3.1

Electric Circuits 1301ENG

Module 3: Sinusoidal AC Analysis

References[1] R. L. Boylestad, Introductory Circuit Analysis, PrenticeHall, 8th ed., 1997, chapters 13 and 14.[2] A. R. Hambley, Electrical Engineering: Principles andApplications, Prentice Hall, 1997, chapter 5.

3.2

3.1 IntroductionWe have so far considered dc (direct current) signals whoseamplitudes do not change with time

Now consider ac (alternating current) signals whoseamplitudes change with time

• ac means sinusoidally varying voltage or current• ac voltage means a sinusoidally varying voltage(whether or not there is any current)

Sinusoidal ac signals are important because• electrical power is normally generated and distributed in

ac form eg. power is distributed to householdpowerpoints as a 50-Hz 240-V ac voltage

• many practical signals are sinusoids or can be expressedas a sum of sinusoids

• sinusoidal “carriers” are used for modulation in radioand other forms of telecommunications

• circuits are commonly specified and tested usingsinusoidal signals with a range of different frequencies

3.2 Sinusoidal waveform basics

Definitions

Figure 1 shows a sinusoidal voltage waveform as afunction of time t − vertical scaling is in volts (v) − horizontal scaling is always in units of time (eg. second(s))

3.3

VmVp-p

v t( )

t0

T1

T3T2

Vm

Figure 1

1 s 1 s 1 s

T = 0 4. s T = 0 5. s(a) (b) (c)

Figure 2

Peak amplitude Vm: maximum value measured fromaverage or mean. Mean value is zero in Fig. 1.

Peak-to-peak value Vp-p: sum of positive and negative peak

amplitudes. V Vp-p m= 2 in Fig. 1.

Period T: time interval between successive repetitions ofsimilar points. Period T T T1 2 3= = in Fig. 1. Thus,sinusoidal waveforms are periodic, repeating same patternof values in each period.

Cycle: portion of waveform in one period. Frequency f:

3.4

• Number of cycles in one second

• Frequency is 1 cycle per second in Fig. 2(a), 212

cycles

per second in Fig. 2(b) where T = 0 4. s, 2 cycles persecond in Fig. 2(c) where T = 0 5. s.

⇒ Large frequency means large number of cycles per second ⇒ Fig. 2(b) shows a higher frequency waveform thanthose in Figs. 2(a) and (c)

• Frequency has unit of hertz

1 1 hertz (Hz) = cycle per second

• Different frequencies in everyday life applications Sound (human ear): 15 Hz−20 kHz FM radio: 88 MHz−108 MHz AM radio: 540 kHz−1600 kHz TV: 54 MHz−890 MHz

• Frequency f is inversely proportional to period T

fT

= 1

f

T

==

Hz

seconds (s)

Tf

= 1

eg. A 50-Hz ac voltage has a period of 20 ms.

Radians/Degrees

3.5

Unit of measurement for horizontal axis (abscissa) of Figs.1 and 2 are sometimes expressed in radians (rad) or degrees

There are 2π rad around a 360o circle (see Fig. 3)

2 360π rad = o

π ≅ 314.

1 57 3 rad = . o

In sinusoidal analysis, angle is measured as positive from

horizontal axis of a circle (ie. from 0o).

23

45 6

0o

90o

180o

270o 3 2π rad

(4.71 rad)

2π rad

(6.28 rad)

π rad

(3.14 rad)

π 2 rad

(1.57 rad)

1 rad

0 28. rad

Figure 3 Conversion between radians and degrees:

3.6

Radians degrees)=

×π180o

(

Degrees radians)=

×180o

π(

Example 1

90°: Radians )= π180

90o

o( =π2

rad

π3

rad: Degrees =

1803

o

ππ

=60o

Mode in your calculator may be in RAD or DEG.

Angular velocity Angular velocity ω is velocity of radius vector rotatingabout center of circle (see Fig. 4).

Angular velocity = angular distance (radians)

time taken (seconds)ω θ=

t

ω

θ

Figure 4

3.7

Time required to complete one revolution (or cycle) isperiod t T= seconds and angular distance travelled in thistime interval is θ π= 2 radians. Thus,

ωπ= 2

T ( rad s)

Thus, smaller T gives larger ω (ie. rotating vector takesless time to complete one cycle of 2π radians and hence itmust travel at a faster speed).

