### Chapter #11 Solutions - Engineering Electromagnetics and Waves - Aziz Inan, Ryan Said, Umran S Inan - 2nd Edition

1. GPS signal transmission through the ionosphere. The Global Positioning System (GPS) uses frequencies of 1.228 and 1.575 GHz. The highest electron density in the earth’s ionosphere occurs near the so-called F-region (at altitudes of ∼300 km) of the ionosphere, reaching values of Ne = 107 cm−3 during periods of high solar activity. By modeling the ionosphere as a single layer of 50-km thickness with uniform electron density Ne, assess the effect of the ionosphere on the GPS signal. What fraction of a GPS signal vertically incident from space onto the ionospheric layer is reflected? Get solution

2. Space vehicle reentry. The intense friction around a space vehicle reentering the atmosphere generates a plasma sheath which is 1 meter thick and is characterized by an electron density of Ne = 1013 cm−3, and a collision frequency of ν = 1011 s−1. What frequency is required in order to transmit plane waves through the plasma sheath with a minimum power loss of 10 dB? Get solution

3. Reflection from the ionosphere. Consider the propagation of waves by reflection from the ionosphere as shown in Figure (a) On a particular date and time of day, and above a particular geographic location, the highest electron density occurs at an altitude of ∼400 km, and the plasma frequency at this altitude is fp ≃ 5 MHz. Neglecting any ionization below this densest layer (i.e., assume electron density to be zero below 400 km altitude), determine the greatest possible angle of incidence θi at which an electromagnetic wave originating at a ground-based transmitter could possibly strike the layer. (b) How far away from the transmitter will the reflected radiation return to earth? (c) What is the highest frequency at which this obliquely incident radiation will be totally reflected?Figure Reflection from the ionosphere. Problem.... Get solution

4. Attenuation rate in the ionosphere. Show that the effective permittivity of a plasma with collisions can be expressed as...where ν is the collision frequency and ωp is the plasma frequency. Using this expression to find a simple approximation for the attenuation rate (in dB-m−1) for a 30 MHz wave passing through the lower ionosphere (assume ν ≃ 107 s−1, Ne ≃ 108 m−3). Get solution

5. Lossy LH Media. (a) In a lossy LH medium with Re{є}0 and Re{μ}0, show that Im{β2} = Im{ω2μє}0. (b) In a lossy RH medium with Re{є}0 and Re{μ}0, show that Im{β2} = Im{ω2μ_}0. Get solution

6. Lossy LH Media. In a lossy LH medium with Re{_}0 and Re{μ}0, use the result from part(a) of Problem to show that Re{β}Im{β}0.Lossy LH Media. (a) In a lossy LH medium with Re{є}0 and Re{μ}0, show that Im{β2} = Im{ω2μє} 0. (b) In a lossy RH medium with Re{є}0 and Re{μ}0, show that Im{β2} = Im{ω2μє}0 Get solution

7. Lorentz model of a metamaterial. In Section 11.2.1, we developed a model for the relative permittivity by considering the dynamics of a bound electron in the presence of an external electric field. The formula for єr given by (11.28) is called the Lorentz model. Using (11.5), we can rewrite (11.28a) as...where we added the subscript e to associate the characteristic frequencies with the bound electron model. A similar model may be used to approximate the permeability of artificially constructed metamaterials:28...(a) First consider the lossless case, where κe = κm = 0. Using ωpe = 2π × 8.0× 109 rad-s−1, ωpm = 2π × 7.0 × 109 rad-s−1, ω0e = 2π × 2.8 × 109 rad-s−1, and ω0m = 2π × 2.5 × 109 rad-s−1: (i) Plot the real and imaginary components of єr and μr in the frequency range 4 f11 GHz. (ii) Plot the real and imaginary components of the index of refraction n in the same frequency range. Use the convention that Re{β} 0 for backward propagation. (iii) In the frequency range 4 f 11 GHz, which frequencies allow for wave propagation? What is the frequency range of backward wave propagation? At what frequency is Re{n} = −1? (iv) Plot βc versus f over the domain 4 f 11 GHz, and comment on the sign of the phase and group velocities over frequencies where propagating solutions exist. (b) Repeat part (a) using κe = κm = 0.05ωpe . Get solution

### Chapter #10 Solutions - Engineering Electromagnetics and Waves - Aziz Inan, Ryan Said, Umran S Inan - 2nd Edition

1. Parallel-plate waveguide modes. A parallel-plate waveguide consists of two perfectly conducting infinite plates spaced 2 cm apart in air. Find the propagation constant γ for the TM0, TE1, TM1, TE2, and TM2 modes at an operating frequency of (a) 5 GHz, (b) 10 GHz, and (c) 20 GHz. Get solution

2. Parallel-plate waveguide modes. A parallel-plate air waveguide has a plate separation of 6 mm and width of 10 cm. (a) List the cutoff frequencies of the seven lowest-order modes (TEm and TMm) that can propagate in this guide. (b) Find all the propagating modes (TEm and TMm) at 40 GHz. (c) Find all the propagating modes at 60 GHz. (d) Repeat part (c) if the waveguide is filled with polyethylene (assume it is lossless, with ∈r ≃ 2.25, μr = 1). Get solution

3. Parallel-plate waveguide modes. An air-filled parallel-plate waveguide with a plate separation of 1 cm is to be used to connect a 25-GHz microwave transmitter to an antenna. (a) Find all the propagating modes. (b) Repeat part (a) if the waveguide is filled with polyethylene (assume it is lossless, with ∈r ≃ 2.25, μr = 1). Get solution

4. VLF propagation in the Earth–ionosphere waveguide. The height of the terrestrial earth– ionosphere waveguide considered in Example varies for VLF (3–30 kHz) from 70 km during the day to about 90 km during the night.39 (a) Find the total number of propagating modes during the day at 12 kHz. (Assume both the ionosphere and the earth to be flat and perfect conductors.) (b) Repeat part (a) for during the night. (c) Find the total number of propagation modes during the day at 24 kHz. (d) Repeat part (c) for during the night. (e) Does the TM17 mode propagate during the day at 30 kHz? (f) Does the TM17 mode propagate during the night at 30 kHz?Example: ELF propagation in the Earth–ionosphere waveguide. Extremely low frequencies (ELF) are ideal for communicating with deeply submerged submarines, because below 1 kHz, electromagnetic waves penetrate into seawater.14 Propagation at these frequencies takes place in the “waveguide” formed between the earth and the ionosphere (Figure); low propagation losses allow nearly worldwide communication from a single ELF transmitter.In J. R. Wait’s simple model,15 the surface of the earth and the bottom of the ionosphere form the boundaries of a terrestrial “parallel”-plate waveguide with lossy walls. The ionosphere is approximated by an isotropic layer beginning at a given altitude and extending to infinity with no horizontal variations allowed. Energy is lost through the “walls” either into the finitely conducting ionosphere or into the ground, with the former loss being dominant. The important feature of propagation below 1 kHz is that there is a single propagating mode, a so-called quasi-TEM mode. All the other modes are evanescent and are almost undetectable at distances in excess of 1000 km. In the far field, the wave consists of a vertical electric field and a horizontal magnetic field transverse to the direction of propagation (Figure). The leakage of energy from this wave into the ocean (see Section 9.8) gives rise to a plane wave propagating vertically downward, and it is this signal that the submarine receiver detects.Consider an idealized earth–ionosphere waveguide where both the ionosphere and the earth are assumed to be perfect conductors. In addition, neglect the curvature of the waveguide and assume it to be flat. The height of the terrestrial waveguide can vary anywhere from 70 km to 90 km depending on conditions; for our purposes, assume it to be 80 km. Find all the propagating modes at an operating frequency of (a) 100 Hz, (b) 1 kHz, and (c) 10 kHz....Figure ELF propagation and submarine reception. Get solution

5. Waveguide in the earth’s crust. It has been proposed40 that radio waves may propagate in a waveguide deep in the earth’s crust, where the basement rock has very low conductivity and is sandwiched between the conductive layers near the surface and the high-temperature conductive layer far below the surface. The upper boundary of this waveguide is on the order of 1 km to several kilometers below the earth’s surface. The depth of the dielectric layer of the waveguide (the basement rock) can vary anywhere from 2 to 20 km, with a conductivity of 10−6 to 10−11 S-m−1 and a relative dielectric constant of ~6. This waveguide may be used for communication from a shore sending station to an underwater receiving station. Consider such a waveguide and assume it to be an ideal nonmagnetic parallel-plate waveguide. (a) If the depth of the dielectric layer of the guide is 2 km, find all the propagating modes of this guide below an operating frequency of 2 kHz. (b) Repeat part (a) (same depth) below 5 kHz. (c) Repeat parts (a) and (b) for a dielectric layer depth of 20 km. Get solution

6. Single-mode propagation. Consider a parallel-plate air waveguide with plate separation a. (a) Find the maximum plate separation amax that results in single-mode propagation along the guide at 10 GHz. (b) Repeat part (a) for the waveguide filled with a dielectric with∈r ≃ 2.54, μr = 1. Get solution

7. Evanescent wave attenuator design. A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanescent wave attenuator to attenuate all the modes except the TEM mode along the guide. Find the minimum length of this attenuator needed to attenuate each mode by at least 100 dB. Assume perfect conductor plates. Get solution

8. Evanescent wave filter design. Consider a parallel-plate air waveguide to be designed such that no other mode but the TEM mode will propagate along the guide at 6 GHz. If it is required by the design constraints that the lowest-order nonpropagating mode should face a minimum attenuation of 20 dB-(cm)−1 along the guide, (a) find the maximum plate separation amax of the guide needed to satisfy this criterion. (b) Using the maximum value of a found in part (a), find the dB-(cm)−1 attenuation experienced by the TE2 and the TM2 modes. Get solution

9. Cutoff and waveguide wavelengths. Consider a parallel-plate waveguide in air with a plate separation of 3 cm to be used at 8 GHz. (a) Determine the cutoff wavelengths of the three lowest-order nonpropagating TMm modes. (b) Determine the guide wavelengths of all the propagating TMm modes. Get solution

10. Guide wavelength of an unknown mode. The waveguide wavelength of a propagating mode along an air-filled parallel-plate waveguide at 15 GHz is found to be λ = 2.5 cm. (a) Find the cutoff frequency of this mode. (b) Recalculate the guide wavelength of this same mode for a waveguide filled with polyethylene (lossless, with ∈r ≃ 2.25, μr = 1). Get solution

11. Guide wavelength, phase velocity, and wave impedance. Consider a parallel-plate air waveguide having a plate separation of 5 cm. Find the following: (a) The cutoff frequencies of the TM0, TM1, and TM2 modes. (b) The phase velocity ῡpm , waveguide wavelength λm and wave impedance ZTMm of the above modes at 8 GHz. (c) The highest-order TMm mode that can propagate in this guide at 20 GHz. Get solution

12. Parallel-plate waveguide design. Design a parallel-plate air waveguide to operate at 5 GHz such that the cutoff frequency of the TE1 mode is at least 25% less than 5 GHz, the cutoff frequency of the TE2 mode is at least 25% greater than 5 GHz, and the power-carrying capability of the guide is maximized. Get solution

13. TEM decomposition of TEm modes. For an air-filled parallel-plate waveguide with 4 cm plate separation, find the oblique incidence angle θi and sketch TE1, TE2, TE3 , and TE4 modes in terms of two TEM waves at an operating frequency of 20 GHz. Get solution

14. Unknown waveguide mode. The electric field of a particular mode in a parallel-plate air waveguide with a plate separation of 2 cm is given by...(a) What is this mode? Is it a propagating or a nonpropagating mode? (b) What is the operating frequency? (c) What is the similar highest-order mode (TEm or TMm) that can propagate in this waveguide? Get solution

