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P9 SOLUTIONS TO PROBLEMS UNIFORM
PLANE ELECTROMAGNETIC WAVES
Section 9.1 Wave Equations
PROBLEM 9.1 3-D Helmholtz equations from Maxwell’s equations. We start with the source-free version of Eqs.(8.81), that is, the complex-domain version of Eqs.(9.1)-(9.4), ∇ × E = −jωμH , (P9.1) ∇ × H = jωεE , (P9.2) ∇ · E = 0 , (P9.3) ∇ · H = 0. (P9.4) We then take the curl of Eq.(P9.1), substitute ∇ × H on the right-hand side of thus obtained equation by the expression on the right-hand side of Eq.(P9.2), and perform transformations similar to those in Eq.(8.90), which gives
∇ × (∇ × E) = ∇(∇ · E) − ∇^2 E = −jωμ (jωεE) = ω^2 εμE. (P9.5)
Using Eqs.(P9.3) and (8.111), we get
∇^2 E + β^2 E = 0 (β = ω
εμ) , (P9.6)
namely, the 3-D Helmholtz equation for the electric field, Eq.(9.8). Analogously, taking the curl of Eq.(P9.2) and substituting ∇ × E on the right- hand side of the resulting equation by the expression on the right-hand side of Eq.(P9.1), combined with Eq.(P9.4), lead to
∇ × (∇ × H) = ∇(∇ · H) − ∇^2 H = ω^2 εμH −→ ∇^2 H + β^2 H = 0 , (P9.7)
which is Helmholtz equation for H, Eq.(9.9).
Section 9.3 Time-Domain Analysis of Uniform
Plane Waves
PROBLEM 9.2 Field expressions for a different propagation direction. (a) The electric field of a uniform plane (not necessarily time-harmonic) electro- magnetic wave propagating in free space in the negative x direction, with E having only a y-component, Fig.P9.1, is given by
E = Ey (x, t) yˆ = f
t +
x c 0
y ˆ , where c 0 =
ε 0 μ 0
, (P9.8)
with f (·) being an arbitrary twice-differentiable function.
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244 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
z
O
x
y
E
H
nn PPPP
Figure P9.1 Electric field intensity vector (E), magnetic field intensity vector (H), and Poynting vector (P) of a uniform plane electromagnetic wave propagating in free space (in the negative x direction).
With this form of the vector E, so E = Ey (x, t) yˆ, Eq.(9.6) simplifies to the following one-dimensional scalar wave equation [see Eqs.(4.126) and (9.10), and note that ∂^2 Ey /∂y^2 = 0 and ∂^2 Ey /∂z^2 = 0]:
∂^2 Ey ∂x^2
− ε 0 μ 0 ∂^2 Ey ∂t^2
= 0 or ∂^2 Ey ∂x^2
c^20
∂^2 Ey ∂t^2
= 0. (P9.9)
That the function Ey = f (t′), where t′^ = t + x/c 0 , is a solution of this equation is verified by direct substitution, as in Eqs.(8.98)-(8.100). Using the chain rule for taking derivatives, the partial derivatives of Ey with respect to t and x can be expressed as ∂Ey ∂t
df dt′
∂t′ ∂t
df dt′^
∂Ey ∂x
df dt′
∂t′ ∂x
c 0
df dt′^
. (P9.10)
Analogously, the expressions for the second partial derivatives are ∂^2 Ey ∂t^2
∂t
df dt′^
d^2 f dt′^2
∂t′ ∂t
d^2 f dt′^2
∂^2 Ey ∂x^2
∂x
c 0
df dt′
c 0
d^2 f dt′^2
∂t′ ∂x
c^20
d^2 f dt′^2
. (P9.11)
Substituting these expressions into Eq.(P9.9), we see that the governing 1-D wave equation is indeed satisfied for Ey in Eq.(P9.8).
(b) To obtain the accompanying magnetic field intensity vector of the wave – from the expression for E, using Maxwell’s equations, we invoke Eq.(9.1) and note that curl E on the left-hand side of the equation becomes (∂Ey /∂x) ˆz [see Eqs.(P9.8) and (4.81)]. This means that H (on the right-hand side of the equation) must be in the following form: H = Hz (x, t) ˆz , (P9.12) with which Eq.(9.1) becomes ∂Ey ∂x
= −μ 0
∂Hz ∂t
. (P9.13)
From Eqs.(P9.10), ∂Ey /∂x = (1/c 0 ) ∂Ey /∂t, so that ∂Hz ∂t
μ 0
∂Ey ∂x
μ 0 c 0
∂Ey ∂t
μ 0 c 0
∂f ∂t
. (P9.14)
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246 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
−→ H =
Ey η 0
(− ˆz) =
E 0
η 0
ejβx^ (− ˆz)
η 0 =
μ 0 ε 0
, (P9.21)
and, of course, vector relationships in Eqs.(9.22) are satisfied.
