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Sinusoids and Phasors
Instructor: Chia-Ming Tsai
Electronics Engineering
National Chiao Tung University
Hsinchu, Taiwan, R.O.C.
Contents
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Introduction
Sinusoids
Phasors
Phasor Relationships for Circuit Elements
Impedance and Admittance
Kirchhoff’s Laws in the Frequency Domain
Impedance Combinations
Applications
Introduction
• AC is more efficient and economical to
transmit power over long distance.
• A sinusoid is a signal that has the form of the
sine or cosine function.
• Circuits driven by sinusoidal current (ac) 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
Sinusoids
Consider t he sinusoidal voltage
v(t )  Vm sin t
where
Vm  the amplitude of the sinusoid

   the angular frequency (radians/s )
ωt  the argument of the sinusoid

The sinusoid repeats itself every T seconds.
2
ωT  2  T 
(T:period )
ω
v(t  nT )  v(t )
Proof :
v(t  nT )  Vm sin  (t  nT )
2
)
ω
 Vm sin( t  2n )
 Vm sin  (t  n
 v(t )
Sinusoids (Cont’d)
• A period 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

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
Trigonomet ric 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  180)   sin t
cos(t  180)   cos t


sin( t  90)   cos t
cos(t  90)   sin t
The Graphical Approach
A cos t  B sin t  C cos(t   )
C  A2  B 2

where 
1 B
  tan 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.
Considerin g a complex number z ,
their are three ways to represent it.
 x  jy : Rectangula r form
r : the magnitude of z

z   r : Polar form
, where 
 : the phase of z
re j : Exponentia l form

Phasors (Cont’d)
 x  jy : Rectangula r form

z   r : Polar form
re j : Exponentia l form

Given x and y, we can get r and  as
y
r  x  y ,   tan
x
If we know r and  , we can obtain x and y as
2
2
1
x  r cos  , y  r sin 
z  x  jy  r  r (cos   j sin  )
Important Mathematical Properties
z  x  jy  r
z1  x1  jy1  r11
z2  x2  jy2  r22
Addition :
z1  z 2  ( x1  x2 )  j ( y1  y2 )
Substracti on :
z1  z 2  ( x1  x2 )  j ( y1  y2 )
Multiplica tion :
z1 z 2  r1r2 (1  2 )
Division :
z1 r1
 (1  2 )
z 2 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 representa tion
of the sinusoid v(t ).
Phasor Representation (Cont’d)
Phasor Diagram
V  Vm 
I  I m  
Sinusoid-Phasor Transformation
v(t )  Vm cos(t   )  V  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 )
dv
 jV
dt
Similarly,
V
 vdt  j

Phasor Relationships for Resistor
If the current th rough 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
Phasor Relationships for Inductor
If the current th rough 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 Relationships for Capacitor
If the voltage across the capacitor is
v  Vm cos(t   )
 V  Vm 
The current th rough the capacitor is
dv
iC
 CVm cos(t    90)  I  jCV
dt
Time domain Phasor domain
Phasor diagram
Impedance and Admittance
V
1
Impedance : Z 
() , Admittance : Y 
(S)
I
Z
V is the phasor vol tage
whe re 
 I is the phasor current
Element
Impedance
R
ZR
L
Z  j L
C
1
Z
j C
Admittance
1
Y
R
1
Y
j C
Y  jL
Impedance and Admittance (Cont’d)
Z  j L
 0
 
1
Z
jC
 0
 
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  R2  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 : conductanc e
where 
 B : susceptanc e
1
1
R  jX
R  jX
G  jB 


 2
R  jX R  jX R  jX R  X 2
R

G  R 2  X 2

X
B  2
R X2

KVL and KCL in the Phasor Domain
Vm1e j1  Vm 2 e j 2  jt 
e   0
 Re 
j

      Vmne n  
For KVL, let v1 , v2 ,..., vn , be the
voltages around a closed loop.
 v1  v2      vn  0
Let VK  Vmk e j k , then
In the sinusoidal steady state,
each volta ge may be written
in cosine form.
 Vm1 cos(t  1 )  Vm 2 cos(t   2 )
     Vmn cos(t   n )  0
This can be written as
j1
Re(Vm1e e
j t
j 2
)  Re(Vm 2 e e
j t
     Re(Vmne j n e jt )  0
)


Re V1  V2      Vn e jt  0
Since e jt  0 for any t ,
 V1  V2      Vn  0
 KVL holds for phasor!!!
In a similar manner,
KCL holds for phasor.
 I1  I 2      I n  0
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
Zk
V
I
, Vk 
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
Yk
I
 V( 
   )
V
, Ik 
I
Z1 Z 2
Zn
Yeq
Yeq
Y- Transformations
Y  Δ Conversion:
Z1Z 2  Z 2 Z 3  Z 3Z1
Za 
Z1
Δ  Y Conversion:
Zb Zc
Z1 
Z a  Zb  Zc
Z1Z 2  Z 2 Z 3  Z 3Z1
Zb 
Z2
ZcZa
Z2 
Z a  Zb  Zc
Z1Z 2  Z 2 Z 3  Z 3Z1
Zc 
Z3
Z a Zb
Z3 
Z a  Zb  Zc
Example 1
Find Zin for
  50 rad/s.
Z in  Z 2 mF  Z 3 10mF || Z8  0.2 H


1
1
 || 8  j  0.2

  3 
j  2 m 
j 10m 
  j10   3  j 2  || 8  j10 

3  j 28  j10 
  j10 
 3.22  j11.07
11  j8
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)
Z an 
j4  2  j4  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 bn  j3 || 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
Find I.
-Y transformation
Applications: Phase Shifters
Vo 
R
1
R
jC
jRC
Vi 
Vi
1  jRC

RC
1 
1 

 tan 
 Vi
2 2 2
 RC 
 1  R C
Leading
output
Phase Shifters (Cont’d)
1
1
jC
Vo 
Vi 
Vi
1
1  jRC
R
jC


1
1

  tan RC  Vi
2 2 2
 1  R C

Lagging
output
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

20
1






Vo 
V1  
45 V1  
45 
45 Vi  90
20  j 20
3
 2

 2
 3

Applications: AC Bridges
Balanced condition : V1  V2
Zx
Z2
V1 
Vs  V2 
Vs
Z1  Z 2
Z3  Z x
Zx
Z3
Z2

 Z 2 Z 3  Z1Z x  Z x 
Z2
Z1  Z 2 Z 3  Z x
Z1
AC Bridges (Cont’d)
Bridge for measuring L
Bridge for measuring C
R1
Lx 
Ls
R2
R1
Cx 
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
1
Z R  R, Z L  jL, ZC 
jC
• Basic circuit laws apply to ac circuits in the
same manner as they do for dc circuits.