15. Unknown waveguide mode. The magnetic field of a particular mode in a parallel-plate air waveguide with a plate separation of 2.5 cm is given by...where x and y are both in meters. (a) What is this mode? Is it a propagating or a nonpropagating mode? (b) What is the operating frequency? (c) Find the corresponding electric field E(x, y). (d) Find the lowest-order similar mode (TEm or TMm) that does not propagate in this waveguide at the same operating frequency. Get solution

16. Maximum power capacity. (a) In Problem. determine the value of constant C1 (assumed to be real) which maximizes the power carried by all the propagating modes without causing any dielectric breakdown (use 15 kV-m−1 for maximum allowable electric field in air, which is half of the breakdown electric field in air at sea level). (b) Using the value of C1 found in part (a), find the maximum time-average power per unit width carried by the mode found in part (a) of Problem.ProblemUnknown waveguide mode. The magnetic field of a particular mode in a parallel-plate air waveguide with a plate separation of 2.5 cm is given by...where x and y are both in meters. (a) What is this mode? Is it a propagating or a nonpropagating mode? (b) What is the operating frequency? (c) Find the corresponding electric field E(x, y). (d) Find the lowest-order similar mode (TEm or TMm) that does not propagate in this waveguide at the same operating frequency Get solution

17. Power-handling capacity of a parallel-plate waveguide. A parallel-plate air waveguide with a plate separation of 1.5 cm is operated at a frequency of 15 GHz. Determine the maximum time-average power per unit guide width in units of kW-(cm)−1 that can be carried by the TE1 mode in this guide, using a breakdown strength of air of 15 kV-(cm)−1 (safety factor of approximately 2 to 1) at sea level. Get solution

18. Power capacity of a parallel-plate waveguide. Show that the maximum power-handling capability of a TMm mode propagating in a parallel-plate waveguide without dielectric breakdown is determined only by the longitudinal component of the electric field for fcm f √ 2fcm and by the transverse component of the electric field for f >√ 2fcm. Get solution

19. Power capacity of a parallel-plate waveguide. For a parallel-plate waveguide formed of two perfectly conducting plates separated by air at an operating frequency of f = 1.5fcm , find the maximum time-average power per unit area of the waveguide that can be carried without dielectric breakdown [use 15 kV-(cm)−1 for maximum allowable electric field in air, which is half of the breakdown electric field in air at sea level] for the following modes: (a) TEM, (b) TE1, and (c) TM1. Get solution

20. Attenuation in a parallel-plate waveguide. Consider a parallel-plate waveguide with plate separation a having a lossless dielectric medium with properties ∈ and μ. (a) Find the frequency in terms of the cutoff frequency (i.e., find f /fcm ) such that the attenuation constants αc due to conductor losses of the TEM and the TEm modes are equal. (b) Find the attenuation αc for the TEM, TEm, and TMm modes at that frequency. Get solution

21. Attenuation in a parallel-plate waveguide. For a TMm mode propagating in a parallel-plate waveguide, do the following: (a) Show that the attenuation constant αc due to conductor losses for the propagating TMm mode is given by...(b) Find the frequency in terms of fcm (i.e., find f /fcm ) such that the attenuation constant αc found in part (a) is minimum. (c) Find the minimum for αc . (d) For an air-filled waveguide made of copper plates 2.5 cm apart, find αc for TEM, TE1, and TM1 modes at the frequency found in part (b). Get solution

22. Parallel-plate waveguide: phase velocity, wavelength, and attenuation. A parallel-plate waveguide is formed by two parallel brass plates (σ = 2.56 × 107 S-m−1, ∈r = 1, μr = 1) separated by a 1.6-cm thick polyethylene slab (∈′r ≃ 2.25, tan δ≃ 4 × 10−4, μr = 1) to operate at a frequency of 10 GHz. For TEM, TE1, and TM1 modes, find (a) the phase velocity ῡp and the waveguide wavelength λ and (b) the attenuation constants αc due to conductor losses and αd due to dielectric losses. Get solution

23. Losses in a parallel-plate waveguide. Consider TEM wave propagation in a parallel-plate waveguide. Although ideally the only nonzero wave components for this mode are Ex and Hy (as given by equation (10.17)), it is clear that there should be a small Ez component due to the finite (Js = n × H) current that flows in the conductors, as mentioned in Footnote 21 on page 834. Since Jz is equal and oppositely directed at the top (x = a) and the bottom (x = 0) conductors, we expect Ez to also have the same magnitude and the opposite sign at x = a and x = 0. Thus, it is reasonable to assume a linear variation between the two values of Ez as:...where K is a constant. Since we must have ∇ • E = 0 within the waveguide, we expect that the fields Ex and Hy may have to be modified to be consistent with this nonzero Ez . Find the modified expressions for Ex and Hy and determine all components of the time-average Poynting flux Sav at x = a/4 inside the waveguide. Get solution

24. Semi-infinite parallel-plate waveguide. Two perfectly conducting and infinitesimally thin sheets in air form a semi-infinite parallel-plate waveguide, with mouth in the plane z = 0 and sides parallel to the y-z plane as shown in Figure. Two perpendicularly polarized (i.e., electric field in the y direction) uniform plane waves 1 and 2 of equal strength are incident upon the mouth of the guide at angles θi as shown. The two waves are in phase, so their separate surfaces of constant phase intersect in lines lying in the y-z plane. The peak electric field strength for each wave is known to be 1 V-m−1. (a) Find an expression (in terms of θi) for the time-average power flowing down the inside of the guide, per unit meter of the guide in the y direction. (b) If the wavelength λ of the incident waves is such that sin θi = λ/(2a) = √ 3/2 and a = 1 cm, find the numerical value of the time-average power transmitted, per unit guide width in the y direction....Figure Semi-infinite parallel-plate waveguide. Problem. Get solution

25. Reflection and transmission in a parallel plate waveguide. The region z >0 in a parallelplate waveguide is filled with nonmagnetic dielectric material with permittivity ∈2 = 3∈0, as shown in Figure. (a) Assuming that the TM1 wave is incident from the left with an incident magnetic field intensity given by...determine the magnitude of the magnetic fields of the reflected and transmitted TM1 waves. (b) A fellow engineer claims that the TM1 wave can be completely transmitted (i.e., without any reflection) across the interface, as long as ∈2 has a specific value. Comment on whether the engineer is correct and if so, determine the value of ∈2 for which there is no reflection....Figure Parallel-plate waveguide with a dielectric. Problem. Get solution

26. Propagating modes in a dielectric slab waveguide. Consider a dielectric slab waveguide with thickness d and refractive indices of 1.5 (for the guide) and 1.48 (for the cladding). (a) Find all the propagating modes at air wavelength λ = 2μm if d = 5μm. (b) Repeat part (a) if d = 15μm. (c) Repeat part (a) if the refractive index of the cladding is 1.49. Get solution

27. Single TE1 mode dielectric slab design. Consider a dielectric slab waveguide with guide thickness d and refractive index 1.55 sandwiched between two cladding regions each with refractive index 1.5. (a) Design the guide such that only the TE1 mode is guided at air wavelength λ = 1μm. (b) Repeat part (a) for a cladding region with a refractive index of 1.53. Get solution

28. Millimeter-wave dielectric slab waveguide. Consider a dielectric slab waveguide with ∈d = 4∈0 and μd = μ0 surrounded by air to be used to guide millimeter waves. Find the guide thickness d such that only the TE1 mode can propagate at frequencies up to 300 GHz. Get solution

29. TE2 mode in a dielectric slab waveguide. Find the cutoff frequency for the TE2 mode in a dielectric slab waveguide with ∈ = 7∈0 and thickness 2 cm, which is embedded in another dielectric with ∈ = 3∈0. Assume nonmagnetic case. Get solution

30. Dielectric slab waveguide modes. A dielectric slab waveguide in air is used to guide electromagnetic energy along its axis. Assume that the slab is 2 cm in thickness, with ∈ = 5∈0 and μ = μ0. (a) Find all of the propagating modes for an operating frequency of 4 GHz and specify their cutoff frequencies. (b) Find αx (in np-m−1) and βx (in rad-m−1) at 8 GHz for each of the propagating modes. (c) Considering a ray theory analysis of the propagating modes, find the incidence angles θi of the component TEM waves within the slab for all of the propagating modes at 8 GHz. Get solution

31. Dielectric slab waveguide modes. Consider a dielectric slab waveguide surrounded by air made of a dielectric core material of d = 10 μm thickness with a refractive index of nd = 1.5 covered with a cladding material of refractive index nc = 1.45, which is assumed to be of infinite extent. (a) Find all the propagating modes at 1 μm air wavelength. (b) Determine the shortest wavelength allowed in the single-mode transmission. Get solution

32. Dielectric waveguide. Consider a nonmagnetic dielectric slab waveguide consisting of a slab with relative permittivity ∈1r = 2.19 surrounded by another dielectric with relative permittivity ∈2r = 2.13. The frequency of operation is 50 GHz. Determine the thickness of the dielectric slab if the lowest order TE mode propagates at a ray angle of θi = 85°. Get solution

33. Dielectric slab waveguide thickness. A dielectric slab waveguide is made of a dielectric core and cladding materials with refractive indices of nd = 1.5 and nc = 1, respectively. If the number of propagating modes is 100 at a free-space wavelength of 500 nm, calculate the thickness of the core material. Get solution

34. TEM decomposition of TMm modes. A dielectric slab waveguide is designed using a core dielectric material of refractive index nd= 3 and thickness 5 μm covered by a cladding dielectric material of refractive index nc = 2.5. Find all the propagating TMm modes and corresponding angles of incidence with which they are bouncing back and forth between the two boundaries at 3 μm air wavelength. Get solution

35. Dielectric above a ground plane. A planar perfect conductor of infinite dimensions is coated with a dielectric material (∈r = 5, μr = 1) of thickness 5.625 cm. (a) Find the cutoff frequencies of the first four TE and/or TM modes, and specify whether they are odd or even. (b) For an operating frequency of 1 GHz, find all of the propagating TE modes. (c) For each of the TE modes found in part (b), find the corresponding propagation constant β. Assume the medium above the coating to be free space. Get solution

36. Group velocity. Derive the expression (equation (10.73)) for the group velocity for Tem or TMm modes in a parallel-plate waveguide. Get solution

37. Group velocity. Consider the propagation in seawater (σ = 4 S/m, ∈ = 81∈0, μ = μ0) of a uniform plane wave signal consisting of the superposition two frequency components at 19 and 21 kHz. (a) Stating all assumptions, determine the phase and group velocities at 20 kHz. (b) Assuming that the two signals are in phase at position z0 = 0 and at time t0 = 0, determine the minimum distance z1 (and the time t1) at which the two signals are once again in phase, that is, have a phase difference of a multiple of 2π. Is the group velocity vg at 20 kHz equal to z1/t1? If not, why not? Get solution

38. Dielectric waveguide. Consider the TEm mode in a nonmagnetic dielectric slab waveguide consisting of a slab with thickness d and permittivity ∈d surrounded by air. (a) Show that the dispersion relation (i.e., β-ω relation) is given by...(b) For ∈d = 4∈0 and for the TE1 mode, find the value of d such that the waveguide phase velocity ῡp = ω/β is equal to the geometric mean of c = (μ0∈0)−1/2 and vpd= (μd∈d ) −1/2. Get solution

