chapter13_answers2.pdf

296

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 highloss material. For the properties given for water:

σ2

ωε2 = 10−9

2π×108×80×8.85×10−12 = 10−5

2π×80×8.85 = 2.25×10−9 << 1

This is a very low loss dielectric. For a low loss dielectric we can write:

η0 = µ0

ε0 = 377 η2 ≈ µ2

ε2 = µ0

80ε0

= 37780

= 42.15 Ω

Now, using the definitions of the reflection and transmission coefficients we can write:

Γ = η2 − η0

η0 + η2 =

µ2

ε2 − µ0

ε0

µ0

ε0 + µ2

ε2

=

37780

− 377

377 + 37780

= − 0.799

The transmission coefficient isT = 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 theelectric field intensity is in the x direction):

Ei(z) = xEie −jβ0z V/m

where the phase constant is that of free space and equals:

β0 = ω µ0ε0 = 2π×108

3×108 = 2π

3 = 2.089 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:

Et(z) = xΤEie −α2ze−jβ2z = x0.201Eie −2.1×10−8ze −j18.73z V/m

where the phase and attenuation constants in material (2) are:

β2 ≈ ω µ2ε2 = 2πfvp2

= 2π×108× 80

3×108 = 18.73 rad

m

α2 ≈ σ2

η2 = 10−9

2 µ2

ε2 = 10−9

2 377

80 = 2.1×10−8 Np

m

Problem 13.2. Incident and reflected waves at a lossless dielectric interface.Calculate the reflection coefficient:

η0 = µ0

ε0 η2 = µ0

2ε0

= η0

2 Ω

a. Reflection coefficient is:

Γ = η2 − η0

η2 + η0 = − 2 − 1

2 + 1 = − 0.17157

Thus:

E1(z) = E0e −jβ1z 1 + Γe j2β1z V/m

E1max = E0 1+Γ = 11.716, E1min = E0 1−Γ = 8.284 V/m

b. In terms of the magnetic field:

H1max = 11.716377

= 0.031 H1min = 8.284377

= 0.0219 Am

297

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 reflectioncoefficient. We calculate the general form of the electric field intensity and from this we evaluate the maximum andminimum values. The location of the minima and maxima are also evaluated from the general relations.

E1(z) = xEi1 e−jβ0z + Γe +jβ0z V/m (1)

where β0 is the phase constant of free space and:

Γ = η2 − η1

η1 + η2

where: η0 = µ0

ε0 = 377, η2 = jωµ2

σ2 + jωε2

Ω (2)

The reflection coefficient is complex for a lossy dielectric because η2 is complex and may be written as ΓejθΓ. In thisexpression, 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 + ΓeθΓe j2β0z V/m

Thus, since the exponential terms all have a magnitude of 1. Maximum electric field intensity is therefore:

E1max = xEi1 1 + Γ V/m

Similarly, the minimum electric field intensity is:

E1min = xEi1 1 − Γ V/m Thus, the ratio between E1max and E1min is:

Emax

Emin

= SWR = 1 + Γ

1 − Γ =

1 + η2 − η0

η2 + η0

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 intoelectricity. The efficiency is the ratio between the power converted into electricity and input power. If the reflectioncoefficient can be decreased, the efficiency of the cell can be increased

a. The reflection coefficient at the silicon-air interface is:

ηair = 377, ηsilicon = 3771.75

= 284.98 Ω → Γ = ηsilicon − ηair

ηsilicon + ηair

= 284.98 − 377284.98 + 377

= − 0.139

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 electricfield intensity inside the cell and finally the power transmitted. The electric field intensity in air is:

Pin = Ein2

2ηair

Wm2

→ Ein = 2Pinηair Vm

The transmitted electric field intensity is:Et = TEin = T 2Pinηair V/m

Thus, the transmitted power into the silicon is:

Pt = Et2

2ηsilicon

= T 2Pinηair

2

2ηsilicon

= T2Pinηair

ηsilicon = 0.8612×1400×377

284.98 = 1372.97 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 W/m2

298

The overall efficiency of the solar cell is the ratio between converted power and input power or:

eff = Pconverted

Pin

= 343.241,400

= 24.5 %

The overall efficiency is 24.5%.

b. If the reflection coefficient can be eliminated, the transmission coefficient becomes T = 1 and all power availableenters the cell. Then:

Pt = 1,400 Wm2

→ Pconverted = Pt×0.25 = 350 Wm2

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 electricfield intensity, magnetic field intensity and power transmitted into the glass are calculated from the transmissioncoefficient 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:

ηair = 377, ηglass = 3771.8

= 281 Ω → Τ = 2ηglass

ηglass + ηair

= 562281 + 377

= 0.8541

The amplitude of the electric field intensity is found from the power density:

Pin = Ein2

2η0

→ Ein = 2Pinη0 = 2×0.1×377 = 8.68 Vm

The incident magnetic field intensity is:

Hin = Ein

η0 = 8.68

377 = 0.023 A

m

These fields are transmitted into glass through the transmission coefficient. Thus,

Eglass = EinT = 8.68×0.8541 = 7.414 V/m Hglass = EinT

ηglass =7.414

281 = 0.0264 A

m

b. The total incident power equals the incident power density multiplied by the area of the beam.

Pin = Pin×πd2

4 = 0.1×π(0.0001)2

4 = 7.854×10−10 W

Out of this, the transmitted power is:

Pglass = PinT 2 ηair

ηglass = 7.854×10−10×(0.8541)2 377

281 = 7.686×10−10 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 isdissipated, causing both the warmth we feel and blistering if exposure is excessive. The conductivity given is onlyimportant 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:

σω ε

= 0.012×π×5×1014×24×8.854×10−12

= 1.5×10−8 << 1

The skin is clearly a low loss material and for the purpose of this computation may be viewed as a dielectric. Thus, theintrinsic impedance of free space and skin are:

η0 = 377, ηs = 37724

= 76.955 Ω

The transmission coefficient is:

Τ = 2ηs

ηs + η0

= 153.9176.955 + 377

= 0.339

299

The power density penetrating into the body is:In free space we can write:

Pin = Ein2

2η0

→ Ein2 = 2η0Pin

Since:

Es = TEin → Ps = T2Ein

2

2ηs

Ps = Pin×Τ 2η0

ηs = 1,300×(0.34)2× 377

76.955 = 731.89 W

m2

Thus, the total power absorbed by the body, assuming a 1 m2 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, theultraviolet portion only penetrates the skin. Thus, even though the power is not that large it can cause considerabledamage 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 andthe body. In this case the body is a low loss dielectric. The power penetrating into the body equals the incident powerdensity, multiplied by the area of the skin, by the transmission coefficient squared and by the ratio of the intrinsicimpedances of the air and body. First we need to make sure that the body is a low loss dielectric at 2.54 GHz:

σω ε

= 0.012×π×2.54×109×24×8.854×10−12

= 0.00295 << 1

Thus, the body may be viewed as a low loss dielectric and the intrinsic impedance of the body is:

ηb = 37724

= 77 Ω

The transmission coefficient at the air-body interface is:

Τ = 2ηb

ηb + η0

= 15477 + 377

= 0.34

The power density allowed by the standard is:

Pin = 10×10−3

10−4 = 100 W

m2

Thus, the total allowable power over the surface of the body is 150 W. The power transmitted into the body is (fromprevious example):

Pb = Pin×T 2η0

ηb = 150×(0.34)2×377

77 = 84.9 W

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 thewall 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)ejωt = Re − yj2Ei1sin(β1z)e jωt = Re y2Ei1sin(β1z)e jωte−jπ/2

= y 2Ei1sin(β1z) cos ω t − π2

= y2Ei1sin(β1z) sin(ωt) V/m

300

a. The first peak is positive and occurs at a distance λ/4 from the wall. The behavior of the standing wave for the electricfield 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 butthis 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 alocation z = 0. The amplitude of the wave at the antenna is 100 V. Now a forward propagating wave propagates towardsthe interface. At the location of the first peak from the antenna, this wave is:

E1st+ = 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:

Eint+ = 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:

Eint− = − 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 antennawe get:

E1st− = − 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 distancepropagated. The total field at this location is the sum of the incident field in (2) and the reflected field in (5):

E1st = E 1st+ + E 1st

− = y100e−αλe−j2π − 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:

E1st = y 100 e−12αe−j2π − e−66αe−jβ66 V/m (7)

Since β = 2π/λ = 2π/12, we can write:

E1st = y 100 e−12αe−j2π − e−66αe−j2πe −j2π×66

1 2 = 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:

E1st = 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

µ1

ε1 = 10−5

2

µ0

4ε0

= 10−5

2 ×377

2 = 94.25×10−5 Np

m (11)

Substituting this in Eq. (10) gives:

E1st = y 100 e−12×94.25×10−5 + e−66×94.25×10−5

= y192.8 V/m

Notes: Without the attenuation constant, the electric field intensity would have been 200 V/m. Also, because of theattenuation constant we had to "follow" the wave and inspect its behavior continuously as it propagates.

y

z

z=0

λ/4

λ

λ2λ3λ

ante

nna

wal

l

ante

nna

wal

l

d12

3dd

firstpeak

Figure A Figure B

301

Problem 13.9. Reflection of waves from conducting surfaces.The incident and reflected waves (electric fields) are:

Ei = Ei0e−jβ0z Er = Ei0Γe +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 = Ei0

η0e−jβ0z − Ei0

η0Γe+jβ0z = Ei0

η0e−jβ0z + e+jβ0z A/m

The ratio between the two is:

EH

= Ei0 e−jβ0z − e+jβ0z

Ei0

η0e−jβ0z + e+jβ0z

= η0 e−jβ0z − e+jβ0z

e−jβ0z + e+jβ0z Ω

Using the relations above:

EH

= η0 e−jβ0z − e+jβ0z

e−jβ0z + e+jβ0z = η0 −j2sinβz

2cosβz = −jη0tanβz 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 incidentmagnetic field intensity parallel to the surface would penetrate into what was the location of the conductor. To cancel thisfield, we introduce a surface current density (A/m) that will produce an opposing field and therefore produce zero magneticfield intensity in the perfect conductor. This current density is proportional to the incident tangential magnetic fieldintensity at the conductor.

a. We start with the configuration shown in Figure A. Assuming that the reflected fields are correct (if not, we willfind them from the conditions at the surface), then the conditions for H in Figure A can also be obtained from FigureB. With a surface current density in the negative y direction, the conditions in Figure A are satisfied. The incidentmagnetic field intensity is:

Hi1 = x Ei

η0e−jβ0z A

m (1)

Taking z = 0 at the interface and using Figure B:

Hr = Ei

η0 A

m (2)

and:

2Hr = 2Ei

η0 = Js → Js = − y2Ei

η0 A

m (3)

b. Without the current density at the surface, the magnetic field intensity everywhere in space is:

Hi1 = x Ei

η0e−jβ0z 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

η0 e−jβ0z for z > 0 and Hr = x Ei

η0 e jβ0z for z < 0 A

m

Thus, the total magnetic field intensity is:To the right of the interface (z > 0):

H = H i1 + H r = x Ei

η0e−jβ0z − x Ei

η0 e−jβ0z = 0 A

m

To the left of the interface (z < 0):

H = H i1 + H r = x Ei

η0e−jβ0z + x Ei

η0 e jβ0z 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.

302

⊗

H

r

E

i1

H

Ei

r

Js

⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗

x

zy

free space conductor Figure A

H

r

i1

H Js

⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗

H

r

i1

H

free space 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 fieldintensity and the normal magnetic flux densities are continuous while the tangential magnetic field intensity and thenormal electric flux density are discontinuous (see chapter 11, Table 11.3.

a. Since the polarization is perpendicular, the electricfield intensity only has a tangential component.Therefore:

E1t = E2t

However, the magnetic field intensity, which must beperpendicular to E (see Figure A) has both atangential and a normal component as shown. Thenormal component of the magnetic flux density iscontinuous:

E1

H1

ε1,µ1,σ1=0

air conductor

H1n

H1t

E2

H2

H2n

H2t

ε2,µ2,σ2

Js

Figure A

B1n = B2n → µ1H1n = µ2H2n

Note: E1 and H1 are the total fields in material (1). The tangential component of the magnetic field intensity is alsodiscontinuous, the discontinuity giving rise to a surface current density Js:

H1t − H2t = 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 aconductor is given as:

γ2 = α2 + jβ2 = (1 +j) πfµ2σ2 The depth of penetration is:

δ2 = 1α

= 1πfµ2σ2

m

and, for a good conductor is very small.The intrinsic impedance in the conductor is:

η2 = (1 + j) 1δ2σ2

Ω

which is very low. The phase velocity is also low and equals:

vp2 = ωβ

= ωπfµ2σ2

= 2ωµ2σ2

ms

Problem 13.12. Application: Condition of no reflection: stealth principles.There are two ways to solve this problem: the easyway is to simply show that if the wave is reflected at90° to the incident wave (or any angle above 90°) nopart of the wave will reflect back. The second is tocalculate the reflected wave and set it to zero. Fromthis you will find the angle α.1. The easy way: see Figure A: This immediatelysays that the angle must be smaller or equal to 45°:

α ≤ 45°

z

P

ε0,µ0Ei

Hi

free space perfectconductor

90°−.

αα

αα

αα

90°−

αθi

Figure A

303

This implies that under ideal conditions (illumination straight on) a 90° corner will not reflect light back to the sourceand an airplane will not be visible. If your car were made as a 90° wedge it would be invisible to radar when viewedstraight on.2. Now for the more difficult way. We note first that the incidence angle (the angle between the direction of propagationof 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 − 2αi) − 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 − 2αi) = 0 or:

180° − 2αi = 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 configurationshown. 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 = xE0e −jβ0(ysinα + zcosα) = x100e −jβ0(ysinα + zcosα) V/m

From Figure A, the incident magnetic field intensity is:

Hi = E0

η0ycosα − z sinα e−jβ0(ysinα + zcosα) = 0.265 ycosα − z sinα e−jβ0(ysinα + zcosα) A

m

where:

η0 = 377 Ω and: β0 = 2π×100×109

c = 2π×100×109

3×108 = 2094.4 rad

m

b. The reflected fields are shown in their assumed directions in Figure A. These are:

Er = − xEre −jβ0(ysinα − zcosα) V/m

Hr = Hr ycosα + zsinα e−jβ0(ysinα − zcosα) 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 = xE0e −jβ0ysinα − xEre −jβ0ysinα = 0 → Er = E0

That is, our assumption as to the direction of the reflected electric field intensity was correct. Thus, the electric fieldintensity is as shown in the figure and, because we must satisfy the Poynting vector, the electric an magnetic fieldintensities with the values for H0 (given) and η0 and β0 (calculated above) must be:

Er = − x100e −j2094.4(ysinα − zcosα) V/m

Hr = 0.265 ycosα + zsinα e−j2094.4(ysinα − zcosα) V/m

c. To calculate the surface current density we argue that the tangential component of the reflected magnetic field intensityis due to this current density. The surface current density shown in Figure B produces a magnetic field intensity equal tothe reflected field on both sides of the interface. In the conductor, this field cancels the incident magnetic field intensitybut in air, it adds to the incident field. For this to happen, the current density must be in the positive x direction. Thetangential magnetic field intensity at the surface (z = 0) is:

Hrt(z = 0) = y2×0.265cosα A/m From Ampere’s law:

Hrt(z = 0) = Js → Js = 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 onlocation on the interface, may be minimum at other positions (see Figure C).

304

x z

y

Ei

Er

Hr

Hi

ε0,µ0

σ

pr

pi

⊗ =∞

αα

αα ⊗ x z

y

⊗⊗⊗⊗⊗⊗

Hr cosα

Hi cosα

Hr cosα

Hi cosαJs

Figure A Figure B

−π/2 0 π/2

0.53

y=0

y=.05m

Js

θi

A/m

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 fieldintensity at the interface has only a tangential component and therefore we need not worry about the continuity of itsnormal 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 fieldintensity is continuous:

E1t = E2t

The normal component is discontinuous as follows:

D1n − D2n = ρs → ε1E1n − ε2E2n = ρs

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 surfacecurrent density Js:

H1t − H2t = Js

b. The difference between the parallel polarization is merely in the components that need to be transferred across theinterface. 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 therequired 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 configurationshown. 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 thesurface.a. The incident magnetic field intensity is in the x direction:

Hi = xH0e −jβ0(ysinα + zcosα) = x100e −jβ0(ysinα + zcosα) A/m

From Figure A, the electric field intensity is:

Ei = η0H0 − y cosα + zsinα e−jβ0(ysinα + zcosα) = 37,700 − y cosα + zsinα e−jβ0(ysinα + zcosα) V/m

305

where η0 = 377 Ω and: β0 = 2π×100×109

c = 2×π×100×109

3×108 = 2094.4 rad

m

b. The reflected fields are shown in their assumed directions in Figure A. These are:

Hr = xHre −jβ0(ysinα − zcosα) A/m

Er = Er ycosα + zsinα e−jβ0(ysinα − zcosα) 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:

Eit(z = 0) + E rt( = 0) = η0H0 − y cosα e−jβ0ysinα + Er ycosα e−jβ0ysinα = 0 V/m or:

Er = η0H0 V/m

That is, the reflection coefficient equals +1. Thus, the electric field intensity is as shown in the figure and, because wemust satisfy the Poynting vector, the electric an magnetic field intensities must be:

Er = η0H0 ycosα + zsinα e−jβ0(ysinα − zcosα) V/m

Hr = xH0e −jβ0(ysinα − zcosα) A/m

With the values above for H0, η0, and β0, we can also write:

Er = 37,700 ycosα + zsinα e−j2094.4(ysinα − zcosα) V/m

Hr = x100e −j2094.4(ysinα − zcosα) A/m

c. To calculate the surface current density we argue that the reflected magnetic field intensity is due to this currentdensity. The surface current density shown in Figure B produces a magnetic field intensity equal to the reflected field onboth sides of the interface. In the conductor, this field cancels the incident magnetic field intensity but in air, it adds tothe incident field. For this to happen, the current density must be in the negative y direction. From Ampere’s law:

2Hr(z = 0) = Js → Js = −y2Hr(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 angleof incidence. Therefore, the surface current density varies with the incidence angle as 2H0cosα. For the parallel polarizedwave, the surface current density is independent of the angle of incidence.

x z

y

Ei

Er

Hr

Hi

ε0,µ0σ

pr

pi

⊗ =∞

αα

α

α

⊗

⊗

Figure A

H

r

i

H

Js

⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗⊗

H

r

i

Hx

zy

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 fieldis used for this purpose since the magnetic field may always be calculated from the electric field. There are twoconfigurations 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 ysinθi + zcosθi V/m

Er = ycosθi + zsinθi Ere− jβ ysinθi − zcosθi V/m

H i = x Hie− jβ ysinθi + zcosθi A/m

H r = x Hre−jβ ysinθi − zcosθi A/m

306

Now we equate the sum of the tangential components of the electric field intensity at z = 0 to zero to find the reflectedelectric 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 jβzcosθi e− jβysinθi

+ zηHisinθi e− jβzcosθi + e jβzcosθi e− jβysinθi = yj2ηHicosθi sin(βzcosθi)e− jβysinθi

+ zηHisinθi cos(βzcosθi)e− jβysinθi = 2ηHi yjcosθi sin(βzcosθi) + zsinθi cos(βzcosθi) e− jβysinθi V/m

Similarly for the magnetic field intensity:

H 1 = H i + H r = xHi e− jβzcosθi + e jβzcosθi e− jβysinθi = x2Hicos(βzcosθi)e− jβysinθi A/m

Note: The total electric and magnetic fields for parallel polarization for the configuration in Figure A are given inEqs. (13.90) and (13.91) but with the obvious change in coordinates (x coordinate becomes the y coordinate and they 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:

E1(x,y,z) = 2×377×15 yjcos30 sin(βzcos30) + zsin30 cos(βzcos30) e− jβysin30

= 11,310 yj0.866 sin173.2z − z0.5 cos173.2z e− j100y V/m(3)

Similarly, for the magnetic field intensity:

H1(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 tothe surface (z direction). The latter is represented by the following components

E1(x,y,z) = yj9794.46 sin(173.2z) e− j100y V/m (5)

H1(x,y,z) = x30 cos173.2ze− j100y A/m (6)

Configuration in Figure B. It is possible to repeat the process above for this configuration as well. However, a quickinspection will show that the only change is in that the horizontal components of the incident and reflected electric fieldare in the negative z direction. All other components are in the same directions as in Figure A. Thus, it is sufficient toreverse the direction of the z component in the final results. This gives:

E1(x,y,z) = 11,310 yj0.866 sin173.2z − z0.5 cos173.2z e− j100y V/m

H1(x,y,z) = x30 cos173.2ze− j100y A/m

The pattern is as above (the z component is real). Thus:

E1(x,y,z) = yj9794.46 sin(173.2z) e− j100y V/m (7)

H1(x,y,z) = x30 cos173.2ze− j100y A/m (8)

b. The amplitude of the electric field intensity is zero wherever sin(173.2z) = 0 and is maximum whereversin(173.2z) = ±1. Thus, the condition for peaks in the electric field intensity is:

173.2z = mπ2

m = 1,3,5,7...n (odd) → z = mπ246.4

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. Thusthe condition for peak in the magnetic field intensity is

173.2z= mπ m=0,1,2,....n → z = mπ173.2

m=0,1,2,3,.. (10)

The peaks are:Epeak = ± 9794.46 V/m Hpeak = ± 30 A/m (11)

c. To calculate the total time averaged power density we write from the total fields in (5) and (6)

PPPPav(x,z) = 12

Re E1(x,z) ×H1*(x,z) = 1

2−11,310 yj0.866 sin173.2z − z0.5 cos173.2z e−j100y × x30 cos173.2ze j100y

= y84,825cos2(173.2z) W/m2

307

x z

y

Ei

Er

Hr

Hi⊗

θ

α

⊗

⊗

iz

y

Er

Hi

x

Hr

θiEi

Figure A Figure B

Problem 13.17. Propagation of waves in the presence of a conducting surface.We can either use the general expressions for theelectric and magnetic field intensities in Eqs.(13.90) and (13.91) and extract the phase constantin both directions required, or we could rely on ageometric discussion to identify the two velocities.The latter is more fundamental.

a. Consider Figure A in which a wavefrontpropagates at an angle towards a conducting interface.During the time point A travels to A’, the point Btravels to A’ vertically. If the phase velocity of pointA perpendicular to the front is vp, then,

x

z

A

B

CA'

θ

..

.

.vpx

p

pzv

v

90°−

θ

wavefront

Figure A

vp

vpx = cos(90 − θi) =sinθi → vpx =

vp

sinθi

= csinθi

ms

Similarly, point C travels horizontally from C to A’ in the same time point A travels to A’. Thus:

vp

vpz = cosθi → vpz =

vp

cosθi

= ccosθi

ms

Specifically for θi = 30°:

vpx = csin30°

= 3×108

0.5 = 6×108 vpz = c

cos30° = 3×108

0.866 = 3.464×108 m

s

Note that both velocities are larger than the speed of light, c.

b. For a wave parallel to the surface, θi = 90°. Thus:

vpx = csin90°

= 3×108

1 = 3×108 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 thephase 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 frommeasurement of d1 and d3). With the transmission angle known, the thickness d may be calculated in terms of the othermeasurements. Snell's law gives the relation between the incidence and transmission angles:

sinθt

sinθi

= ε1µ1

ε2µ2 = ε0µ0

4ε0µ0

= 12

→ sinθi = 2sinθt (1)

Now we can write the angles in terms of the geometrical distances:

sinθi = d1/2

d12/4 + d3

2 = d1

2 d12/4 + d3

2 (2)

308

sinθt = d4/2

d42/4 + d2

= d4

2 d42/4 + d2

= d2

2 d22/4 + d2

(3)

Now, substituting these in Eq. (1):

d1

2 d12/4 + d3

2 = 2d2

2 d22/4 + d2

Expansion of the terms leads to:

d2 = d22 + 4d2

2d32

d12

− d22

4 = 4d2

2d32

d12

+ 3d22

4 = d2

2 4d32

d12

+ 34

or:

d = d2

2d1

16d32 + 3d1

2 = d2

2d1

0.0016 + 3d12 m

d4

d5

θt

θt

d1/2iθiθ

d3 d

d1

d2

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 currentdensity may be viewed as producing the reflected magnetic field intensity and also is responsible for the fact that themagnetic field intensity is zero inside the perfect conductor. That is, we can view the total magnetic field to the left of theconductor as being the sum of the incident tangential component and an additional component due to the surface currentdensity so that in the conductor the magnetic field is zero. Figure A shows the conventional approach. Figure Bshows 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 reflectedmagnetic field intensity is

Hrt = − Hicosθi = − Ei

η0cosθi z < 0 A/m (3)

where z = 0 is the location of the interface. In vector form:

Hrt = − x Ei

η0 cosθi z < 0 A

m

Instead of looking at the problem as one of incident and reflected waves, we may look at it as one of an incident waveeverywhere in space (by removing the conducting surface) and a current sheet at the location of the surface (z = 0) whichproduces the reflected wave. Under this condition, shown in Figure B, there is a “reflected” wave to the right of theinterface as well:

Hrt = Hicosθi = Ei

η0cosθi z > 0 or: H rt = x Ei

η0 cosθi z > 0 A

m

Applying Ampere’s law to the two components of the reflected wave in Figure B, we get:

2HrtL = 2 Ei

η0 Lcosθi = JL → J = 2 Ei

η0 cosθi A

m (5)

And, from the right hand rule, the current density must be in the positive y direction:

J = y 2Eicosθi

η0 = y 2×10cos30°

377 = y 0.0459 A

m (6)

b. Now the magnetic field intensity is parallel to the interface as shown in Figure C. The incident (tangential)magnetic field intensity is therefore:

Hit = y Ei

η0 z < 0 A

m (7)

Viewing Figure C from above (see Figure D), looking onto the y-z plane we can use again the relation in (a). Nowwe can write directly from Eq. (6), but the current density must be in the x direction to produce the reflected magneticfield intensity required:

J = x 2Ei

η0 = x 2×10

377 = x 0.053 A

m

309

x

zyEi

Hi

ε0,µ0

pi

Ht

30°

30°

H

rt

it

H

Js H

rt

it

Hx

zyL

Figure A Figure B

x

zy

Ei

Hi

ε0,µ0

pi30°

Figure C

H

r

i

H

Js H

r

i

H

xz

y

L

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 magneticfield intensities in air and in the dielectric. For this purpose we write the incident electric and magnetic field intensitiesfrom the data given and then proceed to calculate the reflected and transmitted wave. After we sum the incident andreflected 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 incidentelectric field intensity is:

Ei= xEi1e−jβ0(ysinθi + zcosθi) V/m

The magnetic field intensity is (so that propagation is towards the interface):

Hi = Ei1

η0 (ycosθi − zsinθi)e−jβ0(ysinθi + zcosθi) A

m

The assumed reflected waves are:

Er= xΓ⊥Ei1e−jβ0(ysinθi − zcosθi) V/m

Hr = Γ⊥Ei1

η0 (− ycosθi − zsinθi)e −jβ0(ysinθi − zcosθi) 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) =

xEi1 e−jβ0zcosθi) + Γ⊥e jβ0zcosθi e−jβ0ysinθi V/m

H1= Ei1

η0(ycosθi − zsinθi)e −jβ0(ysinθi + zcosθi) + Γ⊥(− ycosθi − zsinθi)e−jβ0(ysinθi − zcosθi) =

y Ei1cosθi

η0e−jβ0zcosθi− Γ⊥e jβ0zcosθi e−jβ0ysinθi − z Ei1sinθi

η0e−jβ0zcosθi + Γ⊥e jβ0zcosθi e−jβ0ysinθi A

m

The time averaged power density is therefore

PPPP av = Re E1×H1*

2 = Re

Ex×Hy*

2 + E x×Hz

*

2 W

m2

We will evaluate each part of the power separately for simplicity:

Ex×Hy*

2 = x Ei1

2e−jβ0zcosθi + Γ⊥e jβ0zcosθi e−jβ0ysinθi×y Ei1cosθi

η0ejβ0zcosθi− Γ⊥e −jβ0zcosθi ejβ0ysinθi

= z Ei12 cosθi

2η0

1 − Γ⊥2 + Γ⊥ ej2β0zcosθi − e−j2β0zcosθi = z Ei1

2 cosθi

2η0

1 − Γ⊥2 + Γ⊥j2sin(2β0zcosθi) W

m2

310

Ex×Hz*

2 = x Ei1

2e−jβ0zcosθi + Γ⊥e jβ0zcosθi e−jβ0ysinθi×z Ei1sinθi

η0−e jβ0zcosθi− Γ⊥e −jβ0zcosθi ejβ0ysinθi

= y Ei12 sinθi

2η0

1 + Γ⊥2 + Γ⊥ ej2β0zcosθi + e−j2β0zcosθi = y Ei1

2 sinθi

2η0

1 + Γ⊥2 + Γ⊥2cos(2β0zcosθi) W

m2

The total power density is the sum of the two power densities

PPPP av = Re yEi12 cosθi

2η0

1 + Γ⊥2 + Γ⊥2cos(2β0zcosθi) + z Ei1

2 sinθi

2η0

1 − Γ⊥2 + Γ⊥j2sin(2β0zcosθi) =

yEi12 cosθi

2η0

1 + Γ⊥2 + Γ⊥2cos(2β0zcosθi) + z Ei1

2 sinθi

2η0

1 − Γ⊥2 W

m2

In these relations the following are also necessary for numerical evaluation:

Γ⊥ = η2cosθi − η0cosθt

η2cosθi + η0cosθt

sinθi

sinθt

= εr2µr2 → θ t = sin−1 sinθi

εr2µr2 η0 = 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 transmissioncoefficient can be calculated. Then, the transmitted waves are:

Et= xEi1T⊥e−jβ2(ysinθt + zcosθt) V/m

Ht = Ei1T⊥

η2 (ycosθt − zsinθt)e −jβ2(ysinθt + zcosθt) A

m

The time averaged power density is then:

PPPP av = Re Et×Ht*

2 = 1

2xEi1T⊥e −jβ2(ysinθt + zcosθt) × Ei1T⊥

η2 (ycosθt − zsinθt)e jβ2(ysinθt + zcosθt)

= Ei12 T⊥

2

2η2

zcosθt + ysinθt Wm2

where the transmission coefficient is calculated as:

T⊥ = 2η2cosθi


and the intrinsic impedances were given in (a).

c. In air, the power has both real and an imaginaryparts. This means that there is both a propagating anda standing wave term for any two different materialsmeeting at the interface. In the dielectric there can beonly a propagating wave (real power density).

x z

y

Ei

Er

Hr

Hi

⊗

θ

⊗

⊗

i

z=0

θtθi⊗Et

Ht

θi

θi

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 transmissionangle 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

Ey

= 105

= 2 → θi = tan−12 = 63.43° = 63°26'

The transmission angle is found from Snell’s law:

sinθt = n1

n2 sinθi = εr1

εr2 sin (63.43°) = 2

3 0.89443 = 0.7303

or: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-zplane. Therefore this is parallel polarization.

311

c. The reflection and transmission coefficients for parallel polarization are given in Eqs. (13.124) and (13.125):

Γ|| = η2cosθt − η1cosθi

η2cosθt + η1cosθi

= µ0/ε2 cosθt − µ0/ε1 cosθi

µ0/ε2 cosθt + µ0/ε1 cosθi

= 1/ 3cos(46.9°) − 1/ 2cos(63.43°)

1/ 3cos(46.9°) + 1/ 2cos(63.43°) = 0.11

T|| = 2η2cosθi


= 2 µ0/ε2 cosθi

µ0/ε2 cosθt + µ0/ε1 cosθi

= 2/ 3cos(63.43°)

1/ 3cos(46°9) + 1/ 2cos(63.43°) = 0.726

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 wavesat the interface. However, we must be careful since the incident electric field intensity is given in terms of itscomponents and the reflection coefficient was calculated in terms of the amplitude of the incident field. Thus, we firstcalculate the amplitude of the incident field:

Ei(x = 0) = x10 + y 5 → Ei = 102 + 52 = 11.18 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:

Er = ΓEi = 0.11×11.18 = 1.23 V/mThe reflected wave is:

Er(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


, η1=η0=377 Ω

For high loss materials (Eq. (13.116)):

η2 = jωµ2

σ2 = (1 + j) ωµ2

2σ2

Ω

Thus:

T⊥ = 2(1+j) ωµ2

2σ2

cosθi

(1+j) ωµ2

2σ2

cosθi + η0cosθt

Let:

a = ωµ2

2σ2

cosθi b = η0cosθt (1)

T⊥ = 2a(1+j)

(1+j)a + b = 2a+j2a

(a + b) + ja (a + b) − ja

(a + b) − ja = 2a(a+b) − j2a2 + j2a(a+b) + 2a 2

(a + b)2+ a 2 = 4a2 + 2ab + j2ab

(a + b)2+ a 2

The phase angle of the transmission coefficient is therefore:

φT⊥ = tan−1 2ab4a2 + 2ab

= tan−1 b2a + b

312

Substituting for a and b from (1):

φT⊥ = tan−1 η0cosθt

2 ωµ2

2σ2

cosθi + η0cosθt

b. Yes, there is a phase angle associated with the reflection coefficient:



= (1+j) ωµ2

2σ2

cosθi − η0cosθt

(1+j) ωµ2

2σ2

cosθi + η0cosθt

Using the notation for a and b in (a):

Γ⊥ = (1+j)a − b

(1+j)a + b = (a − b) + ja

(a + b) + ja.(a + b) − ja

(a + b) − ja = 2a2 − b2 + j2ab

(a + b)2+ a 2

the angle is:

φΓT = tan−1 2ab 2a2 − b2

= 2η0

ωµ2

2σ2

cosθicosθt

2ωµ2

2σ2

cos2θi − η02cos2θt

Problem 13.23. Phase velocity and its dependence on incidence angle.To see how the waves propagate and their variousvelocities, it is possible either too use the incident,reflected and total fields and extract the phasevelocities form these or to use a geometricalrepresentation of the plane wave. From this we canextract the phase velocity from simple geometricalratios (see Figure A and also Problem 13.19).

a. Consider Figure A in which a wavefrontpropagates at an angle towards a conducting interface.During the time point A travels along to A’, point Btravels to A’ vertically. If the phase velocity of pointA perpendicular to the front is vp, then

x

z

A

B

CA'.

