Lesson 15:
Phasors, Complex Numbers and
Impedance
1
Learning Objectives
•
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•
•
•
•
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Define a phasor and use phasors to represent sinusoidal voltages and
currents.
Determine when a sinusoidal waveform leads or lags another Graph a
phasor diagram that illustrates phase relationships.
Define and graph complex numbers in rectangular and polar form.
Perform addition, subtraction, multiplication and division using complex
numbers and illustrate them using graphical methods.
Represent a sinusoidal voltage or current as a complex number in polar and
rectangular form.
Define time domain and phasor (frequency) domain.
Use the phasor domain to add/subtract AC voltages and currents.
2
Learning Objectives
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•
•
•
•
•
•
For purely resistive, inductive and capacitive elements define the voltage
and current phase differences.
Define inductive reactance.
Understand the variation of inductive reactance as a function of frequency
Define capacitive reactance.
Understand the variation of capacitive reactance as a function of frequency
Define impedance.
Graph impedances of purely resistive, inductive and capacitive elements as
a function of phase.
3
Complex numbers
• A complex number (C) is a number of the form: C  a  jb ,
which is known as the rectangular form.
• where a and b are real and
j  1
• a is the real part of C and jb is the imaginary part.
• Complex numbers are merely an invention designed to allow
us to talk about the quantity j.
• j is used in EE to represent the imaginary component to avoid
confusion with CURRENT (i).
• Solving AC circuits is simplified (no, really) through the use
of phasor transforms, which we will now discuss at length…
4
Geometric Representation
• In the rectangular form (C=a+jb), the x-axis is the real axis and the y-axis
is the imaginary (j) axis.
• The polar form (C=Z  Ѳ), where Z is the distance (magnitude) from the
origin and Ѳ is the angle measured counterclockwise (CCW) from the
positive, x (or real) axis (the y-axis is still the imaginary (j) axis).
C = 6 + j8
(rectangular form)
C = 1053.13º
(polar form)
5
Conversion Between Forms
• To convert between forms where:
C  a  jb
(rectangular form)
C  C 
(polar form)
• apply the following relations:
a  C cos
b  C sin 
C  a 2  b2
b
  tan
a
1
6
Example Problem 1
• Convert (5∠60) to rectangular form.
a  C cos  5cos(60)  2.5
C  a  jb (rectangular form)
C  C 
b  C sin   5sin(60)  4.3
(polar form)
a  C cos
C  2.5  j 4.3 (rectangular form)
b  C sin 
C  a 2  b2
• Convert 6 + j 7 to polar form.
b
  tan
a
1
C  a 2  b 2  (62  7 2 )  9.22
  tan 1
b
7
 tan 1  49.4
a
6
C  9.2249.4 (polar form)
7
Properties of j
j  1
j  ( 1)( 1)  1
2
1 1 
 
j j 
j
j
 2 j

j j
8
Addition and Subtraction
of Complex Numbers
• Easiest to perform in rectangular form.
Example: Given A =6 +j12 and B =7 + j2
ADDITION:
• Add the real and imaginary parts separately.
(6  j12)  (7  j 2) = (6  7)  j (12  2) = 13  j14
SUBTRACTION:
• Subtract the real and imaginary parts separately.
(6  j12)  (7  j 2) = (6  7)  j (12  2) =  1  j10
9
Multiplication and Division
of Complex Numbers
• Multiplication and Division is easiest to perform in polar form:
• Multiplication: multiply magnitudes and add the angles:
(670)  (230)  6  2(70  30)  12100
• Division: Divide the magnitudes and subtract the angles:
(670) 6
 (70  30)  340
(230) 2
• Also: the reciprocal of C  C 
1
1
 ( )
C  C
• The conjugate of C is C* and has the same real
value but the OPPOSITE imaginary part:
C  a  jb  C( )
C *  a  jb  C ( )
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Example Problem 2
Given A =1 +j1 and B =2 – j3
• Determine A+B:
(1  j1)  (2  j 3) = (1  2)  j (1  (3)) = 3  j 2
•
and A-B:
(1  j1)  (2  j 3) = (1  2)  j (1  ( 3)) =  1  j 4
Given A =1.4145° and B =3.61-56°
• Determine A/B:
(1.4145)
1.41

(45  (56))  0.391101
(3.61  56) 3.61
•
and A*B:
(1.4145)  (3.61  56)  (1.41  3.61)(45  (56))  5.09  11
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Example Problem 3
Now, make sure you know how to use your calculator:
1. (3-i4) + (10∠44) then convert to rectangular:
a  C cos  10cos(44)  7.19
b  C sin   10sin(44)  6.95
ANS: 10.6∠16.1 (polar)
ANS: 10.2 + j2.9 (rectangular)
(3  j 4)  (7.19  j 6.95) = (3  7.19)  j (4  6.95) = 10.19  j 2.95
2. (22000+i13)/(3∠-17) then convert to rectangular:
C  a 2  b 2  220002  132 )  22k
  tan 1
b
13
 tan 1
 0.339
a
20000
ANS: 7.3*103∠17.0 (polar)
ANS: 7.01*103 + j2.15 (rectangular)
(220000.034) 22000