39. Dispersion in seawater. Reconsider Problem in view of your knowledge of group velocity. Assuming that the frequency dependence of the phase velocity is as given in Problem. derive an expression for the group velocity in seawater, and plot both vp and vg as a function of frequency between 0.5 and 2.5 kHz.ProblemDispersion in sea water. A uniform plane electromagnetic wave in free space propagates with the speed of light, namely, c ≃ 3 × 108 m-s−1. In a conducting medium, however, the velocity of propagation of a uniform plane wave depends on the signal frequency, leading to the “dispersion” of a signal consisting of a band of frequencies. (a) For sea water (σ = 4 S-m−1, ∈r = 81, and μr = 1), show that for frequencies much less than ~890 MHz, the velocity of propagation is approximately given by vp ≃ k1 √ f, where k1 is a constant. What is the value of k1? (b) Consider two different frequency components of a signal, one at 1 kHz, the other at 2 kHz. If these two signals propagate in the same direction in seawater and are in phase at z = 0, what is the phase delay (in degrees) between them (e.g., between their peak values) at a position 100 m away? Get solution

40. Group velocity in a plasma. A cold ionized gas consisting of equal numbers of electrons and protons behaves (see Section 11.1) as a medium with an effective permittivity ∈eff = ∈0(1 − ω2 p/ω2), where ωp = ... is known as the plasma frequency, with N being the volume density of free electrons, qe ≃ −1.6 × 10−19 C the electronic charge, and me ≃ 9.11 × 10−31 kg the electronic mass. (a) Derive the expression for the group velocity vg in this medium. (b) Evaluate the group velocity (and express it as a fraction of the speed of light in free space) for a 1 MHz radio signal propagating through the earth’s ionosphere, where N ≃ 1011 m−3. (c) Repeat part (b) for 100 kHz. Get solution

41. Group velocity in a dielectric slab waveguide. Derive an expression for the group velocity vg for the odd TMm modes in a dielectric slab waveguide of slab thickness d and permittivity ∈d. Get solution

### Chapter #9 Solutions - Engineering Electromagnetics and Waves - Aziz Inan, Ryan Said, Umran S Inan - 2nd Edition

1. Air–perfect conductor interface. A uniform plane electromagnetic wave traveling in air with its electric field given by...is normally incident on a perfect conductor boundary located at x = 0. (a) Find the phase constant β. (b) Find the corresponding magnetic field Hi(x, t). (c) Find the electric and magnetic fields Ēr(x, t) and Hr(x, t) of the reflected wave. (d) Find the nearest two positions in air away from the boundary where the total electric field is always zero. Get solution

2. Air–perfect conductor interface. A uniform plane wave of time-average power density 75 mW-cm−2 in air is normally incident on the surface of a perfect conductor located at y = 0, as shown in Figure. The total magnetic field phasor in air is given by...(a) What is H0? (b) What is the frequency, f ? (c) Find the total electric field Ē1(y, t) at y = −3.5 cm....Figure Normal incidence on a perfect conductor. Problem. Get solution

3. Air–perfect conductor interface. A uniform plane wave propagating in air given by...is normally incident on a perfectly conducting plane located at z = 0. (a) Find the frequency and the wavelength of this wave. (b) Find the corresponding magnetic field Hi(z). (c) Find the electric and magnetic field vectors of the reflected wave [i.e., Er(z) and Hr(z)]. (d) Find the total electric field in air [i.e., E1(z)], and plot the magnitude of each of its components as a function of z. Get solution

4. Air–perfect conductor interface. A uniform plane wave of frequency 12 GHz traveling in free space having a magnetic field given by...is normally incident on a perfect conductor boundary located at y = 0. (a) Find the real-time expression for the reflected wave, Hr(y, t). (b) Compare the polarizations of the incident and the reflected waves. Is there any difference? (c) Find the maximum value of the total magnetic field at y = 0, −1.25 cm, −2.5 cm, −3.75 cm, and −5 cm, respectively. Get solution

5. Air-perfect conductor interface. A 10 mW-m−2, 3 GHz uniform plane wave traveling in air is normally incident on a perfect conductor boundary located at the x = 0 plane, as shown in Figure. (a) Find the phasor-form electric and magnetic field vectors of the incident and reflected waves: Ei(x), Hi(x), Er(x), and Hr(x). (b) Find the two closest positions to the boundary in air where the total magnetic field is always zero....Figure Normal incidence on a perfect conductor. Problem. Get solution

6. Air-dielectric interface. A 65 mW-m−2, 3 GHz uniform plane wave traveling in the z direction in air is normally incident onto a planar, lossless, nonmagnetic dielectric interface with ∈r = 6 located at z = 0, as shown in Figure (a) Find the phasor-form electric and magnetic fields of the incident, reflected, and transmitted waves. (b) Find the two nearest positions in air with respect to the boundary where the total magnetic field is at a local minimum....Figure Normal incidence on a dielectric. Problem. Get solution

7. Air–GaAs interface. A uniform plane wave having a magnetic field given by...is normally incident from air onto a plane air–gallium arsenide (GaAs) interface located at y = 0. Assume GaAs to be a perfect dielectric with ∈r ≃ 13 and μr = 1. (a) Find the reflected (Hr(y)) and the transmitted (Ht(y)) magnetic fields. (b) Calculate the power density of the incident, reflected, and transmitted waves independently and verify the conservation of energy principle. (c) Find the expression for the total magnetic field (H1(y)) in air and sketch its magnitude as a function of y between y = −2 cm and y = 0. Get solution

8. Air–salty lake interface. Consider a uniform plane wave traveling in air with its electric field given by...normally incident on the surface (z = 0) of a salty lake (σ = 0.88 S-m−1, ∈r = 78.8, and μr = 1 at 1 GHz). (a) Assuming the lake to be perfectly flat and lossless (i.e., assume σ = 0), find the electric fields of the reflected and transmitted waves (i.e., Ēr and Ēt). (b) Modify the Ēt expression found in part (a) by introducing in it the exponential attenuation term due to the nonzero conductivity of lake water, given in part (a), and justify this approximation. (c) Using this expression, find the thickness of the lake over which 90% of the power of the transmitted wave is dissipated. What percentage of the power of the incident wave corresponds to this amount? Get solution

9. Aircraft–submarine communication. A submarine submerged in the ocean is trying to communicate with a Navy airplane, equipped with a VLF transmitter operating at 20 kHz, approximately 10 km immediately overhead from the location of the submarine. If the output power of the VLF transmitter is 200 kW and the receiver sensitivity of the submarine is 1 μV-m−1, calculate the maximum depth of the submarine from the surface of the ocean for it to be able to communicate with the transmitter. Assume the transmitter is radiating its power isotropically and assume normal incidence at the air–ocean boundary. Use σ = 4 S-m−1, ∈r = 81, and μr = 1 for the properties of the ocean. Get solution

10. Vertical incidence on plasma. An ionized gas (plasma) consists of free electrons and ions, the motion of which can be represented in terms of a frequency-dependent dielectric constant given by...where fp is known as the plasma frequency. Consider a plasma with fp = 1 MHz onto which a uniform plane wave at frequency f = 500 kHz is normally (θi = 0) incident from air (∈1 = ∈0, μ1 = μ0). In this case, the dielectric constant for the second medium is ∈2 = 3∈0. (a) Determine the reflection coefficient Γ and the complete time-domain expression for the electric field reflected wave. What fraction of the power carried by the incident wave is reflected? (b) Determine the complete time-domain expression for the electric field of the transmitted wave. (c) If the incident wave is left hand circularly polarized (LHCP), what is the polarization of the reflected and transmitted waves? Specify both the type (linear, circular, elliptical) and sense (RH or LH) of the polarization. Get solution

11. Air–fat interface. Consider a planar interface between air and fat tissue (assume it to be of semi-infinite extent). If a plane wave is normally incident from air at this boundary, find the percentage of the power absorbed by the fat tissue at (a) 100 MHz, (b) 300 MHz, (c) 915 MHz, and (d) 2.45 GHz, and compare your results with the results of Example Use Table 9.1 for the parameters of the fat tissue and assume μr = 1.ExampleAir–muscle interface. Consider a planar interface between air and muscle tissue. If a plane wave is normally incident at this boundary, find the percentage of incident power absorbed by the muscle tissue at (a) 100 MHz, (b) 300 MHz, (c) 915 MHz, and (d) 2.45 GHz. Assume μr = 1.TABLE ∈r AND σ FOR BIOLOGICAL TISSUES Muscle, Skin, and Tissues with High Water ContentFat, Bone, and Tissues with Low Water Contentf(MHz)(∈m)rσm(S-m−1)*(∈f)rσf(mS-m−1) *10071.70.8897.4519.1–75.9300541.375.731.6–107750521.545.649.8–138915511.605.655.6–1471,500491.775.670.8–1712,450472.215.596.4–2135,000443.925.5162–30910,00039.910.34.5324–549 Get solution

12. Air–concrete interface. A uniform plane wave operating at 1 GHz is normally incident from air onto the air–concrete interface. At 1 GHz, the complex relative dielectric constants of wet and dry concrete are measured as ∈wr ≃ 14.8 − j 1.73 and ∈dr ≃ 4.5 − j 0.03, respectively.55 For each case, calculate (a) the percentage of the incident power reflected and (b) the penetration depth in the concrete. Assume concrete to be semi-infinite in extent with μr= 1. Get solution

13. Shielding with a copper foil. A 1-GHz, 1-kW-m−2 microwave beam is incident upon a sheet of copper foil of 10 μm thickness (see Example for the electromagnetic properties of copper). Consider neglecting multiple reflections, if justified. (a) Find the power density of the reflected wave. (b) Find the power density transmitted into the foil. (c) Find the power density of the wave that emerges from the other side of the foil. Comment on the shielding effectiveness of this thin copper foil.Example: Air–copper interface. Consider a uniform plane wave propagating in air incident normally on a large copper block. Find the percentage time-average power absorbed by the copper block at 1, 10, and 100 MHz and at 1 GHz. Get solution

14. Absorbing material. Consider a commercial absorber slab56 made of EHP-48 material of 1 m thickness backed by a perfectly conducting metal plate. A 100-MHz uniform plane wave is normally incident from the air side at the air–absorber–metal interface. Find the percentage of incident power lost in the absorber material. For EHP-48, use ∈r = 6.93 − j 8.29 and μr = 1 at 100 MHz. Get solution

15. Radome design. A common material in dielectric radomes for aeronautical applications is fiberglass. For L-band (1–2 GHz), fiberglass has a typical relative dielectric constant of approximately ∈r ≃ 4.6. (a) Assuming a flat-plane radome, determine the minimum thickness of fiberglass that causes no reflections at the center of the L-band. (b) Using the thickness found in part (a), find the percentage of the incident power which transmits to the other side of the radome at each end of the L-band (i.e., 1 GHz and 2 GHz). Assume nonmagnetic case. Get solution

16. Radome design. A radome is to be designed for the nose of an aircraft to protect an X-band weather radar operating between 8.5 and 10.3 GHz. A new type of foam material with ∈r = 2 (assume lossless) is chosen for the design. (a) Assuming a flat planar radome, determine the minimum thickness of the foam that will give no reflections at the center frequency of the band. Assume μr = 1. (b) Using the thickness found in part (a), what percentage of the incident power is reflected at each end of the operating frequency band? (c) A thin layer of a different material (∈r = 4.1, tan δc = 0.04, thickness 0.25 mm) is added on one side of the radome designed in part (a) to protect the radome from rain erosion. What percent of the incident power is reflected at the center frequency? Get solution

