−+=

−++=

tt

ttti

ωω

ωω

V[ ] [ ]

[ ] 31

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

⎥⎤

⎢⎡

−+−++=−∠+∠=

jjV

[ ]

135.27135.720

23

21150120

−=−+=⎥⎥⎦⎢

⎢⎣

−++=

jj

jj

1313

4.30)13()5.27( 22

⎞⎛

=+=V

3.255.27

135.27

13)( 1 −=⎟⎠⎞

⎜⎝⎛ −

=→−

= −TanTan θθ


)3.25cos(4.30)(2o−= tti ω

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.


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 θ we say that Vrotation of θ , we say that V1leads V2 by θ . Alternatively, we could say that V2 lags V1by θ . (Usually, we take θ as the smaller angle between the two phasors.)


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 θ = (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


2( ) 2( ) gv1(t).

Complex Impedances-Inductorp p

mL titi += )sin()( θω

mL

L

mL

tLidt

tdiLtv

titi

+== )cos()()(

)s ()(

θωω

θω

i

dt

I −∠= 90θ

LLmL

mL

LLjZLjLLi

iIIV

I

∠=∠=∠=

−∠=

90)90(

90ωωθω

θ

L LLjZ ∠== 90ωω


LLL Z IV =

Complex Impedances InductorComplex Impedances-Inductor

LL Lj IV ×= ω

o90∠== LLjZL ωωjL

Z IV LLL Z IV =


Complex Impedances-Capacitor

Z IV

p p p

CCC Z IV =

o90111−∠==−=

CCjCjZC ωωω CCjC ωωω

RIV RR RIV =


Impedances-Resistorp

RR RIV =


Exercise 5.5

o

o

)30(10)(

)30cos(10)(1 −=

tt

ttv ω

oo

o

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

)30cos(10)(

3

2

−=+=

+=

tttv

ttv

ωω

ω

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 °


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(

−∠=∠∠

==

∠====

LL

L

Z

jHjLjZVI

ω

9050∠LL Z


Exercise 5.7

100)200cos(100

==c

FCtv

μ

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

1110100

100

−∠=

=

C

j

FCV

μ voltage. Draw the phasor diagram.

9020100

9050102)10100)(200(

11126

∠∠

−∠==−=−==−−

C

C xj

xj

Cj

CjZ

Vωω

90290500100

+∠=−∠∠

==C

CC Z

VI


Exercise 5.8

50)200cos(100)(

Ω==R

Rttv Find the impedance of the resistor, the

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

0500100

50

∠∠=

Ω=

R

RZ

RV

Draw the phasor diagram.

020100050

∠=∠

==

∠==

CR

R RZV

I 02050

∠∠R

R ZI


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