Eletromagnetismo - Hayt - 7ª Ed - Soluções - chapter13 7th solution

Eletromagnetismo - Hayt - 7ª Ed - Soluções - chapter13 7th solution

(Parte 1 de 2)


13.1. A uniform plane wave in air, E+ copper surface at z =0 . What percentage of the incident power density is transmitted into the copper? We need to find the reflection coefficient. The intrinsic impedance of copper (a good conductor) is

Note that the accuracy here is questionable, since we know the conductivity to only two significant figures. We nevertheless proceed: Using η0 = 376.7288 ohms, we write

find a) ω:W e have β = ω√

5= 169 ohms. Next we apply Eq. (76), Chapter 12, to evaluate the Poynting vector (with no loss and consequently with no phase difference between electric and magnetic fields). We find c) < S−1 >: First, we need to evaluate the reflection coefficient:

13.3. A uniform plane wave in region 1 is normally-incident on the planar boundary separating the energy in the incident wave is reflected at the boundary. There are two possible answers.

13.3. (continued) Therefore

13.4. A 10-MHz uniform plane wave having an initial average power density of 5W/m2 is normally- and µ2 = µ0. Calculate the distance into the lossy medium at which the transmitted wave power density is down by 10dB from the initial 5W/m2:

The reflection coefficient encountered by the incident wave from region 1 is therefore

The fraction of the incident power that is reflected is then |Γ|2 =0 .147, and thus the fraction of the power that is transmitted into region 2 is 1 −| Γ|2 =0 .853. Still using the good dielectric approximation, the attenuation coefficient in region 2 is found from Eq. (60a), Chapter 12:

Now, the power that propagates into region 2 is expressed in terms of the incident power through

in which the last equality indicates a factor of ten reduction from the incident power, as occurs for a 10 dB loss. Solve for z to obtain

13.5. The region z< 0i sc haracterized by ′r = µr =1 and ′′r =0 . The total E field here is given as the sum of the two uniform plane waves, Es = 150e−j10z ax + (50 20◦)ej10z ax V/m.

b) Specify the intrinsic impedance of the region z> 0 that would provide the appropriate reflected wave: Use

Γ= Er


13.5 (continued) Now

( 1+Γ

c) At what value of z (−10cm <z < 0) is the total electric field intensity a maximum amplitude? We found the phase of the reflection coefficient to be φ =2 0◦ = .349rad, and we use

a) α1:A s ′′1 =0 , there is no loss mechanism that is modeled (see Eq. (4), Chapter 12), and so α1 =0 .

b) β1: Since region 1 is lossless, the phase constant for the uniform plane wave will be

c) < S+1 >:T o find the power density, we need the intrinsic impedance of region 1, given by

Then the incident power density will be

d) < S−1 >:T o find the reflected power, we need the intrinsic impedance of region 2. This is found using Eq. (48), Chapter 12:

Then the reflection coefficient at the 1-2 boundary is

13.6e) < S+2 >:W e first need the attenuation coefficient in region 2. This is given by Eq. (4) in Chapter 12, which in our case becomes


and ′′r =0 .A uniform plane wave with ω =4 × 108 rad/s is travelling in the az direction toward the interface at z =0 .

a) Find the standing wave ratio in each of the three regions: First we find the phase constant in the middle region,

Then, with the middle layer thickness of 1 m, β2d =2 .67rad. Also, the intrinsic impedance of the middle layer is η2 = η0/√ ′r = η0/2. We now find the input impedance:

Now, at the first interface,

The standing wave ratio measured in region 1 is thus

In region 2 the standing wave ratio is found by considering the reflection coefficient for waves incident from region 2 on the second interface:


Finally, s3 =1 , since no reflected waves exist in region 3. 4

13.7b. Find the location of the maximum |E| for z< 0 that is nearest to z =0 .W e note that the phase of Γ12 is φ = 123◦ =2 .15 rad. Thus

13.8. A wave starts at point a, propagates 100m through a lossy dielectric for which α =0 .5 Np/m, reflects at normal incidence at a boundary at which Γ = 0.3+ j0.4, and then returns to point a. Calculate the ratio of the final power to the incident power after this round trip: Final power, Pf, and incident power, Pi, are related through

Try measuring that.

wavelengths in the two regions are λ1 =5 cm and λ2 =3 cm. What percentage of the energy incident on the boundary is a) reflected; We first note that

c) What is the standing wave ratio in region 1? Use

a) the amplitude of E−1 is one-half that of E+1 : Since region 2 is free space, the reflection coefficient is

13.1. A 150 MHz uniform plane wave in normally-incident from air onto a material whose intrinsic impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of an electric field minimum at 0.3 wavelengths in front of the interface. Determine the impedance of the unknown material: First, the field minimum is used to find the phase of the reflection coefficient, where

where β =2 π/λ has been used. Next,

So we now have

13.12. A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean.

For seawater, σ =4 S/m, and ′r = 78. a) Determine the fractions of the incident power that are reflected and transmitted: First we find the loss tangent:

This value is sufficiently greater than 1 to enable seawater to be considered a good conductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 1), the intrinsic impedance is ηs = √ πfµ/σ(1 + j), and the reflection coefficient becomesΓ=

reflected is

The transmitted fraction is then

b) Qualitatively, how will these answers change (if at all) as the frequency is increased?

