Download Sinusoids and Phasors

2012/11/13
Sinusoids and Phasors
•Introduction
•Sinusoids
•Phasors
•Phasor Relationships for Circuit Elements
•Impedance and Admittance
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•Impedance Combinations
•Applications
Introduction
•AC is more efficient and economical to transmit
power over long distance. (Transformer is the key.)
•A sinusoid is a signal that has the form of the sine
or cosine function.
•Circuits driven by sinusoidal current or voltage
sources are called ac circuits.
•Why sinusoid is important in circuit analysis?
–Nature itself is characteristically sinusoidal.
–A sinusoidal signal is easy to generate and transmit.
–Easy to handle mathematically.
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Sinusoids
Consider the sinusoidal voltage
v(t nT ) v(t )
v (t ) Vm sin t
where
Proof :
Vm the amplitude of the sinusoid

v(t nT ) Vm sin (t nT )



the
angular
frequency
(radians/s
)

2
Vm sin (t n )

ω
t

the
argument
of
the
sinusoid
ω

Vm sin(t 2n)
The sinusoid repeats itself every T seconds.
v(t )
2
ωT2 T 
(T:period )
ω
Sinusoids (Cont
’
d)
•A periodic function is one that satisfies
f(t) = f(t+nT), for all t and for all integers n.
–The period T is the number of seconds per cycle.
–The cyclic frequency f = 1/T is the number of cycles per
second.
1
f 
T
2f
where
: radians per second (rad/s)


f : hertz (Hz)
A more general expression is given as
v(t ) Vm sin(t )
where
(t ) : Argument


: Phase

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Sinusoids (Cont
’
d)
We say that
We say that
v2 leads v1 by 

v1 lags v2 by 

v1 and v2 are in phase, if 0

v1 and v2 are out of phase, if 0

Sinusoids (Cont
’
d)
•To compare sinusoids.
–Use the trigonometric identities.
–Use the graphical approach.
Trigonometric identities :
sin( A B) sin A 
cos B cos A 
sin B
cos( A B) cos A 
cos B sin A 
sin B
) sin t
sin(t 180

cos(t 180
) cos t


) cos t
sin(t 90

) sin t
cos(t 90
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The Graphical Approach
A cos t B sin t C cos(t )
A C cos θ
,B C sin θ

C  A2 B 2

where 
B
tan 1

A

3 cos t 4 sin t
5 cos(t 53.1
)
Phasors
•Sinusoids are easily expressed by using phasors
•A phasor is a complex number that represents the
amplitude and the phase of a sinusoid.
•Phasors provide a simple means of analyzing linear
circuits excited by sinusoidal sources.
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Phasors (Cont
’
d)
Three ways to represent a complex number z :
x jy : Rectangular form

z r : Polar form
, where
re j : Exponential form

r : magnitude of z


: phase of z

Given x and y, we can get r and as
y
r  x 2 y 2 , tan 1
x
If we know r and , we can obtain x and y as
x r cos , y r sin 
y
r
z

x
z x jy rr (cos j sin ) re j
Important Mathematical Properties
z x jy rre j
j
1
z1 x1 jy1 r1
1 r1e
z 2 x2 jy2 r2 2 r2 e j2
Addition :
z1 z2 ( x1 x2 ) j ( y1 y2 )
Substraction :
z1 z 2 ( x1 x2 ) j ( y1 y2 )
Multiplica tion :
z1 z2 ( x1 jy1 )( x2 jy 2 )
( x1 x2 y1 y2 ) j ( x1 y2 x2 y1 )
r1e j1 r2 e j2 r1r2 e j (1 2 )
r1r2 (
2 )
1
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Important Mathematical Properties
z1 x1 jy1 ( x1 jy1 )( x2 jy2 )


z2 x2 jy2 ( x2 jy 2 )( x2 jy2 )
Division :
x x y y
x y x y
 1 22 12 2 j 2 21 12 2
x2 y2
x2 y2
r e j1 r
r
 1 j2  1 e j (1 2 )  1 (
2 )
1
r2 e
r2
r2
Reciprocal :
Square Root :
1 1
 