Also, angular frequency ω is proportional to frequency f

ω π= 2 f ( rad s)

eg. angular velocity of a 50-Hz sinusoidal ac voltage isω π= =2 50 314( ) rad s.

3.3 Sinusoidal waveforms By convention, cosine (rather than sine) waveform is usedin sinusoidal analysis.

A sinusoidal voltage is generally given by:

( )v t V t( ) cos= +m ω θ

Vm = peak amplitude of voltage (v) ω π π= = =2 2T f angular frequency (rad/s) θ = phase (or displacement) angle (rad or degrees)

3.8

Lowercase letter represents time-varying or ac signal (eg.v t( )). Capital letter represents amplitude (eg. Vm)

Case 1: ( )v t V t1( ) cos= m ω for θ = 0

For one cycle (ie. 0 360≤ ≤ωt o), Fig. 5(a) shows that v t1( )has:

• maxima of Vm at ωt = 0 and ωt = 360o

• minima of −Vm at ωt = 180o

• zero value at ωt = 90o and ωt = 270o

• thus, these maxima, minima and zero value occur at

ωt = 0 90o 180o 270o 360o

↑ ↑ ↑ ↑ ↑ v t1( ) = Vm 0 −Vm 0 Vm

Case 2: ( )v t V t2 90( ) cos= +m ω o for θ = 90o

Similar to Case 1, maxima, minima and zero value of v t2( )for one cycle occur at:

ωt + =90o 0 90o 180o 270o 360o

ωt = − 90o 0 90o 180o 270o

↑ ↑ ↑ ↑ ↑ v t2( ) = Vm 0 −Vm 0 Vm

The dotted curve of Fig. 5(b) shows a plot of v t2( ) .

Compared to v t1( ) (solid curve of Fig. 5(b)), waveform

v t2( ) (dotted curve) is shifted to left of time axis by 90o .

3.9

Thus, v t2( ) leads v t1( ) by 90o (or v t1( ) lags v t2( ) by 90o)because peaks of v t2( ) occur before peaks of v t1( ) .

Note that ( )v t t2( ) sin= − ω and hence

( ) ( )sin cosω ωt V t= − +m 90o

Case 3: ( )v t V t3 90( ) cos= −m ω o for θ = −90o

Compared to v t1( ) (solid curve of Fig. 5(c)), waveform

v t3( ) (dotted curve) is shifted to right of time axis by 90o .

v t3( ) lags v t1( ) by 90o (or v t1( ) leads v t3( ) by 90o)because peaks of v t3( ) occur after peaks of v t1( ) .

Note that ( )v t t3( ) sin= ω and hence

( ) ( )sin cosω ωt V t= −m 90o

We can thus deduce that:

− Cosine wave leads sine wave by 90o

− Sine wave lags cosine wave by 90o

− Cosine and sine waves are 90o out of phaseRemarks• θ > 0 means phase lead• θ < 0 means phase lag• For ( )v t V t( ) cos= +m ω θ and given v t V( ) = we have:

3.10

ω θtV

V=

−−cos 1

m

(a) θ = 0

ωt

v t( )

0 90o

180o

270o 360o− 90o

−Vm

Vm ( )v t V t1( ) cos= m ω

450o

(b) θ = 90o

ωt

v t( )

0 90o 180o 270o 360o− 90o

− Vm

Vm

( )v t V t1( ) cos= m ω

90o

90o

− 180o

90o

450o

( )v t V t2 90( ) cos= +m ω o

Figure 5

(c) θ = −90o

3.11

ωt

v t( )

0 90o 180o 270o 360o− 90o

−Vm

Vm ( )v t V t1( ) cos= m ω

90o

90o90o

450o 540o

( )v t V t3 90( ) cos= −m ω o

Figure 5 continued

Relationship between Cosine and SineFigure 6 shows geometric relationship between variousforms of sine and cosine functions

• Starting at + sinα position, + cosα is an additional 90o

in counterclockwise direction. Thus,

( )cos sinα α= + 90o

• Starting at + sinα position, we must travel 180o in

counterclockwise (+ 180o) or clockwise (− 180o ) to get to− sinα . Thus,

( )− = ±sin sinα α 180o

3.12

• Using Fig. 6, various sine-cosine relationships can beobtained:

( )cos sinα α= + 90o

( )sin cosα α= − 90o

( )− = ±sin sinα α 180o

( ) ( )− = + = −cos sin sinα α α270 90o o

etc.

