Sinusoids and Phasors
Chapter 9
George Westinghoues (1946-1914)
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9.2 Sinusoids
• Consider the sinusoidal voltage
v(t) = Vm sint
where
Vm = the amplitude of the sinusoid
 = the angular frequency in radians/s
t = the argument of the sinusoid
T
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2

3
Fig 9.1
2 

v(t  T )  Vm sin  (t  T )  Vm sin   t 



 Vm sin(t  2 )  Vm sin t  v(t )
Hence,
v(t  T )  v(t )
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Fig 9.2
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Example 9.1
• Find the amplitude, phase, period, and frequency of
the sinusoid
v(t) = 12cos(50t + 10°)
• Solution:
The amplitude is Vm = 12V.
The phase is  = 10°
2
2

 0.1257 s.
The period T 
 50
The frequency is f  1  7.958 Hz.
T
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Example 9.2
• Calculus the phase angle between v1 = -10 cos (t +
50°) and v2 = 12 sin (t - 10°). State which sinusoid is
leading.
• Solution:
v1 = -10 cos (t + 50°) = 10 cos (t + 50° - 180°)
v1 = 10 cos (t - 130°) or v1 = 10 cos (t + 130°)
and
v2 = 12 sin (t - 10°) = 12 cos (t - 10° - 90°)
v2 = 12cos (t - 100°) . The phase difference between
v1 and v2 is 30°. We can write v2 as
v2 = 12 cos (t - 130° + 30°) or v2 = 12 cos (t + 260°)
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9.3 Phasor
• A phasor is a complex number that represents
the amplitude and phase of a sinusoid.
z  x  jy
z  r  re j
z  x  jy Rectangular form
z  r
Polar form
Exponential form
z  re j
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Fig 9.6
z  x  yj  r
 r (cos   j sin  )
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• Addition:
z1  z2  ( x1  x2 )  j ( y1  y2 )
• Subtraction:
z1  z2  ( x1  x2 )  j ( y1  y2 )
• Multiplication:
• Division:
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z1 z2  r1r2(1  2 )
z1 r1
 (1  2 )
z2 r2
10
• Reciprocal:
1 1
 ( )
z r
• Square Root:
z  r( / 2)
• Complex Conjugate:
z  x  jy  r( )  re
*
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 j
11
e j  cos   j sin 
cos  Re(e j )
sin   Im( e j )
v(t )  Vm cos(t   )  Re(Vme j (t  ) )
v(t )  Re(Vme j e jt )
v(t )  Re(Ve jt )
where V  Vme j  Vm
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Fig 9.7
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(Time-domain
representation)
Phasor-domain
representation)
v(t )  Vm cos(t   )  V  Vm
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Table 9.1 Sinusoid-phasor
transformation
Time domain
representation
Phasor domain
representation
Vm cos(t   )
Vm
Vmsin(t   )
Vm(  90)
I m cos(t   )
I m
I m sin(t   )
I m(  90)
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Example 9.3
• Evaluate these complex numbers:
1/ 2
(a) (4050  20  30)
10  30  (3  j 4)
(b)
(2  j 4)(3  j 5) *
• Solution:
(a) 4050  40(cos 50  j sin 50)  25.71  j 30.64
20  30  20[cos( 30)  j sin(30)]  17.32  j10
4050  20  30  43.03  j 20.64  47.7225.63
(4050  20  30)1/ 2  6.9112.81
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Example 9.3
10  30  (3  j 4) 8.66  j 5  (3  4 j )
(b)

(2  j 4)(3  j 5) *
(2  j 4)(3  j 5)
11.33  j 9 14.73  37.66


 14  j 22 26.08122.47
 0.565  160.13
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Example 9.4
• Transform these sinusoid to phasors:
(a) i  6 cos(50t  40) A
(b) v  4 sin(30t  50) V
• Solution:
(a) i  6 cos(50t  40) has the phasor
I  6  40 A
(b) Since  sinA  cos( A  90)
v  4 sin(30t  50)  4 cos(30t  50  90)
 4 cos(30t  140) V
The Phasor form of v is V  4140 V
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Example 9.5
• Find the sinusoid representation by these phasors:
(a) I  3  j 4 A
(b) V  j8e j 20 V
• Solution:
(a) I  3  j 4  5126.87
i (t )  5 cos(t  126.87) A
(b) j  190,
V  j8  20  (190)(8  20)
 890  20  870 V
v(t )  8 cos(t  70) V
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9.4 Phasor Relationships for
Circuit Elements
v  iR  RI m cos(t   )
V  RI m
V  RI
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Fig 9.9
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Fig 9.10
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di
v  L  LI m sin(t   )
dt
  sin A  cos( A  90),
 v  LI m cos(t    90)
V  LI me j ( 90)  LI me j e j 90  LI m(  90)
But I m  I, e j 90 j , then V  jLI.
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Fig 9.11
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Fig 9.12
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dv
iC
dt
I  jCV  V 
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I
jC
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Fig 9.13
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Fig 9.14
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Table 9.2 Summary of voltagecurrent relationships
Element
Time domain
Frequency
domain
R
V = Ri
V =RI
L
di
vL
dt
V = jLI
C
dv
iC
dt
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V
I
jC
29
Example 9.8
• The voltage c = 12 cos(60t + 45°) is applied to a 0.1H inductor. Find the steady-state current through the
inductor.
• Solution:
V  jLI, where   60rad/s and V  1245 V.
Hence,
V
1245 1245
I