(0.034  (17))  733317.034
(3  17)
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3. Convert 95-12j to polar:
12
ANS: 95.8∠-7.2 (polar)
Phasor Transform
• To solve problems that involve sinusoids (such as
AC voltages and currents) we use the phasor
transform:
1. We transform sinusoids into complex numbers in
polar form…
2. solve the problem using complex arithmetic (as
described previously)…
3. Then transform the result back to a sinusoid.
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Phasors
• A phasor is a rotating vector whose projection on
the vertical axis can be used to represent a sinusoid.
• The length of the phasor is the amplitude of the
sinusoid (Vm)
• The angular velocity of the phasor is  .
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Representing AC Signals
with Complex Numbers
• By replacing e(t) with it’s phasor equivalent E, we
have transformed the source from the time domain
to the phasor domain.
• Phasors allow us to convert from differential
equations to simple algebra.
• KVL and KCL still work in the phasor domain.
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Formulas from Trigonometry
• Sometimes signals are expressed in cosines instead of sines.
• Below are some formulas to refresh your memory of some of
the trig functions:
cos(t   )  sin(t    90 )
sin(t   )  cos(t    90 )
cos(t  180 )   cos(t )
sin(t  180 )   sin(t )
i.e. : cos(t  70 )  sin(t  160 )   sin(t  20 )
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Using Phasors to Represent
AC Voltage and Current
• Looking at the sinusoid equation, determine Vm and the phase
offset Ѳ:
v(t )  V sin( t  30 ) V
m
• Using Vm, determine VRMS using the formula:
VRMS 
Vm
• The voltage phasor is then:
V
2

RMS
• The same holds true for current:
i (t )  I m sin( t  30 )
I RMS 
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Im
2
I

RMS
Example Problem 4
1. Express 100 sin (ωt)as a voltage phasor.
VRMS 
Vm
2
VRMS 
100V
2
 70.7V
V
  70.70V
RMS
2. Express 50sin(ωt+45⁰) as a voltage phasor.
VRMS 
Vm
2
VRMS 
50V
2
 35.35V
V
  35.3545V
RMS
3. Express 50cos(ωt+45⁰) as a voltage phasor.
 Remember, we are representing the sinusoid as a sin function,
therefore we need to use one of the trig conversions:
cos(ωt+45⁰) = sin(ωt+45⁰+90⁰)
VRMS 
Vm
2
VRMS 
50V
2
 35.35V
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V
  35.3545  90V
RMS
VRMS   35.35135V
Representing AC Signals
with Complex Numbers
• Phasor representations can be viewed as a complex
number in polar form:
E  VRMS   Em  VRMS * 2  e(t )  Em sin(2 ft   )
• NOTE: Before writing the sinusoid equation after
finding the phasor, you need to convert the phasor
magnitude back to the sinusoid representation (i.e.
multiplying the phasor magnitude by 2.)
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Phase Difference
• Phase difference is angular displacement between
waveforms of same frequency.
• If angular displacement is 0°then the waveforms
are in phase.
• If angular displacement is not 0o, they are out of
phase by amount of displacement.
Example:
If v1 = 5 sin(100t) and v2 = 3 sin(100t - 30°), then v1 leads v2 by 30°
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Phase Difference w/ Phasors
• The waveform generated by the leading phasor leads
the waveform generated by the lagging phasor.
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Example Problem 5
i1 = 20 sin (t) mA.
i2 = 10 sin (t+90˚) mA.
i3 = 30 sin (t - 90˚) mA.
Determine the equation for iT.
Recall that I RMS 
I1 
I3 
Im
2
Im
2


20mA
2
30mA
2
Im
2
I

RMS
 14.10mA
I2 
Im
2

10mA
2
 7.0790mA
Now to get back to polar:
C  a 2  b 2  14.12  14.32 )  20.1
 21.2  90mA
  tan 1
IT  I1  I 2  I 3
b
14.3
 tan 1
 45.4
a
14.1
I RMS  20.1  45.4
IT  14.10mA  7.0790mA  21.2  90mA
Remember, it's easier to add in rectangular form so we need to convert:
Now plug the polar back into the sinusoid,
For the real (a) portion of the rectangular complex number (a  C cos  ):
but remember to get back to I m :
14.1cos(0) mA  7.07 cos(90) mA  21.2 cos( 90) mA  14.1
iT (t )  20.1* 2 sin ( wt  45.4) mA
For the imaginary (jb) portion of the rectangular complex number (b  C sin  ):
iT (t )  28.4sin ( wt  45.4) mA
14.1sin(0) mA  7.07 sin(90) mA  21.2sin( 90) mA  14.3
Put it together in the conversion and you get: IT  14.1  j14.3
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R, L and C circuits
with Sinusoidal Excitation
• R, L, C have very different voltage-current
relationships. Recall:
vR  iR R
dvC
iC  C
dt
diL
vL  L
dt
(Ohm's law)
(Capacitor Current relationship)
(Inductor Voltage relationship)
• Sinusoidal (AC) sources are a special case!
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The Impedance Concept
• Impedance (Z) is the opposition (i.e. resistance) that a circuit
element presents to current in the phasor domain. It is
defined as:
V V
Z     Z  
I
I
• Ohm’s law for AC circuits:
V  IZ
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Impedance
• Impedance is a complex quantity that can be made up of
Resistance (R) (real part) and Reactance (X) (imaginary
part).
Z  R  jX
( )
• Unit of impedance is Ohms ().
Z
X