(c) Based on Eqs.(9.38) and (9.39), the total time-average electromagnetic energy density of the wave comes out to be
(wem)ave = (we)ave + (wm)ave = 2(we)ave = ε 0 E^20. (P9.22)
The associated complex Poynting vector, Eq.(9.40), equals
P = E × H∗^ = E 0 ejβx^
E∗ 0
η 0 e−jβx^ ˆy × (− ˆz) =
E 02
η 0 (− xˆ). (P9.23)
As P is purely real, the time-average Poynting vector of the wave is the same [see Eq.(8.195)],
Pave =
E 02
η 0 (− ˆx). (P9.24)
PROBLEM 9.4 Plane wave in a lossless nonmagnetic medium. (a) The wave propagates in the positive x direction, as shown in Fig.P9.2. From Eq.(8.49), the time period of the wave is T = 1/f = 13.33 ns. Having in mind Eqs.(9.36), the phase coefficient of the wave is β = π rad/m, so that Eq.(8.111) gives the wavelength of λ = 2π/β = 2 m. Using Eq.(9.35), the phase velocity of the wave amounts to vp = c = 2πf /β = λf = 1. 5 × 108 m/s. As the propagation medium is nonmagnetic, Eq.(9.47) applies, and the solution for the relative permittivity of the medium (dielectric) is hence εr = (c 0 /c)^2 = 22 = 4, with c 0 standing for the speed of light in free space, Eq.(9.19). Em- ploying Eq.(9.48), the intrinsic impedance of the dielectric then comes out to be η = η 0 /
εr = η 0 /2 = 60π Ω = 188.5 Ω.
z
O
x
y
E
H
nn
PPPP
Figure P9.2 Electric and magnetic field intensity vectors and Poynting vector of a uniform plane time-harmonic wave traveling through a lossless nonmagnetic medium (in the positive x direction).
(b) Eq.(9.20) tells us that the rms electric field intensity of the wave is E 0 = ηH 0 = 188 .5 V/m (H 0 = 1 A/m, from the given expression for H), and we see in Fig.P9. that the vector E is oriented in the positive y direction – such that the cross product,
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P9. Solutions to Problems: Uniform Plane Electromagnetic Waves 247
E × H, is in the direction of the wave propagation. Therefore, the complex electric field intensity vector of the wave is is
E = 188.5 e−jπx^ ˆy V/m (x in m). (P9.25)
On the other side, the instantaneous electric and magnetic field vectors of the wave, Eqs.(9.32), are
E(t) = 266.6 cos(4. 71 × 108 t − πx) yˆ V/m and
H(t) = 1.414 cos(4. 71 × 108 t − πx) ˆz A/m (t in s ; x in m). (P9.26)
(c) Combining Eqs.(9.25) and (P9.26), the instantaneous electromagnetic energy density of the wave equals
wem(t) = εrε 0 E^2 (t) = 2.52 cos^2 (4. 71 × 108 t − πx) μJ/m^3 , (P9.27)
and, by means of Eqs.(9.39) and (P9.25), its time average amounts to
(wem)ave = εrε 0 E 02 = 1. 26 μJ/m^3. (P9.28)
Finally, the complex Poynting vector of the wave, Eq.(9.40), is (Fig.P9.2)
P =
E^20
η
ˆx = 188. 5 ˆx W/m^2. (P9.29)
PROBLEM 9.5 Plane wave computation in free space. (a) As the propa- gation medium is free space and the phase coefficient of the wave is β = 10π rad/m, Eq.(8.111) tells us that the operating frequency is f = ω/(2π) = βc 0 /(2π) = 1 .5 GHz. The magnetic field intensity of the wave amounting to H = 0.02 A/m at t = 0 and z = 1.15 m, the initial (for t = 0) phase in the plane z = 0 of the electric field intensity (θ 0 ) turns out to be
H = 0.02 A/m =
cos(10π × 1 .15 + θ 0 ) A/m −→ cos(11. 5 π + θ 0 ) =
−→ 11. 5 π + θ 0 = ± π 3
−→ θ 0 = − 11. 5 π ± π 3
(+12π) = π 2
π 3
−→ θ 01 =
5 π 6
= 150◦^ and θ 02 =
π 6
= 30◦^ , (P9.30)
so there are two solutions for θ 0.