17. Transmission through a multilayered dielectric. (a) Find the three lowest frequencies at which all of the incident power would be transmitted through the three-layer structure shown in Figure. The permeability of all three media is μ0. (b) If complete transmission is required for any thickness of the center medium, what is the lowest usable frequency? (c) Find the bandwidth of the transmission, defined as the range between the two lowest percentage values adjacent to and on either side of the frequency found in (b). Also find the lowest percentage values of transmission. (d) Why does the reflection from multiply coated optical lenses tend to be purple in color?...Figure Multilayered dielectric. Problem. Get solution

18. Glass slab. Consider a 1-cm thick slab of crown glass, with index of refraction n = 1.52. (a) If a beam of visible light at 500 nm is normally incident from one side of the slab, what percentage of the incident power transmits to the other side? (b) Repeat for 400 and 600 nm. Get solution

19. Refractive index of a liquid. To measure the refractive index of a liquid, a container is designed as shown in Figure to hold the liquid sample.57 Consider a container made of Teflon (∈r ≃ 2.08 and μr = 1 at 10 GHz) with wall thickness of L1 ≃ 1.04 cm on each side. A liquid with refractive index n is poured inside the container’s compartment with thickness L2 ≃ 1.49 cm. When a 10-GHz plane wave is normally incident from one side of the container, the effective reflection coefficient at that side is measured to be Γeff ≃ −0.39. (a) Find the refractive index of the liquid (assume lossless case). (b) Recalculate Γeff at 20 GHz (assume the same material properties apply). (c) Repeat part (b) at 5 GHz....Figure Refractive index of a liquid. Liquid with unknown index of refraction n in a container of known dielectric constant ∈r. Problem. Get solution

20. Antireflection (AR) coating on a glass slab. A beam of light is normally incident on one side of a 1-cm thick slab of flint glass (assume n = 1.86) at 550 nm. (a) What percentage of the incident power reflects back? (b) To minimize reflections, the glass is coated with a thin layer of antireflection coating material on both sides. The material chosen is magnesium fluoride (MgF2), which has a refractive index around 1.38 at 550 nm. Find the approximate thickness of each coating layer of MgF2 needed. (c) If a beam of light at 400 nm is normally incident on the coated glass with the thickness of the coating layers found in part (b), what percentage of the incident power reflects back? (d) Repeat part (c) for a light beam at 700 nm. Get solution

21. AR coating on a glass slab. A 1-cm thick slab of flint glass (n = 1.86) is to be coated only on one side so that when a beam of light is incident on the uncoated side, a sample of the light beam that reflects back from that side can be used to monitor the power of the incident beam. Assuming the light beam to be normally incident at 550 nm and the coating material used on the other side to be MgF2, calculate the percentage of the incident power that reflects back. Get solution

22. Infrared antireflection coating. To minimize reflections at the air–germanium interface in the infrared frequency spectrum, a coating material with an index of refraction of 2.04 is introduced as shown in Figure. (a) If the thickness of the coating material is adjusted to be a quarter-wavelength in the coating material for a free-space wavelength of 4 μm, find the effective reflection coefficient at free-space wavelengths of 4 and 8 μm. (b) Repeat part (a) if the thickness of the coating material is adjusted to be a quarter-wavelength in the coating material at 8 μm....Figure Infrared antireflection coating. Problem. Get solution

23. Reflection from ferrite-titanate slab. A 30-GHz uniform plane wave is normally incident from air onto an interface consisting of a second section (medium 2) of electrical length d having a complex permittivity of ∈c = ∈0(3 − j 4) and a complex permeability of μc = μ0(3 − j 4) followed by a third section (medium 3) consisting of air, as shown in Figure (a) Calculate the values of the real and imaginary parts of the propagation constant γ2 (i.e., α2 and β2) in medium 2 and determine all of the value(s) of d for which the effective reflection coefficient Γeff is zero. (b) For d = 0.25 cm, determine the fraction of the incident wave power that is transmitted into medium 3 (air)....Figure Reflection from ferritetitanate slab. Problem. Get solution

24. Superwide infrared antireflection coating. A wideband antireflection coating system as shown in Figure is designed to be used between air and germanium at infrared frequencies. If the thicknesses of the coating layers are each quarter wavelength for operation at λ1 = 3.5 μm, find and sketch the magnitude of the effective reflection coefficient over the range from 3 to 12 μm....Figure Infrared antireflection coating. Problem. Get solution

25. A snow-covered glacier. Glaciers are huge masses of ice formed in the cold polar regions and in high mountains. Most glaciers range in thickness from about 100 m to 3000 m. In Antarctica, the deepest ice on the polar plateau is 4.7 km. Consider a large glacier in Alaska covered with a layer of snow of 1 m thickness during late winter. A radar signal operating at 56 MHz is normally incident from air onto the air–snow interface. Assume both the snow and the ice to be lossless and nonmagnetic; for ice, ∈r = 3.2, and for snow, ∈r can vary between 1.2 and 1.8. Assuming both the snow and the ice to be homogeneous and the ice to be semi-infinite in extent, calculate the reflection coefficient at the air–snow interface for three different permittivity values of snow: ∈r = 1.2, 1.5, and 1.8. For which case is the snow layer most transparent (invisible) to the radar signal? Why? Get solution

26. Minimum ice thickness. Consider a 500-MHz uniform plane wave radiated by an aircraft radar normally incident on a freshwater (∈r = 88, μr = 1) lake covered with a layer of ice (∈r = 3.2, μr = 1), as shown in Figure. (a) Find the minimum thickness of the ice such that the reflected wave has maximum strength. Assume the lake water to be very deep. (b) What is the ratio of the amplitudes of the reflected and incident electric fields?...Figure Aircraft radar signal incident on an icy lake surface. Problem. Get solution

27. A snow–ice-covered lake. An interior lake in Alaska can be 30–100 m deep and is covered with ice and snow on the top, each of which can be about a meter deep in late winter.58 Consider a 5-GHz C-band radar signal normally incident from air onto the surface of a lake that is 50 m deep, covered with a layer of snow (assume ∈r = 1.5) of 60 cm thickness over a layer of ice (∈r = 3.2) of 1.35 m. Assume both the snow and the ice to be lossless and nonmagnetic. Also assume the lake water, with ∈cr = 68 − j 35 at 5 GHz at 0◦C, to be slightly brackish (salty), with an approximate conductivity of σ = 0.01 S-m−1, and the bottom of the lake to consist of thick silt with ∈r ≃ 50. (a) Calculate the reflection coefficient at the air interface with and without the snow layer. (b) Repeat part (a) at an X-band radar frequency of 10 GHz. Assume all the other parameters to be the same except for the lake water, ∈cr = 42 − j 41 at 10 GHz and 0◦C. Use any approximations possible, with the condition that sufficient justifications are provided. Get solution

28. Incidence on an air gap between two dielectrics. A 1 GHz uniform plane wave having an incident electric field amplitude of and propagating in a lossless dielectric medium (∈1 = 4∈0, μ1 = μ0) is normally incident on an airgap (i.e., ∈2 = ∈0, μ2 = μ0) of length d between itself and an identical dielectric (∈3 = 4∈0, μ3 = μ0). (a) Determine the magnitude of the reflected wave and the complete timedomain expression for the electric field Er(z , t) of the reflected wave for d = λ2/8 and d = λ2/4. (b) Determine the time-average electromagnetic power density (in W-m−2) transmitted through the airgap and the complete time-domain expression for the transmitted electric field ∈3(z , t) for d = λ2/8 and d = λ2/4. (c) Determine the the complete time-domain expression for the electric field of the electromagnetic wave within the airgap (i.e., E2(z , t)) for d = λ2/8 and d = λ2/4.... Get solution

29. Air–water–air interface. A uniform plane wave operating at 2.45 GHz in air is normally incident onto a planar water boundary at 20◦C (∈r = ∈ʹr − j∈ʺr = 79 − j 11).59 (a) Calculate the percentage of the incident power that is transmitted into the water (assume the water region to be nonmagnetic and semi-infinite in extent). (b) What percentage of the incident power is absorbed in the first 1-cm thick layer of water? (c) If the water layer has a finite thickness of 1 cm with air on the other side (i.e., air–water–air interface), calculate the percentage of the incident power that is absorbed in the water layer and compare it with the result of part (b). Get solution

30. Incidence on a coated perfect conductor. A 1.8 GHz uniform plane wave having an incident electric field amplitude of Ei0 = 1ej0 V-m−1 and propagating in a lossless dielectric medium (∈1, μ0) is incident on a perfect conductor coated with a lossy dielectric as shown in Figure. (a) Determine the total amount of power absorbed in the lossy dielectric. (b) Is there a nonzero surface current induced on the surface of the perfect conductor (at z = 0)? If so, determine its magnitude, phase, and direction and write the real time dependent expression, that is, Īs (z , t)....Figure Normal incidence on a coated perfect conductor. Problem. Get solution

31. Air–concrete wall–air. A 900-MHz wireless communication signal is normally incident from one side onto a reinforced concrete wall of thickness d having air on both sides. (a) Find the percentages of the incident power that is reflected back and that is transmitted to the other side of the wall for three different wall thicknesses: 10, 20, and 30 cm. (See Problem for data on the properties of reinforced concrete wall.) (b) Repeat part (a) at 1.8 GHz.ProblemConcrete wall. The effective complex dielectric constant of walls in buildings are investigated for wireless communication applications.72 The relative dielectric constant of the reinforced concrete wall of a building is found to be ∈r = 6.7 − j 1.2 at 900 MHz and ∈r = 6.2 − j 0.69 at 1.8 GHz, respectively. (a) Find the appropriate thickness of the concrete wall to cause a 10 dB attenuation in the field strength of the 900 MHz signal traveling over its thickness. Assume μr = 1 and neglect the reflections from the surfaces of the wall. (b) Repeat the same calculations at 1.8 GHz. Get solution

32. Reflection from multiple interfaces. A uniform plane wave is normally incident on a multiple dielectric interface consisting of two sections (mediums 2 and 3) of the same electrical length (l1 = d and l2 = 2d) as shown in Figure. Determine the value of d (if any) such that the all of the incident wave power is transmitted to medium 4 (i.e., such that the reflection coefficient in medium 1 is zero)....Figure Multiple dielectric interfaces. Problem. Get solution

33. Oblique incidence on a perfect conductor. A 30W-m−2 uniform plane wave in air is obliquely incident on a perfect conductor boundary located at the y = 0 plane. The electric field of the incident wave is given by...(a) Find E0, f , and θi. (b) Find Er. (c) Find the total electric field E1 and the nearest positions (with respect to the conductor surface) of the minima and maxima of its magnitude. Get solution

34. Oblique incidence on a perfect conductor. A parallel-polarized (with respect to the plane of incidence) 100 μW-(cm)−2, 4-GHz wireless communication signal in air is incident on a perfect conductor surface located at y = 0 at an incidence angle of θi = 30◦ as shown in Figure. The signal can be approximated as a uniform plane wave. (a) Write the instantaneous expressions for Ēi(y, z , t) and Hi(y, z , t). (b) Find Er(y, z) and Hr(y, z) of the reflected wave. (c) Find the magnitude of the total magnetic field phasor H1(y, z) and sketch it as a function of y....Figure Oblique incidence on a perfect conductor. Problem. Get solution

35. Plane wave at 45◦ angle. A plane wave in air is incident at 45◦ upon a perfectly conducting surface located at x = 0. The plane wave consists of two components as follows:...(a) Write the perpendicular and parallel polarization components of the electric fields of the reflected wave, and show that the tangential electric fields satisfy the boundary conditions.(b) What are the polarizations of the incident and the reflected waves? Get solution

