Chapter 13: Solutions to Problems Problem 13.1. Reflection and transmission at air-lossy dielectric interface. The wave coming from the left propagates without attenuation until it hits the wall. Now, part of the wave is reflected, part is transmitted. Free space is lossless but for the material to the right we must first decide if we have a low or high loss material. For the properties given for water:
σ2 = 10−9 10−5 = = 2.25×10−9 << 1 8 −12 ωε 2 2π×10 ×80×8.85×10 2π×80×8.85 This is a very low loss dielectric. For a low loss dielectric we can write:
η0 =
µ 0 = 377 ε0
µ2 = ε2
η2 ≈
µ 0 = 377 = 42.15 80 ε0 80
Ω
Now, using the definitions of the reflection and transmission coefficients we can write:
Γ = η2 − η0 = η0 + η2
µ2 − ε2 µ0 + ε0
µ0 377 − 377 ε 0 = 80 = − 0.799 µ2 377 + 377 ε2 80
The transmission coefficient is T = 1 + Γ = 1 − 0.799 = 0.201 The incident wave at the surface is (assuming the wave is calculated at some distance z from the interface and that the electric field intensity is in the x direction): Ei(z) = xE ie −jβ0z V/m where the phase constant is that of free space and equals: 8 β0 = ω µ 0 ε 0 = 2 π ×108 = 2 π = 2.089 3 3×10
rad m
The reflected wave is: Er(z) = xΓEie jβ0z = x0.799Eie j2.089z
V/m
The transmitted electric field intensity at a distance z from the interface is: −8
Et(z) = xΤEie −α2ze −jβ2z = x0.201Eie −2.1×10 ze −j18.73z
V/m
where the phase and attenuation constants in material (2) are: 8 β 2 ≈ ω µ 2 ε 2 = 2 π f = 2 π ×10 ×8 80 = 18.73 v p2 3×10 −9 α 2 ≈ σ η2 = 10 2 2
rad m
µ 2 = 10−9 377 = 2.1×10−8 ε2 2 80
Np m
Problem 13.2. Incident and reflected waves at a lossless dielectric interface. Calculate the reflection coefficient:
η0 =
µ0 ε0
η2 =
a. Reflection coefficient is:
µ 0 = η0 2ε0 2
Ω
Γ = η 2 − η 0 = − 2 − 1 = − 0.17157 η2 + η0 2+1 Thus: E1(z) = E0e −jβ1z 1 + Γe j2β1z V/m E1max = E0 1+ Γ = 11.716, E1min = E0 1− Γ = 8.284 b. In terms of the magnetic field: H1max = 11.716 = 0.031 377
H1min = 8.284 = 0.0219 377
296
V/m A m
Problem 13.3. Incident and reflected waves at a lossy dielectric interface. The electric field intensity to the left of the interface is the sum of the incident and reflected waves and is given in Eq. (13.25). The only difference between reflection at a lossless and lossy interface is in the value of the reflection coefficient. We calculate the general form of the electric field intensity and from this we evaluate the maximum and minimum values. The location of the minima and maxima are also evaluated from the general relations. E1(z) = xEi1 e −jβ0z + Γe +jβ0z where β0 is the phase constant of free space and: Γ = η 2 − η1 where: η0 = η 1 + η2
V/m
µ 0 = 377, ε0
η2 =
(1) j ωµ 2 σ 2 + jωε 2
Ω
(2)
The reflection coefficient is complex for a lossy dielectric because η2 is complex and may be written as Γe jθ Γ . In this expression, θΓ is the phase angle of the reflection coefficient. To calculate the maxima and minima, we write Eq. (1) as: E1(z) = xEi1e −jβ0z 1 + Γ e θΓe j2β0z
V/m
Thus, since the exponential terms all have a magnitude of 1. Maximum electric field intensity is therefore: E 1max = xEi1 1 + Γ
V/m
Similarly, the minimum electric field intensity is: E 1min = xEi1 1 − Γ Thus, the ratio between E1max and E1min is: Emax Emin
V/m
1 + η2 − η0 1+ Γ η2 + η0 = SWR = = 1 − Γ 1 − η2 - η0 η2 + η0
This ratio is called the standing wave ratio of the wave and is an important parameter in electromagnetic design.
Problem 13.4. Application: Transmission of power into solar cells. Part of the incident wave is reflected and part transmitted into the cell. Of the transmitted power, 25% is converted into electricity. The efficiency is the ratio between the power converted into electricity and input power. If the reflection coefficient can be decreased, the efficiency of the cell can be increased a. The reflection coefficient at the silicon-air interface is:
η air = 377,
ηsilicon = 377 = 284.98 1.75
Ω
→
Γ = η silicon − ηair = 284.98 − 377 = − 0.139 η silicon + ηair 284.98 + 377
The transmission coefficient is:
Τ = 1 + Γ = 1 − 0.139 = 0.861 To calculate the transmitted power per unit area of the cell we first find the electric field intensity in air, then the electric field intensity inside the cell and finally the power transmitted. The electric field intensity in air is: 2 W Pin = Ein → Ein = 2ηair m2 The transmitted electric field intensity is: Et = TEin = T 2Pinηair
2Pinηair
V m
V/m
Thus, the transmitted power into the silicon is: Pt =
Et2 = T 2Pinηair 2ηsilicon 2ηsilicon
2
2 2 = T P inη air = 0.861 ×1400×377 = 1372.97 η silicon 284.98
W m2
where Pin is the available solar power density. Out of this power, 25% is converted into electricity or: Pconverted = Pt×0.25 = 1372.97×0.25 = 343.24
297
W/m2
The overall efficiency of the solar cell is the ratio between converted power and input power or: eff = Pconverted = 343.24 = 24.5 % Pin 1,400 The overall efficiency is 24.5%. b. If the reflection coefficient can be eliminated, the transmission coefficient becomes T = 1 and all power available enters the cell. Then: W W P t = 1,400 → Pconverted = Pt×0.25 = 350 m2 m2 the efficiency is therefore 25%, a slight improvement.
Problem 13.5. Power transmitted into glass at normal incidence. The incident electric and magnetic field intensity in the beam are calculated from the beam power density. The electric field intensity, magnetic field intensity and power transmitted into the glass are calculated from the transmission coefficient and incident values. All values are peak values. a. First we calculate the intrinsic impedances in air and glass and the transmission coefficient at the interface:
η air = 377,
ηglass = 377 = 281 1.8
Ω
→
Τ=
2 η glass 562 = = 0.8541 η glass + ηair 281 + 377
The amplitude of the electric field intensity is found from the power density: 2 Pin = Ein 2η0
→
Ein =
2Pinη0 = 2×0.1×377 = 8.68
V m
The incident magnetic field intensity is: Hin = E in = 8.68 = 0.023 η0 377
A m
These fields are transmitted into glass through the transmission coefficient. Thus, Eglass = EinT = 8.68×0.8541 = 7.414 H glass = E inT =7.414 = 0.0264 η glass 281
V/m A m
b. The total incident power equals the incident power density multiplied by the area of the beam. 2
2 Pin = P in×πd = 0.1×π(0.0001) = 7.854×10 −10 W 4 4 Out of this, the transmitted power is: P glass = PinT 2 η air = 7.854×10−10×(0.8541)2 377 = 7.686×10−10 η glass 281
W
Problem 13.6. Application: The sun at the beach or: why do we get sunburns? The solar power impinges on the surface of the skin. Some is reflected, some is transmitted into the skin where it is dissipated, causing both the warmth we feel and blistering if exposure is excessive. The conductivity given is only important to verify that the low loss approximation applies here so that solution is simplified. The average frequency in sunlight (visible range) is about 5×1014 Hz. The ratio σ/ωε is:
σ = 0.01 = 1.5×10 −8 << 1 ω ε 2×π×5×1014×24×8.854×10−12 The skin is clearly a low loss material and for the purpose of this computation may be viewed as a dielectric. Thus, the intrinsic impedance of free space and skin are: η 0 = 377, η s = 377 = 76.955 Ω 24 The transmission coefficient is: 153.91 Τ = 2ηs = = 0.339 η s + η0 76.955 + 377
298
The power density penetrating into the body is: In free space we can write: 2 Pin = Ein 2η0 Since:
→
E i2n = 2η0Pin
2 2 P s = T E in 2ηs 2 η0 2 377 Ps = Pin×Τ = 1,300×(0.34) × = 731.89 ηs 76.955
→
E s = TEin
W m2
Thus, the total power absorbed by the body, assuming a 1 m 2 of exposed skin is 731.89 W. One reason why we "burn" is because this power is dissipated in a very thin layer of the skin. In particular, the ultraviolet portion only penetrates the skin. Thus, even though the power is not that large it can cause considerable damage because the volume power density is large. Note: A more accurate result, using the exact value of η0 gives 731.46 W.
Problem 13.7. Application: Radiation exposure. The power reaching the body is only partially absorbed as defined by the transmission coefficient between free space and the body. In this case the body is a low loss dielectric. The power penetrating into the body equals the incident power density, multiplied by the area of the skin, by the transmission coefficient squared and by the ratio of the intrinsic impedances of the air and body. First we need to make sure that the body is a low loss dielectric at 2.54 GHz:
σ = 0.01 = 0.00295 << 1 ω ε 2×π×2.54×109×24×8.854×10−12 Thus, the body may be viewed as a low loss dielectric and the intrinsic impedance of the body is:
ηb = 377 = 77 24
Ω
The transmission coefficient at the air-body interface is: Τ = 2ηb = 154 = 0.34 η b + η0 77 + 377 The power density allowed by the standard is: Pin = 10×10 10−4
−3
= 100
W m2
Thus, the total allowable power over the surface of the body is 150 W. The power transmitted into the body is (from previous example): P b = Pin×T 2 η0 = 150×(0.34)2×377 = 84.9 W ηb 77 The total energy absorbed in six hours: W = Pbt = 84.9×6 = 509.4
W.h
This is 509.4×3,600 = 1.83×106 W.s or 1.83×106 Joules. Note: A more accurate result using the exact value of η0 and not rounding-off intermediate calculations gives 506.73 W.