(b) Based on Eqs.(9.32), (9.36), and (9.22), the complex magnetic field intensity vector of the wave is given by
H = 28.2 ejθ^0 ej10πz^ ˆx mA/m (z in m). (P9.31)
(c) Having in mind that the wave propagates in the negative z direction, Eq.(9.40) results in the following expression for the time-average Poynting vector of the wave:
Pave =
2 × 377
(− ˆz) W/m^2 = − 0. 3 ˆz W/m^2. (P9.32)
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P9. Solutions to Problems: Uniform Plane Electromagnetic Waves 249
(a, b ≪ λ 0 ) , (P9.36) and this is, of course, the complex equivalent of the expression for eind(t) in Eq.(9.59). To obtain this same complex emf from the complex electric field intensity of the wave (left-hand side of Faraday’s law in the complex domain), we start with the corresponding result for an arbitrarily large contour (previous problem), and transform it into an approximate form, given that the contour is now electrically small, βa ≪ 1 −→ eind = E 0 b e−jβz^
e−jβa^ − 1
≈ −jβE 0 ab e−jβz ( e−jβa^ ≈ 1 − jβa
. (P9.37)
PROBLEM 9.9 Large contour positioned obliquely w.r.t. the wave travel. We first note that, since the free-space wavelength at the frequency of the wave amounts to λ 0 = c 0 /f = 60 cm (c 0 = 3 × 108 m/s) and a = 1 m, the contour in Fig.9.22 cannot be treated as an electrically small one, that is, we cannot use the simplifications pertinent to small contours.
z O
x
y S
E
H
nn
+^ a
H
a cosa H
a
z z + a^ cosa
Figure P9.3 Finding the induced emf in an electrically large square contour po- sitioned obliquely with respect to the direction of propagation of a uniform plane wave (Fig.9.22).
(a) Let us perform the analysis in the complex domain and let us adopt a Cartesian coordinate system in the same way, relative to the field vectors of the wave, as in Fig.9.5, so that the complex rms field intensities of the wave, Ex and Hy , are those in Eqs.(9.36). With this, the coordinates measured along the wave travel (along the z-axis) of the two edges of the contour, in Fig.9.22, to which the x-directed electric field vector of the wave is parallel are z and z + a cos α, respectively, as indicated in Fig.P9.3. Following the procedure of evaluating the induced emf in the contour based on the left-hand side of Faraday’s law of electromagnetic induction in integral form (in the complex domain) given by Eq.(9.55), we thus have
eind =
C
E · dl = −E 1 a + E 2 a = −Ex(z) a + Ex(z + a cos α) a
= E 0 a
[
e−jβ(z+a^ cos^ α)^ − e−jβz
]
= E 0 a e−jβz^
e−jβa^ cos^ α^ − 1
. (P9.38)
To find the magnitude of eind, i.e., the rms value of eind(t), we use Euler’s identity, Eq.(8.61), or the first identity in Eqs.(10.7), and perform the following transforma- tions: eind = E 0 a e−jβz^ e−j(βa^ cos^ α)/^2
[
e−j(βa^ cos^ α)/^2 − ej(βa^ cos^ α)/^2
]
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250 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
= −2jE 0 a e−jβz^ e−j(βa^ cos^ α)/^2 sin
βa cos α 2
−→ |eind| = 2E 0 a
∣sin^
βa cos α 2
= 2E 0 a
∣sin^
πa cos α λ 0
∣ = 9.84 V^.^ (P9.39)
(b) In the alternative way of calculating the emf in the contour, from the right- hand side of Faraday’s law, in the complex domain, we adopt a “broken” surface bounded by the contour (rather than the flat one spanned over the contour) such that the magnetic field vector of the wave is perpendicular to one part of the surface, Fig.P9.3, while being tangential to the remaining three (vertical) parts. Hence, a similar flux integration as in Eq.(9.56) leads to
eind = −jωΦ = −jω
S
B · dS = −jωμ 0
∫ (^) z+a cos α
z′=z
Hy (z′) a dz′
jωμ 0 E 0 a η 0
∫ (^) z+a cos α
z
e−jβz
′ dz′^ = ωμ 0 E 0 a η 0 β
[
e−jβ(z+a^ cos^ α)^ − e−jβz^
]
= E 0 a e−jβz^
e−jβa^ cos^ α^ − 1
, (P9.40)
which is the same result as in Eq.(P9.38).