36. Oblique incidence on a dielectric medium. Consider oblique incidence of a uniform plane wave on the interface between two nonmagnetic dielectric media with permittivities ∈A and ∈B. When a perpendicularly polarized plane wave is incident from medium A (i.e., ∈1= ∈A) onto medium B (i.e., ∈2= ∈B) at an incidence angle θi = θ1, the transmitted angle is θt = θ2, and the reflection coefficient is.... If the propagation direction is reversed, that is, if a new incident wave with the same polarization is now incident from medium B (i.e., ∈1= ∈B) onto medium A (i.e., ∈2= ∈A) at an incidence angle θi = θ2, what is the numerical value of the reflection coefficient...? Get solution

37. Oblique incidence. A uniform plane wave is obliquely incident at an angle θi at the interface between two nonmagnetic (μ1 = μ2 = μ0) dielectric media as shown in Figure. The relative permittivity of the second medium is known to be ∈2r = 3, and the electric field of the incident wave is given by...(a) Calculate the relative dielectric constant ∈1r and the angle of incidence θi. (b) Write the corresponding expression for the magnetic field of the incident wave (i.e., Hi(x, z , t)).(c) Determine the percentage of the incident power that will be transmitted across the interface....Figure Oblique incidence. Problem. Get solution

38. Reflection from ground. A 8-GHz and 200-μW-m−2 microwave communication signal in air is obliquely incident at θi = 60◦ onto the ground (assume lossless and ∈r = 15) located at z = 0. (a) If the incident wave is perpendicularly polarized, write the complete expressions for Ei, Er, and Et. (b) Repeat part (a) for a parallel-polarized wave. Get solution

39. Air–dielectric interface. A uniform plane wave propagating in air has an electric field given by...where E0 is a real constant. The wave is incident on the planar interface (located at y = 0) of a dielectric with μr = 1, ∈r = 3, as shown in Figure. (a) What are the values of the wave frequency and the angle of incidence? (b) What is the polarization of the incident wave (i.e., linear, circular, elliptical, right-handed or left-handed)? (c) Write the complete expression for the electric field of the reflected wave in a simplified form. (d) What is the polarization of the reflected wave?...Figure Air–dielectric interface. Figure for Problem. Get solution

40. Oblique incidence on a dielectric. An elliptically polarized uniform plane wave is incident from free space (∈0, μ0) onto on a lossless nonmagnetic dielectric (∈2, μ0) at an incidence angle θi = 70◦. Assuming the coordinate system and the placement of the interface to be as in Section 9.6 (e.g., Figure), the electric field vector of the incident wave can be written as:...(a) Determine the dielectric constant ∈2 for which the reflected wave is right-hand (RH) circularly polarized. (b) For the value of ∈2 found in part (a), determine the fraction of the incident wave power that is transmitted into the second medium. (c) Specify the polarization (the type, i.e., linear/circular/elliptical, and the sense, i.e., RH or LH) of the wave transmitted into the second medium....Figure Oblique incidence at a dielectric boundary. Uniform plane wave obliquely incident at a dielectric boundary located at the z = 0 plane. Dashed lines AC, BD, and EB are the wavefronts (planar surfaces of constant phase). Get solution

41. Brewster angle at the air–water interface. A perpendicularly polarized uniform plane wave is obliquely incident from air onto the surface of a smooth freshwater lake (assume lossless and nonmagnetic with ∈r ≃ 81) at the Brewster angle (i.e., θi = θiB). Calculate the reflection and transmission coefficients. Get solution

42. Air–ice interface. A 1-W-m−2, 1-GHz radar signal is obliquely incident at an angle θi = 30◦ from air onto an air–ice interface. (a) Assuming the ice to be lossless, nonmagnetic and semiinfinite in extent, with ∈r ≃ 3.17, calculate the reflection and the transmission coefficients and the average power densities of the reflected and the transmitted waves if the incident wave is perpendicularly polarized. (b) Repeat part (a) for an incident wave that is parallelpolarized. (c) Find the Brewster angle and repeat parts (a) and (b) for a wave incident at the Brewster angle. Get solution

43. Communication over a lake. Consider a ground-to-air communication system as shown in Figure. The receiver antenna is on an aircraft over a huge lake circling at a horizontal distance of ∼10 km from the transmitter antenna as it waits for a landing time. The transmitter antenna is located right at the shore mounted on top of a 100-m tower above the lake surface overlooking the lake and transmits a parallel polarized (with respect to the plane of incidence) signal. The transmitter operates in the VHF band. The pilot of the aircraft experiences noise (sometimes called ghosting effect) in his receiver due to the destructive interference between the direct wave and the ground-reflected wave and needs to adjust his altitude to minimize this interference. Assuming the lake to be flat and lossless with ∈r ≃ 80, calculate the critical height of the aircraft in order to achieve clear transmission between the transmitter and the receiver....Figure Communication over a lake. Problem. Get solution

44. Air-to-air communication. Consider two helicopters flying in air separated by a horizontal distance of 2 km over a flat terrain as shown in Figure. The pilot of one of the helicopters, located at an altitude of 100 m, transmits a parallel-polarized (with respect to the plane of incidence) VHF-band signal (assume 200 MHz) to communicate with the other helicopter. The pilot of the other helicopter needs to adjust her altitude to eliminate the noise on her receiver due to the interference of the ground-reflected wave with the direct wave. (a) Assuming the ground to be homogeneous, nonmagnetic and lossless with ∈r ≃ 16, find the critical altitude of the receiver helicopter in order to minimize this interference. (b) Consider another scenario when both helicopters are at 250 m altitude. In this case, what should be the horizontal separation distance between the helicopters in order to achieve clear signal transmission? (c) Repeat both (a) and (b) for the case in which the helicopters try to land at a remote site in Alaska where the ground is covered with permafrost (assume ∈r ≃ 4)....Figure Air-to-air communication. Problem. Get solution

45. Signal to Interference Ratio (SIR). You have just launched your start-up company, with your headquarters located at a distance of d = 140 m from a cellular tower of height h1 = 50 m. As the CEO you choose to have your office on the top floor of your building, at a height of h2 = 20 m, as shown in Figure. (a) Determine the signal-to-interference ratio (SIR):...assuming that the permittivity of the ground is ∈ = 5.55∈0. Provide separate results for both parallel and perpendicular polarization. (b) After a few dropped calls, you feel that you are not satisfied with the SIR you have and decide that you can move your office to a lower floor to maximize SIR. Calculate the height h2 that results in maximum SIR and the value of maximum SIR. Provide separate results for both parallel and perpendicular polarization....Figure Cell phone reception. Problem. Get solution

46. Dry asphalt roads acting as mayfly traps? Adult mayflies have only a few hours in which to find a mate and reproduce. Direct sunlight reflected off the surface of water is strongly polarized in the horizontal plane (i.e., parallel polarized), and many water-dwelling insects, including mayflies, use this reflected polarized light to identify open stretches of water where they can lay their eggs during their brief mating period. However, researchers discovered that light reflected from dry asphalt roads is also horizontally polarized and visually deceives mayflies into laying their eggs on roads instead of in rivers.60 The higher the degree of polarization of the reflected light, the more attractive it is for mayflies. Hence, mayflies swarming, mating, and egg-laying on asphalt roads are predominantly deceived by and attracted to the asphalt surface because the largely horizontally polarized reflected light imitates a water surface. Note that although sunlight has mixed polarization and is incident on the asphalt over a range of angles, the polarization of the reflected light is predominantly horizontal because of the deep minimum for ρ|| in the vicinity of the Brewster angle (see Figure). (a) If the reflected light from asphalt is almost 100% horizontally polarized when the light is incident on the asphalt surface at an incidence angle of about 57.5◦, calculate the effective refractive index of asphalt. (b) Assuming the refractive index of water to be nw ≃ 1.33, find the angle at which the reflected light from the water surface is 100% horizontally polarized....Figure Reflection coefficient versus the angle of incidence. (a) Magnitude and phase of reflection coefficient for perpendicular (Γ⊥) and parallel (Γ||) polarization versus angle of incidence θi for distilled water (∈2r = 81), flint glass (∈2r = 10), and paraffin (∈2r = 2), all assumed to be lossless. For clarity, the complement of the phase angle φ|| (i.e., π − φ||) is sketched, rather than φ|| itself. In all cases, φ|| = π for θiθiB and φ|| = 0 for θi > θiB, while φ⊥ = π for all θi. Get solution

47. Total internal reflection. A uniform plane wave with a magnetic field given by...is obliquely incident at an interface at z = 0 separating two nonmagnetic lossless media as shown in Figure. (a) Calculate the relative dielectric constant ∈1r of medium 1 and the angle of incidence θi. (b) Find the maximum value of the relative dielectric constant ∈2r of medium 2 for total internal reflection to occur. (c) Is it possible to achieve total transmission by adjusting the incidence angle? If yes, use the maximum value of ∈2r found in part (b) to determine the incidence angle at which total transmission would occur....Figure Total internal reflection. Problem. Get solution

48. Reflection from prisms. Consider the various right-angled prisms shown in Figure. (a) What is the minimum index of refraction n1 of the prism necessary in each case for there to be no time-average power transmitted across the hypotenuse when the prisms are (i) in free space, (ii) in water (assume n2 ≃ 1.33). (b) At these refractive index values (found in (a)), what are the exit angles θte?...Figure Reflection from prisms. Problem. Get solution

49. MgF2 prism. A 45◦–90◦–45◦ prism is constructed from MgF2 (n = 1.38) to be used to turn a light beam around by 90◦ by internal reflection at its hypotenuse. Does the light beam exit at the hypotenuse, and if so, what is its exit angle? Get solution

50. Refractive index of a prism. An experiment is designed to measure the refractive index of a prism using the principle of total internal reflection.61 In this experiment, a plane-polarized, collimated, monochromatic beam of light is obliquely incident from one side of the prism at an incidence angle of θi as shown in Figure. The incidence angle θi of the beam is adjusted to a critical value ψc such that the incidence angle on the other side of the prism is equal to the critical angle of incidence θic. Thus, by measuring the refracting angle A of the prism and the critical incidence angle ψc, the refractive index np of the prism can be calculated. (a) Show that...(b) For a prism under test, the refracting angle of the prism and the critical incidence angle adjusted are measured to be A = 60◦ and ψc = 42◦, respectively. Calculate the refractive index np of this prism.... Get solution

51. Right-angle prism. A new technique is proposed to measure small angles in optical systems using right-angle prisms.62 Consider a 45°–90°–45° prism as shown in Figure made of glass having a refractive index of n ≃ 1.515 to be used in an experiment. A light beam that is obliquely incident at an incidence angle θ1 on the entrance face undergoes reflection and refraction at the entrance face, the hypotenuse face, and the exit face of the prism. (a) For an incidence angle of θ1 = 30◦, find the exit angles θ4 and θ6. (b) Find the critical incidence angle θ1 of the incident beam on the entrance face that results in no transmission at the hypotenuse face. (c) What happens to the critical angle found in part (b) when the hypotenuse face of the prism is coated with an antireflection coating (single or multiple layers)? (d) Repeat parts (a) and (b) if the prism is made of a different type of glass with n ≃ 1.845....Figure Right-angle prism. Problem. Get solution

52. A V-shaped prism. A V-shaped right-angled prism is designed to measure accurately the refractive index of liquids63 as shown in Figure. A laser beam normally incident from one side is refracted at the inclined faces, between which the liquid sample under test is placed, and leaves the prism on the other side at a deflection angle ψ. (a) A liquid of known refractive index n ≃ 1.512 is placed in the top compartment of the prism with np ≃ 1.628. Find the deflection angle ψd. (b) Using the same prism, the deflection angle for a different liquid with unknown refractive index is approximately measured to be ψ≃ 55.5°. Calculate the refractive index n of this liquid....Figure A V-shaped prism. Problem. Get solution

