Download Electromagnetic Plane Waves in Free Space

Electromagnetic Plane Waves in Free Space
Brian A. Lail
1
Introduction
Solutions to Maxwell’s equations produce the electromagnetic fields of boundary-value problems.
Maxwell’s equations are coupled differential equations. A single equation may depend on both the
electric field and the magnetic field. Uncoupling these equations leads to a second-order partial
differential equation which is commonly referred to as the wave equation. Here this wave equation
is developed for time-harmonic fields, and solutions are characterized.
2
Maxwell’s Equations in Free Space
In free space (J = 0, ρV = 0), the point form of Maxwell’s equations are given as:
∇ × H = ²o
∂E
∂t
∇ × E = −µo
∂H
∂t
(1)
(2)
∇·E =0
(3)
∇·H =0
(4)
These equations are valid for time-varying E and H, and lead to coupled electromagnetic waves.
3
Time-Harmonic Fields
For the case of sinusoidal (ejωt ) time variation, Maxwell’s equations are written as phasors.
Taking the curl of (6) gives
˜ = jω² E
˜
∇×H
o
(5)
˜ = −jωµ H
˜
∇×E
o
(6)
˜ =0
∇·E
(7)
˜ =0
∇·H
(8)
˜ = −jωµ ∇ × H
˜
∇×∇×E
o
(9)
Substituting Equation (5) into (9) gives
˜
˜ = −jωµ (jω² E)
∇×∇×E
o
o
1
(10)
A vector identity is applied to the left-hand-side of Equation (10)
which produces
˜ − ∇2 E
˜
˜ = ∇(∇ · E)
∇×∇×E
(11)
˜ = ω2µ ² E
˜
˜ − ∇2 E
∇(∇ · E)
o o
(12)
Substituting Equation (7) into (12), the resulting equation may be written
where
˜ + k2 E
˜ =0
∇2 E
o
(13)
√
ko2 = ω µo ²o
(14)
is the square of the free space wave number, ko .
˜
A similar procedure produces an equation in H:
˜ + k2 H
˜ =0
∇2 H
o
(15)
Equations (13) and (15) represent Maxwell’s equation, and may be solved, with appropriate
˜ and H.
˜
boundary conditions, for the fields E
4
Wave Solutions
˜ may be written
Generally, E
˜ = x̂Ẽ + ŷ Ẽ + ẑ Ẽ
E
x
y
z
(16)
From Equation (13), this means that each vector component must satisfy the following:
∂2
∂2
∂2
( 2 + 2 + 2 + ko2 )Ẽi = 0
∂x
∂y
∂z
(17)
where i = x, y, z denotes the component.
We will assume a solution that does not vary in x or y, such that
∂ Ẽi
∂ Ẽi
=
=0
∂x
∂y
(18)
leaving
∂ 2 Ẽi
+ ko2 Ẽi = 0
∂z 2
A general solution to Equation (19) is given by
(19)
Ẽi (z) = Ei+ e−jko z + Ei− e+jko z
(20)
Ei (z, t) = Ei+ cos(ωt − ko z) + Ei− cos(ωt + ko z)
(21)
or, in the time-domain as
In both (20) and (21), the quantities Ei+ and Ei− are generally complex, with both a magnitude
and phase.
A description of the properties of these solutions will be discussed in the next section.
2
5
Properties of Plane Waves in Free Space
Consider the solution given in Equation (21), and note that Hi has a similar form. The two terms
represent traveling waves, with the symbols + and − denoting a direction of propagation in the +z
and −z direction, respectively.
The first term in (21) represents a +ẑ traveling wave. Consider the phase of this term,
ωt − ko z = constant
(22)
represents a fixed point on the waveform. Choose value of the phase to be zero, rearrange and
differentiate to get a velocity.
dz
ω
= vph
(23)
=
dt
ko
This is known as the phase velocity. In free space the phase velocity is equal to the speed of light,c.
vph =
ω
ω
1
= √
=√
= c ≈ 3 × 108 m/s
ko
ω µo ²o
µo ²o
(24)
A plane wave in free space propagates with phase velocity equal to the speed of light.
Consider the ẑ-component of equation (5).
ẑjω²o Ẽz = ẑ(
∂ H̃y
∂ H̃x
−
)
∂x
∂y
(25)
Since we assumed a solution that doesn’t vary in x or y, then
∂ H̃y
∂ H̃x
=
=0
∂x
∂y
(26)
Ẽz = 0
(27)
leaving
A plane wave has no component along its direction of propagation.
