45792-UlabyISMCh07

45792-UlabyISMCh07 - 317 Chapter 7 Plane-Wave Propagation...

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Unformatted text preview: 317 Chapter 7: Plane-Wave Propagation Lesson #43 Chapter -- Section: 7-1 Topics: Time-harmonic fields Highlights: Phasors Complex permittivity Wave equations Special Illustrations: 318 Lesson #44 Chapter -- Section: 7-2 Topics: Waves in lossless media Highlights: Uniform plane waves Intrinsic impedance Wave properties Special Illustrations: Example 7-1 CD-ROM Modules 7.3 and 7.4 319 Lesson #45 and 46 Chapter -- Section: 7-3 Topics: Wave polarization Highlights: Definition of polarization Linear, circular, elliptical Special Illustrations: CD-ROM Demos 7.1-7.5 Liquid Crystal Display Liquid Crystal Display (LCD) LCDs are used in digital clocks, cellular phones, desktop and laptop computers, and some televisions and other electronic systems. They offer a decided advantage over other display technologies, such as cathode ray tubes, in that they are much lighter and thinner and consume a lot less power to operate. LCD technology relies on special electrical and optical properties of a class of materials known as liquid crystals, first discovered in the 1880s by botanist Friedrich Reinitzer. Physical Principle Liquid crystals are neither a pure solid nor a pure liquid, but rather a hybrid of both. One particular variety of interest is the twisted nematic liquid crystal whose molecules have a natural tendency to assume a twisted spiral structure when the material is sandwiched between finely grooved glass substrates with orthogonal orientations (A). Note that the molecules in contact with the grooved surfaces align themselves in parallel along the grooves. The molecular spiral causes the crystal to behave like a wave polarizer; unpolarized light incident upon the entrance substrate follows the orientation of the spiral, emerging through the exit substrate with its polarization (direction of electric field) parallel to the groove's direction. 320 Lesson #47 Chapter -- Section: 7-4 Topics: Waves in lossy media Highlights: Attenuation and skin depth Low loss medium Good conductor Special Illustrations: CD-ROM Demos 7.6-7.8 321 Lesson #48 Chapter -- Section: 7-5 Topics: Current flow in conductors Highlights: Skin depth dependence on frequency Surface impedance Special Illustrations: 322 Lesson #49 Chapter -- Section: 7-6 Topics: EM power density Highlights: Power density in a lossless medium Power density in a lossy medium Time-average power Special Illustrations: CD-ROM Module 7.5 CHAPTER 7 323 Chapter 7 Section 7-2: Propagation in Lossless Media Problem 7.1 The magnetic field of a wave propagating through a certain nonmagnetic material is given by Find (a) the direction of wave propagation, (b) the phase velocity, (c) the wavelength in the material, (d) the relative permittivity of the material, and (e) the electric field phasor. Solution: (a) Positive y-direction. (b) 108 rad/s, k 0 5 rad/m. Hence, and Problem 7.2 Write general expressions