Problem 13.8. Application: Standing waves and reflectometry. Since the wall is a conductor, the electric field intensity at the wall is zero. The electric field intensity to the left of the wall is (Eq. (13.47)): E1(z) = − yj2Ei1sin(β1z) V/m (1) From Eq. (13.50) we have the time dependent field intensity which, in this case is better suited for our calculations. That is: E1(z,t) = Re E1(z)e jωt = Re − yj2Ei1sin(β1z)e jωt = Re y2Ei1sin(β1z)e jωte −jπ/2 = y 2Ei1sin(β1z) cos ω t − π = y2Ei1sin(β1z) sin(ωt) V/m 2
299
a. The first peak is positive and occurs at a distance λ/4 from the wall. The behavior of the standing wave for the electric field intensity is shown in Figure A. Thus, the first positive maximum occurs at λ/4 and then repeats at intervals of λ. The third positive peak is at 2.25λ and the antenna is located at 3.25λ. Note that the field changes with time also but this is of no concern here. b. To calculate the amplitude we assume the antenna generates a wave at time t = 0, and the antenna is placed at a location z = 0. The amplitude of the wave at the antenna is 100 V. Now a forward propagating wave propagates towards the interface. At the location of the first peak from the antenna, this wave is: E +1st = y100e −αλe −jβλ = y100e −αλe −j2π
V/m
(2)
The wave continues to propagate towards the interface. At the interface, the electric field intensity is: E +int = y100e −αd2e −jβd2
V/m
(3)
where d2 is shown in Figure B and equals the distance between antenna and wall (3.25λ). Now the reflected wave at the interface is generated. Since Γ = −1 we have at the interface: E −int = − y100e −αd2e −jβd2
V/m
(4)
This wave propagates now in the negative z direction a distance equal to d3. At the first peak to the right of the antenna we get: E −1st = − y100e −αd2e −jβd2 e −αd3e −jβd3 = − y100e −α(d2+d3)e −jβ(d2+d3) V/m (5) Note that both distances d2 and d3 are taken as positive since we want the total change in phase over the entire distance propagated. The total field at this location is the sum of the incident field in (2) and the reflected field in (5): E 1st = E +1st + E −1st = y100e −αλe −j2 π − y100e −α(d2+d3)e −jβ(d2+d3) = y100 e −αλe −j2π − e −α(2d3+d1)e −jβ(2d3+d1) V/m
(6)
Since λ = 12 m, d3 = 2.25λ or 27 m, and d2 = 3.25λ or 39 m, we can also write: E 1st = y 100 e −12αe −j2π − e −66αe −jβ 66
V/m
(7)
Since β = 2π/λ = 2π/12, we can write: E 1st = y 100 e −12αe −j2π − e −66αe −j2πe −j
2 π ×66 12
= y100 e −12α − e −66αe −j11π e −j2π
V/m
(8)
Since: e −j2π = cos(2π ) − jsin(2π ) = 1
and
e −j11π = cos(11π) − jsin(11π) = −1
(9)
we have the final form of the electric field intensity as: E 1st = y 100 e −12α + e −66α
V/m
(10)
The two waves are in phase at this location (hence the maximum). The attenuation constant for a low loss dielectric is:
α = σ1 2
µ 1 = 10−5 ε1 2
µ 0 = 10−5 ×377 = 94.25×10−5 4ε0 2 2
Np m
(11)
Substituting this in Eq. (10) gives: −5
E 1st = y 100 e −12×94.25×10 + e −66×94.25×10
−5
= y192.8
V/m
λ/4
λ
z
z=0 Figure A
300
antenna
2λ
wall
3λ
first peak d1
d2 Figure B
d3
wall
antenna
Notes: Without the attenuation constant, the electric field intensity would have been 200 V/m. Also, because of the attenuation constant we had to "follow" the wave and inspect its behavior continuously as it propagates. y λ
Problem 13.9. Reflection of waves from conducting surfaces. The incident and reflected waves (electric fields) are: Ei = Ei0e −jβ 0z
Er = Ei0 Γ e +jβ0z
V/m
The total electric field intensity is E = Ei + Er = Ei0e −jβ0z + Ei0Γe +jβ0z = Ei0 e −jβ0z − e +jβ0z
V/m
where the fact that Γ = −1 at a conductor interface was used. The magnetic field is: H = Hi − Hr = E i0 e −jβ0z − E i0 Γe +jβ0z = E i0 e −jβ0z + e +jβ0z η0 η0 η0 The ratio between the two is: −jβ 0z − e +jβ0z e −jβ0z − e +jβ0z E = Ei0 e = η0 −jβ z H Ei0 e −jβ0z + e +jβ0z e 0 + e +jβ0z η0 Using the relations above: −jβ 0z − e +jβ0z E = η0 e = η0 −j2sinβz = −jη0tanβz −jβ0z +jβ0z H e +e 2cosβz
A/m
Ω
Ω
j is equivalent to a phase angle of 90° (in this case the electric field lags behind the magnetic field).
Problem 13.10. Surface current generated by incident waves. To calculate the surface current density we argue as follows: If the conductor were to be removed, then the incident magnetic field intensity parallel to the surface would penetrate into what was the location of the conductor. To cancel this field, we introduce a surface current density (A/m) that will produce an opposing field and therefore produce zero magnetic field intensity in the perfect conductor. This current density is proportional to the incident tangential magnetic field intensity at the conductor. a. We start with the configuration shown in Figure A. Assuming that the reflected fields are correct (if not, we will find them from the conditions at the surface), then the conditions for H in Figure A can also be obtained from Figure B. With a surface current density in the negative y direction, the conditions in Figure A are satisfied. The incident magnetic field intensity is: A Hi1 = x Ei e −jβ0z (1) η0 m Taking z = 0 at the interface and using Figure B: Hr = Ei η0
A m
(2)
and: 2Hr = 2 Ei = Js η0
J s = − y 2Ei η0
→
A m
(3)
b. Without the current density at the surface, the magnetic field intensity everywhere in space is: Hi1 = x Ei e −jβ0z η0
A m
The surface current density produces a magnetic field intensity for z < 0 and for z > 0. These fields are found from Eq. (3) and Ampere’s law: Hr = Ei η0
→
H r = − x Ei e −jβ0z η0
for z > 0
and
H r = x Ei e jβ0z for z < 0 η0
A m
Thus, the total magnetic field intensity is: To the right of the interface (z > 0): H = H i1 + H r = x Ei e −jβ0z − x Ei e −jβ0z = 0 η0 η0 To the left of the interface (z < 0): H = H i1 + H r = x Ei e −jβ0z + x Ei e jβ0z η0 η0
A m A m
That is, the reflected field to the left of the interface may be viewed as a direct consequence of the surface current density.
301
⊗ ⊗ ⊗ ⊗ x Ei ⊗ ⊗ ⊗ Hr Js⊗ z y ⊗ ⊗ ⊗ Er ⊗ free space ⊗ conductor
⊗ ⊗ ⊗ ⊗ ⊗ ⊗ Js⊗ Hr ⊗ ⊗ ⊗ ⊗ free space ⊗
Hi1
Hi1
Hi1
Figure A
Hr free space
Figure B
Problem 13.11. Interface conditions at a conductor interface. The boundary conditions at a conductor's interface are the same as for any general material: The tangential electric field intensity and the normal magnetic flux densities are continuous while the tangential magnetic field intensity and the normal electric flux density are discontinuous (see chapter 11, Table 11.3. air conductor
a. Since the polarization is perpendicular, the electric field intensity only has a tangential component. Therefore:
Js
H1n E2
E1
E1t = E2t However, the magnetic field intensity, which must be perpendicular to E (see Figure A) has both a tangential and a normal component as shown. The normal component of the magnetic flux density is continuous: B 1n = B2n
H1t
H2n
H1 H2t
H2
ε 1 , µ1 , σ1 =0 ε 2 , µ2 , σ2 Figure A
µ 1 H 1n = µ 2 H 2n
→
Note: E 1 and H 1 are the total fields in material (1). The tangential component of the magnetic field intensity is also discontinuous, the discontinuity giving rise to a surface current density Js: H 1t − H 2t = Js
A/m
where Js is the surface current density. b. The wave in the conductor decays very rapidly and its phase changes also rapidly. The propagation constant in a conductor is given as: γ 2 = α 2 + jβ 2 = (1 +j) π fµ 2σ 2 The depth of penetration is: 1 δ2 = 1 = m α π fµ 2σ 2 and, for a good conductor is very small. The intrinsic impedance in the conductor is: η 2 = (1 + j) 1 Ω δ 2σ 2 which is very low. The phase velocity is also low and equals: ω 2ω m v p2 = ω = = µ 2σ 2 s β π fµ 2σ 2
Problem 13.12. Application: Condition of no reflection: stealth principles. z
free space ε 0 , µ0
Ei P Hi
α ≤ 45°
α α − 0°
θi 9
Figure A
302
perfect conductor
α α
.
90°−α
There are two ways to solve this problem: the easy way is to simply show that if the wave is reflected at 90° to the incident wave (or any angle above 90°) no part of the wave will reflect back. The second is to calculate the reflected wave and set it to zero. From this you will find the angle α. 1. The easy way: see Figure A: This immediately says that the angle must be smaller or equal to 45°:
α α
This implies that under ideal conditions (illumination straight on) a 90° corner will not reflect light back to the source and an airplane will not be visible. If your car were made as a 90° wedge it would be invisible to radar when viewed straight on. 2. Now for the more difficult way. We note first that the incidence angle (the angle between the direction of propagation of the incident wave and the normal to the surface is θi = 90° − α. The incident field is Ei = Ei0e −jβ0z V/m The reflected wave is Er = Ei(xsin(180 − 2α i) − zcos(180 − 2αi))Γe −jβ0(xcos(180 − 2αi) − zcos(180 − 2αi)) V/m Only the z component may return to the source therefore, the z component must be zero: Ei(− zcos(180 − 2αi))Γe −jβ0(xcos(180 − 2αi) − zcos(180 − 2αi)) = 0
⇒
cos(180 − 2α i) = 0
or:
→
180° − 2α i = 90°
θi = 45°
Problem 13.13. Oblique incidence on a conducting surface: perpendicular polarization. Referring to Figure A, we write the incident and reflected electric and magnetic fields directly from the configuration shown. The actual directions of the reflected waves are found by matching the incident and reflected fields on the interface. The current density on the conductor’s surface is found from the magnetic field intensity which is parallel to the surface. a. The incident electric field intensity is in the x direction: Ei = xE 0e −jβ0(ysinα + zcosα) = x100e −jβ0(ysinα + zcosα)
V/m
From Figure A, the incident magnetic field intensity is: Hi = E 0 ycos α − z sinα e −jβ0(ysinα + zcosα) = 0.265 ycos α − z sinα e −jβ0(ysinα + zcosα) η0
A m
where:
η0 = 377
Ω
and:
9 9 β 0 = 2π×100×10 = 2π×100×10 = 2094.4 c 3×108
rad m
b. The reflected fields are shown in their assumed directions in Figure A. These are: E r = − xE re −jβ0(ysinα − zcosα) V/m −jβ 0(ysinα − zcosα ) Hr = Hr ycos α + zsin α e A/m Now we match the incident and reflected fields at the interface. For the electric field intensity, the tangential components (the x-directed fields) must add up to zero: E i + E r = xE 0e −jβ0ysinα − xE re −jβ 0ysinα = 0
→
E r = E0
That is, our assumption as to the direction of the reflected electric field intensity was correct. Thus, the electric field intensity is as shown in the figure and, because we must satisfy the Poynting vector, the electric an magnetic field intensities with the values for H0 (given) and η0 and β0 (calculated above) must be: E r = − x100e −j2094.4(ysinα − zcosα ) V/m −j2094.4(ysinα − zcosα ) H r = 0.265 ycos α + zsin α e
V/m
c. To calculate the surface current density we argue that the tangential component of the reflected magnetic field intensity is due to this current density. The surface current density shown in Figure B produces a magnetic field intensity equal to the reflected field on both sides of the interface. In the conductor, this field cancels the incident magnetic field intensity but in air, it adds to the incident field. For this to happen, the current density must be in the positive x direction. The tangential magnetic field intensity at the surface (z = 0) is: Hrt(z = 0) = y2×0.265cosα From Ampere’s law: H rt(z = 0) = Js
→
A/m
J s = x 2Hi(z = 0) = x 0.530cosα e −jβ0ysinα
A/m
The surface current density is angle dependent and is shown in Figure C. It is minimum at α = ±90° but, depending on location on the interface, may be minimum at other positions (see Figure C).