...vpx

p

pzv

v

60°

wavefront

30°

free space dielectric

Figure A

vp

vpx = cos(90 − θi) =sinθi → vpx =

vp

sinθi

= csinθi

ms

Similarly, point C travels horizontally from C to A’ in the same time point A travels to A’. Thus:

vp

vpz = cosθi → vpz =

vp

cosθi

= ccosθi

ms

Specifically for θi = 30°:

vpx = csin30°

= 3×108

0.5 = 6×108 vpz = c

cos30° = 3×108

0.866 = 3.464×108 m

s

Note that both velocities are larger than c.

b. For a wave parallel to the surface, θi = 90°. Thus:

vpx = csin90°

= 3×108

1 = 3×108 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 thephase 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 hereis a dielectric and in Problem 13.19 it is a perfect conductor. There is a significant difference in the magnitudes ofthe fields to the left of the interface but the phase velocities parallel and perpendicular to the interface only depend onthe material to the left of the interface and the angle of incidence.

313

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 tothe surface for all incidence angles if it is perpendicularly polarized. If the electric field intensity has both a parallel andnormal component, the polarization is parallel. Thus, we calculate and plot the reflection coefficient for perpendicular andfor parallel polarization as a function of incidence angle.1. For perpendicular polarization (from Eq. (13.109)):



(1)

From Snell’s law, we re-write the transmission angle as a function of the incidence angle:

sinθt

sinθi

= ε1µ1

ε2µ2 = ε0µ0

2.1ε0µ0

= 12.1

→ sinθ t = sinθi

2.1(2)

However, we need cosθt rather than sinθt:

sinθt = sinθi

2.1 → cosθt = 1 − sin2θt = 1 − sin2θi

2.1 = 1

2.12.1 − sin2θi (3)

The intrinsic impedances η1 and η2 are:η1 = 377 η2 = 377/ 2.1 Ω (4)

Substituting these into the reflection coefficient we get (for θi = θ):

Γ⊥ = 377/ 2.1 cosθ − 377/ 2.1 2.1 − sin2θ

377/ 2.1 cosθ + 377/ 2.1 2.1 − sin2θ = cosθ − 2.1 − sin2θ

cosθ + 2.1 − sin2θ (5)

This is plotted in Figure A for 0 < θi < π/2.2. For parallel polarization (from Eq. (13.124)):

Γ|| = η2cosθt − η1cosθη2cosθt + η1cosθ

(6)

With the results in Eqs. (3) and (4) we get:

Γ|| = sinθt − cosθsinθt + cosθ

=

3772.1

2.1 − sin2θ2.1

− 377cosθ

3772.1

2.1 − sin2θ2.1

+ 377cosθ

-0.22

-0.20

-0.18

-0.16

Γ

Γ||

Γ⊥

0.00 0.20 0.40θ i [rad]

Figure A

or: Γ|| = 2.1 − sin2θ − 2.1cosθ

2.1 − sin2θ + 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:



T|| = 2η2cosθi


Setting θi = θt = 0: Γ|| = η2 − η1

η2 + η1 T|| = 2η2

η2 + η1

These are identical to those for perpendicular incidence and are now independent of polarization. This is because when thewave is perpendicularly incident to a surface the electric field has only a tangential component with either polarization.For parallel incidence θi = 90°:

Γ|| = η2cosθt − η1(0)

η2cosθt + η1(0) = 1 T|| = 2η2(0)

η2cosθt + η1(0) = 0

There is no transmission but total reflection.

314

Problem 13.26. Brewster angle in dielectrics.Using the general relation for Brewster angle for two dielectrics with the same permeability:

θb = sin−1 ε2

ε1 + ε2

a. θb = sin−1 81ε0

81 + 1 ε0

= sin−1 8182

= 83.66° = 83°39'

b. θb = sin−1 4ε0

4 + 1 ε0

= sin−1 45

= 63.43° = 63°26'

c. θb = sin−1 2.25ε0

2.25 + 1 ε0

= sin−1 2.253.25

= 56.39° = 56°18'

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 propagatesin free space and is incident on a perfect dielectric. Therefore we can write for the Brewster angle θb:

sinθb = ε2

ε0 + ε2 → θb = sin−1 ε2

ε0 + ε2 (1)

because µ1 = µ2 = µ0 and ε2 = ε0. Using the first form in Eq. (1):

sin2θb = ε2

ε0 + ε2 → sin2θb ε0 + ε2 = ε2 → ε2 = ε0sin2θb

1 −sin2θb

For the given angle:

ε2 = ε0sin2θb

1 −sin2θb

= ε0 sin262°

1 − sin262° = 3.537ε0 F

m

The permittivity of the dielectric is 3.537ε0.

Problem 13.28. Critical angles in dielectrics. Critical angles in dielectrics.Use the general formula for critical angle between two materials:

a. θc = sin−1 µ2ε2

µ1ε1 = sin−1 µ0ε0

µ0×81×ε0

= sin−1 181

= 6.38° = 6°23'

b. θc = sin−1 µ0×1.75×ε0

µ0×4×ε0

= sin−1 1.754

= 41.41° = 41°25'

c. θc = sin−1 µ0×ε0

µ0×2.25×ε0

= sin−1 12.25

= 41.81° = 41°49'

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, thepermittivity of the material on which the wave impinges can be calculated.

θc = sin−1 µ2ε2

µ1ε1 → sinθc = µ0ε0

µ0ε1

or: sin2θc = ε0

ε1 = 1

εr1 → εr1 = 1

sin2θc

= 1sin236

= 2.894

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 reflectioncoefficients for parallel polarization are used to determine the coefficients. The critical angle is computed from thematerial properties since the critical angle does not depend on the polarization of the wave.

315

a. From Snell's law (Eq. (13.99):

sinθt

sinθi

= 41

= 2 → sinθ t = 2sinθ i → θ t = sin−1 2sinθi

This gives:θt = 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) :



T|| = 2η2cosθi


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 − 188.5cos28377cos69.87 + 188.5cos28

= 0.6883 − 0.882950.6883 + 0.88295

= − 0.12388

T|| = 2×377cos28377cos69.78 + 188.5cos28

= 2.2478

Note: for parallel polarization, the following holds (see exercise 13.12(a)):

1 + Γ||= T||cosθt

cosθi

0≤θi<π/2

which is satisfied for the values above. That is:

1 + Γ|| = 1 − 0.12388 = 0.876 = T||cosθt

cosθi

= 2.2478 cos69.87cos28

= 0.876

b. The critical angle is found from Eq. (13.142) and is not dependent on polarization.

sin θc = µ2ε2

µ1ε1 = 1

4 = 1

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 criticalangle and the material cannot serve as a shield. Thus the plastic with relative permittivity εr = 2 must be used.

b. The critical angle at the interface between glass and coating is:

θc = sin−1 εr2

εr1 = sin−1 2.0

2.25 = sin−1(0.9428) = 70.528° = 70°32'

The critical angle is 70°32'. For any angle larger than this, there will be no penetration of light into the shield. Forθ < θc, there is penetration into the shield.

c. Now we allow propagation in the shield as well. At the interface between shield (εr = 2) and air (εr = 1), the criticalangle is:

θc = sin−1 εr2

εr1 = sin−1 1.0

2.0 = sin−1(0.7071) = 45°

For any angle below this, there will be propagation in air (waves will leak out of the fiber and shield). Thus, tosummarize:

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 constitutelosses).

316

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 thevarious amplitudes (Er0, E2

+, E2− and E3

+) are given in Eqs. (13.156) through (13.159) but here we need to replace thegeneral propagation constant by γ2 = jβ2. With these we get the amplitudes as

Er0 = Γ12 + Γ23e −j2β2d

1 + Γ12Γ23e −j2β2d Ei0 E2

− = T12Γ23e −j2β2d

1 + Γ12Γ23e −j2β2d Ei0 V

m

E2+ = T12

1 + Γ12Γ23e −j2β2d Ei0 E3

+ = T12T23e −jβ2de jβ3d

1 + Γ12Γ23e −j2β2dEi0 V

m

The directions of propagation of these fields is shown in Figure A. In addition we note that:

Γ12 = η2 − η0

η2 + η0, Γ23 = η0 − η2

η0 + η2, T12 = 2η2

η2 + η0, T23 = 2η0

η2 + η0

From this we have Γ23 = − Γ12, T12 = 1 + Γ12 and T23 = 1 + Γ23 = 1 − Γ12. We also have γ1 = γ3 = jβ0.With these we can write:

Er0 = Γ12 1 − e−j2β2d

1 − Γ122 e−j2β2d

Ei0 Vm

E2− =

1 + Γ12 − Γ12 e−j2β2d

1 − Γ122 e−j2β2d

Ei0 = − Γ12 + Γ12

2 e−j2β2d

1 − Γ122 e−j2β2d

Ei0 Vm

E2+ = 1 + Γ12

1 − Γ122 e−j2β2d

Ei0 Vm

E3+ =

1 + Γ12 1 − Γ12 e−jβ2de jβ3d

1 − Γ122 e−j2β2d

Ei0 = 1 − Γ12

2 e−jβ2de jβ3d

1 − Γ122 e−j2β2d

Ei0 Vm

The total fields are as follows

E1 = x Ei0e−jβ0z + Er0e jβ0z = xEi0 e−jβ0z + Γ12 1 − e−j2β2d

1 − Γ122 e−j2β2d

ejβ0z Vm

E2 = x E2+e−jβ2z + E2

−ejβ2z = xEi01 + Γ12 e−jβ2z − Γ12 + Γ12

2 e−j2β2de jβ2z

1 − Γ122 e−j2β2d

Vm

E3 = xE3+e−jβ0z = xEi0

1 − Γ122 e−jβ2de jβ0d

1 − Γ122 e−j2β2d

e−jβ0z Vm

The slab reflection and transmission coefficients are:

Γslab = Er0

Ei0

= Γ12 1 − e−j2β2d

1 − Γ122 e−j2β2d

, Tslab = E3+

Ei0

= 1 − Γ12

2 e−jβ2de jβ0d

1 − Γ122 e−j2β2d

Er1

Ei 1 Ei 2

Er2

Ei 3

Ep

21 3

dε ,µ ,σ2 2 2ε0,µ0 ε0,µ0

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 onthe 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 intrinsicimpedance of free space. Thus:

317

f = 109 Hz , η1=η3=η0 = 377 Ω , η2 = η0

εr

= 3772

= 188.5 Ω

b. The various components of the electric field intensity are as follows (see Eqs. (13.156) through (13.159) byreplacing γ1 = γ3 = jβ0, γ2 = jβ2 or by using the result of Problem 13.32)

Er0 = Γ12 + Γ23e −j2β2d

1 + Γ12Γ23e −j2β2d Ei0, E2

− = T12Γ23e −j2β2d

1 + Γ12Γ23e −j2β2d Ei0 V

m

E2+ = T12

1 + Γ12Γ23e −j2β2d Ei0, E3

+ = T12T23e −jβ2de jβ0d

1 + Γ12Γ23e −j2β2dEi0 V

m

where

Γ12 = η2 − η1

η2 + η1 = 188.5 − 377

188.5 + 377 = − 1

3, Γ23 = η3 − η2

η3 + η2 = 377 − 188.5

377 + 188.5 = 1

3

T12 = 2×188.5

377 + 188.5 = 2

3, T23 = 2×377

377 + 188.5 = 4

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 V/m z < 0

E2 = E2+e−jβ2z + E2

−ejβ2z V/m 0 < z < dE3 = E3

+e−jβ0z V/m z > d

Substituting the values above gives

E1 = Ei0 e−jβ0z + Γ12 + Γ23e −j2β2d

1 + Γ12Γ23e −j2β2dejβ0z V

m z < 0

E2 = Ei0T12

1 + Γ12Γ23e −j2β2de−jβ2z + T12Γ23e −j2β2d

1 + Γ12Γ23e −j2β2dejβ2z = Ei0

T12e −jβ2z + T12Γ23e −j2β2de jβ2z

1 + Γ12Γ23e −j2β2d V/m 0 < z < d

E3 = Ei0T12T23e −jβ2de jβ0d

1 + Γ12Γ23e −j2β2de −jβ0z V

m z > d

The numerical values of the phase constants in free space and in the dielectric are:

β0 = 2πf c

= 2π×109

3×108 = 20π

3 rad

m, β2 = εrβ0 = 40π

3 rad

m

Substitution of these values gives the electric field intensities in the three materials. These now only depend on theincident electric field intensity Ei0, on the thickness d and, of course, on the position z.

E1 = Ei0 e−j20πz/3 + −1/3 + 1/3e −j80πd/3

1 − (1/9)e−j80πd/3ej20πz/3 = Ei0 e−j20πz/3 + 3 −1 + e−j80πd/3

9 − e−j80πd/3ej20πz/3 V

m z < 0

E2 = Ei0(2/3)e−j40πz/3 + (2/3)(1/3)e−j80πd/3e j40πz/3

1 − (1/9)e−j80πd/3 = Ei0

6e−j40πz/3 + 2e−j80πd/3e j40πz/3

9 − e−j80πd/3 V

m 0 < z < d

E3 = Ei0(2/3)(4/3)e−j40πd/3e j20πd/3

1 − (1/9)e−j80πd/3e−j20πz/3 = Ei0

8e−j20πd/3

9 − e−j80πd/3e−j20πz/3 V

m z > d

To calculate wave impedance we need to calculate the magnetic field intensities. These are calculated from the followingrelations:

H1 = 1η1

Ei0e−jβ0z − Er0e jβ0z = Ei0

η1e−jβ0z − Γ12 + Γ23e −j2β2d

1 + Γ12Γ23e −j2β2dejβ0z A

m z < 0

H2 = 1η2

E2+e−jβ2z − E2

−ejβ2z = Ei0

η2 T12e −jβ2z − T12Γ23e −j2β2de jβ2z

1 + Γ12Γ23e −j2β2d A

m 0 < z < d

H3 = 1η3

E3+e−jβ0z = Ei0

η3 T12T23e −jβ2de jβ0d

1 + Γ12Γ23e −j2β2de −jβ0z A

m z > d

With the numerical values above, we have:

318

H1 = Ei0

377e−j20πz/3 − − (1/3) + (1/3)e−j80πd/3

1 − (1/9)e−j80πd/3ej20πz/3 = Ei0

377e−j20πz/3 − 3− 1 + e−j80πd/3

9 − e−j80πd/3ej20πz/3 A

m z < 0

H2 = Ei0

188.5 (2/3)e−j40πz/3 − (2/3)(1/3)e −j80πd/3e j40πz/3

1 − (1/9)e−j80πd/3 = Ei0

188.5 6e−j40πz/3 − 2e −j80πd/3e j40πz/3

9 − e−j80πd/3 A

m 0 < z < d

H3 = Ei0

377 (2/3)(4/3)e−j40πd/3e j20πd/3

1 − (1/9)e−j80πd/3e−j20πz/3 = Ei0

377 8e−j20πd/3

9 − e−j80πd/3e−j20πz/3 A

m z > d

c. By substituting the values foe z and Ei0 = 1 V/m, we get:in material (1), at z = 0:

E1(z=0) = 1 1 + 3 −1 + e−j80π0.01/3

9 − e−j80π0.01/3 = 0.8581 - j0.255 V

m

H1(z=0) = 1377

1 − 3− 1 + e−j0.80π/3

9 − e−j0.80π/3 = 3.03×10−3 + j6.76×10−4 A

mIn material (3), at z = d = 0.01 m:

E3(z=0.01m) = 1 8e−j0.20π/3

9 − e−j0.80π/3 = 0.914 − j0.281 V

m

H3(z=0.01m) = 1377

8e−j0.20π/3

9 − e−j0.80π/3e−j0.20π/3 = E3

377 = 0.914 − j0.281

377 = 2.42×10−3 − j7.46×10−4 A

m

In material (2), at z = d/2 = 0.005 m:

E2(z=d/2) = 1 6e−j40π0.005/3 + 2e −j80π0.01/3e j40π0.005/3

9 − e−j80π0.01/3 = 0.866 − j0.368 V

m

H2(z=d/2) = 1188.5

6e−j40π0.005/3 − 2e −j80π0.01/3e j40π0.005/3

9 − e−j80π0.01/3 = 2.68×10−3 − j2.85×10−4 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 = Γ12 1 − e−j2β2d

1 − Γ122 e−j2β2d

For the reflection coefficient to be zero, we must have:

1 − e−j2β2d = 0 → e −j2β2d = cos2β2d − jsin2β2d = 1 This gives:

cos2β2d = 1 → 2β2d = 2nπ n = 0,1,2... or:

d = nπβ2

= nπω µ2ε2

= nλ2

m

In this case, the wavelength is:

λ = c2f

= 3×108

2×109 = 0.15 m

Thus:d = nπ

β2

= 0.075n m

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, thecondition of no reflection is:

Γslab = Γ12 − e−2γ2d

1 − Γ12Γ23e −2γ2d = 0

First, it is convenient to rewrite this as follows:

319

Γslab = e−γ2d Γ12e γ2d − e−γ2d

e−γ2d eγ2d − Γ12Γ23e −γ2d = Γ12e γ2d − e−γ2d

eγ2d − Γ12Γ23e −γ2d = 0

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 = Γ12 cosh γ2d + sinh γ2d − cosh γ2d + sinh γ2d = 0 The reflection coefficient between the dielectric and free space is

Γ12 = η2 − η0

η2 + η0

With this we haveη2 − η0 cosh γ2d + sinh γ2d − η2 + η0 cosh γ2d + sinh γ2d

= − 2η0cosh γ2d − 2η2sinh γ2d = 0

Now we can write the condition for no reflection as

2η0 + 2η2tanh γ2d = 0 → d = tanh−1 η0

η2

γ 2 m

If the dielectric is lossless, we have:

η0 + η2tanh jβ2d = η0 + jη2tan β2d = 0 → d = tan−1 η0e−jπ/2

η2

β2

m

and, in either case there is no solution since that would imply a complex value for d.