PROBLEM 9.10 Electrically small oblique contour. Now, a = 2 cm is small relative to λ 0 = c 0 /f = 60 cm, namely, the contour in Fig.9.22 is electrically small.
(a) To find the emf induced in the contour from the left-hand side of Faraday’s law of electromagnetic induction, i.e., using the electric field of the wave, we specialize the result from the previous problem (arbitrarily large contour) to obtain
βa ≪ 1 −→ eind = E 0 a e−jβz^
e−jβa^ cos^ α^ − 1
≈ −jβE 0 a^2 cos α e−jβz ( e−jβa^ cos^ α^ ≈ 1 − jβa cos α
. (P9.41)
The magnitude of eind, that is, the rms value of eind(t), equals
|eind| = βE 0 a^2 cos α =
2 πE 0 a^2 cos α λ 0
= 18.1 mV. (P9.42)
(b) Alternatively, to compute eind from the right-hand side of Faraday’s law (using the field H in Fig.9.22), we assume that H is uniform over the surface spanned over the contour, shown in Fig.P9.4, and directly compute the induced emf as follows:
eind = −jωΦ ≈ −jωB · S = −jωBS cos α = −jωμ 0 Hy (z) a^2 cos α
jωμ 0 E 0 a^2 cos α η 0
e−jβz^ = −jβE 0 a^2 cos α e−jβz^ (a ≪ λ 0 ) , (P9.43)
and this, of course, is the same result as in Eq.(P9.41).
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252 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
(d) The integral in Eq.(P9.47), as well as that in Eq.(P9.48), would be maximum if the contour were positioned such that its plane is perpendicular to the vector E in Fig.9.5.
PROBLEM 9.13 Displacement current bounded by a large contour. (a) Let us adopt a Cartesian coordinate system as shown in Fig.P9.5, such that the complex rms field intensities of the wave, Ex and Hy , are those in Eqs.(9.36). Analogously to the computation in Eq.(9.55) and Fig.9.5, or to that in Problem 9.7, the line integral of the complex magnetic field intensity vector of the wave along the contour, Fig.P9.5, is evaluated as follows: ∮
C
H · dl = H 1 a − H 2 a = Hy (z) a − Hy (z + a) a = H 0 a
[
e−jβz^ − e−jβ(z+a)
]
= H 0 a e−jβz^
1 − e−jβa
, where β = ω
εμ. (P9.50)
x z
y
z
z + a
a
C
a
S
H 1 H 2
J d E
Figure P9.5 Computation of the line integral of the complex magnetic field inten- sity vector of a uniform plane time-harmonic electromagnetic wave along a square contour of an arbitrary electrical size and of the associated total displacement cur- rent intensity through a surface spanned over the contour.
(b) Having in mind Eq.(8.5), the total complex rms displacement current intensity through a surface (S) spanned over the contour in Fig.P9.5 is given by
Id =
S
Jd · dS = jωε
S
E · dS , (P9.51)
and this flux integral of E can be computed in the same way the flux of H is found in Eq.(9.56) and Fig.9.5 (or in Problem 9.7). Alternatively, we can use the corrected generalized Amp`ere’s law in Eqs.(8.80), where J = 0 (lossless medium), and the result in Eq.(P9.50) to obtain
Id = jωε
S
E · dS =
C
H · dl = H 0 a e−jβz^
1 − e−jβa
. (P9.52)
PROBLEM 9.14 Displacement current bounded by a small contour. (a)-(b) From the corrected generalized Amp`ere’s law, Eqs.(8.80), with J = 0,
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P9. Solutions to Problems: Uniform Plane Electromagnetic Waves 253
the circulation of H along the contour [part (a) of the problem] equals the total complex rms displacement current intensity through a surface spanned over the contour [part (b) of the problem], and the latter quantity is here (electrically small circular contour) simpler to compute, as follows:
Id =
S
Jd · dS = jωε
S
E · dS ≈ jωεE · S = ±jωεηHS = ±jβH 0 S e−jβz^ (P9.53)
(ωεη = ωε
μ/ε = ω
εμ = β).