303
Hr pr Er
Hr cosα α α
Hi α
⊗
Ei
y x⊗
pi
α
z
Hi cosα
σ=∞
ε 0 , µ0
Figure A
⊗ ⊗ ⊗ ⊗Js ⊗ ⊗
Hr cosα y Hi cosα
x
z
Figure B Js
0.53 A/m y=0 y=.05m
−π/2
0
π/2
θi
Figure C
Problem 13.14. Oblique incidence on a conductor: parallel polarization. The interface conditions given in Chapter 11, Table 3 apply here as well. However, by its nature, the magnetic field intensity at the interface has only a tangential component and therefore we need not worry about the continuity of its normal component. However, the electric field intensity has both a tangential and normal component. a. The electric field intensity has a tangential and a normal component. The tangential component of the electric field intensity is continuous: E1t = E2t The normal component is discontinuous as follows: D 1n − D2n = ρ s
→
ε 1 E 1n − ε 2 E 2n = ρ s
The surface charge density is, of course, a time dependent quantity exactly like the fields that produce it. The tangential component of the magnetic field intensity is also discontinuous, the discontinuity giving rise to a surface current density Js: H 1t − H 2t = Js b. The difference between the parallel polarization is merely in the components that need to be transferred across the interface. In problem 13.11, the normal component of E was zero while here the normal component of H is zero. Note also that for nonconducting media, the current density is zero and the charge density may also be zero, changing the required conditions.
Problem 13.15. Oblique incidence on a conductor: parallel polarization. Referring to Figure A, we write the incident and reflected electric and magnetic fields directly from the configuration shown. The actual directions of the reflected waves are found by matching the incident and reflected fields on the interface. The surface current density on the conductor’s surface is found from the magnetic field intensity which is parallel to the surface. a. The incident magnetic field intensity is in the x direction: Hi = xH 0e −jβ0(ysinα + zcosα) = x100e −jβ0(ysinα + zcosα)
A/m
From Figure A, the electric field intensity is: Ei = η0H0 − y cosα + zsin α e −jβ0(ysinα + zcosα) = 37,700 − y cosα + zsin α e −jβ0(ysinα + zcosα)
304
V/m
η0 = 377
where
Ω
and:
9 9 β 0 = 2π×100×10 = 2×π×100×10 = 2094.4 8 c 3×10
rad m
b. The reflected fields are shown in their assumed directions in Figure A. These are: Hr = xH re −jβ0(ysinα − zcosα) A/m −jβ 0(ysinα − zcosα ) Er = Er ycos α + zsin α e
V/m
Now we match the incident and reflected fields at the interface. For the electric field intensity, the tangential components (the y-directed fields) must add up to zero: E it(z = 0) + E rt( = 0) = η0H0 − y cosα e −jβ0ysinα + Er ycos α e −jβ0ysinα = 0
V/m
or: E r = η 0H 0
V/m
That is, the reflection coefficient equals +1. Thus, the electric field intensity is as shown in the figure and, because we must satisfy the Poynting vector, the electric an magnetic field intensities must be: Er = η0H 0 ycos α + zsin α e −jβ0(ysinα − zcosα) Hr = xH 0e −jβ0(ysinα − zcosα) A/m
V/m
With the values above for H0, η0, and β0, we can also write: E r = 37,700 ycos α + zsin α e −j2094.4(ysinα − Hr = x100e −j2094.4(ysinα − zcosα )
zcosα )
V/m
A/m
c. To calculate the surface current density we argue that the reflected magnetic field intensity is due to this current density. The surface current density shown in Figure B produces a magnetic field intensity equal to the reflected field on both sides of the interface. In the conductor, this field cancels the incident magnetic field intensity but in air, it adds to the incident field. For this to happen, the current density must be in the negative y direction. From Ampere’s law: →
2H r(z = 0) = Js
J s = −y2H r(z = 0) = −y200
A/m
The surface current density is independent of the angle of incidence. d. For perpendicular incidence, the magnetic field intensity has a tangential component which is dependent on the angle of incidence. Therefore, the surface current density varies with the incidence angle as 2H0cosα. For the parallel polarized wave, the surface current density is independent of the angle of incidence. Er pr
⊗
Hr
Hr α α
Hi
⊗ α
y
α
ε 0 , µ0 Ei
x⊗
pi
z
Hi
σ=∞ Figure A
⊗ ⊗ ⊗ ⊗ ⊗ ⊗ Js⊗ ⊗ ⊗ ⊗ ⊗ ⊗
Hr x Hi
y
z
Figure B
Problem 13.16. Standing waves for oblique incidence on a conductor. The standing wave pattern is calculated from the total electric and magnetic field intensity. Usually only the electric field is used for this purpose since the magnetic field may always be calculated from the electric field. There are two configurations possible, depending on how the system of coordinates is arranged. These are shown in Figures A and B. a. Using the system in Figure A, the incident and reflected fields are: Ei = −ycosθi + zsinθi Ei1e − jβ ysinθ i + zcosθ i E r = ycosθi + zsinθi Ere − jβ ysinθ i − zcosθ i H i = x Hie − jβ ysinθ i + zcosθ i A/m −jβ ysin θ i − zcos θ i H r = x H re A/m
305
V/m V/m
Now we equate the sum of the tangential components of the electric field intensity at z = 0 to zero to find the reflected electric field intensity: E it + E rt = −ycosθiEi1e − jβysinθi + ycosθiEre − jβysinθi = 0 V/m This gives Er = Ei. Using this and the fact that Hi is known (Ei = Hiη, Er = Hiη, Hr = Hi), we get the total fields in air: E 1 = E i + E r = yηHicosθi − e − jβzcosθi + e jβ zcosθi e − jβysinθi + zηHisinθi e − jβzcosθi + e jβ zcosθi e − jβysinθi = yj2ηHicosθi sin(β zcosθi)e − jβ ysinθ i + zηHisinθi cos(β zcosθi)e − jβ ysinθ i = 2ηHi yjcosθi sin(β zcosθi) + zsinθi cos(β zcosθi) e − jβysinθi
V/m
Similarly for the magnetic field intensity: H 1 = H i + H r = xHi e − jβzcosθi + e jβ zcosθi e − jβysinθi = x2Hicos(β zcosθi)e − jβysinθi
A/m
Note: The total electric and magnetic fields for parallel polarization for the configuration in Figure A are given in Eqs. (13.90) and (13.91) but with the obvious change in coordinates (x coordinate becomes the y coordinate and the y coordinate becomes the −x coordinate). Substituting now Hi = 15 A/m, β1 = 200 rad/m , η = η0 = 377 Ω, and θi = 30° , we can write: E 1(x,y,z) = 2×377×15 yjcos30 sin(β zcos30) + zsin30 cos(β zcos30) e − jβ ysin30 = 11,310 yj0.866 sin173.2z − z0.5 cos173.2z e − j100y V/m
(3)
Similarly, for the magnetic field intensity: H 1(x,y,z)= x30 cos173.2ze − j100y
A/m
(4)
These clearly represent a wave propagating in the y direction, parallel to the surface and a standing wave perpendicular to the surface (z direction). The latter is represented by the following components E 1(x,y,z) = yj9794.46 sin(173.2z) e − j100y H 1(x,y,z) = x30 cos173.2ze − j100y
V/m A/m
(5) (6)
Configuration in Figure B. It is possible to repeat the process above for this configuration as well. However, a quick inspection will show that the only change is in that the horizontal components of the incident and reflected electric field are in the negative z direction. All other components are in the same directions as in Figure A. Thus, it is sufficient to reverse the direction of the z component in the final results. This gives: E 1(x,y,z) = 11,310 yj0.866 sin173.2z − z0.5 cos173.2z e − j100y H 1(x,y,z) = x30 cos173.2ze − j100y A/m
V/m
The pattern is as above (the z component is real). Thus: E 1(x,y,z) = yj9794.46 sin(173.2z) e − j100y H 1(x,y,z) = x30 cos173.2ze − j100y
V/m A/m
(7) (8)
b. The amplitude of the electric field intensity is zero wherever sin(173.2z) = 0 and is maximum wherever sin(173.2z) = ±1. Thus, the condition for peaks in the electric field intensity is: 173.2z = mπ 2
m = 1,3,5,7...n (odd)
→
z = mπ 246.4
m = 1,3,5,7...n (odd)
(9)
The amplitude of the magnetic field intensity is zero when sin(173.2z) = 0 and is maximum at sin(173.2z) = ±1. Thus the condition for peak in the magnetic field intensity is 173.2z= mπ
m=0,1,2,....n
→
The peaks are:
z = mπ 173.2
m=0,1,2,3,..
Epeak = ± 9794.46 V/m Hpeak = ± 30 A/m c. To calculate the total time averaged power density we write from the total fields in (5) and (6) P av(x,z) = 1 Re E1(x,z) ×H*1(x,z) = 1 −11,310 yj0.866 sin173.2z − z0.5 cos173.2z e −j100y × x30 cos173.2ze j100y 2 2 = y84,825cos 2(173.2z) W/m2
306
(10) (11)
Er
⊗
Hr
Hr
θi Hi
Er
y x⊗
x
θi
Ei
z
⊗
z
y
Hi
α
Ei Figure A
Figure B
Problem 13.17. Propagation of waves in the presence of a conducting surface.
.
C
. . .
vpz
z
A'
vθ
A
90°
p
a. Consider Figure A in which a wavefront propagates at an angle towards a conducting interface. During the time point A travels to A’, the point B travels to A’ vertically. If the phase velocity of point A perpendicular to the front is v p , then, vp = cos(90 − θ i) =sinθ i v px
x
−θ
We can either use the general expressions for the electric and magnetic field intensities in Eqs. (13.90) and (13.91) and extract the phase constant in both directions required, or we could rely on a geometric discussion to identify the two velocities. The latter is more fundamental.
vpx
B
wa
ve
fro
nt
Figure A →
v px =
vp = c sinθi sinθi
m s
Similarly, point C travels horizontally from C to A’ in the same time point A travels to A’. Thus: vp = cosθ i vp z
→
v pz =
vp = c cosθi cosθi
m s
Specifically for θi = 30°: 8 8 c = 3×10 = 3.464×108 vpx = c = 3×10 = 6×10 8 vpz = sin30° 0.5 cos30° 0.866 Note that both velocities are larger than the speed of light, c.
m s
b. For a wave parallel to the surface, θi = 90°. Thus: 8
vpx =
c = 3×10 = 3×10 8 sin90° 1
vpz =
c =∞ cos90°
m s
c. The tangential and normal velocities are always larger or equal to the phase velocity in free space. They approach the phase velocities at normal incidence (for the normal component) and for parallel incidence for the parallel component.