Also, sine is an odd function and cosine is an evenfunction:

sin( ) sin− = −α α

cos( ) cos− =α α

+ sinα− sinα

− cosα

+ cosα

Figure 6

Example 2

Given α and using ( )sin cosα α= − 90o and

cos( ) cos− =α α we have:

3.13

α = 0: sin cos( ) cos( )0 90 90 0o o o= − = =

α = 30o : sin cos( ) cos( ) .30 60 60 0 5o o o= − = =

α = 45o: sin cos( ) cos( ) .45 45 45 1 2 0 707o o o= − = = =

α = 60o : sin cos( ) cos( ) .60 30 30 3 2 0 866o o o= − = = =

α = 90o : sin cos( )90 0 1o = =

3.4 Average valueAverage value of any current or voltage is value indicatedon a dc meter

• average value is equivalent dc value over one cycle

When both dc and ac voltage sources are applied to anelectrical circuit

• need to know dc and ac components of voltage orcurrent in various parts of circuit

dc (or average value)Sum of areas over one period

one period =

T

dc value of rectangular waveDetermine dc components of waveforms of Fig. 7.

In Fig. 7(a):By inspection, area above axis equals area below over oneperiod giving dc = 0 V. Also, using the above dc equation:

3.14

dc(10 V)(1 ms) (10 V)(1 ms)

2 ms V

= −

= 0

In Fig. 7(b):

dc(14 V)(1 ms) (6 V)(1 ms)

2 ms V

= −

= 4

In fact, waveform of Fig. 7(b) is square wave of Fig. 7(a)with a dc shift of 4 V:

v v2 1 4= + V

(a)

0 1 2 3 4

10 V

−10 V

(Square wave)

t (ms)

v1(V)

1 period

Figure 7(a)

3.15

0 2 4 t (ms)1 3

14 V

− 6 V

v2(V)

4 V

dc value = 4 V

1 period

Figure 7(b)

dc value of sinusoidal waveformsArea under one positive (or negative) half of a sine wave(see Fig. 8) is:

[ ]

[ ][ ]

Area

m

m

m

m

m

= ∫

= −

= − −

= − − − +

=

A d

A

A

A

A

sin

cos

cos cos

( )

α α

α

π

π

π0

0

0

1 1

2

o

o

average value mm= =2

0 637A

Aπ

.

dc or average value of a pure sinusoidal waveform overone period is zero

3.16

Am

0 πFigure 8

Example 3Determine dc value of sinusoidal waveform of Fig. 9.

Peak-to-peak value = 16 mV + 2 mV = 18 mVPeak value = 18 mV/2 = 9 mVThus, dc value is 9 mV down from 2 mV (or 9 mV up from-16 mV) ⇒ dc = -7 mV.

0

v

t

+ 2 mV

−16 mV

dc mV= −7

Figure 9

3.17

Instrumentation to measure dc or average valuedc (or average value) of any waveform can be measuredusing a digital multimeter (DMM) or an oscilloscope.• DMM can measure voltage or current levels of anywaveform (dc or ac signal) by setting to dc mode• Oscilloscope can only measure voltage levels

Oscilloscope to measure dc levels of dc circuits1. Choose GND (ground) from DC-GND-AC mode− GND mode blocks any signal into oscilloscope− A horizontal line is shown− Set horizontal line in middle of screen (see Fig. 10(a))

2. Apply oscilloscope probe to voltage to be measured andswitched to DC mode− If dc voltage is present, horizontal line will shift up(positive voltage) or down (negative voltage). (Fig. 10(b))− Positive voltage ≡ higher potential at red or positive lead− Negative voltage ≡ lower potential at red or positive lead

3. Measure voltage using:

Vdc vertical shift in div. vertical sensitivity in V div.= ×( ) ( )

For waveform of Fig. 10(b),

Vdc div. mV div. mV= =( . )( )2 4 50 120

3.18

Vertical sensitivity=50 mVdiv.(a) (b)

Shift=2.4 div.