for the electric and magnetic fields of a 1-GHz sinusoidal plane wave traveling in the y-direction in a lossless nonmagnetic medium with relative permittivity r 9. The electric field is polarized along the x-direction, its peak value is 6 V/m and its intensity is 4 V/m at t 0 and y 2 cm. Eyt Ee jt ^ x7 54 cos 108 t 0 5y (V/m) E 251 33^ y ^ z30e j0 5y 10 3 ^ x7 54e j0 5y ^ k ^ y and H ^ z30e j0 5y 10 3 (A/m) 120 r E ^ k H (c) 2 k 2 0 5 12 6 m. 2 c 2 3 108 (d) r 2 25. up 2 108 (e) From Eq. (7.39b), 120 15 251 33 up k 108 05 2 108 m/s () (V/m) H ^ z 30 cos 108 t 0 5y (mA/m) 324 CHAPTER 7 Hence, which gives and Problem 7.3 The electric field phasor of a uniform plane wave is given by ^ E y 10e j0 2z (V/m). If the phase velocity of the wave is 1 5 10 8 m/s and the relative permeability of the medium is r 2 4, find (a) the wavelength, (b) the frequency f of the wave, (c) the relative permittivity of the medium, and (d) the magnetic field Hzt . (b) (c) From up c r r r 1 r c up 2 1 24 3 15 2 1 67 f up 1 5 108 31 42 4 77 106 Hz 4 77 MHz 2 k 2 02 10 31 42 m Solution: (a) From E ^ y10e j0 2z (V/m), we deduce that k 0 2 rad/m. Hence, Eyt ^ x 6 cos 2 109 t 20y 120 19 0 2 1 rad 120 19 0 0 4 cos 1 4 6 0 84 rad (V/m) 4 6 cos 20 2 10 2 0 6 cos At t 0 and y 2 cm, E 4 V/m: Eyt 0 4 0 k 2 2 2 f 2 109 r r 9 0 c 3 108 ^ x6 cos 2 109 t 20y 0 (V/m) 2 f 2 109 rad/s Solution: For f 1 GHz, r 1, and r 9, 20 rad/m CHAPTER 7 (d) 325 Problem 7.4 The electric field of a plane wave propagating in a nonmagnetic material is given by Determine (a) the wavelength, (b) r , and (c) H. (b) But Hence, (c) H E 1^ k 1 ^ ^ ^ x y3 sin 107t 0 2x z4 cos 107 t 0 2x 3 4 ^ ^ z sin 107 t 0 2x y cos 107t 0 2x (A/m) r c up 2 3 5 108 107 up c r 2 36 up k 107 0 2 5 107 m/s Solution: (a) Since k 0 2, 2 k 2 0 2 10 m E ^ y 3 sin 107t 0 2x ^ z 4 cos with 2 f 9 54 106 rad/s. Hzt 107 t 0 2x (V/m) H ^ x 22 13e j0 2z (mA/m) r 24 120 r 1 67 1 1 ^ ^ ^ z E z y10e j0 2z ^ x 22 13 cos t 0 2z (mA/m) 120 451 94 () 326 with CHAPTER 7 Problem 7.5 A wave radiated by a source in air is incident upon a soil surface, whereupon a part of the wave is transmitted into the soil medium. If the wavelength of the wave is 60 cm in air and 20 cm in the soil medium, what is the soil's relative permittivity? Assume the soil to be a very low loss medium. Problem 7.6 The electric field of a plane wave propagating in a lossless, nonmagnetic, dielectric material with r 2 56 is given by Determine: (a) f , up , , k, and , and (b) the magnetic field H. Solution: (a) k up f 2 0 r up c r 108 1 875 108 m/s 2 56 1 875 108 3 12 cm 6 109 2 201 4 rad/m 3 12 10 2 377 377 235 62 16 2 56 3 f 3 109 Hz 3 GHz 2 f 6 109 rad/s E ^ y 20 cos 6 109 t kz (V/m) r 0 2 Solution: From 0 r , 60 20 2 9 0 r 120 6 20 62 83 () CHAPTER 7 (b) 327 Section 7-3: Wave Polarization Problem 7.7 An RHC-polarized wave with a modulus of 2 (V/m) is traveling in free space in the negative z-direction. Write down the expression for the wave's electric field vector, given that the wavelength is 6 cm. y t=0 z x t=/2 Figure P7.7: Locus of E versus time. Solution: For an RHC wave traveling in ^ z, let us try the following: a Modulus E a 2 a2 a2 2 (V/m). Hence, 2 2 2 E ^ x a cos t kz ^ ya sin t kz H 20 cos 6 109 t kz 20 ^ cos 6 109 t 201 4 z x 235 62 ^ x 8 49 10 2 cos 6 109 t 201 4 z ^ x (A/m) 328 CHAPTER 7 ^ Next, we need to check the sign of the y-component relative to that of the ^ x-component. We do this by examining the locus of E versus t at z 0: Since ^ ^ the wave is traveling along z, when the thumb of the right hand is along z (into the page), the other four fingers point in the direction shown (clockwise as seen from ^ above). Hence, we should reverse the sign of the y-component: with k and Problem 7.8 For a wave characterized by the electric field identify the polarization state, determine the polarization angles , and sketch the locus of E 0 t for each of the following cases: (a) ax 3 V/m, ay 4 V/m, and 0, (b) ax 3 V/m, ay 4 V/m, and 180 , (c) ax 3 V/m, ay 3 V/m, and 45 , (d) ax 3 V/m, ay 4 V/m, and 135 . Solution: 0 sin 2 Case (a) (b) (c) (d) (a) E z t (b) E z t (c) E z t (d) E z t ax 3 3 3 3 ay 4 4 3 4 0 180 45 135 1 ^ x3 cos t ^ x3 cos t ^ x3 cos t ^ x3 cos t kz kz kz kz ^ y4 cos t ^ y4 cos t ^ y3 cos t ^ y4 cos t kz . kz . kz 45 . kz 135 . 0 53 13 53 13 45 53 13 53 13 53 13 45 56 2 0 0 22 5 21 37 sin 20 sin [Eq. (7.59b)] tan 2 tan 20 cos [Eq. (7.59a)] tan ay ax [Eq. (7.60)] Polarization State Linear Linear Left elliptical Right elliptical Ezt ^ xax cos t kz ^ yay cos t kz kc 2 3 6 2 108 1010 (rad/s) 2 2 10 104 72 (rad/m) E kz ^ x 2 cos t ^ y 2 sin t kz (V/m) CHAPTER 7 y 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 -4 1 2 3 4 4 3 2 1 329 y x -4 -3 -2 -1 -1 -2 -3 -4 1 2 3 4 x (a) y 4 3 2 1 -4 -2 -1 -2 -3 -4 2 4 (b) y 4 3 2 1 x -4 -2 -1 -2 -3 -4 2 4 x (c) (d) Figure P7.8: Plots of the locus of E 0 t . Problem 7.9 The electric field of a uniform plane wave propagating in free space ^ ^ x jy 20e jz 6 (V/m). Specify the modulus and direction of the is given by E electric field intensity at the z 0 plane at t 0, 5 and 10 ns. 330 Solution: CHAPTER 7 From Therefore, the wave is LHC polarized. ^ Problem 7.10 A linearly polarized plane wave of the form E xax e jkz can be expressed as the sum of an RHC polarized wave with magnitude a R and an LHC polarized wave with magnitude aL . Prove this statement by finding expressions for aR and aL in terms of ax . Solution: jkz By equating real and imaginary parts, ax aR ax 2. aR aL , 0 ^ xax ^ aR x ^ jy ^ aL x E ER EL ^ jy LHC wave: EL ^ aL x ^ ye j 2 e jkz ^ aL x ^ jy e jkz aR aL , or aL RHC wave: ER ^ aR x ^ ye j 2 e jkz ^ aR x E ^ xax e ^ jy e jkz t 5 10 t 7 0 25 0 5 45 90 At z 0, 0 at t at t at t 0 5 ns 10 ns f c kc 6 3 108 2 2 7 2 f 5 10 rad/s tan 1 Ey Ex t z 6 E 2 Ex 2 1 2 Ey 20 (V/m) 25 107 Hz ^ x20 cos t z 6 ^ y20 sin t z 6 (V/m) ^ x20 cos t z 6 ^ y20 cos t ^ x ^ ye j 2 20e jz 6 jt e ^ x ^ jy 20e jz 6 jt Ezt Ee jt e z 6 2 ax 2, CHAPTER 7 331 Problem 7.11 Ezt The electric field of an elliptically polarized plane wave is given by Determine (a) the polarization