Problem 13.18. Measurement of thickness of dielectrics. Snell's law provides a relation between the incident and transmission angles (the incident angle is known from measurement of d1 and d3). With the transmission angle known, the thickness d may be calculated in terms of the other measurements. Snell's law gives the relation between the incidence and transmission angles: sinθt = ε 1 µ 1 = ε2µ 2 sinθi
ε0µ 0 = 1 4 ε0µ 0 2
→
sin θ i = 2sin θ t
Now we can write the angles in terms of the geometrical distances: d1/2 d1 sinθi = = d12/4 + d32 2 d12/4 + d32
307
(1)
(2)
sinθt =
d4/2 d4 d2 = = d42/4 + d 2 2 d42/4 + d 2 2 d22/4 + d 2
(3)
Now, substituting these in Eq. (1): 2
d1 2 d1 /4 +
d32
= 2
d2
2d2 + d2
d5
d22/4
θt
Expansion of the terms leads to: d2 = d22 +
4d22d32 d12
−
d22 4
=
4d22d32 d12
+
3d22 4
=
2 d22 4d3 d12
θi θi
d1
+3 4
d1/2
or: d = d2 16d32 + 3d12 = d2 0.0016 + 3d12 2d1 2d1
θt
d
d3
m
Figure A
Problem 13.19. Surface currents induced by an obliquely incident wave. The surface current density is produced by the tangential component of the incident magnetic field intensity. This current density may be viewed as producing the reflected magnetic field intensity and also is responsible for the fact that the magnetic field intensity is zero inside the perfect conductor. That is, we can view the total magnetic field to the left of the conductor as being the sum of the incident tangential component and an additional component due to the surface current density so that in the conductor the magnetic field is zero. Figure A shows the conventional approach. Figure B shows the surface current approach. For the magnetic field intensity to cancel to the right of the interface we must have: Hrt + Hit = 0
(1)
a. The incident tangential magnetic field intensity in Figure A is Hicosθi. The tangential component of the reflected magnetic field intensity is Hrt = − Hicosθi = − Ei cos θ i z<0 A/m (3) η0 where z = 0 is the location of the interface. In vector form: A H rt = − x Ei cosθ i z<0 η0 m Instead of looking at the problem as one of incident and reflected waves, we may look at it as one of an incident wave everywhere in space (by removing the conducting surface) and a current sheet at the location of the surface (z = 0) which produces the reflected wave. Under this condition, shown in Figure B, there is a “reflected” wave to the right of the interface as well: A Hrt = Hicosθi = Ei cos θ i z>0 or: H rt = x Ei cosθ i z > 0 η0 η0 m Applying Ampere’s law to the two components of the reflected wave in Figure B, we get: 2HrtL = 2 Ei Lcosθ i = JL η0
→
J = 2 Ei cosθi η0
A m
(5)
And, from the right hand rule, the current density must be in the positive y direction: J = y 2Eicosθi = y 2×10cos30° = y 0.0459 η0 377
A m
b. Now the magnetic field intensity is parallel to the interface as shown in Figure C. The incident (tangential) magnetic field intensity is therefore: A Hit = y Ei z<0 η0 m
(6)
(7)
Viewing Figure C from above (see Figure D), looking onto the y-z plane we can use again the relation in (a). Now we can write directly from Eq. (6), but the current density must be in the x direction to produce the reflected magnetic field intensity required: J = x 2Ei = x 2×10 = x 0.053 η0 377
308
A m
d4
H rt
H rt
x 30°
x pi y
Ei
30°
Hi Ht Figure A
L
Js
Hit
z
Hit
z
y
ε 0 , µ0 Figure B Hr
Hr 30°
x pi
Hi
y
x
L Hi
z
Js
z
Hi y
ε 0 , µ0 Ei
Figure C
Figure D
Problem 13.20. Oblique incidence on a dielectric: perpendicular polarization. To calculate the time averaged power densities in air and in the dielectric we must calculate the total electric and magnetic field intensities in air and in the dielectric. For this purpose we write the incident electric and magnetic field intensities from the data given and then proceed to calculate the reflected and transmitted wave. After we sum the incident and reflected waves left of the interface (in free space) we can calculate the time averaged power density in both materials. a. Referring to Figure A, the incident electric field intensity is (see Eqs. 13.101 through 13.114), the incident electric field intensity is: Ei= xEi1e −jβ0(ysinθi + zcosθi) V/m The magnetic field intensity is (so that propagation is towards the interface): Hi = E i1 (ycosθi − zsinθi)e −jβ0(ysinθi + zcosθi) η0 The assumed reflected waves are: E r= xΓ⊥Ei1e −jβ0(ysinθi − zcosθi) V/m Γ E −j β (ysin θ − zcos θ i) i H r = ⊥ i1 (− ycosθ i − zsinθi)e 0 η0
A m
A m
The total electric and magnetic field intensities in air are: E 1 = E i + E r = xEi1 e −jβ0(ysinθi + zcosθi) + Γ ⊥e −jβ0(ysinθi − zcosθi) = xE i1 e −jβ0zcosθi) + Γ ⊥e jβ0zcosθi e −jβ0ysinθi V/m H 1= Ei1 (ycosθi − zsinθi)e −jβ0(ysinθi + zcosθi) + Γ⊥(− ycosθ i − zsinθi)e −jβ0(ysinθi − zcosθi) = η0 E cos θ i1 i y e −jβ0zcosθi− Γ ⊥e jβ0zcosθi e −jβ0ysinθi − z Ei1sinθi e −jβ0zcosθi + Γ ⊥e jβ0zcosθi e −jβ0ysinθi η0 η0 The time averaged power density is therefore * * E ×H * P av = Re E 1×H 1 = Re x y + E x×H z 2 2 2
A m
W m2
We will evaluate each part of the power separately for simplicity: Ex×Hy* = x Ei1 e −jβ0zcosθi + Γ ⊥e jβ0zcosθi e −jβ0ysinθi× y Ei1cosθi e jβ0zcosθi− Γ ⊥e −jβ0zcosθi e jβ0ysinθi η0 2 2 2 2 2 E cos θ j2 β zcos θ −j2 β zcos θ E cos θ i 1 − Γ + Γ e 0 i − e 0 i = z i1 i 1 − Γ 2 + Γ j2sin(2β zcosθ ) W = z i1 ⊥ ⊥ ⊥ ⊥ 0 i 2η0 2η0 m2
309
E x×H*z = x Ei1 e −jβ0zcosθi + Γ e jβ0zcosθi e −jβ0ysinθi× z E i1sinθ i −e jβ0zcosθi− Γ e −jβ0zcosθi e jβ0ysinθi ⊥ ⊥ η0 2 2 2 2 W = y Ei1sinθi 1 + Γ ⊥2 + Γ ⊥ e j2β0zcosθi + e −j2β0zcosθi = y Ei1sinθi 1 + Γ ⊥2 + Γ⊥2cos(2β0zcosθi) 2η0 2η0 m2 The total power density is the sum of the two power densities 2 2 P av = Re y Ei1cosθi 1 + Γ ⊥2 + Γ⊥2cos(2β0zcosθi) + z Ei1sinθi 1 − Γ ⊥2 + Γ⊥j2sin(2β0zcosθi) 2η0 2η0 2 2 2 E cos θ E i 1 + Γ + Γ 2cos(2β zcosθ ) + z i1sinθi 1 − Γ 2 W y i1 ⊥ ⊥ 0 i ⊥ 2η0 2η0 m2
In these relations the following are also necessary for numerical evaluation: Γ⊥ = η2cosθi − η0cosθt η2cosθi + η0cosθt sinθi = ε r2 µ r2 → θ t = sin −1 sinθi η0 = ε r2 µ r2 sinθt
µ0 ε0
η2 =
=
µ2 ε2
These are known numerical values for any given materials and angle of incidence. b. To find the power density in the dielectric, we assume that the dielectric properties are known and the transmission coefficient can be calculated. Then, the transmitted waves are: Et= xEi1T⊥e −jβ2(ysinθt + zcosθt) V/m E T −jβ 2(ysinθ t + zcosθ t) i1 ⊥ Ht = (ycosθt − zsinθt)e η2 The time averaged power density is then:
A m
* P av = Re E t×H t = 1 xE i1T ⊥e −jβ2(ysinθt + zcosθt) × E i1T ⊥ (ycosθt − zsinθt)e jβ2(ysinθt + zcosθt) η2 2 2 2 2 E T i1 ⊥ W = zcosθt + ysinθt 2η2 m2 where the transmission coefficient is calculated as:
T⊥ =
2η2cosθi η2cosθi + η1cosθt
Ht
⊗ Er Hr
θi
and the intrinsic impedances were given in (a). Hi θi
c. In air, the power has both real and an imaginary parts. This means that there is both a propagating and a standing wave term for any two different materials meeting at the interface. In the dielectric there can be only a propagating wave (real power density).
⊗ Ei
θi θi
θt
⊗Et
y z=0
x⊗
z
Figure A
Problem 13.21. Oblique incidence on a dielectric: parallel polarization. The angle of incidence is found from the tangential and normal components of the incident field. Then, the transmission angle is found through use of Snell’s law. The polarization is identified from the tangential and normal components. a. The angle of incidence is: tan θi = Ex = 10 = 2 → Ey 5 The transmission angle is found from Snell’s law: sinθt = n 1 sinθi = n2
θ i = tan −12 = 63.43° = 63°26'
ε r1 sin (63.43°) = ε r2
or:
2 0.89443 = 0.7303 3
θt = sin −10.7303 = 46.91° = 46°55' b. The incident electric field intensity has components in the x and y directions while the incidence plane is on the y-z plane. Therefore this is parallel polarization.
310
c. The reflection and transmission coefficients for parallel polarization are given in Eqs. (13.124) and (13.125):
Γ|| = η2cosθt − η1cosθi = µ 0 /ε 2 cosθt − µ 0 /ε 1 cosθi = 1/ 3cos(46.9°) − 1/ 2cos(63.43°) = 0.11 η2cosθt + η1cosθi µ 0 /ε 2 cosθt + µ 0 /ε 1 cosθi 1/ 3cos(46.9°) + 1/ 2cos(63.43°) T|| =
2η2cosθi 2 µ 0 /ε 2 cosθi 2/ 3cos(63.43°) = = = 0.726 η2cosθt + η1cosθi µ 0 /ε 2 cosθt + µ 0 /ε 1 cosθi 1/ 3cos(46°9) + 1/ 2cos(63.43°)
Thus:
Γ || = 0.11
T|| = 0.726
d. The scalar components of the transmitted and reflected waves are found by writing the transmitted and reflected waves at the interface. However, we must be careful since the incident electric field intensity is given in terms of its components and the reflection coefficient was calculated in terms of the amplitude of the incident field. Thus, we first calculate the amplitude of the incident field: Ei = 10 2 + 52 = 11.18
→
Ei(x = 0) = x10 + y 5
V/m
The amplitude of the transmitted wave is:
→
Ei(x = 0) = x10 + y 5
Et = T Ei = 0.726×11.18 = 8.12
V/m
The transmitted waves is therefore: Et(x = 0) = x8.12sin(46.91°) + y 8.12cos(46.91°) = x 5.93 + y 5.55
V/m
The amplitude of the reflected wave is: E r = Γ Ei = 0.11×11.18 = 1.23
V/m
The reflected wave is: E r(x = 0) = − x 1.23sin(63.43°) + y 1.23cos(63.43°) = − x 1.1 + y 0.55
V/m
The scalar components of the reflected field are − 1.1 V/m in the x direction and 0.55 V/m in the y direction. The scalar components of the transmitted field are 5.93 V/m in the x direction and 5.55 V/m in the y direction.
