22 Radio Engineering for Wireless Communication and Sensor Applications
In radio engineering we often consider a case where we have an interface with a good conductor. Fields penetrate only a short distance into a good conductor such as metal and not at all into a perfect conductor (s = ∞). We often approximate a good conductor with a perfect conductor, which is lossless. The boundary conditions at the interface of a dielectric and a perfect conductor are:
n × E = 0 |
(2.42) |
n × H = Js |
(2.43) |
n ? D = r s |
(2.44) |
n ? B = 0 |
(2.45) |
Such a boundary is also called an electric wall. Dual to the electric wall is the magnetic wall, where the tangential component of H vanishes.
2.4 Helmholtz Equation and Its Plane Wave Solution
In a source-free ( r = 0, J = 0), linear, and isotropic medium, Maxwell’s equations are simplified into the following forms:
= ? E = 0 |
(2.46) |
= ? H = 0 |
(2.47) |
= × E = −jvm H |
(2.48) |
= × H = jve E |
(2.49) |
When the = × operator is applied on both sides of (2.48), we obtain in a homogeneous medium
= × = × E = −jvm = × H |
(2.50) |
which leads, after utilizing vector identity
= × = × A = =(= ? A) − =2A |
(2.51) |
and (2.46), to
=2E = −v2meE = −k 2E |
(2.52) |
Fundamentals of Electromagnetic Fields |
23 |
This equation is called the Helmholtz equation, which is a special case of the wave equation
=2E − me |
∂2E |
= 0 |
(2.53) |
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∂t 2 |
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The constant k = v √me is called the wave number [1/m].
Let us first consider propagation of a wave in a lossless medium, where er and m r are real. Then k is also real. Let us assume that the electric field has only the x component, that the field is uniform in the x and y directions, and that the wave propagates in the z direction. The Helmholtz equation reduces to
∂2E x |
+ k |
2 |
E x = 0 |
(2.54) |
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∂z 2 |
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The solution of this equation is |
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E x (z ) = E +e −jkz + E −e jkz |
(2.55) |
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where E + and E − are arbitrary amplitudes of waves propagating into the +z and −z directions, respectively. The exact values of E + and E − are determined by the sources and the boundary conditions. In the time domain, (2.55) can be rewritten as
E x (z , t ) = E + cos (vt − kz ) + E − cos (vt + kz ) |
(2.56) |
where E + and E − are now real constants.
The magnetic field of a plane wave can be solved from (2.49). The
result is |
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Hy = |
1 |
(E +e −jkz − E −e jkz ) |
(2.57) |
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h |
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that is, the magnetic field has a component that is perpendicular to the electric field and to the direction of propagation. The ratio of the electric and magnetic fields is called the wave impedance, and it is h = √m /e . In vacuum h0 = √m0 /e0 ≈ 120pV ≈ 377V. Figure 2.5 illustrates a plane
24 Radio Engineering for Wireless Communication and Sensor Applications
Figure 2.5 Plane wave propagating into the +z direction.
wave propagating into the +z direction. The fields of a plane wave repeat themselves periodically in the z direction; the wavelength is
l = |
2p |
= |
2p |
= |
1 |
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(2.58) |
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k |
v√ |
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f √ |
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me |
me |
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The propagation velocity of the wave is
v = f l = |
1 |
(2.59) |
√me |
In a vacuum, the propagation velocity is the speed of light:
v = c = |
1 |
≈ 2.998 |
× 108 m/s |
(2.60) |
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√m0 e0 |
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In a lossy medium having conductivity s, Maxwell’s III and IV equations (the curl equations) are
= × E = −jvmH |
(2.61) |
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= × H = sE + jveE |
(2.62) |
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Now the Helmholtz equation gets the following form: |
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=2E + v 2me S1 − j |
s |
DE = 0 |
(2.63) |
ve |
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Fundamentals of Electromagnetic Fields |
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Compared to (2.52), here jk is replaced by a complex propagation constant,
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s |
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g = a + jb = jv √me √1 − j |
(2.64) |
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ve |
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where a is the attenuation constant and b is the phase constant. In the case of a plane wave propagating into the z direction, we have
∂2E x |
− g |
2 |
E x = 0 |
(2.65) |
∂z 2 |
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leading to
E x (z ) = E +e −gz + E −e gz |
(2.66) |
In the time domain
E x (z , t ) = E +e −az cos (v t − bz ) + E −e az cos (vt + b z )
In the case of a good conductor, that is, when s >> ve , we the propagation constant as
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s |
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vms |
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g = a + jb ≈ jv √me √ |
= (1 + j ) √ |
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jve |
2 |
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(2.67)
obtain
(2.68)
When a plane wave meets a surface of a lossy medium in the perpendicular direction and penetrates into it, its field is damped into 1/e part over a distance called the skin depth:
ds = |
1 |
= |
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(2.69) |
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√vms |
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Example 2.1
Find the attenuation of a 4 mm-thick copper layer at 10 GHz.
Solution
At a frequency of 10 GHz the skin depth in pure copper (s = 5.8 × 107 S/m, m = m 0 ) is only 6.6 × 10−7 m. Therefore, at this frequency a uniform