Within the limits of our good conductor approximation (loss tangent greater than about ten), the reflected power fraction, using the formula derived in part a,i s found to decrease with increasing frequency. The transmitted power fraction thus increases.

13.13. A right-circularly-polarized plane wave is normally incident from air onto a semi-infinite slab of plexiglas ( ′r =3 .45, ′′r = 0). Calculate the fractions of the incident power that are reflected and transmitted. Also, describe the polarizations of the reflected and transmitted waves. First,

The reflected power fraction is thus |Γ|2 =0 .09. The total electric field in the plane of the interface must rotate in the same direction as the incident field, in order to continually satisfy the boundary condition of tangential electric field continuity across the interface. Therefore, the reflected wave will have to be left circularly polarized in order to make this happen. The transmitted power fraction is now 1 −| Γ|2 =0 .91. The transmitted field will be right circularly polarized (as the incident field) for the same reasons.

13.14. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect conductor. a) Construct the superposition of the incident and reflected waves in phasor form: Assume positive z travel for the incident electric field. Then, with reflection coefficient, Γ = −1, the incident and reflected fields will add to give the total field:

b) Determine the real instantaneous form of the result of part a:

E(z, t)=R e{

c) Describe the wave that is formed: This is a standing wave exhibiting circular polarization in time. At each location along the z axis, the field vector rotates clockwise in the xy plane, and has amplitude (constant with time) given by 2E0 sin(βz).

the boundary at z =0 will have no reflection? This frequency gives the condition β2d = π,

The reflection coefficient is now

The standing wave ratio is now

13.16. A uniform plane wave in air is normally-incident onto a lossless dielectric plate of thickness λ/8, and of intrinsic impedance η = 260 Ω. Determine the standing wave ratio in front of the plate. Also find the fraction of the incident power that is transmitted to the other side of the plate: With the a thickness of λ/8, we have βd = π/4, and so cos(βd)= sin(βd)=1 √ 2. The input impedance thus becomes

The reflection coefficient is then


13.17. Repeat Problem 13.16 for the cases in which the frequency is a) doubled: If this is true, then d = λ/4, and thus ηin = (260)2/377 = 179. The reflection coefficient becomes

b) quadrupled: Now, d = λ/2, and so we have a half-wave section surrounded by air. Transmission will be total, and so s =1 and 1 −| Γ|2 =1 .

13.18. A uniform plane wave is normally-incident onto a slab of glass (n =1 .45) whose back surface is in contact with a perfect conductor. Determine the reflective phase shift at the front surface of the glass if the glass thickness is: (a) λ/2; (b) λ/4; (c) λ/8.

With region 3 being a perfect conductor, η3 =0 , and Eq. (36) gives the input impedance to the structure as ηin = jη2 tanβ . The reflection coefficient is then

=Γ r + jΓi

where the last equality occurs by multiplying the numerator and denominator of the middle term by the complex conjugate of its denominator. The reflective phase is now

13.19. You are given four slabs of lossless dielectric, all with the same intrinsic impedance, η, known to be different from that of free space. The thickness of each slab is λ/4, where λ is the wavelength as measured in the slab material. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally-incident. The slabs are to be arranged such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and air spaces such that a) the wave is totally transmitted through the stack: In this case, we look for a combination of half-wave sections. Let the inter-slab distances be d1, d2, and d3 (from left to right). Two possibilities are i.) d1 = d2 = d3 =0 ,t hus creating a single section of thickness λ, or i.) d1 = d3 =0 ,d 2 = λ/2,t hus yielding two half-wave sections separated by a halfwavelength.

b) the stack presents the highest reflectivity to the incident wave: The best choice here is to make d1 = d2 = d3 = λ/4.T huse very thickness is one-quarter wavelength. The impedances transform as follows: First, the input impedance at the front surface of the last slab (slab 4) is ηin,1 = η2/η0.W e transform this back to the back surface of slab 3, moving through a distance of λ/4i n free space: ηin,2 = η20/ηin,1 = η30/η2.W e next transform this impedance to the front surface of slab 3, producing ηin,3 = η2/ηin,2 = η4/η30.W e continue in this manner until reaching the front surface of slab 1, where we find ηin,7 = η8/η70.