Figure 10

Oscilloscope to measure dc or average level of anywaveform1. Using GND, reset horizontal line to middle of screen

2. Switch to AC− only passes ac (alternating or changing) components ofsignal (see Fig. 11(a))− blocks dc components of signal

3. Then switch to DC− passes both dc and ac components of signal− ac waveform of Fig. 11(a) is shifted up to give waveformconsisting of dc and ac components of signal (see Fig.11(b)− Average value of waveform of Fig. 11(b) is:

V Vaverage dc div. V div. V= = =( . )( )08 5 4

3.19

(a)

Shift=0.8 div.Reference

level

(b)

Figure 11

3.5 Effective values (or root mean square (rms))To determine amplitude of a sinusoidal ac current requiredto deliver same power as a dc current using Fig. 12.

ac generator

Switch 2

dc source

Switch 1 R

vV

Water bath

Thermometer

Idc

i I tac m= sinω

Figure 12

If switch 1 only is closed, dc power is delivered to resistorR by dc current Idc

If switch 2 only is closed, ac power is delivered to resistorby ac peak current Im

3.20

For same temperature, average power delivered to resistorby ac source is same as that delivered by dc source.

The ac power delivered to resistor at any instant of time is:

( )P i R I t R I t Rac ac m m2= = =( ) sin ( sin )2 2 2ω ω

sin ( cos )2 12

1 2ω ωt t= − (trigonometric identity)

PI R I R

tacm2

m2

= −2 2

2cos ω

Average power delivered by ac source is just first termsince average value of a sine wave is zero.

Equating average power delivered by ac source to thatdelivered by dc source:

P Pav(ac) dc=

I R

I Rm2

dc2

2= and I Im dc= 2

II

Idcm

m2= = 0 707.

The equivalent dc value of a sinusoidal current or voltageis 1 2 or 0.707 of its maximum value regardless of signalfrequency

3.21

Equivalent dc value = effective value = root mean square(rms) value

In summary: I I II

Irms eff eq dcm

m2= = = = 0 707.

I I Im rms eff= =2 2

Example 4An ac current with peak value of 2 10 14 14( ) .= A isrequired to deliver same power to resistor in Fig. 12 as a dccurrent of 10 A.

rms or effective value of any time-varying waveform (notnecessarily a sinusoidal wave) is:

I I

i t

T

T

rms eff= =∫ 2

0( )

I Ii t

Trms effarea( )= =

2( )

Note that the above rms or effective value equations arealso applicable to time-varying voltages.

Example 5Find the rms or effective value of waveform of Fig. 13(a).

3.22

v t2( ) curve is shown in Fig. 13(b).

V Vrms eff V= = + + =( )( ) ( )( ) ( )( ).

100 2 16 2 4 210

4 899

02 4

(a)

− 10

1 Period

t (s)

v V)(

86 10− 2

4

0 2 4

100

t (s)

v2 V)(

86 10

41 6

(b)

Figure 13

3.6 ac meters and instruments

Digital Multimeters (DMMs)dc meters can be used to measure sinusoidal ac voltagesand currents by using a bridge rectifier which consists offour diodes or electronic switches (see Fig. 14).

DMMs operate using this technique.

3.23

D2

D3

D4

+−

+

−

vi

vmovementD1

Figure 14

vi

Vm

−Vm

0 π 2π α

(a)

vmovement

Vm

0 π 2π α

(b)

V Vdc m= 0 637.

Figure 15

vi > 0 for 0 ≤ ≤α π (see Fig. 15(a))• Diodes D1 and D2 are conducting (ie. they are equivalentto short circuits) and D3 and D4 are not conducting (ie.open circuits)• v vimovement = for 0 ≤ ≤α π (see Fig. 15(b)).

vi < 0 for π α π≤ ≤ 2 (see Fig. 15(a))• D1 and D2 are open circuits and D3 and D4 short circuits

3.24

• v vimovement = − for π α π≤ ≤ 2 ; negative portion ofinput is “flipped over” (see Fig. 15(b)).

Zero average value of Fig. 15(a) has been replaced by awaveform of Fig. 15(b) having an average or dc value of:

VV V

Vdcm m

m= + =2 22

0 637π

.

Movement of pointer is directly related to peak value ofsignal by factor 0.637.

Ratio between rms and dc levels:

V

V

V

Vrms

dc

m

m0.637= ≅0 707

111.

.

V Vrms dc = 111.

