12
AND
PHASORS
COMPLEX
NUMBERS
12-1
12-2
12-3
"12-4
"12-5
Introductionto Phasors
The ComplexNumberSystem
Rectangular
and PolarForms
MathematicalOperations
TechnologyTheoryinto Practice
at
PSpiceTutorial
http:/iwww.prenhalLcom/floyd
r INTRODUCTION
In this chapter,two important tools for the analysisof
ac circuits are introduced. These are phasors and complex numbers.You will see how phasors are a convenient, graphic way to representsinusoidal voltages
and cunents in terms of their magnitude and phase
angle. In later chapters,you will see how phasorscan
also representother ac circuit quantities.
The complex number system is a means for
expressingphasor quantities and for performing mathematical operationswith those quantities.
Phasor diagrams are an abstract method of representing quantities that have both magnitude and
direction. In the case of sinusoidal voltages and currents, the magnitude is the amplitude of the sine wave
and the direction is its phase angle. Phasorsprovide a
way to diagram sine waves and their phase relationships with other sine waves. The complex number
system provides a way to mathematically expressa
phasor quantity and allows phasor quantities to be
added, subtracted,multiplied, or divided. Use of a calculator in complex number conversionsand arithmetic
is covered thoroughly.
The phaserelationship of two sine waves canbe
graphically representedby phasors and it can also be
measuredon an oscilloscope, as you will see in the
TECH TIP in Section 12-5.
r T
T
lr
P
rc
trU
D U
B R
r TECHnoloev
ut
Theory
Into
Practice
r CHAPTER
OBIECTIVES
0 Usea phasorto representa sine wave
0 Usecomplexnumbers to expressphasor quantities
0 Represent
phasorsin two complex forms
tr Do mathematicaloperationswith complex numbers
both manually and with a calculator
432 T
PHASORS
AND COMPLEXNUMBERS
12_1 . INTRODUCTIONTO PHASORS
Phasors provide a graphic meansfor representing quantities that huve both magnitude and direction (angular position). Phasors are especially useful for representing
sine waves in terms of their magnitude and phase angle and also for analysis of reactive circaits studied in later chapters.
After completing this section, you should be uble to
I Use a phasor to represent a sine wave
. Define phasor
. Explain how phasorsare related to the sine wave formula
. Draw a phasor diagram
. Discussangularvelocity
You may already be familiar with vectors. In math and science, a vector is any quantity
with both magnitude and direction. Examples of vectors are force, velocity, and acceleration. The simplest way to describe a vector is to assign a magnitude and an angle to a
quantity.
In electronics, a phasor is similar to a vector but generally refers to quantitiesthat
vary with time such as sine waves. Examples of phasors are shown in Figure 12-1. The
length of the phasor "arrow" representsthe magnitude of a quantity. The angle, 0 (relative
to 0'), representsthe angular position, as shown in parl (a) for a positive angle. The specific phasor example in part (b) has a magnitude of 2 and a phase angle of 45'. The phasor in part (c) has a magnitude of 3 and a phase angle of 180". The phasor in part (d) has
a magnitudeof 1 and a phaseangle of -45o (or +315"). Notice that positive anglesare
measuredcounterclockwise (CCW) from the reference (0") and negative angles are measured clockwise (CW) from the reference.
(c.)
F I G U R E1 2 - 1
Examplesof phasors.
270"
270"
270"
270"
INTRODUCTION
TO PHASORS.
433
PhasorRepresentation
of a SineWave
A full cycle of a sine wave can be representedby rotation of a phasor through 360 degrees.
The instantaneous value of the sine wave at any point is equal to the vertical distance from the tip of the phasor to the horizontal axis.
Figure 12-2 shows how the phasor traces out the sine wave as it goes from 0o to 360o.
You can relate this concept to the rotation in an ac generator (refer to chapter I l).
/
rr.s" z7o"3rs. 360.
2'70"
F I G U R E1 2 - 2
Sine waverepresentedby rotationalphasor motion.
Notice in Figure l2-2 that the length of the phasor is equal to the peak value of the
sine wave (observethe 90' and the 270' points). The angle of the phasor measuredfrom
0' is the corresponding angular point on the sine wave.
Phasors
and the SineWaveFormula
Let's examine a phasor representationat one specific angle. Figure 12-3 shows a voltage
phasor at an angular position of 45o and the correspondingpoint on the sine wave.
The
instantaneousvalue of the sine wave at this point is related to both the position and the
length of the phasor.As previously mentioned, the veftical distance from the phasor
tip
down to the horizontal axis representsthe instantaneousvalue of the sine wave at that ooint.
u = Vrsin 0
F I C U R E1 2 - 3
Right triangle derivation of sine waveformula,
Notice that when a vertical line is drawn from the phasor tip down to the horizontal axis, a right triangle is formed, as shown shadedin Figure r2-3.Thelength of the pha_
sor is the hypotenuseofthe triangle, and the vertical projection is the opposite side. From
trigonometry,
The opposite side of a right triangle is equal to the hypotenuse times the
sine of the angle d.
434 I
PHASORS
AND COMPLEXNUMBERS
The length of the phasor is the peak value of the sinusoidal voltage, Vo. Thus, the
site side of the triangle, which is the instantaneousvalue, can be expressedas
v = Vpsin0
Recall that this formula is the one statedin Chapter 11 for calculating instantaneous
soidal voltage. A similar formula applies to a sinusoidal current.
i = Ipsin 0
Positiveand NegativePhasorAngles
The position of a phasor at any instant can be expressedas a positive angle, as you
seen, or as an equivalent negative angle. Positive angles are measured
from 0o. Negative angles are measuredclockwise from 0o. For a given positive angle
the correspondingnegative angle is 0 * 360", as illustrated in Figure 124(a).In part
a specific example is shown. The angle of the phasor in this case can be expressed
+225" or -135'.
F I G U R E1 2 - 4
Positiveand negativephasor angles,
270"
EXAMPLE
12-1
270"
For the phasor in each part of Figure 12-5, determine the instantaneoussine
value. Also express each positive angle shown as an equivalent negative angle.
length of each phasor representsthe peak value of the sine wave.