angles and (b) the direction of rotation. Solution: (a) Phasor form: Since is defined as the phase of E y relative to that of Ex , Solution: (a) E1 ^ x 2e jkz ^ y 2e jkz e j 2 ^ x 2 cos t kz ^ y 2 cos t kz E1 ^ x 2 cos t kz ^ y 2 sin t Problem 7.12 Compare the polarization states plane waves: ^ ^ (a) wave 1: E1 x2 cos t kz y2 sin t ^ ^ wave 2: E2 x2 cos t kz y2 sin t ^ ^ (b) wave 1: E1 x2 cos t kz y2 sin t ^ 2 cos t kz y2 sin t ^ wave 2: E2 x (b) Since 0, the wave is right-hand elliptically polarized. of each of the following pairs of kz kz kz kz , , , . kz sin 2 sin 20 sin 0 40 or 8 73 2 tan 2 tan 20 cos 0 65 0 tan 1 30 30 10 71 56 or E ^ x10e j30 ^ y30 e jkz 73 5 ^ x10 cos t kz 30 ^ y30 cos t kz Ezt ^ x10 sin t kz 60 ^ y30 cos t kz (V/m) ^ x 10 sin t kz 60 ^ y 30 cos t kz (V/m) 332 CHAPTER 7 Wave 2 has the same magnitude and phases as wave 1 except that its direction is ^ ^ along z instead of z. Hence, the locus of rotation of E will match the left hand instead of the right hand. Thus, wave 2 is LHC. (b) Reversal of direction of propagation (relative to wave 1) makes wave 2 RHC. Determine the polarization state from your plot. Tip of E rotates in accordance with right hand (with thumb pointing along Hence, wave state is RHE. Wave direction is ^ z. At z 0, E ^ x sin t ^ y 2 cos t E ^ x sin t kz ^ y 2 cos t Solution: kz Ezt ^ x sin t kz ^ y 2 cos t Problem 7.13 Plot the locus of E 0 t for a plane wave with E2 ^ x 2e jkz ^ y 2e jkz e j Wave 1 is LHC. E1 ^ x 2e jkz ^ y 2e jkz j 2 e 2 kz E1 ^ x 2 cos t kz ^ y 2 sin t E2 ^ x 2e jkz ^ y 2e jkz e j 2 Hence, wave 1 is RHC. Similarly, kz ^ z). 2 0 tan 1 ay ax tan 1 1 45 CHAPTER 7 y t=/2 2 333 -1 z 1 t=0 x -2 Figure P7.13: Locus of E versus time. Sections 7-4: Propagation in a Lossy Medium Problem 7.14 For each of the following combination of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate , , , up , and c : Solution: Using equations given in Table 7-1: Case (a) Case (b) 4.5 quasi-conductor 9.75 Np/m 12.16 rad/m 51.69 cm 0 52 108 m/s 39 54 j31 72 Case (c) 600 good conductor 6 3 10 4 Np/m 6 3 10 4 rad/m 10 km 0 1 108 m/s 6 28 1 j Type up c 3 6 10 13 low-loss dielectric 8 42 10 11 Np/m 468 3 rad/m 1.34 cm 1 34 108 m/s 168 5 (c) wood with r 1, r 3, and 10 4 S/m at 1 kHz. (b) animal tissue with r 1, r 12, and (a) glass with r 1, r 5, and 10 12 S/m at 10 GHz, 0 3 S/m at 100 MHz, 334 CHAPTER 7 Problem 7.15 Dry soil is characterized by r 2 5, r 1, and 10 4 (S/m). At each of the following frequencies, determine if dry soil may be considered a good conductor, a quasi-conductor, or a low-loss dielectric, and then calculate , , , p , and c : (a) 60 Hz, (b) 1 kHz, (c) 1 MHz, (d) 1 GHz. Good conductor 1 54 Good conductor Quasi-conductor Low-loss dielectric 33.14 0.19 238.27 (m) up (m/s) c () 4 08 2 45 104 j 104 107 180 Problem 7.16 In a medium characterized by r 9, r 1, and 0 1 S/m, determine the phase angle by which the magnetic field leads the electric field at 100 MHz. Hence, quasi-conductor. 