53. Turning a perfect corner. An interesting optical phenomenon is the invariance of the angle between the incoming and outgoing light rays passing through a right-angle isosceles prism with a silver-coated hypotenuse.64 For the isosceles right-angle prism shown in Figure, the incident ray enters at side AB and exits at side AC after being reflected twice and refracted twice. Show that the total deviation angle ψd between the incident and the exiting rays is exactly 90°....Figure Turning a perfect corner. Problem. Get solution

54. Air–oil–water. Consider a layer of oil (assume n ≃ 1.6) about 5 mm thick floating over a body of water (n ≃ 1.33). (a) If a light ray is obliquely incident from air onto the oil surface, find the range of incidence angles (if any) that results in total internal reflection at the oil–water interface. (b) If a light ray is obliquely incident from water onto the oil surface, find the range of incidence angles (if any) that results in total internal reflection at the oil–air interface. Get solution

55. An in-line Brewster angle prism. Brewster angle prisms are optical elements that use light at the polarizing angle to obtain perfect transmission of parallel polarized light. An inline Brewster angle polarizing prism is designed65 as shown in Figure, to polarize a light beam without changing its direction. Consider an unpolarized light beam incident on this prism at point A at an incidence angle of θi = 31.639◦. Given the prism angles to be α = 63.278° and β = 103.939° and the prism refractive index to be n = 1.623, (a) determine whether light exits the prism at points B, C, and/or D, (b) find the exit angles θB, θC, and/or θD at these exit points, (c) specify the polarization of each exiting beam, and (d) find the angle between each exiting beam and the incident beam. Discuss your results in terms of the stated purpose of this particular prism design....Figure In-line polarizing prism. Problem. Get solution

56. V-shaped prism polarizer. A symmetrical three-reflection silicon (Si) polarizer prism is designed as shown in Figure, based on the Brewster-angle internal reflection that occurs at the base of the prism.66 (a) In Figure, the prism angle A is adjusted for use at 1.3 μm light wave communication wavelength (the refractive index of silicon at 1.30 μm is nSi ≃ 3.5053), to A ≃ 52.9613°. Assuming the incident light beam entering the prism on the left side to be unpolarized, find the polarization of the beam exiting on the right side. (b) Find the new value of A for the prism to be used as a polarizer at 1.55-μm wavelength (the refractive index of silicon at 1.55 μm is nSi ≃ 3.4777). (c) Another interesting design is to coat the base of the silicon prism with silicon dioxide (SiO2) as shown in Figure. At 1.3 μm, the prism angle is adjusted to A ≃ 56.217◦. Find the refractive index of SiO2 at 1.3 μm. (d) Repeat part (c) at 1.55 μm when the prism angle is adjusted to A ≃ 56.277°....Figure V-shaped prism polarizer. Problem. Get solution

57. Limestone wall versus brick wall. Consider two buildings, one with limestone (∈r = 7.51, σ = 0.03 S-m−1) exterior walls and the other with brick (∈r = 4.44, σ = 0.01 S-m−1) exterior walls, with the material properties cited measured67 at 4 GHz. These walls represent some of the typical building surfaces that affect the propagation of mobile radio signals. Assuming the walls to be lossless, nonmagnetic, semi-infinite in extent, and neglecting the roughness of their surfaces, calculate the reflection coefficients at the surface of each building for both perpendicular and parallel polarizations at three different angles of incidence of θi = 30°, 45°, and 60° and compare the results. Get solution

58. Refractive index of concrete. Knowledge of the dielectric properties of construction material is important because the reflection and transmission characteristics of buildings and rooms are governed by these properties. The complex refractive index of a plain concrete plate mixed from Portland cement, gravel, sand, and water was measured at 57.5 GHz for use in designing and testing millimeter-wave communication systems.68 The measured refractive index of the concrete 14 months after concreting was n = 2.55 − j 0.084. (a) Using the measured values for concrete, calculate and sketch the magnitude of the reflection coefficient at the air–concrete interface at 57.5 GHz as a function of the incidence angle varying between 0◦ and 90◦ for both perpendicular and parallel polarization cases. Assume the concrete to be semi-infinite in extent. (b) For a 5-cm thick concrete wall having air on both sides, calculate the magnitude of the normal incidence reflection coefficient at 57.5 GHz and compare it with the result of part (a). (c) Repeat part (b) for a thickness of 10 cm. Get solution

59. Oblique incidence on a multiple dielectric interface. Suppose that a parallel polarized uniform plane wave is incident, in air, on a nonmagnetic glass slab (assume ∈r = 2) of thickness d, as shown in Figure. The angle of incidence θi is chosen to be the air-glass Brewster angle so that there is no reflection at the first interface (z = 0). Find the complete expression for the electric field phasor of the wave transmitted into air (z > d), that is, Et(x, z). Hint: First find the reflection coefficient at the z = d interface....Figure Oblique incidence on a multiple interface. Problem. Get solution

60. Oblique incidence on a multiple dielectric interface. A perpendicularly polarized uniform plane wave propagating in air is incident obliquely (at an angle θi) on a structure consisting of two lossless and nonmagnetic dielectrics: a coating layer with permittivity ∈1r and thickness d coated on another dielectric with permittivity ∈2r and of infinite thickness, as shown in Figure. Derive expressions for the effective reflection and transmission coefficients (Γeff and Jeff) in terms of parameters of the media (∈1r , ∈2r), the angle of incidence θi, and the slab thickness d....Figure Oblique incidence on a multiple interface. Problem. Get solution

### Chapter #8 Solutions - Engineering Electromagnetics and Waves - Aziz Inan, Ryan Said, Umran S Inan - 2nd Edition

1. Uniform plane wave. The electric field of a uniform plane wave in air is given by...(a) Find the phase constant β and the wavelength λ. (b) Sketch Ex (z , t ) as a function of t at z = 0 and z = λ/4. (c) Sketch Ex (z , t ) as a function of z at t = 0 and t = π/ω. Get solution

2. Uniform plane wave. The electric field phasor of an 18 GHz uniform plane wave propagating in free space is given by...(a) Find the phase constant, β, and wavelength, λ. (b) Find the corresponding magnetic field phasor H(y). Get solution

3. Uniform plane wave. A uniform plane wave is traveling in the x direction in air with its magnetic field oriented in the z direction. At the instant t = 0, the wave magnetic field has two adjacent zero values, observed at locations x = 2.5 cm and x = 7.5 cm, with a maximum value of 70 mA-m−1 at x = 5 cm. (a) Find the wave magnetic field H(x, t ) and its phasor H(x). (b) Find the corresponding wave electric field E(x, t ) and its phasor E(x). Get solution

4. Broadcast signal. The magnetic field of a TV broadcast signal propagating in air is given as...(a) Find the wave frequency f = ω/(2π). (b) Find the corresponding E(y, t ). Get solution

5. Uniform plane wave. A NASA spacecraft orbiting around Mars receives a radio signal transmitted by the UHF antenna of Curiosity Rover, with an electric field given by...(a) Determine the frequency f and the wavelength λ of this radio signal. (b) Find the corresponding magnetic field H(z , t ). Get solution

6. Lossless nonmagnetic medium. The magnetic field component of a uniform plane wave propagating in a lossless simple nonmagnetic medium (μ = μ0) is given by...(a) Find the frequency, wavelength, and the phase velocity. (b) Find the relative permittivity, єr , and the intrinsic impedance, η, of the medium. (c) Find the corresponding E. (d) Find the time-average power density carried by this wave. Get solution

7. A wireless communication signal. The electric field of a wireless communication signal traveling in air is given in phasor form as...(a) Find the frequency f and wavelength λ. (b) Find the corresponding phasor-form magnetic field H(x). (c) Find the total time-average power density carried by this wave. Get solution

8. Uniform plane wave. An 8 GHz uniform plane wave traveling in air is represented by a magnetic field vector given in phasor form as follows:...(a) Find β and frequency f . (b) Find the corresponding electric field vector in phasor form. (c) Find the total time-average power density carried by this wave. Get solution

9. Cellular phones. The electric field component of a uniform plane wave in air emitted by a mobile communication system is given by...(a) Find the frequency f and wavelength λ. (b) Find θ if E(x, z , t ) ≃ −ŷ25 mV-m−1 at t = 0 and at x = 3 m, z = 2 m. (c) Find the corresponding magnetic field H(x, z , t ). (d) Find the time-average power density carried by this wave. Get solution

10. Superposition of two waves. The sum of the electric fields of two time-harmonic (sinusoidal) electromagnetic waves propagating in opposite directions in air is given as...(a) Find the constant β. (b) Find the corresponding H. (c) Assuming that this wave may be regarded as a sum of two uniform plane waves, determine the direction of propagation of the two component waves. Get solution

11. Unknown material. The intrinsic impedance and the wavelength of a uniform plane wave traveling in an unknown dielectric at 900 MHz are measured to be ∼42Ω and ∼3.7 cm, respectively. Determine the constitutive parameters (i.e., єr and μr) of the material. Get solution

12. Uniform plane wave. A 10 GHz uniform plane wave with maximum electric field of 100 V-m−1 propagates in air in the direction of a unit vector given by û= 0.8x − 0.6ẑ. The wave magnetic field has only a y component, which is approximately equal to 40 mA-m−1 at x = z = t = 0. Find E and H. Get solution

13. Standing waves. The electric field phasor of an electromagnetic wave in air is given by...(a) Find the wavelength, λ. (b) Find the corresponding magnetic field H(x). (c) Is this a traveling wave? Get solution

14. Propagation through wet versus dry earth. Assume the conductivities of wet and dry earth to be σwet = 10−2 S-m−1 and σdry = 10−4 S-m−1, respectively, and the corresponding permittivities to be єwet = 10є0 and єdry = 3є0. Both media are known to be nonmagnetic (i.e., μr = 1). Determine the attenuation constant, phase constant, wavelength, phase velocity, penetration depth, and intrinsic impedance for a uniform plane wave of 20 MHz propagating in (a) wet earth, (b) dry earth. Use approximate expressions whenever possible. Get solution

15. 15 Propagation in lossy media. (a) Show that the penetration depth (i.e., the depth at which the field amplitude drops to 1/e of its value at the surface) in a lossy medium with μ = μ0 is approximately given by...where tan δc is the loss tangent.70 (b) For tan δc≪1, show that the above equation can be further approximated as d ≃ [0.318(c/f )]/[tan δcr...]. (c) Assuming the properties of fat tissue at 2.45 GHz to be σ = 0.12 S-m−1, єr = 5.5, and μr = 1, find the penetration depth of a 2.45 GHz plane wave in fat tissue using both expressions, and compare the results. Get solution

16. Glacier ice. A glacier is a huge mass of ice that, unlike sea ice, sits over land. Glaciers are formed in the cold polar regions and in high mountains. They spawn billions of tons of icebergs each year from tongues that reach the sea. The icebergs drift over an area of 70,000,000 km2, which is more than 20% of the ocean area, and pose a serious threat to navigation and offshore activity in many areas of the world. Glacier ice is a low-loss dielectric material that permits significant microwave penetration.71 The depth of penetration of an electromagnetic wave into glacier ice with loss tangent of tan δc≃ 0.001 at X-band (assume 3 cm air wavelength) is found to be ∼5.41 m. (a) Calculate the dielectric constant and the effective conductivity of the glacier ice. (Note that μr = 1.) (b) Calculate the attenuation of the signal measured in dB-m−1. Get solution