F I G U R E1 2 - 5
(b)
(c)
t0v
(d)
(f)
INTRODUCTION
TO PI-IASORSr
435
Solution
(a) v = 10sin0'= 10(0)= 0 V
0=0"
(b) v = l0 sin30'= 10(0.5)
= 5V
=
=
=
0 30o 0 360" 30o- 360' = -330'
(c) v = 10sin90'= 10(l)= 10V
0=90"=-270o
(d) v = 10 sin 135"= 10(0.707)
=7.07Y
0=I35"=-225o
(e) v = 10 sin270' = 10(-1)= -10 V
0 =2J0o= -90'
(f) y = 10 sin330' = 10(-0.5)= -5 V
0=330'= -30o
The equivalentnegativeanglesareshownin Figure 12-5.
RelatedProblem If a phasoris at 45' and its lengthrepresents15 v, what is the
instantaneous
sinewavevalue?
PhasorDiagrams
A phasor diagram can be used to show the relative relationship of two or more sine waves
of the same frequency.A phasor in a fxed position is used to represent a complete sine
wave because once the phase angle between two or more sine waves of the same frequency or between the sine wave and a reference is established,the phase angle remains
constantthroughout the cycles. For example, the two sine waves in Figure I2-6(a) can be
representedby a phasor diagram, as shown in pafi (b). As you can see,sine wave B leads
sine wave A by 30" and has less amplitude than sine wave A, as indicated by the lengths
of the phasors.
12-6
of a phasor diagram.
270"
12-2
Use a phasor diagram to representthe sine waves in Figure l2-7.
F I G U R E1 2 - 7
yv(v)
(v)
436 T
PHASORS
AND COMPLEXNUMBERS
Solution The phasor diagram representing the sine waves is shown in
12-8. The length of each phasor representsthe peak value of the sine wave.
F I G U R E1 2 - B
45"a l.sv
Related Problem Describe a phasor to representa 5 V peak sine wave that lags sine
wave C in Figure L2-7 by 25'.
AngularVelocityof a Phasor
As you have seenoone cycle of a sine wave is traced out when a phasor is
through 360 degrees.The faster it is rotated, the faster the sine wave cycle is traced
Thus, the period and frequency are related to the velocity of rotation of the phasor.
velocity of rotation is called the angular velocity and is designatedc,r(the small
letter omega).
When a phasor rotates through 360 degreesor 2n radians, one complete
traced oul Therefore, the time required for the phasor to go through 2n radians is
period of the sine wave. Becausethe phasor rotates through 2n radransin a time
the period f, the angular velocity can be expressedas
"
f t l = -
2n
T
Since/= 177
a =2nf,
When a phasor is rotated at an angular velocity c,r,then or is the angle
which the phasor has passedat any instant. Therefore, the following relationship can
stated:
'1$,+,eoI.,
(1
Substituting 2nf for @,you get 0 = 2nft.With this relationship between angle and
the equation for the instantaneousvalue of a sinusoidal voltage, y = Vrsin 0, canbe
ten as
i = VpsinZnfl
The instantaneousvalue can be calculated at any point in time along the sine wave
if the frequency and peak value are known. The unit of 2nft is the radian.
T H EC O M P L E X
N U M B E RS Y S T E M.
EXAMPLE
12-3
437
Whatis thevalueof a sinusoidal
voltageat 3 ptsfrom thepositive-going
zerocrossing
whenV, = 10V andf = 50 kHz?
Solution
v=VosinZnft
= 10 sin[2n(50
kHz)13x 10 o s)] = 8.09V
Related Problem What is the valueof a sinusoidalvoltageat 12 ptsfrom the positive-goingzerocrossingwhenV, = 50 V andf = 10 kHz?
12-1
1. What is a phasor?
2. Whatis theangularvelocityof a phasorrepresenting
a sinewavewith a frequency
of 1500Hz?
3. A cerlain phasor has an angular velocity of 628 radls. To what frequency does this
correspond?
4. Sketch a phasor diagram to representthe two sine waves in Figure 12-9. Use peak
values.
F I G U R E1 2 - 9
v(v)
2-2T THE COMPLEXNUMBERSYSTEM
Complex numhers qllow mathematical operations with phasor qaantities snd are useful in analysis of ac circuits. With the complex number system,you can add, subtract,
mukiply, und divide quantities that have both magnitude and angle, such as sine
waves und other uc circuit quantities that will be studied later.
After completing this section, you should be able to
I Use complex numbers to express phasor quantities
. Describe the complex plane
. Representa point on the complex plane
. Discuss real and imasinarv numbers
Positiveand NegativeNumbers
Positive numbers can be representedby points to the right of the origin on the horizontal
axis of a graph, and negative numbers can be representedby points to the left of the origin, as illustratedin Figure l2-10(a). Also, positivenumberscan be representedby points
on the vertical axis above the origin, and negative numbers can be representedby points
below the origin, as shown in Figure 12-10(b).
438 I
PHASORS
AND COMPLEXNUMBERS
(a)
(b)
FtcuRE12-10
Graphic representation of positive and negative numbers.
T h e C o mp l e xP l ane
To distinguish between values on the horizontal axis and values on the vefiical axis,a
complex plane is used. In the complex plane, the horizontal axis is called the real axis,
and the vertical axis is called the imaginary axis, as shown in Figure l2-ll.
F I G U R E1 2 - 11
The complexplane.
Negative real axis
Negativej axis
In electrical circuit work, a t7 prefix is used to designatenumbers that lie on the
imaginary axis in order to distinguish them from numbers lying on the real axis. This prefix is known as the j operator. In mathematics,an I is used instead of a j, but in electric
circuits, the i can be confused with instantaneouscurrent, so I is used.
AngularPositionon the ComplexPlane
Angular positions can be representedon the complex plane, as shown in Frgure 12-12.
The positive real axis representszero degrees.Proceeding counterclockwise, the +j axis
represents90o, the negative real axis represents180', the -7 axis is the 270" point, and,
after a full rotation of 360", we are back to the positive real axis. Notice that the planeis
sectionedinto four quadrants.