125 67 1 j2 71 49 j44 18 84 04 1 2 c 1 j j 1 2 120 1 r 0 r 1 2 2 9 0 1 36 108 10 9 2 31 72 Solution: The phase angle by which the magnetic field leads the electric field is where is the phase angle of c . 1 54 1 6 28 1 j 204 28 j65 89 106 18 108 (rad/m) 10 4 6 28 10 4 3 49 10 2 19 108 (Np/m) 1 54 10 4 6 28 10 4 1 13 10 2 1 19 10 2 f r 0 Type of medium f 60 Hz 12 104 Solution: r 2 5, r 1, 10 4 S/m. 1 kHz 720 1 MHz 0.72 1 GHz 72 10 4 2 CHAPTER 7 Therefore 31 72 . ^ Since H 1 c k E, H leads E by E by 31 72 . 335 Problem 7.17 Generate a plot for the skin depth s versus frequency for seawater for the range from 1 kHz to 10 GHz (use log-log scales). The constitutive parameters of seawater are r 1, r 80 and 4 S/m. Solution: See Fig. P7.17 for plot of s versus frequency. 10 1 Skin depth vs. frequency for seawater 10 Skin depth (m) 0 10 -1 10 -2 10 -3 10 -2 10 -1 10 10 Frequency (MHz) 0 1 10 2 10 3 10 Figure P7.17: Skin depth versus frequency for seawater. 3 0 0 r 2 f r c2 0 r 80 c2 80 108 2 4 36 2 f 10 9 80 72 80 f s 1 1 2 1 2 1 2 1 , or by 31 72 . In other words, H lags 109 4 336 CHAPTER 7 Problem 7.18 Ignoring reflection at the air-soil boundary, if the amplitude of a 3-GHz incident wave is 10 V/m at the surface of a wet soil medium, at what depth will it be down to 1 mV/m? Wet soil is characterized by r 1, r 9, and 5 10 4 S/m. Solution: 9 Hence, medium is a low-loss dielectric. Problem 7.19 The skin depth of a certain nonmagnetic conducting material is 3 m at 5 GHz. Determine the phase velocity in the material. Problem 7.20 Based on wave attenuation and reflection measurements conducted at 1 MHz, it was determined that the intrinsic impedance of a certain medium is 28 1 45 and the skin depth is 2 m. Determine (a) the conductivity of the material, (b) the wavelength in the medium, and (c) the phase velocity. Solution: (a) Since the phase angle of c is 45 , the material is a good conductor. Hence, or 28 1 cos 45 19 87 1 j c 28 1e j45 28 1 cos 45 j28 1 sin 45 up 2 f 2 f s 2 5 109 3 10 6 9 42 104 Solution: For a good conductor, z 287 82 m , and for any material s 10 3 10e 0 032z ln 10 4 1 . Hence, (m/s) 2 120 2 r 5 10 4 120 2 9 0 032z 0 032 5 10 4 36 2 3 109 10 9 3 32 E z E0 e z 10e z 10 4 (Np/m) CHAPTER 7 337 Problem 7.21 The electric field of a plane wave propagating in a nonmagnetic medium is given by Obtain the corresponding expression for H. Solution: From the given expression for E, From (7.65a) and (7.65b), Using the above values for , , and , we obtain the following: r 1 j 1 j 0 r r r 1 2 377 16 5 47 16 1 2 1 j c 1 2 5 47 r 16 2 2 2 c2 r 157 9 e j36 85 2 2 2 2 0 0 r 40 (rad/m) 30 (Np/m) 2 109 (rad/s) 2 c2 r E ^ z 25e 30x cos 2 (c) up f 106 12 57 1 26 107 m/s. 109 t 40x (V/m) 2 2 05 4 12 57 m () (b) Since Therefore, for a good conductor, and 0 5, it follows that 0 5. Since 1 s 1 2 0 5 Np/m, 19 87 05 19 87 2 52 10 2 S/m 338 CHAPTER 7 ^ z 25e 30x e j40x 1 ^ 1 ^ x k E c 157 9 e j36 85 Section 7-5: Current Flow in Conductors Problem 7.22 In a nonmagnetic, lossy, dielectric medium, a 300-MHz plane wave is characterized by the magnetic field phasor