PROBLEM 9.15 Power flow through a large rectangular aperture. The energy delivered by the wave through the aperture (opening) to the other side of the screen in one hour, Waperture, can be computed based on Eq.(9.63) or (9.64), and Fig.9.6.
(a) If the wave is incident normally onto the screen, Eq.(9.63) gives
Waperture = (Paperture)ave∆t = PaveS∆t =
E^20
η 0
ab ∆t = Em^2 2 η 0
ab ∆t = 95.69 mJ ( ∆t = 1 h = 3,600 s ≫ T =
f = 33.3 ps
. (P9.54)
(b) In the case of the wave propagation direction making the angle α = 60◦^ with the normal to the screen (or the aperture), we have [see Eq.(9.61) and Fig.9.6]
Waperture = PaveS cos α∆t =
E^2 m 2 η 0
ab cos α ∆t = 47.84 mJ. (P9.55)
PROBLEM 9.16 Safety limits for human exposure to electromagnetic radiation. Based on Eqs.(9.40) and (8.195), the maximum permissible levels for the rms intensities of the electric and magnetic fields, Emax and Hmax, in air corre- sponding to the maximum permissible time-average intensity of the Poynting vector, (Pave)max, at a given frequency, f , are given by
Pave =
E 02
η 0
= η 0 H^20 −→ Emax =
η 0 (Pave)max and Hmax =
Emax η 0
(P9.56)
where η 0 = 377 Ω. Taking the values for (Pave)max according to the IEEE standard (for human exposure safety limits), the computed maximum field levels at frequen- cies f 1 = 150 MHz, f 2 = 1.5 GHz, and f 3 = 15 GHz using these expressions are given in Table P9.1.
Table P9.1 Maximum permissible rms field levels for human exposure at different frequencies frequency (Pave)max Emax (rms) Hmax (rms) 150 MHz 2 W/m^2 27 .46 V/m 72 .8 mA/m 1 .5 GHz 10 W/m^2 61 .4 V/m 163 mA/m 15 GHz 100 W/m^2 194 .2 V/m 515 mA/m
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P9. Solutions to Problems: Uniform Plane Electromagnetic Waves 255
H = 9.19 e−j
◦ ( ˆx − ˆy) mA/m. (P9.60) Having in mind Eq.(9.76), the complex Poynting vector of the wave is
P = E × H∗^ =
|E 0 |^2
η 0
n ˆ = 36.8( ˆx + ˆy + ˆz) mW/m^2 (P9.61)
(| − xˆ − ˆy + 2 ˆz| =
Section 9.7 Theory of Time-Harmonic Waves in
Lossy Media
PROBLEM 9.20 More on Helmholtz equations for a lossy medium. (a) Governing Maxwell’s equations [Eqs.(8.81)] for this case are given by
∇ × E = −jωμH , (P9.62)
∇ × H = σE + jωεE , (P9.63) ∇ · E = 0 , (P9.64) ∇ · H = 0 , (P9.65) where ρ = 0 in the third equation because the propagation medium is homogeneous (see Example 8.10). Taking the curl of Eq.(P9.62), substituting ∇ × H on the right- hand side of thus obtained equation by the expression on the right-hand side of Eq.(P9.63), and performing transformations similar to those in Eq.(8.90) yield
∇ × (∇ × E) = ∇(∇ · E) − ∇^2 E = −jωμ∇ × H = −jωμ (σE + jωεE)
=
ω^2 εμ − jωμσ
E. (P9.66)
Then, using Eqs.(P9.64), (9.82), and (9.78), we get
∇^2 E − γ^2 E = 0
[
γ^2 = −ω^2 εeμ = −ω^2
ε − j
σ ω
μ = −ω^2 εμ + jωμσ
]
, (P9.67)
with εe standing for the equivalent complex permittivity in Eq.(9.78), and the result is exactly the E-field 3-D Helmholtz equation for a lossy medium, in Eqs.(9.86).
(b) To verify that the field Ex(z) = E 0 e−γz^ , in Eq.(9.81), is a solution of Eq.(P9.67), we first realize that, with this form of the vector E, so E = Ex(z) ˆx, Eq.(P9.67) simplifies to the following 1-D scalar Helmholtz equation [see Eqs.(4.126) and (9.10), and note that ∂^2 Ex/∂x^2 = 0 and ∂^2 Ex/∂y^2 = 0]:
d^2 Ex dz^2
− γ^2 Ex = 0. (P9.68)
We then take the second derivative of Ex(z),
d^2 Ex dz^2
= E 0
d^2 dz^2
e−γz^ = (−γ)^2 E 0 e−γz^ = γ^2 Ex , (P9.69)
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256 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
and what we obtain is indeed Eq.(P9.68), that is, Eq.(P9.67).