THE COMPLEXNUMBERSYSTEM f
439
2nd, quadrant//.-
FIGURE
12-12
Angleson the complexplane.
Representing
a Pointon the ComplexPlane
A point located on the complex plane can be classified as real, imaginary (t7), or a combination of the two. For example, a point located 4 units from the origin on the positive
real axis is the positive real number, +4, as shown in Figure 12-13(a). A point 2 units
from the origin on the negativereal axis is the negativereal number, -2, as shown in part
(b). A point on the +7 axis 6 units from the origin, as in part (c), is the positive imaginary
number, +76. Finally, a point 5 units along the -7 axis is the negative imaginary number,
j5, as in part (d).
(a) Real number, +4
(b) Real number, - 2
(c) Imaginary number, +76
(d) Imaginary number, -j5
FIGURE
12-13
Real and imaginary Q) numberson the complexplane.
440 r
AND COMPLEXNUMBERS
PHASORS
When a point lies not on any axis but somewherein one of the four quadrants,it is
a complex number and can be defined by its coordinates.For example, in Figure 12-14,
the point located in the first quadrant has a real value of +4 and a7 value of +74 andis
expressedas +4, +j4. The point located in the second quadrant has coordinates -3 and
+j2.The point located in the third quadranthas coordinates-3 and -j5. The point located
in the fourth quadrant has coordinatesof+6 and-j4.
H G U R E1 2 - 1 4
Coordinatepoints on the complexplnne.
12-4
EXAMPLE
(a) Locate the following points on the complex plane:7, j5;5,
-j2; -3.5, jt; and
-5.s,-j6.5.
(b) Determine the coordinatesfor each point in Figure 12-15.
+J
i7,js
_3;s1jl
---i-;'r
t
t
l
-
l
:
t
,,.i,-.,t,ial
i
_J
FtcuRE12-'t6
FtcuRE12-1s
Solution
(a) SeeFigure12-16.
(b) l:2,j6
2:ll, jl
5: -1, -j9
6: -5, -j5
3:6, -j2
4:10, -jl0
7: -6, j5
8: -2, jll
Related Problem In what quadrant is each of the following points located?
(d) -11, {6.8
(a) +2.5,+jr
(b) 7,-j5
(c) -10, -r5
t2
RECTANCULAR
AND POLARFORMS .
441
Valueof i
If we multiply the positive real value of +2 by j, the result is +72. This multiplication has
effectively moved the +2 through a 90' angle to the +7 axis. Similarly, multiplying +2 by
-/ rotates it -90' to the -7 axis. Thus,7 is considereda rotational operator.
Mathematically, the 7 operator has a value of V-1. If +72 is multiplied by j we get
izz = 1{-gf{ -UfU = eD(2)= -2
This calculation effectively places the value on the negative real axis. Therefore, multiplying a positive real number by 7' converts it to a negative real number, which, in effect,
is a rotation of 180' on the complex plane. This operation is illustrated in Figure 12-17.
FIGURE
12-17
Effect of thej operatoron location of a number on the complexplane.
12-2
1. Locate the following points on the complex plane:
(a) +3
(b) -4
(c) +71
What is the angular difference between the following numbers:
(b) {6 and -6
(a) +4 and +j4
(c) +j2 and -j2
I RECTANGULAR
AND POLARFORMS
The rectangular form und the polar form are two forms of complex numbers that are
used to representphasor quantities. Esch hus certuin advantages when used in circuit
analysis, depending on the particular application.
After completing this section, you should be sble to
I Represent phasors in two complex forms
. Show how to representa phasor in rectangular form
. Show how to representa phasor in polar form
. Convert between rectangular and polar forms
As you know, a phasor quantity contains both magnitude and angular position or phase.
In this text, italic letters such as V and I are used to representmagnitude only, and boldface nonitalic letters such as V and I are used to represent complete phasor quantities.
Other circuit quantities that can be expressedin phasor form will be studied in later
chapters.
442 T
PHASORS
AND COMPLEXNUMBERS
Form
Rectangular
A phasor quantity is representedin rectangular form by the algebraic sum of the real
value (A) of the coordinate and the 7 value (B) of the coordinate, expressedin the following general form:
A+jB
An "arrow" drawn from the origin to the coordinate point in the complex plane is used
to represent graphically the phasor quantity. Examples of phasor quantities arc | + j2,
5 - j3, -4 + j4, and-2 -76, which are shown on the complex plane in Figure 12-18.As
you can see, the rectangular coordinates describe the phasor in terms of its values proiected onto the real axis and the I axis.
FIGURE
12-18
Examplesof phasors specifi.edby rectangular
coordinates.
PolarForm
Phasorquantities can also be expressedin polar form, which consistsof the phasormagnitude (C) and the angular position relative to the positive real axis (0), expressedin the
following general form:
CZ!O
Examples are2145o, 52.120",4Lll0", md 82-30' . The first number is the magnitude,
and the symbol lprecedes the value of the angle.Figure 12-19 showsthesephasorsonthe
complex plane. The length of the phasor,of course,representsthe magnitude of the quantity. Keep in mind that for every phasor expressedin polar form, there is also an equivalent
expressionin rectangularform.
F I G U R E1 2 - 1 9
Examples of phasors specifiedby polar
values.
RECTANCULAR
AND POTARFORMS .
443
Most scientific calculators have provisions for conversion between rectangular and
polar forms. The basic conversionmethods are presentedfirst so that you will understand
the mathematical procedure,followed by the calculator methods.
Conversionfrom Rectangular
to PolarForm
A phasor can exist in any of the four quadrantsof the complex plane, as indicated in Figure 12-20. The phase angle 0 in each case is measuredrelative to the positive real axis
(0') as shown.
-NI
e= 180"+0
(a) lst quadrant
(b) 2nd quadrant
(c) 3rd quadrant
(d) 4th quadrant
F I G U R E1 2 - 2 0
All possiblephasor quadrantlocations.0 is the angk of the phasor relative to the positive
real axis in each case,and $ is the angle in the 2nd and 3rd quadrantsrelative to the
negativereal axis.