Problem 13.22. Phase shift of transmitted and reflected waves. a. The transmission coefficient is (Eq. (12.110)): T⊥ =
2η2cosθi , η2cosθi + η1cosθt
η 1= η 0=377
Ω
For high loss materials (Eq. (13.116)): j ωµ 2 = (1 + j) σ2
η2 =
ωµ 2 2σ2
Ω
Thus:
ωµ 2 cosθ i 2σ2
2(1+j) T⊥ = (1+j)
ωµ 2 cosθ + η cosθ i 0 t 2σ2
Let: a= 2a+j2a T⊥ = 2a(1+j) = (1+j)a + b (a + b) + ja
ωµ 2 cosθ i 2σ2
b = η 0cosθ t
(a + b) − ja = 2a(a+b) − j2a 2 + j2a(a+b) + 2a 2 = 4a 2 + 2ab + j2ab (a + b) − ja (a + b)2+ a 2 (a + b)2+ a 2
The phase angle of the transmission coefficient is therefore:
φT⊥ = tan − 1
2ab = tan − 1 b 4a 2 + 2ab 2a + b
311
(1)
Substituting for a and b from (1):
η0cosθt
φT⊥ = tan −1 2
ωµ 2 cosθ + η cosθ i 0 t 2σ2
b. Yes, there is a phase angle associated with the reflection coefficient: (1+j)
Γ⊥ = η2cosθi − η1cosθt = η2cosθi + η1cosθt (1+j)
ωµ 2 cosθ − η cosθ i 0 t 2σ2 ωµ 2 cosθ + η cosθ i 0 t 2σ2
Using the notation for a and b in (a): 2 2 Γ⊥ = (1+j)a − b = (a − b) + ja .(a + b) − ja = 2a − b 2+ j2ab (1+j)a + b (a + b) + ja (a + b) − ja (a + b) + a 2
the angle is: 2η0 ωµ 2 cosθicosθt 2σ2 2ab φΓT = tan = 2 2 ωµ 2 2a − b 2 cos 2θ i − η 02cos 2θt 2σ2 −1
Problem 13.23. Phase velocity and its dependence on incidence angle. To see how the waves propagate and their various velocities, it is possible either too use the incident, reflected and total fields and extract the phase velocities form these or to use a geometrical representation of the plane wave. From this we can extract the phase velocity from simple geometrical ratios (see Figure A and also Problem 13.19).
. . .
.
vpz
dielectric z
A'
vp
60°
30°
C
vpx
A
a. Consider Figure A in which a wavefront propagates at an angle towards a conducting interface. During the time point A travels along to A’, point B travels to A’ vertically. If the phase velocity of point A perpendicular to the front is v p , then vp = cos(90 − θ i) =sinθ i v px
x
free space
B
wa
ve
fro
nt
Figure A →
v px =
vp = c sinθi sinθi
m s
Similarly, point C travels horizontally from C to A’ in the same time point A travels to A’. Thus: vp = cosθ i vp z
→
Specifically for θi = 30°: 8 vpx = c = 3×10 = 6×10 8 sin30° 0.5
v pz =
vpz =
vp = c cosθi cosθi
m s
8 c = 3×10 = 3.464×108 cos30° 0.866
m s
Note that both velocities are larger than c. b. For a wave parallel to the surface, θi = 90°. Thus: 8
vpx =
c = 3×10 = 3×10 8 sin90° 1
vpz =
c =∞ cos90°
m s
c. The tangential and normal velocities are always larger or equal to the phase velocity in free space. They approach the phase velocities at normal incidence (for the normal component) and for parallel incidence for the parallel component. d. There is no difference between the phase velocities here and those in Problem 13.19, even though the material here is a dielectric and in Problem 13.19 it is a perfect conductor. There is a significant difference in the magnitudes of the fields to the left of the interface but the phase velocities parallel and perpendicular to the interface only depend on the material to the left of the interface and the angle of incidence.
312
Problem 13.24. Reflection coefficient and its dependency on angle of incidence. For a wave propagating from free space onto a perfect dielectric, the electric field intensity can only remain tangential to the surface for all incidence angles if it is perpendicularly polarized. If the electric field intensity has both a parallel and normal component, the polarization is parallel. Thus, we calculate and plot the reflection coefficient for perpendicular and for parallel polarization as a function of incidence angle. 1. For perpendicular polarization (from Eq. (13.109)):
Γ⊥ = η2cosθi − η1cosθt η2cosθi + η1cosθt
(1)
From Snell’s law, we re-write the transmission angle as a function of the incidence angle:
ε0µ 0 = 1 sinθt = ε 1 µ 1 = ε2µ 2 sinθi 2.1 ε 0µ 0 2.1 However, we need cosθt rather than sinθt: sinθt = sinθi → cos θ t = 2.1 The intrinsic impedances η1 and η2 are: η 1 = 377
sin θ t = sinθi 2.1
→
(2)
2 1 − sin θi = 1 2.1 − sin 2θi 2.1 2.1
1 − sin 2θt =
η 2 = 377/ 2.1
(3)
Ω
(4)
Substituting these into the reflection coefficient we get (for θi = θ):
Γ⊥ =
377/ 2.1 cosθ − 377/ 2.1 377/ 2.1 cosθ + 377/ 2.1
2.1 − sin 2θ
= cosθ − 2.1 − sin 2θ cosθ +
This is plotted in Figure A for 0 < θ i < π /2. 2. For parallel polarization (from Eq. (13.124)):
Γ|| = η2cosθt − η1cosθ η2cosθt + η1cosθ
(5)
-0.16
Γ||
(6)
-0.18
Γ
With the results in Eqs. (3) and (4) we get:
Γ⊥ -0.20
377 2.1 − sin 2θ − 377cosθ 2.1 Γ|| = sinθt − cosθ = 2.1 sinθt + cosθ 377 2.1 − sin 2θ + 377cosθ 2.1 2.1 or:
2.1 − sin 2θ 2.1 − sin 2θ
-0.22
0.00
0.20
θ i [rad] Figure A
0.40
2 Γ|| = 2.1 − sin θ − 2.1cosθ 2.1 − sin 2θ + 2.1cosθ
This is plotted in Figure A for angles between zero and π/2.
Problem 13.25. Parallel and perpendicular incidence: Reflection & transmission coefficients. The reflection and transmission coefficients (for parallel polarization) are:
Γ|| = η2cosθt − η1cosθi η2cosθt + η1cosθi Setting θi = θt = 0:
T|| =
Γ|| = η 2 − η 1 η2 + η1
2η2cosθi η2cosθt + η1cosθi T|| =
2η2 η2 + η1
These are identical to those for perpendicular incidence and are now independent of polarization. This is because when the wave is perpendicularly incident to a surface the electric field has only a tangential component with either polarization. For parallel incidence θi = 90°: 2η2(0) Γ|| = η2cosθt − η1(0) = 1 T|| = =0 η2cosθt + η1(0) η2cosθt + η1(0) There is no transmission but total reflection.
313
Problem 13.26. Brewster angle in dielectrics. Using the general relation for Brewster angle for two dielectrics with the same permeability:
ε2 ε1 + ε2
θb = sin −1 θb = sin −1
a.
θb = sin −1
b.
θb = sin −1
c.
81 ε0 = sin −1 81 + 1 ε 0 4ε0 = sin −1 4 + 1 ε0 2.25ε0 = sin −1 2.25 + 1 ε 0
81 = 83.66° = 83°39' 82 4 = 63.43° = 63°26' 5 2.25 = 56.39° = 56°18' 3.25
Problem 13.27. Calculation of permittivity from the Brewster angle. Since the Brewster angle only depends on permeability and permittivity of the materials on the two sides of the interface, it is possible to calculate the permittivity of the dielectric if its permeability is known. In this case, the wave propagates in free space and is incident on a perfect dielectric. Therefore we can write for the Brewster angle θb: sinθ b =
ε2 ε 0 + ε2
ε2 ε 0 + ε2
θ b = sin −1
→
(1)
because µ 1 = µ 2 = µ 0 and ε 2 = ε 0 . Using the first form in Eq. (1): sin 2 θ b =
ε2 ε 0 + ε2
sin 2θ b ε 0 + ε2 = ε2
→
2 ε 2 = ε 0sin 2θ b 1 −sin θ b
→
For the given angle: 2 2 ε 2 = ε 0sin 2θ b = ε0 sin 62° = 3.537 ε 0 1 −sin θ b 1 − sin 262°
F m
The permittivity of the dielectric is 3.537ε 0 .
Problem 13.28. Critical angles in dielectrics. Critical angles in dielectrics. Use the general formula for critical angle between two materials: a. b. c.
µ 2 ε 2 = sin −1 µ 0ε0 1 = 6.38° = 6°23' = sin −1 µ 1ε1 µ 0×81× ε 0 81 µ 0×1.75× ε 0 = sin −1 1.75 = 41.41° = 41°25' θc = sin −1 µ 0×4× ε 0 4 −1 µ × ε −1 0 0 1 = 41.81° = 41°49' θc = sin = sin µ 0×2.25× ε 0 2.25
θc = sin −1
Problem 13.29. Application: Use of critical angle to measure permittivity. With the known properties of the material through which the wave propagates (free space), and the critical angle, the permittivity of the material on which the wave impinges can be calculated.
θc = sin −1 or:
µ 2ε2 µ 1ε1
sin 2θc = ε 0 = 1 ε 1 ε r1
sinθ c =
→
ε r1 =
→
µ 0ε0 µ 0ε1
1 = 1 = 2.894 sin 2θc sin 236
Relative permittivity of the material is 2.894.
Problem 13.30. Critical angles in dielectrics. The reflection and transmission angles are calculated from Snell's law. Then, the formulas for transmission and reflection coefficients for parallel polarization are used to determine the coefficients. The critical angle is computed from the material properties since the critical angle does not depend on the polarization of the wave.
314
a. From Snell's law (Eq. (13.99):
This gives:
sinθt = 4 = 2 sinθi 1
sin θ t = 2sin θ i
→
θ t = sin −1 2sinθi
→
θ t = sin −1 2sin28° = 69.87° The reflection coefficient for parallel polarization is given in Eq. (13.124) and the transmission coefficient in Eq. (13.125): 2η2cosθi Γ|| = η2cosθt − η1cosθi T|| = η2cosθt + η1cosθi η2cosθt + η1cosθi To evaluate these we need the intrinsic impedances. η2 is that of free space (377 Ω) and η1 is half that much (188.5 Ω). Thus:
Γ|| = 377cos69.87 − 188.5cos28 = 0.6883 − 0.88295 = − 0.12388 377cos69.87 + 188.5cos28 0.6883 + 0.88295 T|| =
2×377cos28 = 2.2478 377cos69.78 + 188.5cos28
Note: for parallel polarization, the following holds (see exercise 13.12(a)): 1 + Γ||= T|| cosθt cosθi which is satisfied for the values above. That is:
0≤ θ i< π /2
1 + Γ|| = 1 − 0.12388 = 0.876 = T|| cosθt = 2.2478 cos69.87 = 0.876 cosθi cos28 b. The critical angle is found from Eq. (13.142) and is not dependent on polarization. sin θc =
µ 2ε2 = µ 1ε1
1 =1 4 2
→
θc = 30°
Problem 13.31. Application: Design of sheathing for optical fibers. a. For a coating to be an effective shield its permittivity must be lower than that of glass. Otherwise there is no critical angle and the material cannot serve as a shield. Thus the plastic with relative permittivity ε r = 2 must be used. b. The critical angle at the interface between glass and coating is:
θc = sin −1
ε r2 = sin −1 ε r1
2.0 = sin −1(0.9428) = 70.528° = 70°32' 2.25
The critical angle is 70°32'. For any angle larger than this, there will be no penetration of light into the shield. For θ < θc, there is penetration into the shield. c. Now we allow propagation in the shield as well. At the interface between shield (ε r = 2) and air (ε r = 1), the critical angle is: θc = sin −1 ε r2 = sin −1 1.0 = sin −1(0.7071) = 45° ε r1 2.0 For any angle below this, there will be propagation in air (waves will leak out of the fiber and shield). Thus, to summarize: For θ > 70°32', waves can only propagate in the glass core. For 45° < θ < 70°32', waves can propagate in glass and shield but not in air. For θ < 45°, waves can also leak out into air. The critical angle for the fiber is now 45°, since, in general, we cannot allow propagation in air (this would constitute losses).
