Download Complex Numbers, Phasors and Circuits

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Complex Numbers, Phasors and Circuits
Complex numbers are defined by points or vectors in the complex
plane, and can be represented in Cartesian coordinates
z = a + jb
j = −1
or in polar (exponential) form
z = A exp( jφ) = A cos ( φ ) + jA sin ( φ )
a = A cos ( φ ) real part
b = A sin ( φ ) imaginary part
where
2
A= a + b
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φ = tan
−1  b 
 
 a
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Im
z
b
A
φ
a
Note :
Re
z = A exp( jφ) = A exp( jφ ± j 2 nπ)
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Every complex number has a complex conjugate
z* = ( a + jb ) * = a − jb
so that
z ⋅ z* = ( a + jb) ⋅ ( a − jb)
2
2
=a +b = z
2
= A2
In polar form we have
z* = ( A exp( jφ) ) * = A exp(− jφ)
= A exp ( j 2 π − jφ )
= A cos ( φ ) − jA sin ( φ )
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The polar form is more useful in some cases. For instance, when
raising a complex number to a power, the Cartesian form
z n = ( a + jb) ⋅ ( a + jb)… ( a + jb)
is cumbersome, and impractical for non−integer exponents.
polar form, instead, the result is immediate
In
n
z = [ A exp( jφ)] = An exp ( jnφ )
n
In the case of roots, one should remember to consider φ + 2kπ as
argument of the exponential, with k = integer, otherwise possible
roots are skipped:
n z = n A exp jφ + j 2 kπ = n A exp  j φ + j 2 kπ 
(
)


n 
 n
The results corresponding to angles up to 2π are solutions of the
root operation.
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Transmission Lines
In electromagnetic problems it is often convenient to keep in mind
the following simple identities
 π
j = exp  j 
 2
 π
− j = exp  − j 
 2
It is also useful to remember the following expressions for
trigonometric functions
exp( jz) + exp(− jz)
exp( jz) − exp(− jz)
cos ( z ) =
; sin ( z ) =
2
2j
resulting from Euler’s identity
exp(± jz) = cos( z) ± j sin( z)
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Complex representation is very useful for time-harmonic functions
of the form
A cos ( ω t + φ ) = Re [ A exp ( jω t + jφ )]
= Re [ A exp ( jφ ) exp ( jω t )]
= Re [ A exp ( jω t )]
The complex quantity
A = A exp ( jφ )
contains all the information about amplitude and phase of the
signal and is called the phasor of
A cos ( ω t + φ )
If it is known that the signal is time-harmonic with frequency
phasor completely characterizes its behavior.
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ω, the
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Often, a time-harmonic signal may be of the form:
A sin ( ω t + φ )
and we have the following complex representation
A sin ( ω t + φ ) = Re  − jA ( cos ( ω t + φ ) + j sin ( ω t + φ ) ) 
= Re [ − jA exp ( jω t + jφ )]
= Re [ A exp ( − jπ / 2 ) exp ( jφ ) exp ( jω t )]
= Re  A exp ( j ( φ − π / 2 ) ) exp ( jω t )
= Re [ A exp ( jω t )]
with phasor
A = A exp ( j ( φ − π / 2 ) )
This result is not surprising, since
cos(ω t + φ − π / 2) = sin(ω t + φ)
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Transmission Lines
Time differentiation can be greatly simplified by the use of phasors.
Consider for instance the signal
V ( t ) = V0 cos ( ω t + φ )
with phasor
V = V0 exp ( jφ )
The time derivative can be expressed as
∂ V ( t)
= −ωV0 sin ( ω t + φ )
∂t
= Re { jωV0 exp ( jφ ) exp ( jω t )}
⇒
jωV0 exp ( jφ ) = jω V
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is the phasor of
∂ V ( t)
∂t
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Transmission Lines
With phasors, time-differential equations for time harmonic signals
can be transformed into algebraic equations. Consider the simple
circuit below, realized with lumped elements
R
L
i (t)
v (t)
C
This circuit is described by the integro-differential equation
d i(t)
1 t
v( t ) = L
+ R i + ∫ i( t ) dt
dt
C −∞
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Transmission Lines
Upon time-differentiation we can eliminate the integral as
d2 i( t)
d v( t )
di 1
=L
+ R + i( t )
dt
dt C
dt 2
If we assume a time-harmonic excitation, we know that voltage and
current should have the form
v( t ) = V0 cos(ω t + αV )
phasor ⇒ V = V0 exp( jαV )
i( t ) = I0 cos(ω t + α I )
phasor ⇒
If
I = I0 exp( jα I )
V0 and αV are given,
⇒
I0 and αI are the unknowns of the problem.
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Transmission Lines
The differential equation can be rewritten using phasors
{
}
L Re −ω2 I exp ( jω t ) + R Re { jω I exp ( jω t )}
1
+ Re { I exp ( jω t )} = Re { jωV exp ( jω t )}
C
Finally, the transform phasor equation is obtained as
1 

V =  R + jω L − j
I=ZI

ωC 

where
Z
=
Impedance
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R
Resistance
+
1 

j ωL −

C
ω


Reactance
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Transmission Lines
The result for the phasor current is simply obtained as
V
V
I= =
= I0 exp ( jα I )
1 
Z 
 R + jω L − j ω C 


which readily yields the unknowns I0 and αI .
The time dependent current is then obtained from
i( t ) = Re { I0 exp ( jα I ) exp ( jω t )}
= I0 cos ( ω t + α I )
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Transmission Lines
The phasor formalism provides a convenient way to solve timeharmonic problems in steady state, without having to solve directly
a differential equation. The key to the success of phasors is that
with the exponential representation one can immediately separate
frequency and phase information. Direct solution of the timedependent differential equation is only necessary for transients.
Integro-differential
equations
Transform
Algebraic equations
based on phasors
I=?
i(t)=?
Direct Solution
( Transients )
i(t)
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Solution
AntiTransform
I
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Transmission Lines
The phasor representation of the circuit example above has
introduced the concept of impedance. Note that the resistance is
not explicitly a function of frequency. The reactance components
are instead linear functions of frequency:
Inductive component
⇒ proportional to ω
Capacitive component ⇒ inversely proportional to ω
Because of this frequency dependence, for specified values of L
and C , one can always find a frequency at which the magnitudes of
the inductive and capacitive terms are equal
ωr L =
1
ωr C
⇒
ωr =
1
LC
This is a resonance condition. The reactance cancels out and the
impedance becomes purely resistive.
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Transmission Lines
The peak value of the current phasor is maximum at resonance
I0 =
| I0|
V0
1 

R + ωL −

ω
C


2
2
IM
ωr
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ω
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Consider now the circuit below where an inductor and a capacitor
are in parallel
I
L
R
C
V
The input impedance of the circuit is
 1

Zin = R + 
+ jω C 
 jω L

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−1
= R+
jω L
1 − ω2 LC
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When
ω=0
Zin = R
1
ω=
LC
ω→∞
Zin → ∞
Zin = R
At the resonance condition
ωr =
1
LC
the part of the circuit containing the reactance components
behaves like an open circuit, and no current can flow. The voltage
at the terminals of the parallel circuit is the same as the input
voltage V.
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