The first step to convert from rectangular form to polar form is to determine the
magnitude of the phasor. A phasor can be visualized as forming a right triangle in the
complex plane, as indicated in Figure 12-21, for each quadrant location. The horizontal
side of the triangle is the real value, A, and the vertical side is the 7 value, B. The
hypotenuseof the triangle is the length of the phasor, C, representingthe magnitude, and
can be expressed,using the Pythagoreantheorem, as
c=fA'+E
('t2-4)
Next, the angle d indicated in parts (a) and (d) of Figure 12-21 is expressedas an
inversetangentfunction.
-'l tB\
d=tan-l
I
\ A /
lst quadrant
(b) 2nd quadrant
12-21
angle relationships in the complex plane.
(12-s)
(d) 4th quadrant
4 4 4 T P H A S O RA
SN D C O M P L E X
NUMBERS
The angle d indicated in parts (b) and (c) of Figure 12-21 is
d=+180'Td
which includes both conditions as indicated by the dual signs.
o = +180o + tun-'/4)
\A/
(t2-6)
In each case the appropriate signs must be used in the calculation.
The general formula for converting from rectangular to polar is as follows:
-rA ljB = C./.+0
Example l2-5 illustrates the conversionprocedure.
EXAMPLE
12-5
Convert the following complex numbers from rectangular form to polar form by determining the magnitude and angle:
(a) 8 +76
(b) 10 - j5
(c) -r2 * j18
(d) -7 +r10
Solution
(a) The magnitude of the phasor representedby 8 + j6 is
c=f*+*=f{\6r=Vloo=to
Since the phasoris in the flrst quadrant,use Equation (12-5). The angle is
s= tan-'(!!
) =,* '19)= ro.s'
A
\
/
\8/
0 is the angle relative to the positive real axis. The complete polar expressionfor
this phasor is
C = 10236.9'
Boldface, nonitalic letters representphasor quantities.
(b) The magnitude of the phasor representedby l0 -75 is
c =f rd\ esf =f tzs=1L.2
Since the phasor is in the fourth quadrant, use Equation (I2-5). The angle is
o = r a n' f - l \ = - 2 6 . 6 .
\ 10/
0 is the angle relative to the positive real axis. The complete polar expressionfor
this phasoris
C = ll.2Z-26.6
(c) The magnitudeof the phasorrepresentedby -12 - 718 is
s =f en)u+ (-tlf =f +ea=21.6
Since the phasor is in the third quadrant,use Equation (12-6). The angle is
d = -180'. t*-'(#)
= -180o+ 56.3'= -123.7"
The complete polar expressionfor this phasor is
C = 21.62-123.7"
RECTANCULAR
AND POLARFORMS .
445
(d) The magnitude of the phasor representedby -7 + 710 is
c = f ( J ) z+ t e = f u g = 1 2 . 2
Since the angle is in the secondquadrant,use Equation (12-6). The angle is
d = l g o o - , u n - t ( - 1 9 ) = l g o o * 5 5 o= 1 2 5 o
\ 1I
The completepolar expression
for this phasoris
C = 72.21125"
RelatedProblem Convertl8 + j23 to polar form.
Conversionfrom Polarto RectangularForm
The polar form gives the magnitude and angle of a phasor quantity, as indicated in Figure
1a aa
F I G U R E1 2 - 2 2
Polar componentsof a phasor.
I B = c s i n0
A=Ccosd
To get the rectangular form, sidesA and B of the triangle must be found, using the rules
from trigonometry statedbelow:
A=Ccos0
(12-8)
B=Csin0
(12*9)
The polar-to-rectangular
conversionformulais asfollows:
CZ A =Ccos 0 +7C sinA = A + jB
The following exampledemonstrates
this conversion.
E 12-6
Convert the following polar quantities to rectangular form:
(a) 102.30'
(b) 2002-45"
(c) 42135'
Solution
(a) The real part of the phasor representedby 10230' is
A = C c o s d = 1 0 c o s 3 0 " = 1 0 ( 0 . 8 6 6=) 8 . 6 6
The j part of this phasor is
jB = jC titl I = j10 sin 30' - j10(0.5) =75
The complete rectangular expressionis
A + jB = 8.66*"t5
(12-10)
446 I
PHASORS
AND COMPLEXNUMBERS
(b) The real part of the phasor representedby 20Ol-45" is
A = 200 cos(-45") = 200(0.101)= I4l
The 7 part is
jB = j2O0 sin(-45') = j200(-0.101) = -jr41
The complete rectangular expressionis
A+jB=141 -jl4l
(c) The real part of the phasor representedby 42135" is
A = 4 cos 135o= 4(-0.70'7)= -2.83
The j part is
jB = j4 sin 135' = 4(0.101)= 2.83
The complete rectangular expressionis
A+jB=-2.83+j2.83
Related Prohlem
Convert 782-26'to
rectangular form.
CalculatorConversionof ComplexNumbers
The following discussion and examples apply to the TI-85 calculator. The procedureon
other calculators may differ.
Entering Complex Numbers All complex numbers must begin and end with parentheses,and each component must be separatedby either a comma for rectangular or an
angle sign for polar. The angle sign is a secondary function. Complex numbers are
entered on the TI-85 as follows. A number in rectangular form is entered in this format:
(real,imaginary)
A number in polar form is enteredin this format:
(magnitudeZangle)
EXAMPLE 12-7
Enter the following complex numberson your calculator:
(a) 3 + j4
(b) 10 - j8
(c) 4/30"
(d) 252-60'
Solution
(a) 3 + j4 is entered as (3,4).
(b) 10 -78 is enteredas (10,-8).
(c) 4130" is entered as (4130).
(d) 252-60" is enteredas (24/-60).
Related Problem
and 130245'.
Enter each of these complex numbers on your calculator: 15 + jl8
Form Conversions To convert one form of complex number to the other, you will
use the MODE key on your calculator to select the form. To get into the mode sc
press the @ *"y and then the fffi]
key. The mode screenwill appear as shown in
we 12-23. The highlighted (dark) items indicate the default selections.Of course,we
RECTANCULAR
AND POLARFORMS .