Let us now consider how the field solutions are coupled. We have determined that a plane wave
traveling in the ẑ direction has no ẑ component. Let’s assume that the solution for Ẽ has only a x̂
component, or Ẽ = x̂Ẽx and that the wave is travelling in the +ẑ direction.
Ẽ(z) = x̂Ẽx+ (z) = Ex+ e−jko z
(28)
From Equation (6) we get the corresponding solution for H̃.
H̃y =
H̃x = 0
(29)
ko + −jko z
E e
ωµo x
(30)
H̃z = 0
(31)
By defining the intrinsic impedance of free space, ηo , as the ratio of Ẽx to H̃y we get
r
µo
ηo =
≈ 377Ω
²o
3
(32)
and the coupled field solutions may be written as
Ẽ(z) = Ex+ e−jko z
H̃(z) = ŷ
1 + −jko z
E e
ηo x
(33)
(34)
Equations (33) and (34) represent coupled electric and magnetic fields, which form a plane wave
with the electric field along x̂, the magnetic field along ŷ, and the direction of propagation along
ẑ. This is known as a transverse electromagnetic wave (TEM), where Ẽ, H̃, and the direction of
propagation are perpendicular to each other.
We may think of the wave number as a vector, with the unit vector k̂ denoting the direction of
propagation. Then the three vectors E, H, and k̂ form a right-handed system, with the direction
of E × H being k̂.
The corresponding time-domain (instantaneous) fields are given by
E(z, t) = x̂|Ex+ |cos(ωt − ko z + φ+ )
(35)
|Ex+ |
cos(ωt − ko z + φ+ )
ηo
(36)
H(z, t) = ŷ
6
EXAMPLE
Consider a plane wave in free space given by
E = 250x̂cos(106 t − ko z)V /m
6.1
(37)
angular frequency
The angular (radian) frequency, ω, is the coefficient of t. Therefore,
ω = 2πf = 106 rad/s
(38)
f = 159kHz
(39)
which gives a frequency
6.2
period
The period of the wave is the inverse of the frequency.
T =
6.3
1
= 6.28µs
f
(40)
wavelength
The wavelength λ , frequency f , and phase velocity vph are related as
λ=
vph
c
= = 1.88km
f
f
4
(41)
6.4
propagation constant or wave number
ω
√
ko = ω µo ²o = = 0.0033m−1
c
6.5
(42)
amplitude of H
The amplitude or magnitude of the coupled fields are related by
Hy =
6.6
250
Ex
=
= 0.663A/m
ηo
377
(43)
direction of propagation
The form of this solution represents propagation in the +ẑ direction.
7
Poynting Vector
For a wave with electric field E and magnetic field H, the power per unit area carried by the wave
is given by
P = E × H W/m2
(44)
Equation (44) is known as the Poyting vector. Since E, H, and k̂ form a right-handed system, the
power flows in the direction of wave propagation.
For example, a plane wave with fields:
E = x̂Ex cos(ωt − ko z)
(45)
Ex
cos(ωt − ko z)
ηo
(46)
H = ŷ
has power flow given by the Poynting vector
P = ẑ
Ex2
cos2 (ωt − ko z)W/m2
η
(47)
The time-average power density can be found by integrating over one period and dividing by the
period
Z
1 T Ex2
cos2 (ωt − ko z)dt
(48)
Pz,ave =
T 0 η
or, in the phasor representation as
1
P̃z,ave = <e(Ẽ × H̃ ∗ )
2
5
(49)
8
Wave Polarization
So far we’ve seen Ex and Hy propagating along ẑ. Generally, for ẑ propagation, the direction of E
and H in the plane perpendicular to ẑ may vary in both time and position.
Define wave polarization −→ electric field vector orientation as a function of time, at a fixed
position in space.
The polarization describes the shape and locus of the E vector in the plane perpendicular to
the direction of propagation, at a given point in space as a function of time.