(c) Finally, the lossy version of Eq.(9.43) gives
Hy =
jE 0 ωμ
d dz e−γz^ =
−jγ ωμ E 0 e−γz^ =
εe μ Ex =
Ex η
γ = jω
εeμ
, (P9.70)
which is simply the H-field expression for the wave via the complex intrinsic impedance of the medium, Eq.(9.90).
PROBLEM 9.21 Different propagation direction in a lossy medium. (a) The expression for Ez is (Fig.P9.6)
Ez = E 0 eγy^ , where γ = jω
εeμ and εe = ε − j σ ω
, (P9.71)
and, since d^2 Ez dy^2
= E 0
d^2 dy^2
eγy^ = γ^2 E 0 eγy^ = γ^2 Ez −→
d^2 Ez dy^2
− γ^2 Ez = 0 , (P9.72)
we see that the corresponding 1-D Helmholtz equation is indeed satisfied. z
O
x
y
E
nn H PPPP
Figure P9.6 Electric and magnetic field intensity vectors and Poynting vector of a uniform plane time-harmonic electromagnetic wave propagating in the negative y direction through a lossy medium of parameters ε, μ, and σ.
(b) Invoking the first equation in Eqs.(8.81), where, because E = Ez (y) ˆz, ∇ × E [Eq.(4.81)] becomes ( dEz / dy) ˆx, and thus H = Hx(y) ˆx, we substitute in it the expression for Ez from Eq.(P9.71), which yields dEz dy
= −jωμHx −→ Hx =
jE 0 ωμ
d dy
eγy^ =
jγ ωμ
E 0 eγy^ = −
εe μ
Ez = −
Ez η
−→ H =
Ez η (− ˆx) =
E 0
η eγy^ (− ˆx)
η =
μ εe
, (P9.73)
with η standing for the complex intrinsic impedance of the medium [Eq.(9.91)], and the vector H is shown in Fig.P9.6. Obviously, the vector relationships in Eqs.(9.22) hold true for the results for E and H.
(c) From Eq.(9.96), the time-average Poynting vector of the wave (having in mind that it propagates in the negative y direction – Fig.P9.6) is given by
Pave =
E^20
|η|
e^2 αy^ cos φ (− yˆ) , (P9.74)
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258 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
(b) Conversely, from Eq.(9.98), the power and field intensity ratios, P 1 /P 2 and E 1 /E 2 , can be expressed in terms of AdB as follows:
AdB = 10 log
P 1
P 2
= 20 log
E 1
E 2
P 1
P 2
= 10AdB/^10 and
E 1
E 2
= 10AdB/^20 , (P9.79) and thus computed P 1 /P 2 and E 1 /E 2 for the given values of AdB are presented in Table P9.3.
Table P9.3 Power and field intensity ratios for the given values of AdB AdB 60 dB 14 dB 6 dB 1 dB 0 dB −3 dB −14 dB −100 dB P 1 /P 2 106 25 4 1.26 1 0.5 0.04 10 −^10 E 1 /E 2 103 5 2 1.122 1 0.707 0.2 10 −^5
Section 9.8 Explicit Expressions for Basic
Propagation Parameters
PROBLEM 9.24 Finding parameters of a lossy medium from wave travel. (a) As the phase lag of the magnetic field behind the electric field of the wave is φ = 21◦, we can compute the ratio of coefficients α and β, and then find α from the known β (β = 204 rad/m), using Eqs.(9.111), as follows:
σ ωε
= tan 2φ = tan 42◦^ = 0. 9 −→ u =
1 + tan^2 2 φ = 1. 345
α β
u − 1 u + 1 = 0. 384 −→ α = 0. 384 β = 78.3 Np/m. (P9.80)
Next, Eqs.(9.113) and (9.114), with the angular frequency of the wave amounting to ω = 6. 28 × 109 rad/s, give the relative permittivity and conductivity of the (nonmagnetic) medium,
εr =
2 α^2 c^20 ω^2 (u − 1) = 81 and σ = ωεrε 0 tan 2φ = 4 S/m (μ = μ 0 ) , (P9.81)
where c 0 = 3 × 108 m/s (free-space wave velocity). Note that material parameters εr = 81 and σ = 4 S/m indicate that the propagation medium might be seawater.