447
only interested in the Radian/Degree and the RectC/PolarC selections. Degree is the
angular default, and PolarC is the complex form default. To choose a nonselecteditem,
move 'he flashing highlight box to the desireditem using the arrow keys and pr"s M@.
Press lG8iffi]to exit the mode screen.
F I G U R E1 2 - 2 3
Defoult modescreen.
With PolarC highlighted, you can enter a number in rectangular form and convert it
to polar form. To do this, simply enter the rectangular number and press the [Sffi$] key.
(real,imaginary) ENTER ---)
(magnitudeZangle)
To convert from polar to rectangular,use the directional arrow key and move the flashing
highlight box from Normal down to RectC and press llxiEil. Exit the mode screen,enter
the polar number, and press WF1 .
(magnitudeZangle) ENTER ---------) (real,imaginary)
(a) Convert l0 - j5 to polar form.
(b) Convert l0Z3O" to rectansular form.
Solution
(a) Select PolarC on the mode screen, exit, enter the retangular number, and press
ti.
The resultis shownin Figure l2-24(a). To view rhe part of rhe number
not originally displayed, press the right arrow key.
(b) Select RectC on the mode screen,exit, enter the polar number, and press tENTTn
The result is shown in Figure l2-24(b).
Related Problem
tansular form.
(a) Convert 8 + j6 to polar form. (b) Convert 2002_45" to rec-
The Complex (CPLX) Mode Another way of polar-to-rectangularand rectangular-topolar conversionsis the CPLX mode, which is selectedby pressing [ffiJ and @.
trt
this mode you will see the screen in Figure l2*25(a). Press the @
t"y and you get
the screenin Figure l2-25(b).
448 T
P H A S O RA
SN D C O M P L E X
NUMBERS
(b)Press@
{a)Press
Gndl rhenGNTEEI
(c) Enterpolarnumber,pressGTl.
Enterrectangular
number,pressGl.
F I G U R E1 2 - 2 5
Using the CPLX modefor conversions.
To convert a polar number to rectangular form, enter the number in the properf,
mat (magnitude/.angle) and press @ then t$Xffiil . To convert a rectangular number
polarform, enterthe numberin the properformatandpressfffi] then ffi&ffiS]
. fne
are shownfor two examplenumbersin Figure 12-25(c).
SECTION 12-3
REVIEW
1. Namethetwo partsof a complexnumberin rectangular
form.
2. Namethetwo p*r, of a complexnumberin potarform.
3. Convert2 + j2 topolarform. In whichquuar*t i*t *t, phasorlie?
4. Convert52-45'to rectangular
[orm.In whichquadrantdoesthisphasorlie?
12_4 . MATHEMATICAT
OPERATIONS
Complex numbers can be added, subtracted, multiplied, and divided.
After completing this section, you should be able to
I Do mathematical operations with complex numbers
. Add complex numbers
. Subtract complex numbers
. Multiply complex numbers
. Divide complex numbers
. Apply complex numbers to sine waves
Addition
Complex numbers must be in rectangular form in order to add them. The rule is
Add the real parts of each complex number to get the real part of the
sum. Then add thej parts of each complex number to get theT part of the
sum.
EXAMPLE 12-9
Add thefollowingsetsof complexnumbers:
(a) 8 + j5 and2 + jl
(b) 20 -710 and12+ j6
Solution
(a) (8 +r5) + (2 + jl)= (8 + 2) +j(5 + l) = 10 +16
( b ) ( 2 0 - 7 1 0 )+ ( r 2 + j 6 ) = ( 2 0 + 1 2 ) +j ( - 1 0 + 6 ) = 3 2 + j ( - 4 ) = 3 2 - j 4
RelatedProhlem Add 5 -j11 and-6 + j3.
MATHEMATICAL
OPERATIONS.
449
Subtraction
As in addition, the numbers must be in rectangular form to be subtracted.The rule is
Subtract the real parts of the numbers to get the real part of the difference. Then subtract theT parts of the numbers to get theT part of the difference.
EXAMPLE12-10
Performthe following subtracrions:
(a) Subtract| + j2 from3 + j4.
(b) Subtract10- j8 from 15+715.
Solution
(a) Q + jD - Q + j2)= (3 - 1)+ j(4 - 2) =2 + jz
(b) (ls +jl5)-(10-j8)=(ls -10)+jlls -(-8)] = s + j23
RelatedProblem Subrract3.5- j4.5 from -10 - 79.
Multiplication
Multiplication of two complex numbers in rectangularform is accomplishedby multiplying, in turn, each term in one number by both terms in the other number and then combining the resulting real terms and the resulting/ terms (recall that j x j = -1). As an
examole.
r-f=l
j3)(2-
(5 +
l
u
-J
j4) = l0 - j20 + j6 + 12= 22
t
Multiplication of two complex numbers is easier when both numbers are in polar
form. The rule is
Multiply the magnitudes, and add the angles algebraically.
EXAMPLE
12_11 Performthe following multiplications:
(a) 10/45' times5220o
(b) 2260" times4l-30o
Solution
(a) (10245")(5220')= (r0)(5)z(45'+ 20")= 50265"
(b) (2260")(42-30') = (2)(4)2160'+ (-30")l = 8230"
RelatedProblem Multiply 50270' times302-60'.
Division
Division of two complex numbers in rectangular form is accomplished by multiplying
both the numerator and the denominator by the complex conjugate of the denominator
and then combining terms and simplifying. The complex conjugate of a number is found
by changing the sign of theT term. As an example,
10+/5 _ (r0 + js)(z- j4) _ 20- j30+ 20 _ 40- j30
(2+j4)(2-j4)
2+j4
4+16
20
450 I
P H A S O RA
SN D C O M P L E X
NUMBERS
Like multiplication, division is easier when the numbers are in polar form. The
rule is
Divide the magnitude of the numerator by the magnitude of the denominator to get the magnitude of the quotient. Then subtract the denominator angle from the numerator angle to get the angle of the quotient.