z r
z  r  2
Complex Conjugate : z x jy r
Phasor Representation
 e j cos j sin 

cos Re(e j)

j
sin Im(e )
v(t ) Vm cos(t )
Re(Vm e j (t ) ) Re(Vm e je jt )
 v(t ) Re( Ve jt )
V Vm e j Vm 
V is the phasor representation
of the sinusoid v(t ).
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Phasor Representation (Cont
’
d)
Phasor Diagram
V Vm 
I I m 
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Sinusoid-Phasor Transformation
v(t ) Vm cos(t ) Re(Ve jt ) where V Vm e j Vm 
dv(t )
Vm sin(t ) Vm cos(t 90
)
dt
Re(Vm e jt e je j 90) Re e j 90(Vm e j)e jt

Re( jVe
jt

) Re(V
e )
jt
Finally we have
Sinusoids
v(t ) Vm cos(t )
Phasors
V Vm 
dv
Vm cos(t 90
) jV Vm (90
)
dt
V
V Vm
vdt  m cos(t 90
)
 (90
)


j 
Phasor Relationship for Resistor
If the current through the resistor is
i I m cos(t )
 I I m 
By Ohm' s law,
v iR RI m cos(t )  V RI m RI
Time domain Phasor domain
Phasor diagram
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Phasor Relationship for Inductor
If the current through the inductor is
i I m cos(t )
 I I m 
The voltage across the inductor is
di
v L LI m cos(t 90
)  V jLI
dt
Time domain Phasor domain
Phasor diagram
Phasor Relationship for Capacitor
If the voltage across the capacitor is
v Vm cos(t )
 V Vm 
The current through the capacitor is
i C
dv
CVm cos(t 90
)  I jCV
dt
Time domain Phasor domain
Phasor diagram
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Impedance and Admittance
V
1
Impedance : Z  () , Admittance : Y  (S)
I
Z
V is the phasor voltage

where 
I is the phasor current
Element Impedance Admittance
1
R
Z R
Y
R
1
L
Z jL
Y
jL
1
C
Z
Y jC
jC
Impedance and Admittance (Cont
’
d)
Z jL
 0
 
Z
1
jC
 0
 
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Impedance and Admittance (Cont
’
d)
Z R jX
R : resistance
where 
X : reactance
The impedance is said to be
inductive when X is positive


capacitive when X is negative

Z R jX Z 
Z  R 2 X 2

where 
1 X
tan
R

R Z cos 
and 
X Z sin 
If X is positive, then
Z R jX : inductive or lagging



since current lags voltage
Z R jX : capacitive or leading


since current leads voltage

Impedance and Admittance (Cont
’
d)
1
Y  G jB
Z
G : conductance

where 
B : susceptance
1
1
R jX
R jX
G jB 


 2
R jX R jX R jX R X 2
R

G 2
2

  R X
X
B  2
 R X 2
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KVL and KCL in the Phasor Domain
For KVL, let v1 , v2 ,..., vn , be the voltages around a closed loop.
 v1 v2 


vn 0
In the sinusoidal steady state, each voltage may be written in
cosine form.
 Vm1 cos(t 1 ) Vm 2 cos(t 2 ) 


Vmn cos(t n ) 0
This can be rewritten as
Re(Vm1e j1 e jt ) Re(Vm 2 e j2 e jt ) 


Re(Vmn e jn e jt ) 0


Vm1e j1 Vm 2 e j2 jt 


 Re 
e 0
jn 






V
e


mn

 

for any t
KVL and KCL (Cont
’
d)
Let Vk Vmk e jk , then

Re
Ve
 
Re 
V1 V2 


Vn 
e jt Re VT e jt
T
j (t T )
V
T

cos(t 
T ) 0
where VT VT e jT V1 V2 


Vn
1

Possible 
(1) cos(t 
T ) ()
T ) 0  t  ( 2 k


solutions 
(2) VT 0 ()

 V1 V2 


Vn VT e jT 0 (KVL holds for phasor.)
In a similar manner, KCL holds for phasor.
 I1 I 2 