Consider the electric field
˜
E(z)
= x̂Ẽx (z) + ŷ Ẽy (z)
(50)
with
Ẽx (z) = Exo e−jkz
(51)
Ẽy (z) = Eyo e−jkz
(52)
Exo and Eyo are generally complex −→ magnitude and phase angle.Wave polarization depends on
the phase of Eyo relative to the phase of Exo
Consider a phase difference δ between Exo and Eyo
Exo = |Exo |
(53)
Eyo = |Eyo |ejδ
(54)
˜
E(z)
= (x̂|Exo | + ŷ|Eyo |ejδ )e−jkz
(55)
E(z, t) = x̂|Exo |cos(ωt − kz) + ŷ|Eyo |cos(ωt − kz + δ)
(56)
such that
and the instantaneous field is
In order to describe the polarization, two quantities are considered: intensity and direction.
intensity:
£
¤1/2 £
¤1/2
|E(z, t)| = Ex2 (z, t) + Ey2 (z, t)
= |Exo |2 cos2 (ωt − kz) + |Eyo |2 cos2 (ωt − kz + δ)
direction:
ψ = tan−1 (
Ey (z, t)
Ex (z, t)
(57)
(58)
is the angle of inclination from the x̂-axis, for a given z.
8.1
Linear Polarization
Let δ = 0 or π. For convenience select z = 0. Therefore,
E(0, t) = (x̂|Exo | + ŷ|Eyo |)cos(ωt) =⇒ δ = 0
(59)
E(0, t) = (x̂|Exo | − ŷ|Eyo |)cos(ωt) =⇒ δ = π
(60)
and
For both cases, the intensity is found to be
£
¤1/2
|E(0, t)| = |Exo |2 + |Eyo |2
cos(ωt)
6
(61)
which shows that at z = 0 the magnitude of E oscillates as cos(ωt).
In order to determine the direction, the angle of incidence is computed. For the in-phase (δ = 0)
case:
|Eyo |
ψ = tan−1 (
)
(62)
|Exo |
and for the out-of-phase (δ = π) case:
ψ = tan−1 (
−|Eyo |
)
|Exo |
(63)
In both cases, ψ is independent of both z and t.
Figure (1) depicts linear polarization. For a fixed z =constant plane, the electric field oscillates
as cos(ωt) at a fixed angle of inclination. Note that if |Ey,o | = 0 then ψ = 0 giving an x-polarized
wave. Similarly, if |Exo | = 0 then ψ = π giving a y-polarized wave.
Figure 1: Linear Polarization
7
8.2
Circular Polarization
Let |Exo | = |Eyo | = Eo and δ = ±π/2.
8.2.1
Left-hand-circular, LHC
Now specify that δ = +π/2. This gives intensity
£
¤1/2
|E(z, t)| = Eo2 cos2 (ωt − kz) + Eo2 sin2 (ωt − kz)
= Eo
and direction
ψ = tan−1 (
−Eo sin(ωt − kz)
= −(ωt − kz)
Eo cos(ωt − kz)
(64)
(65)
The magnitude of the electric field is constant, while the direction varies in both z and t. For fixed
position, select z = 0, this gives ψ = −ωt decreasing with time. Therefore, the tip of E traces out
a circle in the xy-plane, in the clockwise direction when viewing the wave approaching. The left
hand thumb in the direction of propagation puts the left hand fingers in the direction of rotation
of E. For that reason, this is called left-hand circular polarization.
Figure 2: Linear Polarization
8
8.2.2
Right-hand-circular, RHR
Again, here |Exo | = |Eyo | = Eo , but now specify that δ = −π/2. This gives intensity
|E(z, t)| = Eo
(66)
ψ = (ωt − kz)
(67)
and direction
Therefore, for a fixed z-plane, say z=0 for convenience, the magnitude of the electric field is
constant while the direction is such that the tip of E traces out a circle in the xy-plane in the
counterclockwise direction when viewing the wave approaching. The right hand thumb in the
direction of propagation puts the right hand fingers in the direction of rotation of E. For that
reason, this is called right-hand-circular polarization.
8.3
EXAMPLE
˜ = 10(ŷ + j ẑ)e−j25x V /m.
The electric field of a uniform plane wave in free space is given by E
˜ and the polarization.
Determine the frequency, the magnetic field phasor, H,
8.3.1
frequency
Since k =
8.3.2
ω
c
then f =
kc
2π
=
(25)(3×108 )
2π
= 1.2GHz
magnetic field
˜ requires a positive ẑ-component of H,
˜
With propagation in the +x direction, the ŷ-component of E
˜ requires a negative ŷ-component of H,
˜ in order to form the right-handed
and the ẑ-component of E
system. Therefore,
˜ = 10 (ẑ − j ŷ)e−j25x A/m
H
(68)
ηo
8.3.3
polarization
The electric field phasor may be rewritten as
˜ = 10(ŷ + ẑejπ/2 )e−j25x V /m
E
(69)
˜ is a constant, while the phase angle δ is π/2. Therefore, the
Here we see that the magnitude of E
polarization of this wave is left-hand-circular.
9