EXAMPLE12_12 Performthe following divisions:
(a) Divide 100250"by 25220'.
(b) Dividel5ZlO' by 32-30'.
Solution
100250'= / 100\
20')= 4/30"
,u*
\;)t(sO't 5 z t 0 "= \/;1) z5l\t }.'.( - 3 0 " 1=;5 t 4 0 "
tot
uas"
ta)
Related
Problem Divide242-30"bv 6112".
UsingComplexNumberswith SineWaves
Since sine waves can be representedby phasors,they can be described in terms of complex numbers either in rectangularor polar form. For example, four seriessinusoidalvolt
age sourcesall with the same frequency are shown in Figure 12-26. The sine wavesare
graphed in Figure l2-27(a), and the phasor representationis shown in part (b). The total
voltage acrossthe load in Figure 12-26 can be determined by first converting eachphasor to rectangular form and then adding as follows:
Y,o,=Yr + V2 + V, + Vo
= l0ll20" Y + 4230" Y + 8l-30" Y + 61-t30' Y
= ( - 5 V + j 8 . 6 6 V ) + ( 3 . 4 6Y + j 2 V ) + ( 6 . 9 3v - j 4 V ) + ( - 3 . 8 6 V - j 4 . 6 0 V )
= 1 . 5 3V + j 2 . 0 6 Y = 2 . 5 72 5 3 . 4 " Y
The total voltagehas a magnitudeof 2.51 V and a phaseangle of 53.4".
As you have seen, sine waves can be representedin complex form and can be
added, subtracted,multiplied, and divided using the rules that we have discussed.Also, as
you will learn later, other electrical quantities, such as capacitive and inductive reactances, impedance, and power, can be described in complex form to ease many circuit
analysis problems.
F I G U R E1 2 - 2 6
Superimpos
ed sinusoidalsources,
r 0 t 1 2 0 "v
4130'v
Y,o,
8Z 30"V
6Z-130'V
2.51253.4"
V
Frc
ap'
four
Figt
MATHEMATICAL
OPERATIONST
451
12-27
diagram representing the
out-of-phasesine waves in
12-26.
(a)
10v
4V
ComplexNumberOperationswith the Calculator
Addition, subtraction, multiplication, and division of complex numbers can be easily
done with the calculator.All operationsare accomplishedby simply entering the numbers
in the proper format along with the arithmetic sign, just as you would for real numbers.
The TI-85 is used to illustrate the procedures.
Addition
To add two or more complex numbers, enter the numbers with the plus sign
and press
format, or there can be a mixture of both formats. The form of the result is determinedby
the selection on the mode screen.
EXAMPLE
12-13
Add the following sets of complex numbers with the results in rectangular form:
(c) 12120" and 8 + j6
(a) 8 + j5 and 2 + jI
(b) 5245" and 10130"
Solution For all of these additions, selectRectC on the mode screenso that the sums
are in rectangular form; then exit the mode screen. The calculator screens should
.
appear as follows in Figure 12-28 after entering the numbers and pressing W
Add 100 +750 and 15 - j45 and 35 +765. Obtain the sum in polar
452 I
P H A S O RA
SN D C O M P L E X
NUMBERS
Subtraction
To subtract complex numbers, enter the numbers with the minus sign and
press [EitLTTF),
as illusrrared in rhe following example.
EXAMPLE12_14 Subtract the following sets of complex numbers with the results in rectangular form:
(a) 1 + j2 from 3 + j4
(b) 8235' from t2Zt0.
(c) 22260. from 100 +750
Solution For all of these subtractions,select RectC on the mode screen so that the
differences are in rectangular form; then exit the mode screen.The calculator screens
rl,g!{ appear as follows in Figure 12-29 after entering the numbers and pressing
{Efrf,,:'EEl.
F I G U R E1 2 - 2 9
RelatedProblem subtracrl0 + j50 from 25 - 730. obtain rhe differencein polar
form.
Multiplication Tomultiply complex numbers, enter the numbers with the multiplication sign andpress(ffiFfr], as illustrated in the following example.
EXAMPTE
12_15 Multiply the following sets of complex numbers with the results in polar form:
(a) 1 + j2 and 3 + j4
(b) 8235" and t2lt0"
(c) 22160" and 100 +750
Solution For all of these multiplications, select PolarC on the mode screen so that
the products are in polar form; then exit the mode screen. The calculator screens
+qql{ appear as follows in Figure 12-30 after entering rhe numbers and pressing
GnTqil.
RelatedProblem Multiply 10 +/50 and25-730. obtain the productin rectangular
form.
TECHNOLOCYTHEORYINTO PRACTICE.
453
Division To divide complex numbers, enter the numbers with the division sisn and
press [FF-,,GFI,
as illustrated in the following example.
EXAMPLE 12-16
Divide the following sets of complex numbers with the results in polar form:
(a) 5+j2by4+j4
(b) t6Z45"byt0Z50.
( c ) 2 5 1 5 0 "a n d 8 5 + j 5 5
Solution For all of thesedivisions, selectPolarC on the mode screenso that the quotients are in polar form; then exit the mode screen. The calculator screens should
appear as follows in Figure 12-37 after entering the numbers and pressing GNTERI
(b)
(a)
(c)
RelatedProblem Divide 15 + j20 by 15 +710. obrainthe quotienrin rectangular
form.
12-4
1. Add I + j2 and 3 - jl. Verify your result with a calculator.
2. Subtract12 + j18 from 15 + j25. Verify your resulrwith a calculator.
I
I
j
i
3. Muttiply 8145" times2165". Verify your resultwith a calcularor.
4. Divide 302'15" by 6260". Verify your result with a calculator.
2-5 r TECHnology
TheoryInto Practice
Phusors and complex numbers are mathematical concepts thst are usedfor the purpose of ac circuit analysis. Although this Tech TIP does not deul directty with phasors
or complex numbers, the angular relationships thqt are meusured cun be represented
by phasors.