17. Good dielectric. Alumina (Al2O3) is a low-loss ceramic material that is commonly used as a substrate for printed circuit boards. At 10 GHz, the relative permittivity and loss tangent of alumina are approximately equal to єr = 9.7 and tan δc = 2 × 10−4. Assume μr = 1. For a 10 GHz uniform plane wave propagating in a sufficiently large sample of alumina, determine the following: (a) attenuation constant, α, in np-m−1; (b) penetration depth, d; (c) total attenuation in dB over thicknesses of 1 cm and 1 m. Get solution

18. Concrete wall. The effective complex dielectric constant of walls in buildings are investigated for wireless communication applications.72 The relative dielectric constant of the reinforced concrete wall of a building is found to be єr = 6.7 − j 1.2 at 900 MHz and єr = 6.2 − j 0.69 at 1.8 GHz, respectively. (a) Find the appropriate thickness of the concrete wall to cause a 10 dB attenuation in the field strength of the 900 MHz signal traveling over its thickness. Assume μr = 1 and neglect the reflections from the surfaces of the wall. (b) Repeat the same calculations at 1.8 GHz. Get solution

19. Unknown medium. The magnetic field phasor of a 100 MHz uniform plane wave in a nonmagnetic medium is given by...(a) Find the conductivity σ and relative permittivity єr of the medium. (b) Find the corresponding time-domain electric field E(z , t ). Get solution

20. Unknown biological tissue. The electric field component of a uniform plane wave propagating in a biological tissue with relative dielectric constant is given by...(a) Find ...and ... Assume σ = 0 and μr = 1. (b) Write the corresponding expression for the wave magnetic field. (c) Write the mathematical expression for the time-average Poynting vector Sav and sketch its magnitude from y = 0 to 5 cm. Get solution

21. Propagation in seawater. Transmission of electromagnetic energy through the ocean is practically impossible at high frequencies because of the high attenuation rates encountered. For seawater, take σ = 4 S-m−1, єr = 81, and μr = 1, respectively. (a) Show that seawater is a good conductor for frequencies much less than ∼890 MHz. (b) For frequencies less than 100 MHz, calculate, as a function of frequency (in Hz), the approximate distance over which the amplitude of the electric field is reduced by a factor of 10. Get solution

22. Wavelength in seawater. Find and sketch the wavelength in seawater as a function of frequency. Calculate λsw at the following frequencies: 1 Hz, 1 kHz, 1 MHz, and 1 GHz. Sketch log λsw vs. log f . Use the following properties for seawater: σ = 4 S-m−1, єr = 81, and μr = 1. Get solution

23. Communication in seawater. ELF communication signals (f ≤ 3 kHz) can more effectively penetrate seawater than VLF signals (3 kHz ≤ f ≤ 30 kHz). In practice, an ELF signal used for communication can penetrate and be received at a depth of up to 80 m below the ocean surface.73 (a) Find the ELF frequency at which the skin depth in seawater is equal to 80 m. For seawater, use σ = 4 S-m−1, єr = 81, and μr = 1. (b) Find the ELF frequency at which the skin depth is equal to half of 80 m. (c) At 100 Hz, find the depth at which the peak value of the electric field propagating vertically downward in seawater is 40 dB less than its value immediately below the surface of the sea. (d) A surface vehicle-based transmitter operating at 1 kHz generates an electromagnetic signal of peak value 1 V-m−1 immediately below the sea surface. If the antenna and the receiver system of the submerged vehicle can measure a signal with a peak value of as low as 1 μV-m−1, calculate the maximum depth beyond which the two vehicles cannot communicate. Get solution

24. Submarine communication near a river delta. A submarine submerged in the sea (σ = 4 S-m−1, єr = 81, μr = 1) wants to receive the signal from a VLF transmitter operating at 20 kHz. (a) How close must the submarine be to the surface in order to receive 0.1% of the signal amplitude immediately below the sea surface? (b) Repeat part (a) if the submarine is submerged near a river delta, where the average conductivity of seawater is ten times smaller. Get solution

25. Human brain tissue. Consider a 1.9 GHz electromagnetic wave produced by a wireless communication telephone inside a human brain tissue74 (єr = 43.2, μr = 1, and σ = 1.29 S-m−1) such that the peak electric field magnitude at the point of entry (z = 0) inside the tissue is about 100 V-m−1. Assuming plane wave approximation, do the following: (a) Calculate the electric field magnitudes at points z = 1 cm, 2 cm, 3 cm, 4 cm, and 5 cm inside the brain tissue and sketch it with respect to z . (b) Calculate the time-average power density at the same points and sketch it with respect to z . (c) Calculate the time-average power absorbed in the first 1 cm thickness of a tissue sample having a cross-sectional area of 1 cm2. Get solution

26. Dispersion in sea water. A uniform plane electromagnetic wave in free space propagates with the speed of light, namely, c ≃ 3 × 108 m-s−1. In a conducting medium, however, the velocity of propagation of a uniform plane wave depends on the signal frequency, leading to the “dispersion” of a signal consisting of a band of frequencies. (a) For sea water (σ = 4 S-m−1, єr = 81, and μr = 1), show that for frequencies much less than ∼890 MHz, the velocity of propagation is approximately given by vp ≃ k1√f, where k1 is a constant. What is the value of k1? (b) Consider two different frequency components of a signal, one at 1 kHz, the other at 2 kHz. If these two signals propagate in the same direction in seawater and are in phase at z = 0, what is the phase delay (in degrees) between them (e.g., between their peak values) at a position 100 m away? Get solution

27. Electromagnetic earthquake precursor. A group of Stanford scientists measured75 mysterious electromagnetic waves varying with ultralow frequencies in the range of 0.01–10 Hz during two different earthquakes which occurred in Santa Cruz, California, in 1989 and in Parkfield, California, in 1994. A member of the group speculates that these waves may result from a local disturbance in the earth’s magnetic field caused by charged particles carried by water streams that flow along the fault lines deep in the earth’s crust as a result of the shifts that led to the quake. These low-frequency waves can penetrate rock much more easily than those of higher frequencies but can still travel only about 15 km through the ground. Since this low-frequency electromagnetic activity was recorded close to a month before the quake and lasted about a month after, this phenomenon has a potential use as an earthquake predictor. Consider three plane waves of equal amplitudes with frequencies of 0.1 Hz, 1 Hz, and 10 Hz, all produced at a depth of 15 km below the earth’s surface during an earthquake. Assuming each of these waves to be propagating vertically up toward the surface of the earth, (a) calculate the percentage time-average power of each wave that reaches the surface of the earth and (b) using the results of part (a), comment on which one of the three signals is more likely to be picked up by a receiver located on the earth’s surface, based on their signal strengths. For simplicity, assume the earth’s crust to be homogeneous, isotropic, and nonmagnetic with properties σ = 10−3 S-m−1 and єr = 10, respectively. Get solution

28. Phantom muscle tissue. In order to develop radiofrequency (RF) heating techniques for treating tumors at various locations and depths in patients, it is necessary to carry out experiments to determine the energy absorbed by an object exposed to electromagnetic fields over a wide range of RF frequencies. An artificial muscle tissue (“muscle phantom”) is designed to be used in these experiments to simulate actual muscle tissue for applications in the frequency range used for RF hyperthermia.76 (a) Given the relative dielectric constant and the conductivity of the muscle phantom at 915 MHz and 22◦C to be єr ≃ 51.1 and σ≃1.27 S-m−1, calculate the depth of penetration in the phantom. Note that μr = 1. (b) Repeat part (a) at 2.45 GHz when єr ≃ 47.4 and σ ≃ 2.17 S-m−1. Which frequency can penetrate deeper into the muscle phantom? (c) Calculate the total dB attenuation over a muscle phantom of 1.5 cm thickness at both frequencies. Get solution

29. Unknown medium. The skin depth and the loss tangent of a nonmagnetic conducting medium at 21.4 kHz are approximately equal to 1.72 m and 4.15 × 104, respectively. (a) Find the conductivity σ and the relative dielectric constant єr of the medium. What medium is this? (b) Write the mathematical expressions for the electric and magnetic field components of a 21.4 kHz uniform plane wave propagating in this medium, assuming the maximum peak value of the electric field to be 10 V-m−1. (c) Repeat part (b) at 2.14 MHz. Assume the properties of the medium to be the same at both frequencies. Get solution

30. Unknown medium. A uniform plane wave propagates in the x direction in a certain type of material with unknown properties. At t = 0, the wave electric field is measured to vary with x as shown in Figure At x = 40 m, the temporal variation of the wave electric field is measured to be in the form shown in Figure Using the data in these two Figure, find (a) σ and єr (assume nonmagnetic case), (b) the depth of penetration and the attenuation in dB-m−1, and (c) the total dB attenuation and the phase shift over a distance of 100 m through this medium.Figure Unknown medium. Problem.... Get solution

31. Uniform plane waves. The electric field of a 1 GHz uniform plane wave propagating in a low-loss dielectric is given by...(a) Stating all assumptions, determine the conductivity σ and permittivity є of this dielectric and calculate the depth of penetration d. (b) Write a complete (i.e., with all quantities specified in terms of numerical values) expression for the vector magnetic field intensity H(x, y, t ) of this wave. Get solution

32. Thickness of beef products. Microwave heating is generally uniform over the entire body of the product being heated if the thickness of the product does not exceed about 1–1.5 times its penetration depth.77 (a) Consider a beef product to be heated in a microwave oven operating at 2.45 GHz. The dielectric properties of raw beef at 2.45 GHz and 25◦C are ...= 52.4, μr = 1, and tan δc = 0.33.78 What is the maximum thickness of this beef product for it to be heated uniformly? (b) Microwave ovens operating at 915 MHz are evidently more appropriate for cooking products with large cross sections and high dielectric loss factors. The dielectric properties of raw beef at 915 MHz and 25◦C are ...= 54.5, μr = 1, and tan δc = 0.411. Find the maximum thickness of the beef product at 915 MHz and compare it with the results of part (a). Get solution

33. Beef versus bacon. The dielectric properties of cooked beef and smoked bacon at 25◦C are given by єr ≃ 31.1 − j 10.3 at 2.45 GHz and єr ≃ 2.5 − j 0.125 at 3 GHz, respectively (see the references in the preceding problem). Assuming μr = 1, calculate the loss tangent and the penetration depth for each and explain the differences. Get solution

34. Aluminum foil. A sheet of aluminum foil of thickness ∼25 μm is used to shield an electronic instrument at 100 MHz. Find the dB attenuation of a plane wave that travels from one side to the other side of the aluminum foil. (Neglect the effects from the boundaries.) For aluminum, σ = 3.54 × 107 S-m−1 and єr = μr = 1. Get solution

35. Unknown material. Using the results of a reflection measurement technique, the intrinsic impedance of a material at 200 MHz is found to be approximately given by...Assuming that the material is nonmagnetic, determine its conductivity σ and the relative dielectric constant ... Get solution

36. Poynting flux. The electric and magnetic field expressions for a uniform plane wave propagating in a lossy medium are as follows:...The frequency of operation is f = 108 Hz, and the electrical parameters of the medium are є = 18.5є0, μ0, and σ. (a) Find the time-average electromagnetic power density entering a rectangular box-shaped surface like that shown in Figure assuming a = d = 1 m and b = 0.5 m. (b) Determine the power density exiting this region and compare with (a). (c) The difference between your results in (a) and (b) should represent electromagnetic power lost in the region enclosed by the square-box region. Can you calculate this dissipated power by any other method (i.e., without using the Poynting vector)? If yes, carry out this calculation. Hint: You may first need to find σ.Figure Poynting flux. Problem.... Get solution