315
Problem 13.32. Propagation through lossless slab. The electric field intensities in the three domains are given in Eqs. (13.145), (13.147) and (13.149) where the various amplitudes (E r0, E2+, E2− and E3+) are given in Eqs. (13.156) through (13.159) but here we need to replace the general propagation constant by γ 2 = jβ 2 . With these we get the amplitudes as −j2 β 2 d Er0 = Γ 12 + Γ 23e −j2β d Ei0 2 1 + Γ 12Γ 23e
E2+ =
E2− =
T 12Γ 23e −j2β2d Ei0 1 + Γ 12Γ 23e −j2β2d
V m
−jβ2d jβ 3d e E3+ = T 12T 23e Ei0 1 + Γ 12Γ 23e −j2β2d
T12 Ei0 1 + Γ 12Γ 23e −j2β2d
V m
The directions of propagation of these fields is shown in Figure A. In addition we note that:
Γ12 = η 2 − η 0 , η2 + η0
Γ 23 = η 0 − η 2 , η0 + η2
T12 =
2η2 , η2 + η0
T23 =
2η0 η2 + η0
From this we have Γ 23 = − Γ 12, T 12 = 1 + Γ 12 and T 23 = 1 + Γ 23 = 1 − Γ 12. We also have γ 1 = γ 3 = jβ 0 . With these we can write: Γ 1 − e −j2β2d V Er0 = 12 Ei0 2 −j2 β 2d m 1 − Γ 12 e E2− =
1 + Γ 12 − Γ 12 e −j2β2d 1−
E3+ =
2 −j2 β 2 d Γ 12 e E2+ =
1− 1 + Γ 12 Ei0 2 −j2 β 2d 1 − Γ 12 e
1 + Γ 12 1 − Γ 12 e −jβ2de jβ3d 1−
2 Γ 12 + Γ 12 e −j2β2d
Ei0 = −
2 −j2 β 2d Γ 12 e
Ei0 =
2 −j2 β 2d Γ 12 e
V m
2 1 − Γ 12 e −jβ2de jβ3d
1−
V m
Ei0
2 −j2 β 2d Γ 12 e
V m
Ei0
The total fields are as follows E1 = x Ei0e −jβ0z + Er0e jβ0z = xE i0 e −jβ0z + E2 = x E2+e −jβ2z + E2−e jβ2z = xE i0
Γ 12 1 − e −j2β2d 2 −j2 β 2d 1 − Γ 12 e
V m
e jβ 0 z
2 1 + Γ 12 e −jβ2z − Γ 12 + Γ 12 e −j2β2de jβ2z 2 −j2 β 2d 1 − Γ 12 e
E3 = xE 3+e −jβ0z = xE i0
2 1 − Γ 12 e −jβ2de jβ0d
1−
2 −j2 β 2d Γ 12 e
e −jβ0z
V m
V m
The slab reflection and transmission coefficients are:
Γ 1 − e −j2β2d Γslab = E r0 = 12 2 , Ei0 1 − Γ 12e −j2β2d ε 0 , µ0
d ε 2 , µ 2 , σ2
2 + 1 − Γ 12 e −jβ2de jβ0d Tslab = E3 = 2 −j2 β 2d Ei0 1 − Γ 12 e
ε 0 , µ0
E p
Er1 Ei 1 1
Er2 Ei 2
Ei 3
2
3
Figure A
Problem 13.33. Propagation through lossless dielectric slab. Refer to Figure A in Problem 13.32. The characteristic impedance of each material is the same as for any dielectric (material (1) and (3) have characteristic impedance of free space while material (2) has half the impedance of material (1)). The wave impedance is always the ratio between the electric and magnetic field intensities and these vary depending on the reflections that may exist. a. The intrinsic impedance in each material is that of a perfect dielectric. In materials (1) and (3) this is the intrinsic impedance of free space. Thus:
316
f = 109 Hz ,
η 2 = η0 = 377 = 188.5 Ω εr 2
η1=η3=η0 = 377 Ω ,
b. The various components of the electric field intensity are as follows (see Eqs. (13.156) through (13.159) by replacing γ 1 = γ 3 = jβ 0 , γ 2 = jβ 2 or by using the result of Problem 13.32) −j2 β 2 d Er0 = Γ 12 + Γ 23e −j2β d Ei0 , 2 1 + Γ 12Γ 23e
E2+ =
T12 Ei0 , 1 + Γ 12Γ 23e −j2β2d
T 12Γ 23e −j2β2d Ei0 1 + Γ 12Γ 23e −j2β2d
V m
−jβ2d jβ 0d e E3+ = T 12T 23e Ei0 1 + Γ 12Γ 23e −j2β2d
V m
E2− =
where
Γ12 = η 2 − η 1 = 188.5 − 377 = − 1 , η 2 + η 1 188.5 + 377 3 2×188.5 2 T 12 = = , 377 + 188.5 3
Γ 23 = η 3 − η 2 = 377 − 188.5 = 1 η 3 + η 2 377 + 188.5 3 2×377 T23 = =4 377 + 188.5 3
The total electric fields in the three materials are (with z = 0 at the interface between air and the surface of the dielectric): E1 = Ei0e −jβ0z + Er0e jβ0z E2 = E2+e −jβ2z + E2−e jβ2z E3 = E3+e −jβ0z
V/m z<0 V/m 0 d
Substituting the values above gives −j2 β 2d E1 = Ei0 e −jβ0z + Γ 12 + Γ 23e −j2β d e jβ0z 2 1 + Γ 12Γ 23e
E2 = Ei0
V m
z<0
−j2 β 2d −jβ2z + T Γ e −j2 β 2d e jβ2z T12 12 23 e −jβ2z + T 12Γ 23e −j2β d e jβ2z = Ei0 T 12 e −j2 β 2 d 2 1 + Γ 12Γ 23e 1 + Γ 12Γ 23e 1 + Γ 12Γ 23e −j2β2d −jβ2d jβ 0d e E3 = Ei0 T 12T 23e e −jβ0z 1 + Γ 12Γ 23e −j2β2d
V m
V/m
0
z> d
The numerical values of the phase constants in free space and in the dielectric are: 9 β0 = 2 π f = 2 π ×108 = 20π c 3 3×10
rad , m
β 2 = ε r β0 = 40π 3
rad m
Substitution of these values gives the electric field intensities in the three materials. These now only depend on the incident electric field intensity Ei0, on the thickness d and, of course, on the position z. −j80 π d/3 −j80 π d/3 E1 = Ei0 e −j20 π z/3 + −1/3 + 1/3e−j80 π d/3 e j20 π z/3 = Ei0 e −j20 π z/3 + 3 −1 + e−j80 π d/3 e j20 π z/3 1 − (1/9)e 9−e
V m
−j40πz/3 + (2/3)(1/3)e −j80 π d/3 e j40 π z/3 −j40πz/3 + 2e −j80 π d/3 e j40 π z/3 E2 = Ei0 (2/3)e = Ei0 6e −j80 π d/3 1 − (1/9)e 9 − e −j80 π d/3 −j40 π d/3 j20 π d/3 −j20 π d/3 e E3 = Ei0 (2/3)(4/3)e e −j20 π z/3 = Ei0 8e e −j20 π z/3 −j80 π d/3 1 − (1/9)e 9 − e −j80 π d/3
V m V m
z<0 0
z> d
To calculate wave impedance we need to calculate the magnetic field intensities. These are calculated from the following relations: −j2 β 2d A H1 = 1 Ei0e −jβ0z − Er0e jβ0z = E i0 e −jβ0z − Γ 12 + Γ 23e −j2β d e jβ0z z<0 2 η1 η1 m 1 + Γ 12Γ 23e −jβ2z − T Γ e −j2 β 2d e jβ2z 12 23 A H2 = 1 E2+e −jβ2z − E2−e jβ2z = E i0 T 12 e 0 d η3 η3 1 + Γ 12Γ 23e −j2β2d m With the numerical values above, we have:
317
−j80 π d/3 −j80 π d/3 H1 = Ei0 e −j20 π z/3 − − (1/3) + (1/3)e e j20 π z/3 = Ei0 e −j20 π z/3 − 3− 1 + e−j80 π d/3 e j20 π z/3 −j80 π d/3 377 1 − (1/9)e 377 9−e −j40πz/3 − (2/3)(1/3)e −j80 π d/3 e j40 π z/3 −j40 π z/3 (2/3) e E E i0 i0 6e − 2e −j80π d/3e j40 π z/3 H2 = = −j80 π d/3 188.5 1 − (1/9)e 188.5 9 − e −j80 π d/3 −j40 π d/3 j20 π d/3 −j20 π d/3 A H3 = Ei0 (2/3)(4/3)e −j80eπ d/3 e −j20 π z/3 = Ei0 8e e −j20 π z/3 −j80 π d/3 m 377 377 9 − e 1 − (1/9)e
A m
z<0
A m
0
z> d
c. By substituting the values foe z and Ei0 = 1 V/m, we get: in material (1), at z = 0: −j80 π 0.01/3 E1(z=0) = 1 1 + 3 −1 + e−j80 π 0.01/3 = 0.8581 - j0.255 9−e 1 − 1 + e −j0.80π /3 = 3.03×10−3 + j6.76×10−4 H1(z=0) = 1−3 377 9 − e −j0.80π /3 In material (3), at z = d = 0.01 m:
V m A m
−j0.20 π /3 V E3(z=0.01m) = 1 8e = 0.914 − j0.281 m 9 − e −j0.80π /3 8e −j0.20π /3 e −j0.20π /3 = E3 = 0.914 − j0.281 = 2.42×10−3 − j7.46×10−4 H3(z=0.01m) = 1 377 9 − e −j0.80 π /3 377 377
A m
In material (2), at z = d/2 = 0.005 m: −j40π 0.005/3 + 2e −j80 π 0.01/3 e j40 π 0.005/3 V E2(z=d/2) = 1 6e = 0.866 − j0.368 m 9 − e −j80 π 0.01/3 6e −j40π0.005/3 − 2e −j80 π 0.01/3e j40 π 0.005/3 = 2.68×10−3 − j2.85×10−4 H2(z=d/2) = 1 188.5 9 − e −j80 π 0.01/3
A m
Problem 13.34. Conditions for transparency of dielectrics. The input reflection coefficient is first calculated and set to zero. The condition for transparency is found from this: The slab reflection coefficient due to the lossless material is (see Eq. (13.164):
Γslab =
Γ 12 1 − e −j2β2d
2 −j2 β 2d 1 − Γ 12 e For the reflection coefficient to be zero, we must have:
1 − e −j2 β 2 d = 0
e −j2β2d = cos2β2d − jsin2β2d = 1
→
This gives:
or:
cos2 β 2d = 1
2β 2d = 2n π
→
d = n π = nπ = n λ β2 ω µ 2 ε 2 2
n = 0,1,2... m
In this case, the wavelength is: 8
λ = c = 3×10 9 = 0.15 2f 2×10
m
Thus: d = n π = 0.075n m β2 for n = 0,1,..... The thinnest possible dielectric is d = 0.075 m (except of course for d = 0)
Problem 13.35. Conditions for transparency. Here material (3) is replaced with a perfect conductor for which Γ23 = −1 and T23 = 0. For a general dielectric, the condition of no reflection is: −2 γ 2 d Γslab = Γ 12 − e −2γ d = 0 1 − Γ 12Γ 23e 2 First, it is convenient to rewrite this as follows:
318
Γslab =
γ2d − γ2d e −γ2d Γ12e γ2d − e −γ2d = γΓd12e − e −γ d = 0 γ2d − γ2d e e − Γ 12Γ 23e e 2 − Γ 12Γ 23e 2 − γ2d
For this to be satisfied, the nominator must be zero (assuming the denominator is not - this is in general true since Γ12 < 0 for a dielectric in free space) Γ12e γ2d − e −γ2d = Γ 12 cosh γ 2 d + sinh γ 2 d − cosh γ 2 d + sinh γ 2 d = 0 The reflection coefficient between the dielectric and free space is Γ12 = η 2 − η 0 η2 + η0 With this we have η 2 − η 0 cosh γ 2 d + sinh γ 2 d − η 2 + η 0 cosh γ 2 d + sinh γ 2 d = − 2η0cosh γ 2 d − 2 η 2sinh γ 2 d = 0 Now we can write the condition for no reflection as 2η 0 + 2η 2tanh γ 2 d = 0
→
tanh −1 η0 η2 d= γ2
m
If the dielectric is lossless, we have: −jπ /2 tan −1 η0e η2 η 0 + η 2tanh jβ2d = η 0 + jη 2tan β2d = 0 → d= β2 and, in either case there is no solution since that would imply a complex value for d.