(b) The complex propagation coefficient of the wave, Eq.(9.82), equals
γ = α + jβ = (78.3 + j204) m−^1. (P9.82)
(c) From Eqs.(9.118), the magnitude of the complex intrinsic impedance of the medium is |η| =
η 0 √ εru
= 36 Ω (η 0 = 377 Ω) , (P9.83)
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P9. Solutions to Problems: Uniform Plane Electromagnetic Waves 259
and, based on Eq.(9.93), the instantaneous magnetic field intensity vector of the wave comes out to be
H =
Em(0) |η| eαy^ cos(ωt+βy −φ) ˆz = 27.8 e^78.^3 y^ cos(6. 28 × 109 t+204y − 21 ◦) ˆz mA/m
(t in s ; y in m) , (P9.84) with Em(0) = 1 V/m being the amplitude of the electric field in the plane y = 0.
(d) Having in mind Eq.(9.96), the time-average Poynting vector of the wave is
Pave =
E^20
|η|
e^2 αy^ cos φ (− ˆy) = 13 e^156.^6 y^ (− ˆy) mW/m^2 (y in m) , (P9.85)
where E 0 = Em(0)/
2 is the rms electric field intensity of the wave for y = 0.
PROBLEM 9.25 Finding parameters of a biological tissue. (a) As the wave amplitude is reduced by 3.25 dB for every cm traveled, invoking Eq.(9.88) the attenuation coefficient of the wave is obtained as
AdB = 8. 686 αd −→ α =
AdB
- 686 d
8. 686 × 0. 01
Np/m = 37.42 Np/m. (P9.86) The phase coefficient of the wave is β = 260 rad/m, so we now know the ratio α/β, and thus, from Eqs.(9.111), also the temporary parameter u (useful for further computations),
α β
u − 1 u + 1
−→ u = 1 + (α/β)^2 1 − (α/β)^2
= 1. 042. (P9.87)
With the use of Eqs.(9.118), the phase angle of the complex intrinsic impedance of the medium is φ =
arctan
u^2 − 1 = 8. 19 ◦^ , (P9.88)
and hence Eqs.(9.113) and (9.114) result in the following values of the relative per- mittivity and conductivity of the biological tissue (which, of course, is a nonmagnetic medium):
εr =
α^2 c^20 2 π^2 f 2 (u − 1)
= 42. 1 and σ = 2πf εrε 0 tan 2φ = 1.3 S/m. (P9.89)
(b) The magnitude of η, by means of Eqs.(9.118), turns out to be
|η| =
η 0 √ εru
= 56.9 Ω , (P9.90)
so that the time-average Poynting vector of the wave, Eq.(9.96), amounts to
Pave = |η|H 02 e−^2 αz^ cos φ ˆz = 35.2 e−^74.^84 z^ ˆz mW/m^2 (z in m). (P9.91)
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P9. Solutions to Problems: Uniform Plane Electromagnetic Waves 261
and u =
1 + tan^2 2 φ = 1. 08 −→ α = 2πf
εrε 0 μ 0 2
u − 1 = 23.4 Np/m
and η = η 0 √ εru
ejφ^ = 54.2 ej11.^1
◦ Ω (η 0 = 377 Ω). (P9.97)
At the second frequency (f = 1.9 GHz), from the same equations, these parameters turn out to be φ = 8. 16 ◦, u = 1.042, α = 37.4 Np/m and η = 57 ej8.^16
◦ Ω.
(b) By virtue of Poynting’s theorem in complex form, Eq.(8.196), the time-average power absorbed (lost to heat) in the first d = 1 cm of depth into the material per S = 1 cm^2 of cross-sectional area, that is, in the first v = 1 cm × 1 cm × 1 cm of the material past the interface, (PJ)ave, equals the negative of the real part of the outward complex power flow through the surface S 0 enclosing v. Namely,
0 = (PJ)ave + jQ +
S 0
P · dS −→ (PJ)ave = −Re
S 0
P · dS
S 0
Re{P} · dS = −
S 0
Pave · dS = Pave(0)S − Pave(d)S , (P9.98)
where Pave is the time-average Poynting vector of the wave, and, given that it is directed into the material, (PJ)ave turns out to be the difference of fluxes of Pave through the surface of area S at the entry of the wave into the solution and through the surface of the same area d past the interface. Using Eq.(9.96), we thus have
(PJ)ave =
E^20
|η|
cos φ −
E 02
|η|
e−^2 αd^ cos φ =
E 02
|η|
cos φ
1 − e−^2 αd
, (P9.99)
and the result is (PJ)ave = 16.9 W at f = 835 MHz, while (PJ)ave = 22.8 W at f = 1.9 GHz.
