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Chapter 9
Complex Numbers and Phasors
Chapter Objectives:
 Understand the concepts of sinusoids and phasors.
 Apply phasors to circuit elements.
 Introduce the concepts of impedance and admittance.
 Learn about impedance combinations.
 Apply what is learnt to phase-shifters and AC
bridges.
Huseyin Bilgekul
EENG224 Circuit Theory II
Department of Electrical and Electronic Engineering
Eastern Mediterranean University
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Complex Numbers
 A complex number may be written in RECTANGULAR FORM as:
RECTANGULAR FORM
z = x+ jy j= -1, x=Re  z  , y=Im(z)
 A second way of representing the complex number is by specifying the
MAGNITUDE and r and the ANGLE θ in POLAR form.
POLAR FORM
z = x+ jy= z  =r
 The third way of representing the complex number is the EXPONENTIAL form.
EXPONENTIAL FORM
z = x+ jy= z  =re j
• x is the REAL part.
• y is the IMAGINARY part.
• r is the MAGNITUDE.
• φ is the ANGLE.
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Complex Numbers
 A complex number may be written in RECTANGULAR FORM as: forms.
z = x+ jy j= -1 RECTANGULAR FORM
x  r cos 
y  r sin 
z= r 
r  x2  y2
z= re j
r  x2  y2
POLAR FORM
y
 =tan -1
x
EXPONENTIAL FORM
y
 =tan -1
x
z = x + jy= r  = re j
e j =cos +jsin
Euler's Identity
cos  Re e j  Real part
sin  Im e j  Imaginary part
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Complex Number Conversions
 We need to convert COMPLEX numbers from one form to the other form.
z  x  jy  r  re j =r (cos  j sin  )
z  x  jy  r  re j =r (cos  j sin  )
y
r  x 2  y 2 ,   tan 1
Rectangular to Polar
x
x  rcos , y  r sin  Polar to Rectangular
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Mathematical Operations of Complex Numbers
 Mathematical operations on complex numbers may require conversions from one
form to other form.
ADDITION: z1 + z 2 =(x1 + x 2 )+j(y1 + y 2 )
SUBTRACTION: z1 - z 2 =(x1 -x 2 )+j(y1 - y 2 )
MULTIPLICATION: z1z 2 = r1 r2 1 +2
DIVISION:
z1 r1
= 1 -2
z 2 r2
RECIPROCAL:
1 1
= -
z r
SQUARE ROOT: z = r  
2
COMPLEX CONJUGATE: z  x  jy  r     re  j
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Phasors
 A phasor is a complex number that represents the amplitude and phase of a sinusoid.
 Phasor is the mathematical equivalent of a sinusoid with time variable dropped.
 Phasor representation is based on Euler’s identity.
e j =cos  jsin
Euler's Identity
cos  Re e j  Real part
sin  Im e j  Imaginary part
 Given a sinusoid v(t)=Vmcos(ωt+φ).
v(t )  Vm cos(t   )  Re(Vme j (t  ) )  Re(Vm e j e jt )  Re(Ve jt )
V  Vme j  Vm  PHASOR REP.
v(t )  Vm cos(t   )  V  Vm 
(Time Domain Re pr.)
v(t )  Re{Ve jt }
(Phasor Domain Re presentation)
(Converting Phasor back to time)
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Phasors
 Given the sinusoids i(t)=Imcos(ωt+φI) and v(t)=Vmcos(ωt+ φV) we can obtain the
phasor forms as:
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Phasors
 Amplitude and phase difference are two principal
concerns in the study of voltage and current sinusoids.
 Phasor will be defined from the cosine function in all our
proceeding study. If a voltage or current expression is in
the form of a sine, it will be changed to a cosine by
subtracting from the phase.
• Example
• Transform the following sinusoids to phasors:
–
–
i = 6cos(50t – 40o) A
v = –4sin(30t + 50o) V
Solution:
a. I  6  40 A
b. Since –sin(A) = cos(A+90o);
v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V
Transform to phasor => V  4140 V
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Phasors
•
Example 5:
•
Transform the sinusoids corresponding to
phasors:
a)
b)
V  1030 V
I  j(5  j12) A
Solution:
a) v(t) = 10cos(t + 210o) V
5
)  13 22.62
12
b) Since I  12  j5  12 2  52  tan 1 (
i(t) = 13cos(t + 22.62o) A
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Phasor as Rotating Vectors
v(t )  Vm cos(t   )
v(t )  Re Vm e( jt  ) 
v(t )  Re Vm ( jt   ) 
Rotating Phasor
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Phasor Diagrams
 The SINOR
Ve
j t
Rotates on a circle of radius Vm at an angular velocity of ω in the counterclockwise
direction
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Phasor Diagrams
Time Domain Re presentation
Vm cos(t   )
Phasor Domain Re p.
Vm 
Vm sin(t   )
Vm   90
I m cos(t   )
I m 
I m sin(t   )
I m   90
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Time Domain Versus Phasor Domain
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Differentiation and Integration in Phasor Domain
 Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by jω.
v(t )  Vm cos(t   )  Re  Ve jt 
dv(t )
 Vm sin(t   )  Vm cos(t    90)
dt
dv
= Re  j Ve jt 
 J V
dt
 Integrating a sinusoid is equivalent to dividing its corresponding phasor by jω.
(Time Domain)
(Phasor Domain)
v(t )  Vm cos(t   )

V  Vm 
v(t )  Vm sin(t   )

V  Vm   90
dv
dt
 vdt

JV

V
J
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Adding Phasors Graphically
 Adding sinusoids of the same frequency is equivalent to adding
their corresponding phasors.
V=V1+V2
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20cos(5t  30) A
1

5
1
H
10
2F
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Solving AC Circuits
 We can derive the differential equations for the following
circuit in order to solve for vo(t) in phase domain Vo.
d 2vo 5 dv0
400
o


20
v


sin(4
t

15
)
0
2
dt
3 dt
3

However, the derivation may sometimes be very tedious.
Is there any quicker and more systematic methods to do it?
 Instead of first deriving the differential equation and then
transforming it into phasor to solve for Vo, we can transform all the
RLC components into phasor first, then apply the KCL laws and other
theorems to set up a phasor equation involving Vo directly.
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