R
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Resistance and Sinusoidal AC
• For a purely resistive circuit, current and voltage
are in phase.
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Resistors
• For resistors, voltage and current are in phase:
VR VR  VR
ZR 

 0  R0  R
I
I 
I
Z R  R0
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Example Problem 6
Two resistors R1=10 kΩ and R2=12.5 kΩ are in series.
i(t) = 14.7 sin (ωt + 39˚) mA
a)
b)
c)
d)
IT 
Compute VR1 and VR2
Compute VT = VR1 + VR2
Calculate ZT
Compare VT to the results of VT = I ZT
Im
2

14.7 mA
2
 10.3939mA
VR1  IT * Z R1  10.3939mA *10k 0  10439V
VR 2  IT * Z R 2  10.3939mA *12.5k 0  129.939V
VT  VR1  VR 2  10439V  129.939V  23439V
ZT  Z R1  Z R 2  10k 0  12.5k 0  22.5k 0
Note that these are in
phase so we simply sum
the magnitudes. May not
always be this easy…
VT  IT * ZT  10.3939mA * 22.5k 0  23439V  Same value calculated previously.
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Inductance and Sinusoidal AC
• Voltage-Current relationship for an inductor:
diL
d
 L  I m sin t 
dt
dt
  LI m cos t   LI m sin t  90
vL  L

vL  LI m sin t  90 
ZL 

iL
I m sin t
 LI m

90
2
Im
0
2
  L90 ( )
• It should be noted that for a purely inductive circuit voltage
leads current by 90º.
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Inductive Impedance
• Impedance can be written as a complex number (in
rectangular or polar form):
Z L   L90  j L ( )
• Since an ideal inductor has no real resistive
component, this means the reactance (X) of an
inductor is the pure imaginary part:
X L  L
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Inductance and Sinusoidal AC
Again, voltage leads current by 90˚.
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Inductance
• For inductors, voltage leads current by 90º.
VL VL 90 VL
ZL 

 90   L90  j L
I
I 0
I
Z L  jX L  X L 90
X L   L  2 fL
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Impedance and AC Circuits
Solution technique
1.
2.
3.
4.
Transform time domain currents and voltages into phasors.
Calculate impedances for circuit elements.
Perform all calculations using complex math.
Transform resulting phasors back to time domain (if reqd).
34
Example Problem 7
For the inductive circuit:
vL = 40 sin (ωt + 30˚) V
f = 26.53 kHz
L = 2 mH
VL 
40V 90
 28.330 V
2
ZL 
VL
V
 I L  L 90
I
ZL
IL 
28.3V 30 V
 85  60mA
33390 
IL 
Determine VL and IL
Graph vL and iL
Z L  jX L  X L 90
X L   L  2 fL
Z L  2fL90 
Z L  2 (26.3kHz )(2mH )90 
Z L  330.590 
Notice
90°phase
difference!
iL
 iL  I L * 2
2
iL  85 2 sin(t  60)
iL = 120 sin (ωt - 60˚) mA
35
Capacitance and Sinusoidal AC
• Current-voltage relationship for an capacitor:
dvC
d
iC  C
 C Vm sin t 
dt
dt
 CVm cos t  CVm sin t  90
Zc 
vc
Vm sin t

ic CVm sin t  90


Vm
0
1
2


  90 ( )
Vm
C
90 C
2
• It should be noted that, for a purely capacitive circuit current
leads voltage by 90º.
36
Capacitive Impedance
• Impedance can be written as a complex number (in
rectangular or polar form):
 1 
 1 
Zc  
   90   
 j ( )
 C 
 C 
• Since a capacitor has no real resistive component, this
means the reactance of a capacitor is the pure imaginary
part:
 1 
Xc  

 C 
37
Capacitance and Sinusoidal AC
Again, current leads voltage by 90˚.
38
Capacitance
• For capacitors, current leads voltage by 90º.
VC VC 0 VC
1
 1 
ZC 

   90 
  90   
j
I
I 90
I
C
 C 
ZC   jX C  X C   90
 1   1 
XC  


 C   2 fC 
39
ELI the ICE man
Pneumonic
ELI the ICE man
I leads E
E leads I
When voltage is applied to an
inductor, it resists the change
of current. The current builds
up more slowly, lagging in time
and phase.
41
Since the voltage on a capacitor is
directly proportional to the charge
on it, the current must lead the
voltage in time and phase to
conduct charge to the capacitor
plate and raise the voltage
Frequency Dependency
• Inductors:
• Capacitors:
42
QUESTIONS?
43