Meter indication = 1.11 (dc or average value)

dc or average reading multimeter is then scaled 1.11 to readrms value (voltage or current).

Example 6Determine the reading of a DMM placed across a resistorwith Vm=20Vac.Solution:

V Vrms m V) V= = =0 707 0 707 20 1414. . ( .

3.25

OscilloscopesIn general, a signal consists of both dc and ac components

Switch or knob with DC/GND/AC mode (see Fig. 16)

DC (Direct Coupling) mode• DC means Direct Coupling but not measuring dc (oraverage) value of quantity• DC mode directly couples complete signal consisting ofdc and ac components to display

AC (Alternating Current) mode• only passes ac component of signal• dc component of signal is blocked by capacitor

GND (Ground)• Input signal is blocked by direct ground connection

− A single horizontal line will be shown on display

AC

DC

GND

Oscilloscope

Display AC

GND

DC

Capacitor

Input signal

Figure 16

3.26

3.7 Complex Numbers

3.7.1 IntroductionHow do we find the algebraic sum of voltages and currentsthat are varying sinusoidally?

One solution: find algebraic sum on a point-to-point basis − A long and tedious process

Phasor analysis• A quick, direct and accurate technique for findingalgebraic sum of sinusoidal ac waveforms• Allows same techniques used for dc networks to be usedfor sinusoidal ac networks• Makes use of complex numbers− need to consider complex numbers first which provide amathematical tool for phasor analysis (Section 3.9)

Complex number• represents a point in a two-dimensional plane (Fig. 17)

− real numbers are on real (Re) horizontal axis− complex numbers on imaginary (Im) vertical axis

Im( j)

Re

+

+

−

−

Figure 17

3.27

Two forms to represent complex numbers• Rectangular form: represents a point in the plane• Polar form: represents a radius vector from origin to point

Definition

• The operator j is a point on imaginary axis and has a 90o

counterclockwise rotation

j j2 1 1= − = − or

3.7.2 Rectangular formA complex number in rectangular form (see Fig. 18(a)) is:

C A jB= +

where A and B are real numbers. See example in Fig. 18(b)

Im

ReA

B

C A jB= +

(a )

Im

Re

C j= − −10 20( )b

0− 10

− 20

Figure 18

3.28

3.7.3 Polar formA complex number in polar form (see Fig. 19(a)) is:

C C= ∠θ

∠θ = = +e jjθ θ θcos sin

where C is magnitude of C and θ is a positive anglemeasured counterclockwise from positive real axis.

A negative sign in front of polar form gives a complexnumber directly opposite to complex number with apositive sign (see Fig. 19(b)):

− = − ∠θ = ∠θ ±C C C π

Im

Re

C

(a )

C

θ

Im

Re

C

(b)

θπ

− π− C

Figure 19

3.29

Example 7

Im

Re

C= ∠5 30o

(a )

5

θ = 30o

Im

Re

( )bC = ∠−7 120o

7 θ = −120o

Figure 20

3.7.4 Conversion between formsThe two forms are related by the following equations.

Rectangular to polar form (See Fig. 21).

C A B= +2 2

θ = −tan 1 B

A

Polar to rectangular form (See Fig. 21).

A C= cosθ

B C= sinθ

3.30

Im

ReA

B

C C A jB= ∠θ = +

θ

C

Figure 21

Example 8Convert the following from rectangular to polar form:

C j= +3 4 (Fig. 22(a))Solution:

C = + = =3 4 25 52 2

θ =

=−tan .1 4

35313o

Therefore C = ∠5 5313. o

Example 9Convert the following from polar to rectangular form:

C = ∠10 45o (Fig. 22(b))

Solution:

A = = =10 45 10 0 707 7 07cos ( )( . ) .o

B = = =10 45 10 0 707 7 07sin ( )( . ) .o

Therefore C j= +7 07 7 07. .

3.31

Im

Re3

4

C j= +3 4

θ

C

Im

Re

C = ∠10 45o

(b)

10

45o

(a)

Figure 22

If complex number is in 2nd, 3rd, or 4th quadrant• determine angle of vector in that quadrant• then determine angle of vector relative to horizontal realaxis

Example 10Convert the following from rectangular to polar form:

C j= − +6 3 (Fig. 23(a))Solution:

C = + =6 3 6 712 2 .

