CT Lecture 2
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Transcript of CT Lecture 2
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7/31/2019 CT Lecture 2
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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.
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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)
x is the REAL part.
y is the IMAGINARY part.
ris the MAGNITUDE.
is the ANGLE.
A second way of representing the complex number is by specifying the
MAGNITUDE and rand the ANGLE in POLAR form.
z = x+ jy= zPOLAR FORM
=r
The third way of representing the complex number is the EXPONENTIAL form.
z = x+
EXPONENT
jy= z
IAL FORM
= jre
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Complex Numbers
A complex number may be written in RECTANGULAR FORM as: forms.
2 2 -1
j
2 2 -1
j
j
z = x+ jy j= -1
cos y sin
z=
=tan
z= e
RECTANGULAR FORM
POLAR FORM
EXPO
=tan
z = x + jy=
NE
= e
e =cos +j
NTIAL FORM
x r r
r
yr x y
x
r
yr x y
x
r r
j
j
sin
cos Re
Euler's Identity
Real part
Imaginary p
e
sin Im e art
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Complex Number Conversions
We need to convert COMPLEX numbers from one form to the other form.
z = (cos sin )jx jy r re r j
2 2 1 Rectangular to Polar
Polar to Rectangu
z = (cos sin )
, tan
cos , sin lar
jx jy r re r j
yr x y
x
x r y r
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Mathematical Operations of Complex Numbers
Mathematical operations on complex numbers may require conversions from one
form to other form.
1 2 1 2 1 2
1 2 1 2 1 2
1 2 1 2 1 2
1 11 2
2 2
z + z =(x + x )+j(y + y )
z - z =(x -x )+j(y - y )
z z = r r +
ADDITION:
SUBTRACTION:
MULTIPLICATION:
DIVISION:
RECIPROCAL:
SQUARE ROOT:
z r= -
z r
1 1
= -z r
z=
COMPLEX CO
r2
NJUGATE: z
rj
x jy re
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Phasors
(Time Domain Re pr.) (Phasor Domain Re presentation)
( ) Re{ } (Converting Phasor back to time)
( ) cos( )m m
j t
v t V t V
v t e
V
V
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 Eulers identity.
Given a sinusoidv(t)=Vmcos(t+).
j
j
j
e =cos jsin
co
Euler's Identity
Real part
Imaginary pa
s Re e
s rtin Im e
( )( ) cos( ) Re( ) Re( ) Re( )
PHA S .OR REP
j t j t j t
m m
j
m m
j
mv t V t V e e
V
e e
V
V
e
V
V
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Phasors
Given the sinusoids i(t)=Imcos(t+I) and v(t)=Vmcos(t+ V) 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 ourproceeding study. If a voltage or current expression is inthe form of a sine, it will be changed to a cosine bysubtracting from the phase.
Example
Transform the following sinusoids to phasors:
i = 6cos(50t40o) A
v =4sin(30t + 50o) V
Solution:
a. I A
b. Since sin(A) = cos(A+90o);
v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V
Transform to phasor => V V
406
1404
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Phasor as Rotating Vectors
( )
( ) cos( )
( ) Re
( ) Re ( )
Rotating Phasor
m
j t
m
m
v t V t
v t V e
v t V j t
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Phasor Diagrams
The SINOR
Rotates on a circle of radius Vm at an angular velocity of in the counterclockwise
direction
j te
V
Ph D
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Phasor Diagrams
cos( )
sin(
Time
) 90
cos
Domain Representation Phasor Domain Re
( )
sin( ) 0
p.
9
m m
m m
m m
m m
V t V
V t V
I t I
I t I
Ph
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Time Domain Versus Phasor Domain
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Differentiation and Integration in Phasor Domain
(Time Domain) (Phasor Domain)
( ) cos( )
( ) sin( ) 90
V
m m
m m
v t V t V
v t V t V
dvJ
dt
vdtJ
V
V
V
Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by j.
( ) cos( ) Re
( )sin( ) cos( 90 )
= Re
j t
m
m m
j t
v t V t e
dv t
d
V t V t dt
ev
j Jdt
V V
V
Integrating a sinusoid is equivalent to dividing its corresponding phasor by j.
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20cos(5 30 ) At 15
2 F1
H10
S l i AC Ci it
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We can derive the differential equations for the followingcircuit in order to solve for vo(t) in phase domain Vo.
20
02
5 40020 sin(4 15 )
3 3
ood v dv v tdt dt
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 componentsinto phasor first, then apply the KCL laws and other
theorems to set up a phasor equation involving Vo directly.
Solving AC Circuits