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 reflectioncoefficient for a dielectric slab in free space is:


1 − Γ122 e−j2β2d

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−j2β2d = 0 → e −j2β2d = cos2β2d − jsin2β2d = 1

This gives:cos2β2d = 1 → 2β2d = 2nπ n = 0,1,2...

or:

d = nπβ2

= nπω µ2ε2

= nλ2

m

and the minimum thickness is

d = λ2

m

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 veryuseful in protecting the antenna. With the given properties, the wavelength is:

λ = vp

f = 1

f µ2ε2

→ d = λ2

= 12f µ2εr2

= c2f εr2

= 3×108

2×1010 4 = 0.015

2 = 0.0075 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 settingthe reflection coefficient of the dielectric slab in free space to zero (Eq. (13.164)). This gives:


1 − Γ122 e−j2β2d

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:

320

1 − e−j2β2d = 0 → e −j2β2d = cos2β2d − jsin2β2d = 1 This gives:

cos2β2d = 1 → 2β2d = 2nπ n = 0,1,2... or:

d = nπβ2

= nπω µ2ε2

= nλ2

m

where n = 1,2,3,..... Taking n = 1:

d = π2πf µ1ε1

= 12f 3.8µ0ε0

= c2f 3.8

= 3×108

2×2.45×109 3.8 = 0.0314 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 characteristicimpedance of free space while material (2) is now complex because of the loss term). The wave impedance in a planewave is always the ratio between the electric and magnetic field intensities and these vary depending on the reflectionsthat may exist.

a. The intrinsic impedance in each material is that of the dielectric, regardless of any interfaces that may exist. Inmaterials (1) and (3) this is the intrinsic impedance of free space. Thus:

f = 109 Hz , η1 = η3 = η0 = 377 Ω

η2 = jωµ0

jωε2 + σ = j2π×109×4π×10−7

j2π×109×4×8.854×10−12 + 0.001 = 188.37 + j0.42 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))

Er0 = Γ12 + Γ23e −2γ2d

1 + Γ12Γ23e −2γ2d Ei0 V

m

E2− = T12Γ23e −2γ2d

1 + Γ12Γ23e −2γ2d Ei0 E2

+ = T12

1 + Γ12Γ23e −2γ2d Ei0 V

m

E3+ = T12T23e −γ2de γ3d

1 + Γ12Γ23e −2γ2dEi0 V

m

whereγ0 = jβ0, γ2 =α2 + jβ0, γ3 = jβ0

and:

Γ12 = η2 − η1

η2 + η1 = 188.37 + j0.42 − 377

188.37 + j0.42 + 377 = − 0.336 + j0.01 ≈ − 1

3

Γ23 = η3 − η2

η3 + η2 = 377 − 188.37 − j0.42

377 + 188.37 + j0.42 = 0.3336 − j0.01 ≈ 1

3

T12 = 2η2

η3 + η2 =

2× 188.37 + j0.42)

377 + 188.37 + j0.42 = 0.6664 + j0.001 ≈ 2

3

T23 = 2η3

η3 + η2 = 2×377

377 + 188.37 + j0.42 = 0.6664 − j0.001 ≈ 4

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 < z < d E3 = E3

+e−jβ0z V/m z > dSubstituting the values above gives

E1 = Ei0e−jβ0z + Er0e jβ0z = Ei0 e−jβ0z + Γ12 + Γ23e −2(α2+jβ2)d

1 + Γ12Γ23e −2(α2+jβ2)de+jβ0z V

m z < 0

E2 = E2+e−jβ0z + E2

−ejβ0z = Ei0T12e −(α2+jβ2)z + T12Γ23e −2(α2+jβ2)de (α2+jβ2)z

1 + Γ12Γ23e −2(α2+jβ2)d V

m 0 < z < d

321

E3 = E3+e−jβ0z = Ei0 T12T23e − α2 + jβ2 de jβ0d

1 + Γ12Γ23e −2 α2 + jβ2 de−jβ0z 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:

σω ε

= 0.0012×π×109×4×8.854×10-12

= 0.00049 << 1

The parameters are therefore:

f = 109 Hz , η1 = η3 = η0 = 377 Ω , η2 ≈ 188.5 Ω

β0 = 2πf c

= 2π×109

3×108 = 2π

3 = 20.94 rad

m, β2 = εrβ0 = 4π

3 = 41.89 rad

m

α2 ≈ σ2ηn = 0.001

2×188.5 = 0.098 Np

m

The magnetic field intensities are calculated from the following relations (Eqs. 13.146, 13.148 and 13.150):

H1 = 1η0

Ei0e−jβ0z − Er0e jβ0z Am

H2 = E2+

η2e−(α2+jβ2)z − E2

−

η2e(α2+jβ2)z A

m

H3 = 1η0

E3+e−jβ0z A

m

In terms of the electric fields above:

H1 = 1η1

Ei0e−jβ0z − Er0e jβ0z = Ei0

η0e−jβ0z − Γ12 + Γ23e −2(α2+jβ2)d

1 + Γ12Γ23e −2(α2+jβ2)dejβ0z A

m z < 0

H2 = 1η2

E2+e−(α2+jβ2)z − E2

−e (α2+jβ2)z = Ei0

η2 T12e −(α2+jβ2)z − T12Γ23e−2(α2+jβ2)de (α2+jβ2)z

1 + Γ12Γ23e −2(α2+jβ2)d A

m 0 < z < d

H3 = 1η3

E3+e−jβ0z = Ei0

η0 T12T23e −(α2+jβ2)dejβ0d

1 + Γ12Γ23e −2(α2+jβ2)de −jβ0z A

m z > d

Substitution of these values gives the electric field intensities in the three materials. These now only depend on theincident 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,

E1 = 1 1 + − (1/3) + (1/3)e−2(0.098+j40π/3)0.01

1 − (1/9)e−2(0.098+j40π/3)0.01 = 0.864 − j0.254 V

m

H1 = 1377

1 − − (1/3) + (1/3)e−2(0.098+j40π/3)0.01

1 − (1/9)e−2(0.098+j40π/3)0.01 = 3.0×10−3 + j6.72×10−4 A

m

In material (3) at z = 0.01 m,

E3 = 1 (2/3)(4/3)e− 0.098 + j40π/3 0.01e j0.2π/3

1 − (1/9)e−2(0.098+j40π/3)0.01e−j0.2π/3 = 0.887 + j0.32 V

m

H3 = 1377

(2/3)(4/3)e− 0.098 + j40π/3 0.01e j0.2π/3

1 − (1/9)e−2(0.098+j40π/3)0.01e−j0.2π/3 = 2.35×10−3 + j8.5×10−4 A

mIn material (3) at z = 0.005 m,

E2 = 1 (2/3)e−(0.098+j40π/3)0.005 + (2/3)(1/3)e −2(0.098+j40π/3)0.01e (0.098+j40π/3)0.005

1 − (1/9)e−2(0.098+j40π/3)0.01 = 0.679 − j0.197 V

m

H2 = 1188.5

(2/3)e−(0.098+j40π/3)0.005 − (2/3)(1/3)e −2(0.098+j40π/3)0.01e (0.098+j40π/3)0.005

1 − (1/9)e−2(0.098+j40π/3)0.01

= 3.61×10−3 − j1.04×10−3 Am

Problem 13.39. Propagation through a two layer slab.