m
Problem 13.36. Application: Design of radomes. The radome is a dielectric layer in front of the radar antenna. Thus it can be viewed as a dielectric slab. The slab reflection coefficient for a dielectric slab in free space is:
Γslab =
Γ 12 1 − e −j2β2d 2 −j2 β 2d 1 − Γ 12 e
where (2) indicates the radome material. For this to be transparent it must be zero. This can be satisfied either if η2 = η0 or if: 1 − e −j2 β 2 d = 0 → e −j2β2d = cos2β2d − jsin2β2d = 1 This gives: cos2 β 2d = 1 → 2β 2d = 2n π n = 0,1,2... or: d = n π = nπ = n λ m β2 ω µ 2 ε 2 2 and the minimum thickness is d= λ m 2 for n = 1. You can use n = 0 with the simple result that d = 0. This in fact is a correct solution but perhaps not very useful in protecting the antenna. With the given properties, the wavelength is:
λ=
vp = 1 f f µ 2ε2
→
8 1 d=λ = = c = 3×10 = 0.015 = 0.0075 2 2f µ 2 ε r2 2f ε r2 2×1010 4 2
m
That is, d = 7.5 mm
Problem 13.37. Application: Design of a dielectric window. The condition for transparency of the dielectric window in free space is the same as for a radome and is found by setting the reflection coefficient of the dielectric slab in free space to zero (Eq. (13.164)). This gives:
Γslab =
Γ 12 1 − e −j2β2d 2 −j2 β 2d 1 − Γ 12 e
where (2) indicates the window material. For this to be transparent it must be zero. This can be satisfied either if η2 = η0 or if:
319
1 − e −j2 β 2 d = 0
→
e −j2β2d = cos2β2d − jsin2β2d = 1
This gives: cos2 β 2d = 1
→
2β 2d = 2n π
n = 0,1,2...
or: d = n π = nπ = n λ β2 ω µ 2 ε 2 2
m
where n = 1,2,3,..... Taking n = 1: d=
3×108 π 1 = = c = = 0.0314 2πf µ 1 ε 1 2f 3.8 µ 0ε 0 2f 3.8 2×2.45×109 3.8
m
This is 31.4 mm in thickness. Any multiple of this thickness is acceptable.
Problem 13.38. Propagation through a lossy dielectric slab. The characteristic impedance of each material is the same as for any dielectric (material (1) and (3) have characteristic impedance of free space while material (2) is now complex because of the loss term). The wave impedance in a plane wave is always the ratio between the electric and magnetic field intensities and these vary depending on the reflections that may exist. a. The intrinsic impedance in each material is that of the dielectric, regardless of any interfaces that may exist. In materials (1) and (3) this is the intrinsic impedance of free space. Thus: f = 109
η2 =
j ωµ 0 = j ωε 2 + σ
Hz ,
η 1 = η 3 = η 0 = 377
Ω
j2π ×109×4π ×10−7 = 188.37 + j0.42 j2π×109×4×8.854×10−12 + 0.001
Ω
Although the intrinsic impedance is complex, its imaginary part is small and will be approximated as 188.5 Ω. b. The various components of the electric field intensity are as follows (see Eqs. (13.156) through (13.159)) −2 γ 2 d V Er0 = Γ 12 + Γ 23e −2γ d Ei0 m 1 + Γ 12Γ 23e 2 −2 γ 2 d T12 E2− = T 12Γ 23e −2γ d Ei0 E2+ = Ei0 1 + Γ 12Γ 23e 2 1 + Γ 12Γ 23e −2γ2d − γ2d γ 3 d V E3+ = T 12T 23e e−2γ d Ei0 m 1 + Γ 12Γ 23e 2
V m
where
γ 0 = jβ 0 ,
γ 2 = α 2 + jβ 0 ,
γ 3 = jβ 0
and:
Γ12 = η 2 − η 1 = 188.37 + j0.42 − 377 = − 0.336 + j0.01 ≈ − 1 η 2 + η 1 188.37 + j0.42 + 377 3 η 377 − 188.37 − j0.42 3 − η2 1 Γ23 = = = 0.3336 − j0.01 ≈ η 3 + η 2 377 + 188.37 + j0.42 3 2× 188.37 + j0.42) T 12 = 2η2 = = 0.6664 + j0.001 ≈ 2 η 3 + η 2 377 + 188.37 + j0.42 3 2 η 2×377 3 T 23 = = = 0.6664 − j0.001 ≈ 4 η 3 + η 2 377 + 188.37 + j0.42 3 The total electric fields in the three materials are (see Eqs. (13.145, 13.147 and 13.149): E1 = Ei0e −jβ0z + Er0e jβ0z V/m z<0 E2 = E2+e −(α2+jβ2)z + E2−e (α2+jβ2)z V/m 0 d Substituting the values above gives −2(α2+jβ2)d V E1 = Ei0e −jβ0z + Er0e jβ0z = Ei0 e −jβ0z + Γ 12 + Γ 23e −2(α +jβ )d e +jβ0z 2 2 m 1 + Γ 12Γ 23e −(α2+jβ2)z + T12Γ23e −2(α2+jβ2)de (α2+jβ2)z E2 = E2+e −jβ0z + E2−e jβ0z = Ei0 T 12e 1 + Γ 12Γ 23e −2(α2+jβ2)d
320
V m
z<0 0
− α 2 + jβ 2 d j β 0 d e E3 = E3+e −jβ0z = E i0 T 12T 23e e −jβ0z 1 + Γ 12Γ 23e −2 α2 + jβ2 d
V m
z> d
In this case, the dielectric is a low loss dielectric. The numerical values of the various parameters are: The attenuation and phase constants are:
σ = 0.001 = 0.00049 << 1 ω ε 2×π×109×4×8.854×10-12 The parameters are therefore: f = 109
Hz , η 1 = η 3 = η 0 = 377 Ω , η 2 ≈ 188.5 9 β0 = 2 π f = 2 π ×108 = 2 π = 20.94 rad , β 2 = ε r β0 = 4 π = 41.89 c m 3 3 3×10 Np α 2 ≈ σ ηn = 0.001 ×188.5 = 0.098 m 2 2
Ω rad m
The magnetic field intensities are calculated from the following relations (Eqs. 13.146, 13.148 and 13.150): A H1 = 1 Ei0e −jβ0z − Er0e jβ0z η0 m + − A H2 = E2 e −(α2+jβ2)z − E 2 e (α2+jβ2)z η2 η2 m A H3 = 1 E3+e −jβ0z η0 m In terms of the electric fields above: −2(α2+jβ2)d A H1 = 1 Ei0e −jβ0z − Er0e jβ0z = E i0 e −jβ0z − Γ 12 + Γ 23e −2(α +jβ )d e jβ0z z<0 2 2 η1 η0 m 1 + Γ 12Γ 23e −(α2+jβ2)z − T12Γ23e −2(α2+jβ2)de (α2+jβ2)z A H2 = 1 E2+e −(α2+jβ2)z − E2−e (α2+jβ2)z = E i0 T 12e 0 d η3 η0 1 + Γ 12Γ 23e −2(α2+jβ2)d m Substitution of these values gives the electric field intensities in the three materials. These now only depend on the incident electric field intensity Ei0 and on the thickness d and on position z. c. Substituting the various values we obtain the electric and magnetic field intensities. In material (1) at z = 0, −2(0.098+j40 π /3)0.01
E1 = 1 1 + − (1/3) + (1/3)e = 0.864 − j0.254 1 − (1/9)e −2(0.098+j40 π /3)0.01 −2(0.098+j40 π /3)0.01 H1 = 1 1 − − (1/3) + (1/3)e = 3.0×10−3 + j6.72×10−4 −2(0.098+j40 π /3)0.01 377 1 − (1/9)e
V m A m
In material (3) at z = 0.01 m, − 0.098 + j40 π /3 0.01 j0.2 π /3 e V E 3 = 1 (2/3)(4/3)e −2(0.098+j40 π /3)0.01 e −j0.2π/3 = 0.887 + j0.32 m 1 − (1/9)e − 0.098 + j40 π /3 0.01 j0.2 π /3 e H3 = 1 (2/3)(4/3)e −2(0.098+j40 π /3)0.01 e −j0.2π/3 = 2.35×10 −3 + j8.5×10 −4 377 1 − (1/9)e In material (3) at z = 0.005 m,
A m
−(0.098+j40π/3)0.005 + (2/3)(1/3)e −2(0.098+j40π /3)0.01e (0.098+j40 π /3)0.005 = 0.679 − j0.197 E2 = 1 (2/3)e 1 − (1/9)e −2(0.098+j40 π /3)0.01 (2/3)e −(0.098+j40π/3)0.005 − (2/3)(1/3)e −2(0.098+j40π /3)0.01e (0.098+j40 π /3)0.005 H2 = 1 188.5 1 − (1/9)e −2(0.098+j40 π /3)0.01 −3 A = 3.61×10 − j1.04×10−3 m
Problem 13.39. Propagation through a two layer slab. 321
V m
We write the forward and backward electric and magnetic fields in each section using the amplitudes of the electric field intensity as the unknown values. In material (1), only the reflected wave is unknown and in material (4), there is only a forward propagating wave. This will result in a system of six equations in six unknowns which is solved to find the amplitudes of the fields and from these, the reflection and transmission coefficients. a. To write the fields in each material, we use Figure A. In material (1) there is a forward propagating wave made of the incident field and a backward propagating wave made of the reflected wave at the interface: E 1 = x Ei0e −jβ0z + Er0e jβ0z
V/m
The magnetic field intensity is then: H 1 = y Ei0 e −jβ0z − E r0 e jβ0z η0 η0
A m
In these relations Ei0 = E0 and is known. Also, β0 is known based on the material properties and frequency. In material (2) we have similar relations but now both the forward and backward propagating waves are unknown (see section). Denoting the forward and backward propagating electric field amplitudes as E2+ and E2−, we can write: E 2 = x E2+e −jβ2z + E2−e jβ2z
V m
+ − H 2 = y E2 e −jβ2z − E 2 e jβ2z η2 η2
A m
V m
+ − H 3 = y E3 e −jβ3z − E 3 e jβ3z η3 η3
A m
In material (3) we again have: E 3 = x E3+e −jβ3z + E3−e jβ3z
In material (4), there is only a forward propagating wave and we get: E 4 = x E4+e −jβ0z
+ H 4 = y E4 e −jβ0z η0
V m
A m
In all these relations, all values are known except Er and E2−, E2+, E3−, E3+ and E4+. Thus we need six equations to solve for these six unknowns based on the incident field, material properties and dimensions. To do so, we match the electric fields and, separately, the magnetic fields at the three interfaces. Doing so gives: At z = 0: E i0 − E r0 = E 2+ − E 2− η0 η0 η2 η2
E i0 + Er0 = E2+ + E2− At z = d1:
E 2+ e −jβ2d1 − E 2− e jβ2d1 = E 3+ e −jβ3d1 − E 3− e jβ3d1 η2 η2 η3 η3
E2+e −jβ2d1 + E2−e jβ2d1 = E3+e −jβ3d1 + E3−e jβ3d1 At z = d2:
E 3+ e −jβ3d2 − E 3− e jβ3d2 = E 4+ e −jβ0d2 η3 η3 η0
E3+e −jβ3d2 + E3−e jβ3d2 = E4+e −jβ0d2
Before substituting the actual values and solving it is worth rewriting the six equations as follows: − Er0 + E2+ + E2− = Ei0 = E0 E r0 + E 2+ − E 2− = E 0 η0 η2 η2 η0 + −jβ 2d 1 E2 e + E2−e jβ2d1 − E3+e −jβ3d1 − E3−e jβ3d1 = 0 E2+ e −jβ2d1 − E 2− e jβ2d1 − E 3+ e −jβ3d1 + E 3− e jβ3d1 = 0 η2 η2 η3 η3 E3+e −jβ3d2 + E3−e jβ3d2 − E4+e −jβ0d2 = 0 E3+ e −jβ3d2 − E 3− e jβ3d2 − E 4+ e −jβ0d2 = 0 η3 η3 η0 Writing this as a system of equations:
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(1) (2) (3) (4) (5) (6)
−1
1
1
0
0
0
E0 E r0
1 η0
1 η2
− 1 η2
0
0
0
0
e −jβ2d1
e jβ 2 d 1
− e −jβ3d1
− e jβ 3 d 1
0
E2−
0
e −jβ2d1 η2
jβ 2 d 1 −e η2
−jβ3d1 −e η3
e jβ 3 d 1 η3
0
E3+
0
0
0
e −jβ3d2
e jβ 3 d 2
− e −jβ0d2
0
0
0
e −jβ3d2 η3
jβ 3 d 2 −e η3
−jβ0d2 −e η0
E0 η0
E2+ =
0 0
E3−
0
E4+
0
Although this can be solved in general terms (but it is complicated and very tedious) it is a relatively easy matter to solve the equations for particular values of β0, β2, β3, η0, η2, η3, d1, d2, and E0. These are as follows: 6 β 0 = ω µ 0 ε 0 = 2×π×150×10 = π 3×108
In free space:
rad m
6 ×1.5 = 1.5π β 2 = ω µ0×2.25ε0 = 2×π×150×10 8 3×10
In material (2):
6 ×2 = 2 π β 3 = ω µ 0×4 ε 0 = 2×π×150×10 3×108 η 2 = 377 = 251.34 η 3 = 377 = 188.5 2.25 4
In material (3):
η 0 = 377
rad m rad m Ω
With d1 = 0.1 m, d2 = 0.2 m, we get: −1
1
1
0
0
0
1 377 0
1 251.34 e −j0.15π
−
1 251.34 e j0.15 π
0
0
0
− e −j0.2 π
− e j0.2π
0
E2−
j0.15 π − e 251.34 0
−j0.2 π −e 188.5 e −j0.4 π
e j0.2π 188.5 e j0.4 π
0
E3+
0
e −j0.15π 251.34 0
− e −j0.2 π
0
0
0
e −j0.4 π 188.5
j0.4 π − e 188.5
−j0.2 π −e 377
0
1 E r0 E2+ =
1 377 0 0
E3−
0
E4+
0
Expanding the exponents using Euler's equation as follows: e j0.15π = cos(0.15π ) + j sin(0.15π) = 0.891 + j 0.454 e − j0.15π = cos(0.15π ) − j sin(0.15π) = 0.891 − j 0.454 e j0.2π = cos(0.2π ) + j sin(0.2π) = 0.809 + j 0.588 e −j0.2π = cos(0.2π ) − j sin(0.2π) = 0.809 − j 0.588 e j0.4π = cos(0.4π ) + j sin(0.4π) = 0.309 + j 0.951 e −j0.4π = cos(0.4π ) − j sin(0.4π) = 0.309 − j 0.951 Substituting these in the system of equations and solving the system in complex variables, gives: E r0 = − 0.46693 − j 0.12261,
E2+ = 0.75552 − j 0.020435
V/m
E2− = − 0.22246 − j 0.10218,
E3+ = 0.65441 + j 0.055284
V/m
E3− = − 0.16564 − j 0.14313,
E4+ = 0.74918 − j 0.45302
V/m
Substituting these back into the field equations in each section of space, the electric and magnetic field intensities are:
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E 1 = x 1e −jπz + ( − 0.46693 − j 0.12261)e jπz H 1 = y 1 1e −jπz + (0.46693 + j 0.12261)e jπz 377
V/m , A, m
z<0 z<0
E 2 = x (0.75552 − j 0.020435)e −j1.5πz − (0.22246 + j 0.10218)e j1.5π z H2 = y 1 (0.75552 − j 0.020435)e −j1.5πz + (0.22246 + j 0.10218)e j1.5π z 251.34 E 3 = x (0.65441 + j 0.055284)e −j2πz − (0.16564 + j 0.14313)e j2π z H 3 = y 1 (0.65441 + j 0.055284)e −j2πz + (0.16564 + j 0.14313)e j2π z 188.5 E 4 = x (0.74918 − j 0.45302)e −jπz H 4 = y 1 (0.74918 − j 0.45302)e −jπz 377
V/m , A, m
V/m , A, m V/m , A, m
0 < z < 0.1 m 0 < z < 0.1 m
0.1 m < z < 0.2 m 0.1 m < z < 0.2 m
z > 0.2 m z > 0.2 m
b. The slab reflection is the ratio between the amplitude of the reflected wave and that of the incident wave:
Γslab = E r0 = − 0.46693 − j 0.12261 E0 which is complex, with magnitude |Γ| = 0.4828. The transmission coefficient for the slab is the ratio between the amplitude of the electric field in material (4) and the incident electric field: + T slab = E 4 = 0.74918 − j 0.45302 E0 This has a magnitude |T| = 0.8755.
ε 0 , µ0
2.25ε0 , µ 0, σ =0
Er0
E 2−
E3−
E2+
E3+
E4
3
z= d2
E p
Ei 0 1
z=0
2
4ε0 , µ0 , σ =0
z= d1
ε 0 , µ0
4
Figure A.
Problem 13.40. Transmission of power through an interface. The idea here is to calculate the transmission coefficient between air and material 1. This gives the electric field and magnetic field intensities entering the material. The Poynting vector is calculated. Since the attenuation constant is very high, all power is dissipated within a short distance. Thus, the Poynting vector gives the total power dissipated per unit area. In air, the incident electric and magnetic field intensities are: V m
E = xE0cos(ω t − βz)
and:
H = y E0 cos(ωt − βz) η0
A m
where the direction of E was chosen arbitrarily but so that propagation is in the positive z direction. Note that η0 is real but η1 is complex. Since the material can be assumed to be a good conductor (σ/ωε = 898 >> 1), the intrinsic impedance of material 1 is given by
η1 = (1 + j)
π fµ = (1 + j) σ
π×108×4π×10−7×50 = (1 + j)44.43 = 62.833∠45° 10
Ω
The high loss in the material also means that the material can be assumed to be infinitely thick (no field reaches the opposite surface). Thus, only reflection from the interface needs to be considered. The reflection and transmission coefficients at the interface are:
Γ1 = η 1 − η 0 η0 + η1
and:
In particular, the transmission coefficient is:
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T1 =
2η1 η0 + η1
T1 =
2η1 = 2×(1 + j)44.43 = 0.23 + j0.18655 = 0.296∠39° η 0 + η 1 377 + (1 + j)44.43
The electric and magnetic field intensities in material 1, immediately next to the interface is equal to H 1 = y E0 T 1 η1
V, m
E1 = xE 0 T 1
A m
Now, we can calculate the magnitude of the time averaged Poynting vector as P av = 1 E 1×H 1* 2
W m2
This is the time averaged power density entering material (1) per unit area of the interface. The phasor forms of the transmission coefficient and the intrinsic impedance are: T1 = 0.296e j39°,
η 1 = 62.833e j45°
Now we can write: 2 P av = E 1×H 1* = 1 xE 0T 1 e −jβ1z × y E0 T1* e jβ1z = 1 xE 0T 1 e −jβ1z × y E0 T * e jβ1z = 1 z E0 T 1 T 1 * ∗ ∗ 2 2 η1 2 η1 2 η1∗ 2 Since T1T1* = T we can write:
P av = z
2 2 E02 T 1 2 = z (100) (0.296) = 6.972e j45° = 6.972∠45° ∗ −j45° 2η1 2×62.833e
W m2
W m2
The real part of this is:
Pav = 6.972cos45° = 4.93
W m2
Since only real power can be dissipated. Thus for each m 2 of the material there are 4.93 W of power dissipated. Note that the phase angle of the power density is due to the intrinsic impedance of the material.
Problem 13.41. Transmission/reflection through conductor backed slabs. The reflection coefficient for a lossless dielectric slab backed by a perfect conductor is given in Eq. (13.162) after substituting γ 2 = jβ 2 and Γ 23 = − 1. For the reflection coefficient to be maximum, the thickness must be such that the magnitude of the reflection coefficient is 1. −j2 β 2d Γslab = Γ 12 − e −j2β d 2 1 − Γ 12e
a. For the given quantities:
η0 =
µ 0 = 377 ε0
η2 =
µ 0 = 377 = 168.6 5ε0 5
β2 = ω 5 ε 0µ 0 = 2π×900×106 5×8.854×10−12×4π×10−7 = 42.178
Ω rad m
With these, Γ12 is
and the reflection coefficient is:
Γ 12 = 377 − 168.6 = − 0.382 377 + 168.6
−j84.356 Γslab = − 0.382 − e = − 0.382 − cos(0.8436rad) + jsin(0.8436rad) = − 1.0468 + j0.74704 1.254 − j0.28537 1 + 0.382e −j84.356 1 + 0.382cos(0.8436rad) − j0.382sin(0.8436rad) = − 0.9226 + j 0.3858
The magnitude of the reflection coefficient is:
Γ = 0.999 b. To see what the condition for maximum reflection, we One possibility for the reflection coefficient to be maximum, is to set Γslab = −1. This gives:
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−j2 β 2 d − 1 = Γ 12 − e −j2β d 2 1 − Γ 12e
Γ 12 − e −j2β2d = − 1 + Γ 12e −j2β2d
→
Or:
Γ 12 + 1 = − e −j2β2d = Γ12e −j2β2d + e −j2β2d = e −j2β2d Γ 12 + 1
→
Γ 12 + 1 = Γ 12 + 1 e −j2β2d
Or: e −j2β2d = cos2β2d − jsin2β2d = 1
→
cos2β2d = 1
→
2β2d = n2π
The required thickness is then: nπ n d = nπ = = = 0.0745n, β2 2πf 5 ε 0µ 0 2×900×106 5×8.854×10−12×4π×10−7
n = 0,1,2,...
Thus, the minimum thickness is zero (no dielectric). The next solution is for n = 1 and the thickness d should be 0.0745 m (74.5 mm).
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