 2  45 A
jL j 60  0.1 690
Converting ,
i (t )  2 cos(60t  45) A
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9.5 Impedance and Admittance
V
Z
I
or
V  ZI
• The impedance Z of a circuit is the ratio of the
phasor voltage V to the phasor current I,
measured in ohms (Ω).
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Table 9.3 Impedance and
admittances of passive elements
Element
R
L
C
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Impedance
Admittances
Z=R
1
Y
R
Z = jL
Z
I
jC
Y
I
jL
Y = jC
32
Fig 9.15
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Z  R  jX  Z 
where
X
Z  R  X ,   tan
R
2
2
1
and
R  Z cos ,
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X  Z sin 
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Example 9.9
• Find v(t) and i(t) in the circuit shown in Fig. 9.16.
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Example 9.9
• From the voltage source 10 cos 4t,  = 4,
Vs  100 V
• The impedance is
1
1
Z  5
 5
 5  j 2.5
jC
j 4  0.1
• Hence the current
Vs 100 10(5  j 2.5)
I

 2
Z 5  j 2.5
5  2.52
 1.6  j 0.8  1.78926.57 A
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Example 9.9
• The voltage across the capacitor is
1.78926.57
V  IZ C 

jC
j 4  0.1
1.78926.57

 4.47  63.43 V
0.490
• Converting I and V, we get
I
i (t )  1.789 cos(4t  26.57) A
v()  4.47 cos(4t  63.43) V
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9.6 Kirchhoff’s Law in the
Frequency Domain
V1  V2   Vn  0
I1  I 2   I n  0
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9.7 Impedance Combinations
V  V1  V2    VN  I (Z1  Z 2    Z N )
V
Z eq   Z1  Z 2    Z N
I
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Fig 9.19
V
I
Z1  Z 2
Since V1  Z1I and V2  Z 2I, then
Z1
Z2
V1 
V, V2 
V
Z1  Z 2
Z1  Z 2
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Fig 9.20
1
1
1 
I  I1  I 2    I N  V 
 

ZN 
 Z1 Z 2
1
I
1
1
1
 

 
Z eq V Z1 Z 2
ZN
Yeq  Y1  Y2    YN
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Fig 9.21
1
1
1
Z1Z 2
Z eq 



Yeq Y1  Y2 1/Z1  1 / Z 2 Z1  Z 2
V  IZ eq  I1Z1  I 2 Z 2
Z2
Z1
I1 
I, I 2 
I
Z1  Z 2
Z1  Z 2
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Fig 9.22
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• Y-Δ Conversion
Z1Z 2  Z 2 Z3  Z3Z1
Za 
Z1
Z1Z 2  Z 2 Z3  Z3Z1
Zb 
Z2
Z1Z 2  Z 2 Z3  Z3Z1
Zc 
Z3
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• Δ-Y Conversion
Ζb Z c
Z1 
Z a  Zb  Z c
Ζc Za
Z2 
Z a  Zb  Z c
Ζ a Zb
Z3 
Z a  Zb  Z c
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Example 9.10
• Find the input impedance of the circuit in Fig. 9.23.
Assume that the circuit operations at  = 50 red/s.
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Example 9.10
Let
Z1 = Impedance of the 2-mF capacitor.
Z2 = Impedance of the 3-Ω resistor in series with the
10-mF capacitor
Z3 = Impedance of the 0.2-H inductor in series with
the 8-Ω resisotr
1
1
Z


  j10 
Then
1
3
jC j 50  2  10
1
1
Z2  3 
 3
 (3  j 2) 
3
jC
j 50  10  10
1
Z3  8 
 8  j 50  0.2  (8  j10) 
jC
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Example 9.10
The input impedance is
(3  j 2)(8  j10)
Zin  Z1  Z 2 Z3   j10 
11  j8
(44  j14)(11  j8)
  j10 
  j10  3.22  j1.07 
2
2
11  8
Thus
Zin  3.22  j11.07 
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Example 9.11
• Determine v0(t) in the circuit of Fig. 9.25.
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Example 9.11
vs  20 cos( 4t  15)  Vs  20  15 V,   4
1
1
10mF 

jC j 4  10  103
5 H  jL  j 4  5  j 20 
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Example 9.11
Let
Z1 = Impedance of the 60-Ω resistor
Z2 = Impedance of the parallel combination of the 10mF capacitor and the 5-H inductor
Then Z1 = 60 Ω and
 j 25  j 20
Z 2   j 25 j 20 
 j100 
 j 25  j 20
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Example 9.11
By the voltage-division principle,
Z2
j100
Vo 
Vz 
(20  15)
Z1  Z 2
60  j100
 (0.857530.96)( 20  15)  17.1515.96 V
We converting this to the time domain and obtain
vo (t )  17.15 cos(4t  15.96) V
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