322

We write the forward and backward electric and magnetic fields in each section using the amplitudes of the electric fieldintensity as the unknown values. In material (1), only the reflected wave is unknown and in material (4), there is only aforward propagating wave. This will result in a system of six equations in six unknowns which is solved to find theamplitudes 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 ofthe 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

η0e−jβ0z − Er0

η0ejβ0z 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 (seesection). Denoting the forward and backward propagating electric field amplitudes as E2

+ and E2−, we can write:

E 2 = x E2+e−jβ2z + E2

−ejβ2z Vm

H 2 = y E2+

η2e−jβ2z − E2

−

η2ejβ2z A

m

In material (3) we again have:

E 3 = x E3+e−jβ3z + E3

−ejβ3z Vm

H 3 = y E3+

η3e−jβ3z − E3

−

η3ejβ3z A

m

In material (4), there is only a forward propagating wave and we get:

E 4 = x E4+e−jβ0z V

m H 4 = y E4

+

η0e−jβ0z 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 electricfields and, separately, the magnetic fields at the three interfaces. Doing so gives:

At z = 0:

Ei0 + Er0 = E2+ + E2

− E i0

η0 − Er0

η0 = E2

+

η2 − E2

−

η2

At z = d1:

E2+e−jβ2d1 + E2

−ejβ2d1 = E3+e−jβ3d1 + E3

−ejβ3d1 E 2+

η2e−jβ2d1 − E2

−

η2ejβ2d1 = E3

+


−

η3ejβ3d1

At z = d2:

E3+e−jβ3d2 + E3

−ejβ3d2 = E4+e−jβ0d2 E 3

+


−

η3ejβ3d2 = E4

+

η0e−jβ0d2

Before substituting the actual values and solving it is worth rewriting the six equations as follows:

− Er0 + E2+ + E2

− = Ei0 = E0 (1)Er0

η0 + E2

+

η2 − E2

−

η2 = E0

η0 (2)

E2+e−jβ2d1 + E2

−ejβ2d1 − E3+e−jβ3d1 − E3

−ejβ3d1 = 0 (3)E2

+


−

η2ejβ2d1 − E3

+

η3e−jβ3d1 + E3

−

η3ejβ3d1 = 0 (4)

E3+e−jβ3d2 + E3

−ejβ3d2 − E4+e−jβ0d2 = 0 (5)

E3+


−

η3ejβ3d2 − E4

+

η0e−jβ0d2 = 0 (6)

Writing this as a system of equations:

323

−1 1 1 0 0 0

1η0

1η2

− 1η2

0 0 0

0 e−jβ2d1 ejβ2d1 − e−jβ3d1 − e jβ3d1 0

0 e−jβ2d1

η2 − e

jβ2d1

η2 − e

−jβ3d1

η3

ejβ3d1

η30

0 0 0 e−jβ3d2 ejβ3d2 − e−jβ0d2

0 0 0 e−jβ3d2

η3 − e

jβ3d2

η3 − e

−jβ0d2

η0

Er0

E2+

E2−

E3+

E3−

E4+

=

E0

E0

η0

0

0

0

0

Although this can be solved in general terms (but it is complicated and very tedious) it is a relatively easy matter to solvethe equations for particular values of β0, β2, β3, η0, η2, η3, d1, d2, and E0. These are as follows:

In free space: β0 = ω µ0ε0 = 2×π×150×106

3×108 = π rad

m

In material (2): β2 = ω µ0×2.25ε0 = 2×π×150×106×1.5

3×108 = 1.5π rad

m

In material (3): β3 = ω µ0×4ε0 = 2×π×150×106×2

3×108 = 2π rad

m

η0 = 377 η2 = 3772.25

= 251.34 η3 = 3774

= 188.5 Ω

With d1 = 0.1 m, d2 = 0.2 m, we get:

−1 1 1 0 0 0

1377

1251.34

− 1251.34

0 0 0

0 e−j0.15π ej0.15π − e−j0.2π − e j0.2π 0

0 e−j0.15π

251.34− ej0.15π

251.34− e

−j0.2π

188.5ej0.2π

188.50

0 0 0 e−j0.4π ej0.4π − e−j0.2π

0 0 0 e−j0.4π

188.5 − ej0.4π

188.5 − e

−j0.2π

377

Er0

E2+

E2−

E3+

E3−

E4+

=

1

1377

0

0

0

0

Expanding the exponents using Euler's equation as follows:

ej0.15π = 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

ej0.2π = 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 ej0.4π = 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:

Er0 = − 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:

324

E 1 = x 1e−jπz + ( − 0.46693 − j 0.12261)e jπz V/m , z < 0

H 1 = y 1377

1e−jπz + (0.46693 + j 0.12261)e jπz Am

, z < 0

E 2 = x (0.75552 − j 0.020435)e−j1.5πz − (0.22246 + j 0.10218)e j1.5πz V/m , 0 < z < 0.1 m

H 2 = y 1251.34

(0.75552 − j 0.020435)e−j1.5πz + (0.22246 + j 0.10218)e j1.5πz Am

, 0 < z < 0.1 m

E 3 = x (0.65441 + j 0.055284)e−j2πz − (0.16564 + j 0.14313)e j2πz V/m , 0.1 m < z < 0.2 m

H 3 = y 1188.5

(0.65441 + j 0.055284)e−j2πz + (0.16564 + j 0.14313)e j2πz Am

, 0.1 m < z < 0.2 m

E 4 = x (0.74918 − j 0.45302)e−jπz V/m , z > 0.2 m

H 4 = y 1377

(0.74918 − j 0.45302)e−jπz Am

, 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 = Er0

E0

= − 0.46693 − j 0.12261

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 theincident electric field:

Tslab = E4+

E0

= 0.74918 − j 0.45302

This has a magnitude |T| = 0.8755.

Er0

Ei 0 E

E2

Ep

21 3d

ε ,µ ,σε0,µ0

E

E

E

4d

ε ,µ ,σ ε0,µ02.25 0 0 0 0=0 =0

1 2

2+

−3

3 4+

−

4

z=0 z= z= 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 andmagnetic field intensities entering the material. The Poynting vector is calculated. Since the attenuation constant is veryhigh, all power is dissipated within a short distance. Thus, the Poynting vector gives the total power dissipated per unitarea. In air, the incident electric and magnetic field intensities are:

E = xE0cos(ωt − βz) Vm

and: H = yE0

η0 cos(ωt − βz) 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) πfµσ

= (1 + j) π×108×4π×10−7×50

10 = (1 + j)44.43 = 62.833∠45° Ω

The high loss in the material also means that the material can be assumed to be infinitely thick (no field reaches theopposite surface). Thus, only reflection from the interface needs to be considered.The reflection and transmission coefficients at the interface are:

Γ1 = η1 − η0

η0 + η1 and: T1 = 2η1

η0 + η1

In particular, the transmission coefficient is:

325

T1 = 2η1

η0 + η1 = 2×(1 + j)44.43

377 + (1 + j)44.43 = 0.23 + j0.18655 = 0.296∠39°

The electric and magnetic field intensities in material 1, immediately next to the interface is equal to

E1 = xE0T1 Vm

, H1 = yE0

η1T1 A

m

Now, we can calculate the magnitude of the time averaged Poynting vector as

PPPPav = 12

E1×H1* Wm2

This is the time averaged power density entering material (1) per unit area of the interface. The phasor forms of thetransmission coefficient and the intrinsic impedance are:

T1 = 0.296e j39°, η1 = 62.833e j45°

Now we can write:

PPPPav = E1×H1*

2 = 1

2xE0T1 e−jβ1z × y E0

η1∗T1* e jβ1z = 1

2xE0T1 e−jβ1z × y E0

η1∗T * e jβ1z = 1

2zE0

2T1T1*η1∗

Wm2

Since T1T1* = T 2 we can write:

PPPP av = zE0

2 T12

2η1∗

= z (100)2(0.296)2

2×62.833e−j45° = 6.972e j45° = 6.972∠45° W

m2

The real part of this is:

Pav = 6.972cos45° = 4.93 Wm2

Since only real power can be dissipated. Thus for each m2 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) aftersubstituting γ2 = jβ2 and Γ23 = − 1. For the reflection coefficient to be maximum, the thickness must be such thatthe magnitude of the reflection coefficient is 1.

Γslab = Γ12 − e−j2β2d

1 − Γ12e −j2β2d

a. For the given quantities:

η0 = µ0

ε0 = 377 η2 = µ0

5ε0

= 3775

= 168.6 Ω

β2 = ω 5ε0µ0 = 2π×900×106 5×8.854×10−12×4π×10−7 = 42.178 radm

With these, Γ12 is

Γ12 = 377 − 168.6377 + 168.6

= − 0.382

and the reflection coefficient is:

Γslab = − 0.382 − e−j84.356

1 + 0.382e−j84.356 = − 0.382 − cos(0.8436rad) + jsin(0.8436rad)

1 + 0.382cos(0.8436rad) − j0.382sin(0.8436rad) = − 1.0468 + j0.74704

1.254 − j0.28537

= − 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:

326

− 1 = Γ12 − e−j2β2d

1 − Γ12e −j2β2d → Γ12 − e−j2β2d = − 1 + Γ12e −j2β2d

Or:Γ12 + 1 = − e−j2β2d = Γ12e −j2β2d + e−j2β2d = e−j2β2d Γ12 + 1 → Γ12 + 1 = Γ12 + 1 e−j2β2d

Or:

e−j2β2d = cos2β2d − jsin2β2d = 1 → cos2β2d = 1 → 2β2d = n2π

The required thickness is then:

d = nπβ2

= nπ2πf 5ε0µ0

= n

2×900×106 5×8.854×10−12×4π×10−7 = 0.0745n, n = 0,1,2,...

Thus, the minimum thickness is zero (no dielectric). The next solution is for n = 1 and the thickness d should be0.0745 m (74.5 mm).

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