I n 0
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Series-Connected Impedance
Applying KVL gives
V V1 V2 


Vn
I (Z1 Z 2 


Z n )
V
 Z eq  Z1 Z 2 


Z n
I
V
Z
I  , Vk  k V
Z eq
Z eq
Parallel-Connected Impedance
Applying KCL gives
I I1 I 2 


I n
I
 Yeq  Y1 Y2 


Yn
V
1
1
1
I
Y
V (  


 )
V  , Ik  k I
Z1 Z 2
Zn
Yeq
Yeq
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Y-Transformations
Y Δ Co
n
v
e
r
s
ion:
ΔY Conversion:
Z Z Z 2 Z 3 Z 3 Z1
Za  1 2
Z1
Zb Zc
Z1 
Z a Z b Z c
Z Z Z 2 Z 3 Z 3Z1
Zb  1 2
Z2
ZcZa
Z2 
Z a Z b Z c
Z Z Z 2 Z 3 Z 3Z1
Zc  1 2
Z3
Z a Zb
Z3 
Z a Z b Z c
Example 1
Find Z in for
50 rad/s.
Z in Z 2 mF Z 310 mF || Z 80.2 H



1
1


3

8 j
0.2
|| 
j
2m 
j


10
m



j10

3 j 2
|| 8 j10


3 j 2
8 j10
3.22 j11.07
j10 
11 j8
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Example 2
Find vo (t).
Sol:
vs 20 cos(4t 15
)
 Vs 2015, 4
j 20 || j 25
Vo 
Vs
60 
j 20 || j 25
j100



2015
60 j100




0.857530.96
2015
17.1515.96
 vo (t ) 17.15 cos(4t 15.96
)
Example 3
Sol:
j 4(2 j 4)
j 4 2 j 4 8
1.6 j 0.8
j 4(8)
Z bn 
j 3.2
10
8(2 j 4)
Z cn 
1.6 j 3.2
10
Z 12 Z an 
Z an 
Find I.
-Y transformation

Z bn j 3
|| Z cn j 6 8
13.6 j1 13.644.204
V
500
I 
Z 13.644.204
3.6664.204
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Applications: Phase Shifters
R
jRC
Vo 
Vi 
Vi
1
1 jRC
R
jC
jRC (1 jRC )
RC (RC j1)

Vi 
Vi
2 2 2
1  R C
1 2 R 2C 2
 RC

1 

tan 1 
Vi


2 2 2
RC 
 1  R C

Output leads input.
Phase Shifters (Cont
’
d)
1
1
1 jRC
jC
Vo 
Vi 
Vi 
Vi
1
1 jRC
1 2 R 2C 2
R
jC


1

tan 1 
RC 
Vi

2 2 2
 1  R C

Output lags input.
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Example
Design an RC circuit
to provide a phase of
90leading.
Sol:
20(20 j 20)
Z 20 || (20 j 20) 
12 j 4
40 j 20
2

Z
12 j 4

V1 
Vi 
Vi 
45
Vi


Z j 20
12 j 24
3

2
 2

2
 1
20


 45

Vo 
V1 
45
V1 
45
Vi  90






20 j 20
2
2
3

 


 3
Applications: AC Bridges
Balanced condition : V1 V2
Z2
Zx
V1 
Vs V2 
Vs
Z1 Z 2
Z 3 Z x
Z2
Zx
Z

 Z 2 Z 3 Z1Z x  Z x  3 Z 2
Z1 Z 2 Z 3 Z x
Z1
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AC Bridges (Cont
’
d)
Bridge for measuring L
Bridge for measuring C
R1 jLx R2 jLs
R1 jC x R2 jCs
R
 Lx  2 Ls
R1
R
 C x  1 Cs
R2
Summary
•Transformation between sinusoid and phasor is
given as
v(t ) Vm cos(t )  V Vm 
•Impedance Z for R, L, and C are given as
Z R R, Z L jL, Z C 
1
jC
•Basic circuit laws apply to ac circuits in the same
manner as they do for dc circuits.
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