37. Laser beams. The electric field component of a laser beam propagating in the z direction is approximated by...where E0 is the amplitude on the axis and a is the effective beam radius, where the electric field amplitude is a factor of e−1 lower than E0. (a) Find the corresponding expression for the magnetic field E. (b) Show that the time-average power density at the center of the laser beam is given by...where η = 377. (c) Find the total power of the laser beam. Consider a typical laboratory helium-neon laser with a total power of 5 mW and an effective radius of a = 400 μm. What is the power density at the center of the beam? (d) The power density of solar electromagnetic radiation at the Earth’s surface is 1400 W-m−2. At what distance from the Sun would its power density be equal to that for the helium-neon laser in part (c)? (e) One of the highest-power lasers built for fusion experiments operates at λ = 1.6 μm, produces 10.2 kJ for 0.2 ns and is designed for focusing on targets of 0.5 mm diameter. Estimate the electric field strength at the center of the beam. Is the field large enough to break down air? What is the radiation pressure of the laser beam? How much weight can be lifted with the pressure of this beam? Get solution

38. Maxwell’s equations. Consider a parallel-plate transmission line with perfectly conducting plates of large extent, separated by a distance of d meters. As shown in Figure, an alternating surface current density Js in the z direction flows on the conductor surface:...Figure Surface current. Problem....(a) Find an expression for the electric field, and determine the voltage between the plates, for d = 0.1 m and J0 = 1 A-m−1. (b) Use the continuity equation to find an expression for the surface charge density ρs (z , t ). Get solution

39. Uniform plane wave. A uniform plane electromagnetic wave propagates in free space with electric and magnetic field components as shown in Figure:...The wave frequency is 300 MHz and the electric field amplitude E0 = 1 V-m−1. A square loop antenna with side length a = 10 cm is placed at z = 2 m as shown. (a) Find the voltage Vind(t ) induced at the terminals of the loop. (b) Repeat (a) for the loop located at a distance of d = 3 m from the x axis instead of 2 m as shown. Compare your answers in (a) and (b).Figure Uniform plane wave. Problem.... Get solution

40. FM radio. An FM radio station operating at 100 MHz radiates a circularly polarized plane wave with a total isotropically radiated power of 200 kW. The transmitter antenna is located on a tower 500 m above the ground. (a) Find the rms value of the electric field 1 km away from the base of the antenna tower. Neglect the effects of reflections from the ground and other boundaries. (b) If the primary coverage radius of this station is ∼100 km, find the approximate time-average power density of the FM wave at this distance. Get solution

41. Mobile phones. Cellular phone antennas installed on cars have a maximum output power of 3 W, set by the Federal Communications Commission (FCC) standards. The incident electromagnetic energy to which the passengers in the car are exposed does not pose any health threats, both because they are some distance away from the antenna and also because the body of the car and glass window shield them from much of the radiation.79 (a) For a car with a synthetic roof, the maximum localized power density in the passenger seat is about 0.3 mW-(cm)−2. For cars with metal roofs, this value reduces to 0.02 mW-(cm)−2 or less. Antennas mounted on the trunk or in the glass of the rear windshield deliver power densities of about 0.35–0.07 mW-(cm)−2 to passengers in the back seat. Compare these values with the IEEE safety limit (IEEE Standard C95.1-1991) in the cellular phone frequency range (which is typically 800–900 MHz) and comment on the safety of the passengers. Note that from 300 MHz to 15 GHz, IEEE safety limits 80 specify a maximum allowable power density that increases linearly with frequency as |Sav|max = f /1500 in mW-(cm)−2, where the frequency f is in MHz. For reference, the maximum allowable power density is 1 mW-(cm)−2 at 1.5 GHz. (b) Calculate the maximum output power of a cellular phone antenna installed on the metal roof of a car such that the localized power density in the passenger compartment is equal to the IEEE safety limit at 850 MHz. Get solution

42. Radar aboard a Navy ship. Some shipboard personnel work daily in an environment where the radio frequency (RF) power density only a few feet above their heads may exceed safe levels. Some areas on the deck of the ship are not even allowed to personnel due to high power densities. Pilots of aircrafts routinely fly through the ship’s radar beams during takeoff and landing operations. On one of the Navy aircraft carriers, the average power density along the axis of the main beam (where the field intensity is the greatest) 100 ft away from a 6 kW missile control radar operating in the C-band is measured 81 to be about 300 mW-(cm)−2. (a) If it is assumed that human exposures in such environments are limited to less than 10 mW-(cm)−2, calculate the approximate distance along the beam axis that can be considered as the hazardous zone. (b) Assuming the operation frequency of the C-band radar to be 5 GHz, recalculate the hazardous zone along the radar’s main beam based on the IEEE safety limit 82 for the average power density given by |Sav|max = f /1500 in mW-(cm)−2 (valid over the frequency range 300 MHz–15 GHz), where f is in MHz. Get solution

43. Radio frequency exposure time. In 1965, the U.S. Army and Air Force amended their use of the prevailing 10 mW-(cm)−2 exposure guideline to include a time limit for exposures, given by the formula...where |Sav| is the average power density [in mW-(cm)−2] of exposure and tmax is the maximum recommended exposure duration, in minutes.83 Consider a person standing on the deck of a Navy ship where the peak rms electric field strength due to a microwave radar transmitter is measured to be 140 V-m−1. Using the above formula, calculate the maximum exposure time allowed (in hours) for this person to stay at that location. Get solution

44. Microwave cataracts in humans. Over 50 cases of human cataract induction have been attributed to microwave exposures, primarily encountered in occupational situations involving acute exposure to presumably relatively high-intensity fields.84 The following are three reported incidents of cataracts caused by microwave radiation: (1) A 22-year-old technician exposed approximately five times to 3 GHz radiation at an estimated average power density of 300 mW-(cm)−2 for 3 min/exposure developed bilateral cataracts. (2) A person was exposed to microwaves for durations of approximately 50 hour/month over a 4-year period at average power densities of less than 10 mW-(cm)−2 in most instances, but with a period of 6 months or more during which the average power density was approximately 1 W-(cm)−2. (3) A 50-year-old woman was intermittently exposed to leakage radiation from a 2.45 GHz microwave oven of approximately 1 mW-(cm)−2 during oven operation, with levels of up to 90 mW-(cm)−2 when the oven door was open, presumably over a period of approximately 6 years prior to developing cataract. For each of these above cases, compare the power densities with the IEEE standards (see Problem ) and comment.Mobile phones. Cellular phone antennas installed on cars have a maximum output power of 3 W, set by the Federal Communications Commission (FCC) standards. The incident electromagnetic energy to which the passengers in the car are exposed does not pose any health threats, both because they are some distance away from the antenna and also because the body of the car and glass window shield them from much of the radiation.79 (a) For a car with a synthetic roof, the maximum localized power density in the passenger seat is about 0.3 mW-(cm)−2. For cars with metal roofs, this value reduces to 0.02 mW-(cm)−2 or less. Antennas mounted on the trunk or in the glass of the rear windshield deliver power densities of about 0.35–0.07 mW-(cm)−2 to passengers in the back seat. Compare these values with the IEEE safety limit (IEEE Standard C95.1-1991) in the cellular phone frequency range (which is typically 800–900 MHz) and comment on the safety of the passengers. Note that from 300 MHz to 15 GHz, IEEE safety limits 80 specify a maximum allowable power density that increases linearly with frequency as |Sav|max = f /1500 in mW-(cm)−2, where the frequency f is in MHz. For reference, the maximum allowable power density is 1 mW-(cm)−2 at 1.5 GHz. (b) Calculate the maximum output power of a cellular phone antenna installed on the metal roof of a car such that the localized power density in the passenger compartment is equal to the IEEE safety limit at 850 MHz. Get solution

45. VHF TV signal. The magnetic field component of a 10 μW-m−2, 200 MHz TV signal in air is given by...(a) What are the values of H0 and a? (b) Find the corresponding electric field E(x, y, t ). What is the polarization of the wave? (c) An observer at z = 0 is equipped with a wire antenna capable of detecting the component of the electric field along its length. Find the maximum value of the measured electric field if the antenna wire is oriented along the (i) x direction, (ii) y direction, (iii) 45◦ line between the x and y directions. Get solution

46. FM polarization. Find the type (linear, circular, elliptical) and sense (right- or left-handed) of the polarization of the FM broadcast signal given in ExampleFM broadcast signal. An FM radio broadcast signal traveling in the y direction in air has a magnetic field given by the phasor...(a) Determine the frequency (in MHz) and wavelength (in m). (b) Find the corresponding E(y).(c) Write the instantaneous expression for E(y, t ) and H(y, t ). Get solution

47. Unknown wave polarization. The magnetic field component of a uniform plane wave in air is given by...where a is a real constant. (a) Find the wavelength λ and frequency f . (b) Find the total time-average power density carried by this wave. (c) Determine the type (linear, circular, elliptical) and sense (right- or left-handed) of the polarization of this wave when a = 1. (d) Repeat part (c) when a = 3. Get solution

48. Linear and circularly polarized waves. Two electromagnetic waves operating at the same frequency and propagating in the same direction (y direction) in air are such that one of them is linearly polarized in the x direction, whereas the other is left-hand circularly polarized (LHCP). However, the electric and magnetic field components of the two waves appear identical at one instant within every 50 ps time interval. The linearly polarized wave carries a time-average power density of 1.4 W-m−2. An observer located at y = 0 uses a receiving antenna to measure the x component of the total electric field only and records a maximum field magnitude of about 65 V-m−1 over every time interval of 0.5 ns. (a) Write the mathematical expressions for the electric field components of each wave, using numerical values of various quantities whenever possible. (b) Find the ratio of the time-average power densities of the LHCP and the linearly polarized waves. Get solution

49. Two circularly polarized waves. Consider two circularly polarized waves traveling in the same direction transmitted by two different satellites operating at the same frequency given by...where E01 and E02 are real constants. (a) If the total time-average power densities of these two waves are equal, find the polarization of the total wave. (b) Repeat part (a) for the case when the total time-average power density of the first wave is four times the total time-average power density of the second wave. Get solution

50. Wave polarization. Consider the following complex phasor expression for a time-harmonic magnetic field in free space:...(a) Is this a uniform plane wave? What is its frequency? (b) What is the direction of propagation and the state of polarization (specify both the type and sense of polarization) of this electromagnetic field? (c) Find the associated electric field phasor and the total time-average power density in the direction of propagation. Get solution

51. Wave polarization. The electric field component of a communication satellite signal travelling in free space is given by...(a) Find the corresponding H(z ). (b) Find the total time-average power density carried by this wave. (c) Determine the polarization (both type and sense) of the wave. Get solution

52. Wave polarization. A fellow engineer makes the following two measurements of the electric field vector of a uniform plane wave propagating in the x direction in a simple, lossless and nonmagnetic (μ = μ0) medium:...(a) Are these two measurements enough to determine the polarization type and sense of the wave? What about the wavelength of the wave? (b) Now the engineer offers two other measurements made at the same time t = 0:...Furthermore, you are informed that the electric field amplitude at t = 0 does not exceed 5 V-m−1 at any point between x = 0 and x = 0.75 m. Given this information, determine the type and sense of polarization of the wave. Get solution

53. Superposition of two waves. The electric field components of two electromagnetic waves at the same frequency and propagating in free space are represented by...Find and sketch the locus of the total electric field measured at the origin (x = y = z = 0) if E0 is equal to (a) 10 V-m−1, (b) 20 V-m−1, (c) 40 V-m−1, respectively. Get solution