Assuming η< η0, the ratio ηn/ηn−10 becomes smaller as n increases (as the number of slabs increases). The reflection coefficient for waves incident on the front slab thus gets close to unity, and approaches 1 as the number of slabs approaches infinity.

13.20. The 50MHz plane wave of Problem 13.12 is incident onto the ocean surface at an angle to the normal of 60◦. Determine the fractions of the incident power that are reflected and transmitted for a) s polarization: To review Problem 12, we first we find the loss tangent:

This value is sufficiently greater than 1 to enable seawater to be considered a good con- ductor at 50MHz. Then, using the approximation (Eq. 65, Chapter 1), and with µ = µ0,

refraction, which means that we need to know the refractive index of seawater at 50MHz. Fora uniform plane wave in a good conductor, the phase constant is

β = nsea ωc

Then, using Snell’s law, the angle of refraction is found:

This angle is small enough so that cosθ2 .=1 . Therefore, for s polarization,

b) p polarization: Again, with the refracted angle close to zero, the relection coefficient for p olarization is

13.21. A right-circularly polarized plane wavein air is incident at Brewster’s angle onto a semi-infinite

the angle of refraction is θ2 =9 0◦−θB (see Example 13.9), or θ2 =2 8.3◦. With incidence at Brewster’s angle, all p-polarized power will be transmitted — only s-polarized power will be reflected. This is found through

nent represents one-half the total incident wave power, and so the fraction of the total power that is reflected is .302/2=0 .15, or 15%. The fraction of the incident power that is transmitted is then the remainder, or 85%.

b) Describe the polarizations of the reflected and transmitted waves: Since all the p-polarized component is transmitted, the reflected wave will be entirely s-polarized (linear). The transmitted wave, while having all the incident p-polarized power, will have a reduced s-component, and so this wave will be right-elliptically polarized.

13.2. A dielectric waveguide is shown in Fig. 13.16 with refractive indices as labeled. Incident light enters the guide at angle φ from the front surface normal as shown. Once inside, the light totally reflects at the upper n1 − n2 interface, where n1 >n 2. All subsequent reflections from the upper an lower boundaries will be total as well, and so the light is confined to the guide.

Express, in terms of n1 and n2, the maximum value of φ such that total confinement will occur, with n0 =1 . The quantity sinφ is known as the numerical aperture of the guide.

From the illustration we see that φ1 maximizes when θ1 is at its minimum value. This minimum will be the critical angle for the n1 − n2 interface, where sinθc = sinθ1 = n2/n1. Let the refracted angle to the right of the vertical interface (not shown) be φ2, where n0 sinφ1 =

is the numerical aperture angle.

13.23. Suppose that φ1 in Fig. 13.16 is Brewster’s angle, and that θ1 is the critical angle. Find n0 in terms of n1 and n2: With the incoming ray at Brewster’s angle, the refracted angle of this ray (measured from the inside normal to the front surface) will be 90◦ − φ1. Therefore, φ1 = θ1, and thus sinφ1 = sinθ1.T hus

Alternatively, we could have used the result of Problem 13.2, in which it was found that

13.24. A Brewster prism is designed to pass p-polarized light without any reflective loss. The prism of Fig. 13.17 is made of glass (n =1 .45), and is in air. Considering the light path shown, determine the apex angle, α: With entrance and exit rays at Brewster’s angle (to eliminate reflective loss), the interior ray must be horizontal, or parallel to the bottom surface of the prism. From the geometry, the angle between the interior ray and the normal to the prism surfaces that it intersects is α/2. Since this angle is also Brewster’s angle, we may write:

13.25. In the Brewster prism of Fig. 13.17, determine for s-polarized light the fraction of the incident power that is transmitted through the prism: We use Γs =( ηs2 − ηs1)/(ηs2 + ηs1), where


Thus, at the first interface, Γ = (1−n2)/(1+n2). At the second interface, Γ will be equal but of opposite sign to the above value. The power transmission coefficient through each interface is 1 −| Γ|2,s o that for both interfaces, we have, with n =1 .45:



13.26. Show how a single block of glass can be used to turn a p-polarized beam of iight through 180◦, with the light suffering, in principle, zero reflective loss. The light is incident from air, and the returning beam (also in air) may be displaced sideways from the incident beam. Specify all pertinent angles and use n =1 .45 for glass. More than one design is possible here.

The prism below is designed such that light enters at Brewster’s angle, and once inside, is turned around using total reflection. Using the result of Example 13.9, we find that with glass, θB =5 5.4◦, which, by the geometry, is also the incident angle for total reflection at the back of the prism. For this to work, the Brewster angle must be greater than or equal to the

(Parte 1 de 2)