One way to measurethe phase angle between two sine waves with the same frequency is
to flrst adjust the amplitude of one waveform on the scope screenso that both waveforms
appearequal. Measure the time (Ar) between correspondingpoints on the waveforms and
measurethe period. Then, using the relationship A,tlT = 0/360", calculate the phase angle
0 as 0 = (LtlT)360'. On most digital oscilloscopes,the cursorsmay be usedto determine
L,t and T.
Another way to measurethe phaseangle between two sine waves with the samefrequency is to use an analog oscilloscope and conveft the horizontal axis into angular divisions. The seconds/divisioncontrol can be switched off of the calibrated position and varied until there is exactly one half-cycle of the sine wave displayed across the screen as
illustrated in Figure 12-32(a). (Generally,the time base of digital oscilloscopescannot be
I
il
{l
tl
454 T
PHASORS
AND COMPLEXNUMBERS
(b)
(a)
F I G U R E1 2 - 3 2
decalibrated.)Since a half-cycle contains 180', each of the ten main horizontal divisions
represents 18' and each of the small divisions represents3.6o, as indicated. A quarter
cycle can be used although it is difficult to establish the exact positive peak when the
waveform is spreadout.
After the phase scale has been establishedusing one of the sine waves, the traces
for both oscilloscope channels must be superimposedon each other and aligned along
the horizontal axis to prevent any vertical offset of the waveforms. This is done by
touching channel I probe to ground (0 V) or by switching the channel input coupling
switch to the ground (GND) position and adjusting the vertical position to bring it to the
center line on the screen,as indicated in Figure l2-32(b). This is repeatedfor the channeI 2 trace.
Next the probes are connected to the two signals and each is ac coupled (channel input switch set to AC) into the scope. The scope should be triggered on the reference channel-do not use composite or vertical-mode triggering. You should set the
amplitudes to an approximately equal height by taking the volts/division controls off
of the calibrate position and adjusting each one. This is illustrated in Figure 12-33,
The phase angle between the two waveforms can be most easily measured from the
zero crossings, as shown. In this example, the angle is 36'.
F I G U R E1 2 - 3 3
FORMULAS r
455
PhaseAngleMeasurement
r
'toN
12-5
EW
Determine the phase angle for each of the scope displays in Figure 12-34. A quarter
cycle is shown in part (c).
1.. When measuringthe phaseangle between two waveforns on an oscilloscope,is the
seconds/divisionsettingcritical?
2. How do you prevent vertical offset of the waveforms when measuring the phase
angle?
3. What triggering methodshould be selectedfor a phaseshift measurement?
SUMMARY
I
The angular position of a phasor representsthe angle of the sine wave with respectto a reference,
and the length of a phasor representsthe amplitude.
I
A complex number representsa phasor quantity.
I
Complex numbers can be added, subtracted,multiplied, and divided.
I
The rectangular form of a complex number consists of a real part and aj part of the form A + jB.
The polar form of a complex number consists of a magnitude and an angle of the form CZX 0.
I
GTOSSARY
These terms are also in the end-of-book glossary.
Angular velocity The rotational velocity of a phasor which is related to the frequency of the sine
wave that the phasor represents.
Complex plane An area consisting of four quadrantson which a quantity containing both magnitude and direction can be represented.
Imaginary number
A number that exists on the vertical axis of the complex plane.
Phasor A representation of a sine wave in terms of its magnitude (amplitude) and direction
(phase angle).
Polar form
One form of a complex number made up of a magnitude and an angle.
Rectangular form
FORMULAS
One form of a complex number made up of a real part and an imaginary part.
(r2-r)
ot=2nf
(r2-2\
(r2-3)
0=at
v = Vpsin2nfl.
456 I
PHASORS
AND COMPLEXN U M B E R S
(r2-4)
C=\,EiE
(r2-s)
A=tar
^
-'l+B\
\ A /
-, /,8 \
(12-6)
d = +180"-r t a n ' l - l
(r2-7)
(r2-8)
lA!
A=Ccos0
0?-9)
(12-10)
B=Csin0
Cl? = Ccos d +7C sin 0 = A + jB
\A/
r SEIF.TEST
jB = CZ+0
l. A phasor represents
(a) the magnitude of a quantity
(c) the phase angle
(b) the magnitude and direction of a quantity
(d) the length of a quantity
2. A positive angle of 20' is equivalent to a negative angle of
(a) -160'
(b) -340"
(c) -70"
(d) -20'
3. In the complex plane, the number 3 + j4is located in the
(a) flrst quadrant
(b) second quadrant
(c) third quadrant
(d) fourth quadranr
4. In the complex plane, 12 -j6
(a) flrst quadrant
is located in the
(b) second quadrant
(c) third quadrant
(d) fourth quadrant
5. The complex number 5 +75 is equivalent to
(a) 52"45"
(b) 2520'
(c) 1.07245'
(d) 1.0'72135"
6. The complex number 35260'is equivalentto
(a) 35 + j35
(b) 35 +i60
(c) 17.5+ j30.3
(d) 30.3 + i1'7.s
,f
(4 + j7) + (-2 + j9) is equalto
(^)2+jr6
( b ) 1 1 + r 1 1 ( c )- 2 + j r 6
8. (16-i8) - (12+ j5) is equalto
(a)28-jr3
(b)4-jt3
@ )a - f i
9. (5245")(2220")is equalto
(a) 1265'
(b) 10225"
(c) 10265"
10. (502rc') I (25230")is equalto
(a) 25240'
(b) 2240'
(c) 252-20"
r PROBLEMS
(d)2-j2
(d)-4+jt3
(d) 1225"
(d) 2l-20"
More dfficub problems are indicated by an asterisk (*).
SECTION12-1 Introductionto Phasors
t.
phasordiagram
to represent
thesinewavesin Figure12-35withrespect
to a 0oref|.f,il.i
2. Sketch the sine waves representedby the phasor diagram in Figure 72-36. The phasor lengths
representpeak values.
F I G U R E1 2 - 3 5
yry)
I
0.5
P R O B L E M S.
