Transcript Phasors/Complex Numbers in AC

Lesson 18
Phasors & Complex Numbers in
AC
Learning Objectives
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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.
Define a phasor and use phasors to represent sinusoidal voltages
and currents.
Define time domain and phasor (frequency) domain
Represent a sinusoidal voltage or current as a complex number in
polar and rectangular form.
Use the phasor domain to add/subtract AC voltages and currents.
Determine when a sinusoidal waveform leads or lags another.
Graph a phasor diagram that illustrates phase relationships.
Complex numbers
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A complex number is a number of the form C =
a + jb where a and b are real and j = 1
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a is the real part of C and b 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)
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Geometric Representation
C = 6 + j8
(rectangular form)
C = 1053.13º
(polar form)
Conversion Between Forms
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To convert between forms where
C  a  jb
C  C
(rectangular form)
(polar form)
apply the following relations
a  C cos
b  C sin 
C  a 2  b2
1 b
  tan
a
Example Problem 1
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Convert (5∠60) to rectangular form.
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Convert 6 + j 7 to polar form.
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Convert -4 + j 4 to polar form.
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Convert (5∠220) to rectangular form.
Properties of j
j  1
j  ( 1)( 1)  1
2
1 1 
 
j j 
j
j
 2 j

j j
Addition and Subtraction of Complex Numbers
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Easiest to perform in rectangular form
Add/subtract real and imaginary parts separately
(6  j12)  (7  j 2) = (6  7)  j (12  2) = 13  j14
(6  j12)  (7  j 2) = (6  7)  j (12  2) =  1  j10
Multiplication and Division of Complex Numbers
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Easiest to perform in polar form
Multiplication: multiply magnitudes and add the
angles
(670)  (230)  6  2(70  30)  12100
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Division: Divide the magnitudes and subtract the
angles
(670) 6
 (70  30)  340
(230) 2
Example Problem 2
Given A =1 +j1 and B =2 – j3
 Determine A+B and A-B.
Given A =1.4145° and B =3.61-56°
 Determine A/B and A*B.
Reciprocals and Conjugates
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The reciprocal of C = C , is
1
1
  
C C
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The conjugate of C is denoted C*,
which has the same real value but
the opposite imaginary part:
C  a  jb  C
C  a  jb  C  
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Example Problem 3
And now you can try with your TI!!
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(3-i4) + (10∠44)
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(22000+i13)/(3∠-17)
Convert 95-12j to polar:
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ANS: 10.6∠16.1
ANS: 10.2 + 2.9i
ANS: 7.3E3∠17.0
ANS: 95.8∠-7.2
Phasor Transform
To solve problems that involve sinusoids
(such as AC voltages and currents) we
use the phasor transform.
 We transform sinusoids into complex
numbers in polar form, solve the problem
using complex arithmetic (as described),
and then transform the result back to a
sinusoid.
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THE SINUSOIDAL WAVEFORM
Generating a sinusoidal waveform through the vertical projection of a
rotating vector.
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 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 phasor domain.
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Using phasors to represent AC voltage and current
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Looking at the sinusoid eqn, determine VPk and phase offset
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v(t )  VPK sin(t  30 )
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Using VPK, determine VRMS using the formula:
“The equivalent dc value of a sinusoidal current or voltage is
0.707 of its peak value”
V
VRMS 
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The phasor is then
V
PK
2

RMS
Representing AC Signals with Complex
Numbers
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Phasor representations can be viewed as
a complex number in polar form.
e(t )  2Em sin(t   )
E = Erms
Example Problem 4
i1 = 20 sin (t) mA.
i2 = 10 sin (t+90˚) mA.
i3 = 30 sin (t - 90˚) mA.
Determine the equation for iT.
Phase Difference
Phase difference is angular
displacement between waveforms of
same frequency.
If angular displacement is 0° then
waveforms are in phase
If angular displacement is not 0o, they
are out of phase by amount of
displacement
Phase Difference
If v1 = 5 sin(100t) and v2 = 3 sin(100t - 30°), v1 leads v2 by
30°
Phase Difference w/ Phasors
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The waveform generated by the leading
phasor leads the waveform generated by
the lagging phasor.
Formulas from Trigonometry
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Sometimes signals are expressed in
cosines instead of sines.
cos(t   )  sin(t    90 )
sin(t   )  cos(t    90 )
cos(t  180 )   cos(t )
sin(t  180 )   sin(t )
 cos(t  70 )  sin(t  160 )   sin(t  20 )
Example Problem 5
Draw the phasor diagram, determine phase
relationship, and sketch the waveform for the
following:
i = 40 sin(t + 80º) and v = -30 sin(t - 70º)