Section 9.10 Wave Propagation in Good Conductors
PROBLEM 9.28 Various combinations of material parameters and fre- quency. We start with the classification into a medium behaving as a good dielec- tric, quasi-conductor, or good conductor of each of the combinations of material parameters and frequency performed in Example 9.16. Then, for good dielectrics, the attenuation and phase coefficients, dB attenuation per meter traveled, wave- length, and phase velocity of a uniform plane wave propagating in the material, and the complex intrinsic impedance of the material are computed using Eqs.(9.123), (9.89), (8.111), (9.35), and (9.124), respectively. For quasi-conductors, these pa- rameters are evaluated based on Eqs.(9.105), (9.106), (9.89), (8.111), (9.35), and (9.109). Finally, we invoke Eqs.(9.135), (9.89), (9.143), and (9.136) if the material at a given frequency is classified as a good conductor. All results are tabulated in
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262 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
Table P9.4.
Table P9.4 Various propagation parameters for different combinations of material parameters and fre- quency. material f (MHz) behaves as α (Np/m) β (rad/m) α in dB/m λ (m) vp (m/s) η (Ω) glass 0. 1 good diel. 8. 42 × 10 −^11 0. 00469 7. 32 × 10 −^10 1341 1. 34 × 108 168 ej fresh water 0. 1 quasi-cond. 0. 016 0. 0247 0. 139 255 2. 55 × 107 26 .9 ej
◦
copper 0. 1 good cond. 4785 4785 4. 16 × 104 0. 00131 131 1. 17 × 10 −^4 ej ◦ rural ground 12,840 good diel. 0. 503 1007 4. 37 0. 00624 8. 01 × 107 101 ej rural ground 12. 84 quasi-cond. 0. 458 1. 11 3. 98 5. 68 7. 29 × 107 84 .7 ej22.^5 ◦ rural ground 0. 01284 good cond. 0. 0225 0. 0225 0. 196 279 3. 58 × 106 3 .18 ej ◦
Section 9.11 Skin Effect
PROBLEM 9.29 1/1000 depth of penetration in seawater. The expression for the ocean depth at which the electric field amplitude of a radio wave decreases to 1 /1000 of its value at the ocean surface is obtained in a complete analogy with the derivation of the expression for the one-percent depth of penetration in Eqs.(9.141) and (9.142), which now become
Em(δ 1 / 1000 ) = Em(0) e−α δ^1 /^1000 = Em(0) e−δ^1 /^1000 /δ^ =
Em(0) 1000 −→ δ 1 / 1000 = δ ln 1000 ≈ 6. 9 δ , (P9.100) where δ is the skin depth of the material (seawater), at the same frequency. At frequencies f 1 = 1 kHz, f 2 = 10 kHz, f 3 = 100 kHz, f 4 = 1 MHz, and f 5 = 10 MHz, σ/(ωε) for seawater with εr = 81 and σ = 4 S/m turns out to be 8. 88 × 105 , 8. 88 × 104 , 8. 88 × 103 , 888, and 88.8, respectively, meaning that the condition in Eq.(9.133) is satisfied and seawater can be assumed to behave as a good conductor in all cases. So, the computed values for the 1/1000 depth of penetration, phase velocity, wavelength, and index of refraction in seawater, using Eqs.(P9.100), (9.139), (9.143), and (9.144), at each of the frequencies f 1 - f 5 , are given in Table P9.5.
Table P9.5 Various wave parameters for seawater at five different frequencies. frequency 1 kHz 10 kHz 100 kHz 1 MHz 10 MHz δ 1 / 1000 (m) 0. 276 0. 873 2. 76 8. 73 27. 6 vp (m/s) 5 × 104 1. 58 × 105 5 × 105 1. 58 × 106 5 × 106 λ (m) 50 15. 8 5 1. 58 0. 5 n 6000 1900 600 190 60
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