457
FtcuRE12-36
3. Determine the frequency for each angular velocity:
(a) 60 rad/s
(b) 360 radls
(c) 2 rad/s
(d) 1256 radls
4. Determine the value of sine wave A in Figure 12-35 ar each of the following times, measured
from the positive-going zero crossing. Assume the frequency is 5 kHz.
(a) 30 ps
(b) 75 ps
(c) 125 ps
*5. In Figure 12-35, how many microseconds after the zero crossing does sine wave
A reach
0.8 V? Assume the frequencyis 5 kHz.
SECTION12-2 The ComplexNumberSystem
6. Locate the following numbers on the complex plane:
(a) +6
(b) -2
(c) +73
(d) -j8
7. Locate the points representedby each of the following coordinateson the complex plane:
(a) 3,j5
(b) -7,j1
(c) -10, -j10
*8. Determine the coordinates of each point having the same magnitude but located
180' away
from each point in Problem 7.
*9. Determine the coordinates of each point having the same magnitude but located
90o away
from those in Problem 7.
SECTION12-3 Rectangular
and PolarForms
10. Points on the complex plane are describedbelow. Express each point as a complex number in
rectangular form:
(a) 3 units to the right of the origin on the real axis, and up 5 units on the axis.
7
(b) 2 units to the left of the origin on the real axis, and 1.5 units up on rhej axis.
(c) 10 units to the left of the origin on the real axis, and down 14 units on the -l axis.
1 1 . What is the value of the hypotenuseof a right triangle whose sides are l0 and 15?
12. Convert each of the following rectangular numbers to polar form:
@) a0 - j40
(b) s0 -i200
(c) 35 - j20
(d) 98 +r45
13. Convert each of the following polar numbers to rectangular form:
(a) 10002-50"
(b) 15Z160'
(c) 252-135"
(d) 32180"
14. Express each of the following polar numbers using a negative angle to replace the positive
angre:
(a) l0Zl20'
(b) 32285"
(c) 52310"
15. Identify the quadrant in which each of the points in Problem 12 is located.
16. Identify the quadrant in which each point in Problem 14 is located.
17. Write the polar expressionsusing positive angles for each phasor inFigure 12-37.
458 r
NUMBERS
P H A S O RA
SN D C O M P L E X
SECTION12-4 MathematicalOperations
18. Add the following sets of complex numbers:
(b) 3.5- j4 and2.2+ j6
( a )9 + j 3 a n d 5 + 7 8
(d) 12145"and20Z32'
(c) -18 + j23 and30 -il5
(f) 50 - j39 and60Z-30"
(e\ 3.8275'and1 +/1.8
1.9. Performthe following subtractions:
(b) (-4s - j23) - (36+ jr2)
(a) (2.5+.ir.2) - (1.4+ j0.s)
(c) (8 - j4) - 3t25'
fi) 481135'-332-60'
20. Multiply the followingnumbers:
(b) l20Z-220" and95Z200'
(a) 4.5248" and3.2Z90'
(d) 67 + j84 and 102240"
(c) -3Zl5O" and4 - j3
(ft 0.8+10.5andL2 - jr.5
(e) 15-rl0 and-25 - j30
21. Performthe following divisions:
9,./5o.4
ra)ffi;
6?./-910
tntffi
., 28/30'
( c )' , *
A -.l rz
(d)+#
22. Pe{orm the following operations:
(10021s'x8s-jl50)
2.5265"- l.8Z-23"
(b)
25 + j45
t.2237"
-j100)
(250290"
+ 1752'75'\(50
.
,( d" ), { 1 . 5 ) 2 t +3t.l ;8 t , / 8 _ _ , 4 \
tct
-, . ,
(r25+ j9o\B5z5o\
" 2)
(a)
*23. Three sinusoidal voltage sourcesare connectedin seriesas shown in Figure 12-38. Determine
the total voltage and current expressedas polar quantities. Resistancealways has a zero phase
angle as you will learn later, so Rr = 2.220' kdt.
*24, What is the magnitude and phase of the voltages acrosseach resistor in Figure 12-38?
F I G U R E1 2 - 3 8
1.0ko
2.520"v
4.2230"
V
R2
2.2ko.
5.rz-45"V
R3
1.8kO
I
ANSWERS
TO SECTION
REVIEWS
Section12-1
1.
2.
3.
4.
of the magnitudeandangularpositionof a time-varyingquantity
A graphicrepresentation
9425rails
100Hz
SeeFisure12-39.
F l c u R E1 2 - 3 9
r 459
ANSWERS
TO SELF-TEST
Section12-2
1. (a) 3 units right of the origin on real axis
(c) I unit aboveorigin on 7 axis
(b) 90"
(c) 180'
2. (a) 90'
(b) 4 units left of the origin on real axis
Section12-3
1.
2.
3.
4.
Realpart andj (imaginary)part
Magnitudeandangle
2.828145";first
3.54-73.54, fourth
Section 12-4
r . 4 +j l
2 . 3 +j 7
3. t6tlt0"
4. 5ZI5'
Section12-5
1. No, the seconds/divisionsetting is not critical.
2. Superimposethe traces by adjusting the vertical deflection (position).
3. Reference channel
SWERS
RELATED
PROBLEMS
EXAMPTES
lz-t 10.6v
l2-2 5Vat-85"
t2-i 34.2y
(b) 4th
124 (a) lst
(c) 3rd
(d) 2nd
l2-5 29.2252"
t2-6 70.r - j34.2
r2-1 (16218);(13024s)
-141.421356237)
r2-8 (a) (r0236.86989764s8) (b) (14r.421356237,
-1
-j8
r2-9
l2-r0 -r3.5 - j4.s
lz-tt 15002-50.
12-12 42-42"
12-13 (221.359
436212
2t8.4349488229)
r2-r4 (52.201s32s446
Z7 3.300755766)
12-1s (17s0,9s0)
12-16 (1.307
692307
69,.46I 538461538)
1. (b)
ANSWERS
9
TOSELF.TEST . ( c )
2.(b)
10.(d)
3. (a)
4.(d)
5. (c)
6.(c)
7, la)
8. (b)