β =

=−tan .1 3

626 57o

θ = − =180 26 57 15353o o o. .

Therefore C = ∠6 71 153 43. . o

3.32

Im

Re6

3

C j= − +6 3

β

C

C = ∠10 230o(a)

θ

Im

Bβ

(b)

A

10

θ = 230o

Re

Figure 23

Example 11Convert the following from polar to rectangular form:

C = ∠10 230o (Fig. 23(b))Solution:

A C= = − = =cos cos( ) cos .β 10 230 180 10 50 6 428o o o

B C= = =sin sin .β 10 50 7 66o

Therefore C j= − −6 428 7 66. .

3.8 Mathematical operations with complex numbersSimilar to real numbers, we can also perform addition,subtraction, multiplication and division on complexnumbers.

Consider the following definitions for the symbol jassociated with imaginary part of complex numbers:

j j2 1 1= − = − or

3.33

Thus j j j j j3 2 1= = − = − j j j4 2 2 1 1 1= = − − = +( )( )

j j5 =

Also, 1 1

12j j

j

j

j

j

j=

= =−

Thus 1j

j= −

3.8.1 Complex conjugateConjugate or complex conjugate of a complex number canbe found by:• changing the sign of imaginary part in rectangular form• putting a negative of the angle in polar form

Example 12The conjugate of C j= +2 3 is 2 3− j . See Fig. 24(a).

The conjugate of C = ∠2 30o is 2 30∠ − o . See Fig. 24(b)

Re2

3

C j= +2 3 Im

Re

C

(b)

2

30o

− 3

Conjugate of

C

j2 3−(a)

2

− 30o

Conjugate of C

Im

Figure 24

3.34

3.8.2 AdditionTo add two or more complex numbers, just add real andimaginary parts separately. For example, if

C A jB1 1 1= ± ± and C A jB2 2 2= ± ±

then C C A A j B B1 2 1 2 1 2+ = ± ± + ± ±( ) ( ) (see Fig. 25(a))

Note in Fig. 25(a) that the resultant vector C C1 2+ isobtained by adding the head of one vector (eg. C1) with thetail of another vector (eg. C2).

Example 13(a) Add C j1 2 4= + and C j2 3 1= + (see Fig. 25(a))Solution: C C j j1 2 2 3 4 1 5 5+ = + + + = +( ) ( )(b) Add C j1 3 6= + and C j2 6 3= − +Solution: C C j j1 2 3 6 6 3 3 9+ = − + + = − +( ) ( )

3.8.3 SubtractionIn subtraction, real and imaginary parts are againconsidered. For example, if

C A jB1 1 1= ± ± and C A jB2 2 2= ± ±

then [ ] [ ]C C A A j B B1 2 1 2 1 2− = ± − ± + ± − ±( ) ( )

(see Fig. 25(b))

3.35

Re

C C1 2+

(a)

Im

0 2 4 6

2

4

6C1

C2 ReC C1 2−

(b)

Im

0 2 4 6

2

4

6 C1

C2

− 2

− C2

Figure 25

Example 14(a) Subtract C j2 1 4= + from C j1 4 6= + (see Fig. 25(b))Solution: C C j j1 2 4 1 6 4 3 2− = − + − = +( ) ( )(b) Subtract C j2 2 5= − + from C j1 3 3= +Solution: [ ]C C j j1 2 3 2 3 5 5 2− = − − + − = −( ) ( )

3.8.4 Addition or subtraction in polar form

Addition or subtraction cannot be performed in polar formunless the complex numbers have the same angle θ or

differ only by multiples of 180o .

Example 15

(a) 2 45 3 45 5 45∠ + ∠ = ∠o o o (see Fig. 26(a))

(b) 2 0 4 180 6 0∠ − ∠ = ∠o o o (see Fig. 26(b))

3.36

Im

Re

(a)

45o23 5

Im

Re

(b)

26

− ∠4 1 8 0o

4 1 8 0∠ o

Figure 26

3.8.5 Multiplication

Rectangular formMultiply real and imaginary parts of one in turn by real andimaginary parts of the other. For example, if

C A jB1 1 1= + and C A jB2 2 2= +

then

C C A jB A jB

A A jB jB A jB

A A jA B jB A B B

1 2 1 1 2 2

1 2 2 1 2 2

1 2 1 2 1 2 1 2

⋅ = + += + + += + + −

( )( )