Obtain time-domain expressions for the electric and magnetic field vectors. To find c , we need and . From the given expression for H, whose solution gives Similarly, from (7.65b), which gives r 0 91 2 2 9 2 3 108 2 4 10 7 r 2 2 r 1 95 10 9 36 4 81 2 3 108 2 4 10 7 r 2 2 2 Also, we are given than f 300 MHz 3 108 Hz. From (7.65a), 9 (rad/m) 2 (Np/m) E ^ c k H Solution: ~ H ^ x j4^ e z 2y e j9y (A/m) 10 9 36 H He jt ^ y 0 16 e 30x cos 2 109 t 40x 36 85 (A/m) H E ^ z 25e 30x e j40x ^ y 0 16 e 30x e 40x e j36 85 CHAPTER 7 c 339 Hence, Problem 7.23 A rectangular copper block is 30 cm in height (along z). In response to a wave incident upon the block from above, a current is induced in the block in the positive x-direction. Determine the ratio of the a-c resistance of the block to its d-c resistance at 1 kHz. The relevant properties of copper are given in Appendix B. l w J 30 cm Figure P7.23: Copper block of Problem 7.23. ^ xe 2y cos t 9y ^ z 4e 2y sin t ^ x j4^ e z e e H He jt 2y j9y jt 9y (A/m) ^ z 256 9 e 2y cos t 9y 12 6 ^ x 1 03 103 e 2y cos t E Ee jt 9y 102 6 ^ x 4e j 2 ^ z 256 9 e 2y e j9y j12 6 e ^ x j4 ^ z 256 9 e 2y e j9y j12 6 e E ^ 256 9 e j12 6 y ^ x j4^ e z 2y e j9y (V/m) 1 j 0 r 0 91 1 95 1 2 1 j 1 2 377 0 93 1 95 j0 21 256 9 e j12 6 340 Solution: CHAPTER 7 Problem 7.24 The inner and outer conductors of a coaxial cable have radii of 0.5 cm and 1 cm, respectively. The conductors are made of copper with r 1, r 1 and 5 8 107 S/m, and the outer conductor is 0.5 mm thick. At 10 MHz: (a) Are the conductors thick enough to be considered infinitely thick so far as the flow of current through them is concerned? (b) Determine the surface resistance R s . (c) Determine the a-c resistance per unit length of the cable. Solution: (a) From Eqs. (7.72) and (7.77b), Hence, Hence, conductor is plenty thick. (b) From Eq. (7.92a), 58 107 21 10 5 (c) From Eq. (7.96), 5 3 2 R Rs 2 1 a 1 b 82 10 2 4 1 10 1 10 0 039 Rs 1 s 1 d s 0 5 mm 0 021 mm 25 82 10 4 (/m) s f 1 2 107 4 10 7 58 107 1 2 0 021 mm 03 f 03 103 4 10 7 58 107 1 2 Rac Rdc 03 s a-c resistance Rac 143 55 d-c resistance Rdc l l A 0 3 w l ws CHAPTER 7 341 Section 7-6: EM Power Density Problem 7.25 The magnetic field of a plane wave traveling in air is given by ^ H x 50 sin 2 107 t ky (mA/m). Determine the average power density carried by the wave. Solution: Determine the direction of wave travel and the average power density carried by the wave. Solution: 0 r The wave is traveling in the negative x-direction. Problem 7.27 The electric-field phasor of a uniform plane wave traveling downward in water is given by (a) obtain an expression for the average power density, (b) determine the attenuation rate, and (c) determine the depth at which the power density has been reduced by 40 dB. ^ where z is the downward direction and z 0 is the water surface. If E ^ x 5e 0 2z e j0 2z (V/m) Sav ^ x 32 22 2 ^ x 13 2 40 ^ x0 05 120 9 40 () (W/m2 ) E ^ y 3 cos 107t kx ^ z 2 cos 107t kx (V/m) 4 S/m, Problem 7.26 A wave traveling in a nonmagnetic medium with r