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Slide 1EE40 Spring 2008 Venkat Anantharam
EE40Lecture 14
Venkat Anantharam
2/27/08 Reading: Chap. 5: phasors
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Slide 2EE40 Spring 2008 Venkat Anantharam
Complex Numbers (1)• x is the real part• y is the imaginary part• z is the magnitude• θ is the phase
( 1)j = −
θ
z
x
y
real axis
imaginary axis
• Rectangular Coordinates Z = x + jy
• Polar Coordinates: Z = z ∠ θ
• Exponential Form:
θcoszx = θsinzy =
22 yxz +=xy1tan−=θ
(cos sin )z jθ θ= +Z
j je zeθ θ= =Z Z
0
2
1 1 1 0
1 1 90
j
j
e
j eπ
= = ∠ °
= = ∠ °
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Slide 3EE40 Spring 2008 Venkat Anantharam
Complex Numbers (2)
2 2
cos2
sin2
cos sin
cos sin 1
j j
j j
j
j
e e
e ej
e j
e
θ θ
θ θ
θ
θ
θ
θ
θ θ
θ θ
−
−
+=
−=
= +
= + =
Euler’s Identities
Exponential Form of a complex numberj je ze zθ θ θ= = = ∠Z Z
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Slide 4EE40 Spring 2008 Venkat Anantharam
Arithmetic With Complex Numbers• To compute phasor voltages and currents, we
need to be able to perform computations with complex numbers.– Addition– Subtraction– Multiplication– Division
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Slide 5EE40 Spring 2008 Venkat Anantharam
Addition
• Addition is most easily performed in rectangular coordinates:
A = x + jyB = z + jw
A + B = (x + z) + j(y + w)
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Slide 6EE40 Spring 2008 Venkat Anantharam
Addition
Real Axis
Imaginary Axis
AB
A + B
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Slide 7EE40 Spring 2008 Venkat Anantharam
Subtraction
• Subtraction is most easily performed in rectangular coordinates:
A = x + jyB = z + jw
A - B = (x - z) + j(y - w)
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Slide 8EE40 Spring 2008 Venkat Anantharam
Subtraction
Real Axis
Imaginary Axis
AB
A - B
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Slide 9EE40 Spring 2008 Venkat Anantharam
Multiplication
• Multiplication is most easily performed in polar coordinates:
A = AM ∠ θB = BM ∠ φ
A × B = (AM × BM) ∠ (θ + φ)
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Slide 10EE40 Spring 2008 Venkat Anantharam
Multiplication
Imaginary Axis
Real Axis
A
A × BB
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Slide 11EE40 Spring 2008 Venkat Anantharam
Division
• Division is most easily performed in polar coordinates:
A = AM ∠ θB = BM ∠ φ
A / B = (AM / BM) ∠ (θ − φ)
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Slide 12EE40 Spring 2008 Venkat Anantharam
Division
Imaginary Axis
A
B
Real Axis
A / B
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Slide 13EE40 Spring 2008 Venkat Anantharam
Phasors• Assuming a source voltage is a sinusoidal time-
varying functionv(t) = V cos (ωt + θ)
• We can write:
• Similarly, if the function is v(t) = V sin (ωt + θ)
( ) ( )( ) cos( ) Re Rej t j t
j
v t V t V e Ve
Define Phasor as Ve V
ω θ ω θ
θ
ω θ
θ
+ +⎡ ⎤ ⎡ ⎤= + = =⎣ ⎦ ⎣ ⎦= ∠
( )
( )2
2
( ) sin( ) cos( ) Re2
j tv t V t V t Ve
Phasor V
πω θ
πθ
πω θ ω θ+ −
−
⎡ ⎤= + = + − = ⎢ ⎥
⎣ ⎦
= ∠
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Slide 14EE40 Spring 2008 Venkat Anantharam
Phasor from rotating Complex Vector
{ } )( tjjwtj eeVetVtv ωφφω VReRe)cos()( ==+=
Imaginary Axis
VReal Axis
Rotates at uniform angular velocity ωt
cos(ωt+φ)
The head start angle is φ.
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Slide 15EE40 Spring 2008 Venkat Anantharam
Complex Exponentials• We represent a real-valued sinusoid as the real
part of a complex exponential after multiplying by .
• Complex exponentials – provide the link between time functions and phasors.– Allow derivatives and integrals to be replaced by
multiplying or dividing by jω– make solving for AC steady state simple algebra with
complex numbers.• Phasors allow us to express current-voltage
relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor.
tje ω
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Slide 16EE40 Spring 2008 Venkat Anantharam
I-V Relationship for a Capacitor
C v(t)
+
-
i(t)
dttdvCti )()( =
Suppose that v(t) is a sinusoid:v(t) = Re{VM ej(ωt+θ)}
Find i(t).
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Slide 17EE40 Spring 2008 Venkat Anantharam
Capacitor Impedance (1)
C v(t)
+
-
i(t)dttdvCti )()( =
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( ) cos( )2
( )( )2 2
sin( ) cos( )2 2
(
2
j t j t
j t j t j t j t
j t j t
c
Vv t V t e e
dv t CV d CVi t C e e j e edt dt
CV e e CV t CV tj
V VZCVI
ω θ ω θ
ω θ ω θ ω θ ω θ
ω θ ω θ
ω θ
ω
ω πω ω θ ω ω θ
θ θ θπ ωθ
+ − +
+ − + + − +
+ − +
⎡ ⎤= + = +⎣ ⎦
⎡ ⎤ ⎡ ⎤= = + = −⎣ ⎦ ⎣ ⎦
− ⎡ ⎤= − = − + = + +⎣ ⎦
∠= = = ∠ −
⎛ ⎞∠ +⎜ ⎟⎝ ⎠
VI
1 1 1) ( )2 2
jC C j C
π πω ω ω
− = ∠ − = − =
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Slide 18EE40 Spring 2008 Venkat Anantharam
Capacitor Impedance (2)
C v(t)
+
-
i(t)dttdvCti )()( =
Phasor definition
( )
( )( )
( ) cos( ) Re
( )( ) Re Re
1( )
j t
j tj t
c
v t V t Ve V
dv t dei t C CV j CVe Idt dtV VZI j CV j C
ω θ
ω θω θ
ω θ θ
ω θ
θ θ θθ ω ω
+
++
⇒
⇒
⎡ ⎤= + = = ∠⎣ ⎦⎡ ⎤
⎡ ⎤= = = = ∠⎢ ⎥ ⎣ ⎦⎣ ⎦
∠= = = ∠ − =
∠
V
I
VI
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Slide 19EE40 Spring 2008 Venkat Anantharam
Example
v(t) = 120V cos(377t + 30°)C = 2µF
• What is V?• What is I?
• What is i(t)?
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Slide 20EE40 Spring 2008 Venkat Anantharam
Computing the Current
Note: The differentiation and integration operations become algebraic operations
ωjdt 1
⇒∫ωjdtd
⇒
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Slide 21EE40 Spring 2008 Venkat Anantharam
Inductor Impedance
V = jωL I
L
+
-
i(t)
dttdiLtv )()( =v(t)
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Slide 22EE40 Spring 2008 Venkat Anantharam
Example
i(t) = 1µA cos(2π 9.15 107t + 30°)L = 1µH
• What is I?• What is V?
• What is v(t)?