( ) ( )

C C A A B B j A B B A1 2 1 2 1 2 1 2 1 2⋅ = − + +( ) ( )

Example 16Find C C1 2⋅ if C j1 2 3= − − and C j2 4 7= −

3.37

Solution:C C j

j1 2 2 4 3 7 2 7 3 4

29 2

⋅ = − − − − + − − + −= − +

[( )( ) ( )( )] [( )( ) ( )( )]

Polar formIn polar form, multiply magnitudes and add angles. Forexample, if C C C C1 1 1 2 2 2= ∠θ = ∠θ and then

C C C C1 2 1 2 1 2⋅ = ∠θ + θ

Example 17

Find C C1 2⋅ if C C1 25 30 10 20= − ∠ = ∠ −o o and

Solution: C C1 2 50 10⋅ = − ∠ o

3.8.6 Division

Rectangular formNeed to multiply numerator and denominator by theconjugate of denominator and then collect real andimaginary parts. If C A jB1 1 1= + and C A jB2 2 2= + then

C

C

A jB A jB

A jB A jB

A A B B j A B B A

A B

1

2

1 1 2 2

2 2 2 2

1 2 1 2 1 2 1 2

22

22

= + −+ −

= + + − ++

( )( )( )( )

( ) ( )

C

C

A A B B

A Bj

A B B A

A B1

2

1 2 1 2

22

22

1 2 1 2

22

22

= ++

+ − ++

( ) ( )

3.38

Example 18Find C C1 2 if C j1 1 4= − + and C j2 4 5= −

Solution:

C

Cj

j j

1

22 2 2 2

1 4 4 5

4 5

1 5 4 4

4 524

411141

0585 0 268

= − + −+

+ − − − ++

= − + ≅ − +

( )( ) ( )( ) ( )( ) ( )( )

. .

Polar form• Divide magnitude of numerator by magnitude ofdenominator• Subtract angle of denominator from angle of numerator

If C C C C1 1 1 2 2 2= ∠θ = ∠θ and then

C

C

C

C1

2

1

21 2= ∠θ − θ

Also 1 1

C C∠θ= ∠ − θ

Example 19

Find C C1 2 if C C1 215 10 2 7= ∠ − = − ∠o o and .

Solution: C

C1

2

152

10 7 7 5 17=−

∠ − − = − ∠ −o o o.

3.39

3.9 Phasor Analysis

Having considered complex numbers, we are now ready toconsider phasors and to show how phasors can be used tosolve sinusoidal ac circuit problems.

A sinusoidal waveform (eg. voltage)

v t V V t( ) sin sin= =m mα ω

can be generated through the vertical projection of a radiusvector rotating at an uniform angular velocity ω (see Fig.27).

Similarly, a sinusoidal waveform

v t V t( ) sin( );= + =m ω θ θ 45o

can also be generated (see Fig. 28(b)) through the verticalprojection of a rotating vector (see Fig. 28(a)).

This rotating vector, having a constant magnitude Vm withone end fixed at the origin, is called a phasor in sinusoidalac circuits.

3.40

Figure 27

3.41

θ = 45oVm

ω

45oθ = 45o 135o 225o 315oωt

Vmsin45oVm

0

v t V t( ) sin( );= + =m ω θ θ 45o

t = 0

( )a ( )b

Figure 28

During its rotational development of the sine wave, thephasor will, at the instant t = 0, have the position shown inFig. 28(a) which is called a phasor diagram.

Note that the vertical projection of the rotating vector (ie.

Vm sin45o ) in Fig. 28(a) corresponds to the amplitude of

v t( ) at time t = 0 (ie. v V( ) sin0 45= mo).

Thus, the phasor diagram is a “snapshot” of the rotatingvectors at time t = 0 .

Hence, the phasor representation of a sinusoidal acwaveform v t( ) is given by Vm∠ ± θ . That is,

v t V t( ) sin( )= ±m ω θ ⇒ Vm∠ ± θ

3.42

Example 20Figure 29 shows an example of the addition of two currentwaveforms using phasors.

Note that the vertical heights of the waveforms in Fig.29(b) at t = 0 are determined by the vertical projections ofthe radius vectors in Fig. 29(a).

Imagine trying to add these waveforms “by hand” withoutusing phasors!

Figure 29