characterized by an electric field given by Sav ^ z ^ x 10 6 ^ y 0 50 2 2 120 50 2 2 10 6 ^ y0 48 E ^ 0 y H ^ z0 50 sin 2 107t H ^ x50 sin 2 107t ky (mA/m) ky (mV/m) (W/m2 ) 9 is 342 Solution: (a) Since CHAPTER 7 From Eq. (7.109), (b) A 8 68z 8 68 0 2z 1 74z (dB). (c) 40 dB is equivalent to 10 4 . Hence, Problem 7.28 The amplitudes of an elliptically polarized plane wave traveling in a lossless, nonmagnetic medium with r 4 are Hy0 3 (mA/m) and Hz0 4 (mA/m). Determine the average power flowing through an aperture in the y-z plane if its area is 20 m2 . Solution: 0 120 60 188 5 r 4 2 188 5 2 ^ ^ x Hy0 Hx0 x 9 16 10 6 2 36 2 2 Sav A 2 36 10 3 20 47 13 (mW) Problem 7.29 A wave traveling in a lossless, nonmagnetic medium has an electric field amplitude of 24.56 V/m and an average power density of 2.4 W/m 2 . Determine the phase velocity of the wave. Solution: or 24 56 2 2 24 125 67 Sav E0 2 2 E0 2 2Sav P Sav (mW/m2 ) or z 23 03 m. 10 4 e 2z e 0 4z ln 10 2 4 0 4z Sav ^ z E0 2 e 2 c 2z cos ^ z 25 e 0 0707 0 4z cos 45 ^ z125e 0 4z (W/m2 ) 1 j 1 j 1 j 0 05 c 0 2, the medium is a good conductor. 02 4 0 0707e j45 () CHAPTER 7 But 343 Hence, Problem 7.30 At microwave frequencies, the power density considered safe for human exposure is 1 (mW/cm2 ). A radar radiates a wave with an electric field 3 000 R (V/m), where R is the amplitude E that decays with distance as E R distance in meters. What is the radius of the unsafe region? Solution: Problem 7.31 Consider the imaginary rectangular box shown in Fig. 7-19 (P7.31). (a) Determine the net power flux P t entering the box due to a plane wave in air given by ^ E xE0 cos t ky (V/m) (b) Determine the net time-average power entering the box. Solution: (a) 0 b Pt St Ay S t Ay cos2 t 2 E0 ac cos2 t 0 St E H ^ y 2 E0 cos2 t 0 ky H E ^ x E0 cos t ky E0 ^ z cos t ky 0 R 12 104 10 1 2 2 R2 34 64 m 10 3 103 R 2 1 120 12 104 kb Sav E R 20 2 1 (mW/cm2 ) 10 3 W/cm2 up c r 3 108 3 1 108 m/s 10 W/m2 0 r 377 r r 377 125 67 2 9 344 z CHAPTER 7 b a c y x Figure P7.31: Imaginary rectangular box of Problems 7.31 and 7.32. (b) 0 Net average energy entering the box is zero, which is as expected since the box is in a lossless medium (air). Problem 7.32 Repeat Problem 7.31 for a wave traveling in a lossy medium in which Solution: (a) ^ y 64e 40y cos 2 10 t 9 40y cos 2 109 t ^ z 0 64 e 20y cos 2 109t ^ x 100e 20y cos 2 St E H 109t 40y 40y 36 85 The box has dimensions A 1 cm, b 2 cm, and c 0 5 cm. 40y 36 85 H ^ z 0 64e 20y cos 2 10 t 9 40y 36 85 E ^ x 100e 20y cos 2 109 t 40y (V/m) (A/m) 0 Pav cos2 t cos2 t 2 E0 ac 0 2 where T 2 . 2 kb dt 0 Pav P t dt 1 T T CHAPTER 7 345 0 0 This is the average power absorbed by the lossy material in the box. Problem 7.33 Given a wave with calculate: (a) the time-average electric energy density av 0 0 (b) the time-average magnetic energy density av 0 0 av Solution: (a) 0 av we 2 E0 cos2 t and (c) show that we wm av . 1 2T T kz dt wm wm dt 1 T T 1 2T T H 2 dt we we dt 1 T T 1 2T E ^ xE0 cos t kz Pav 7 05 With a 1 cm, b 2 cm, and c 0 5 cm, 10 4 (W) T Pav 32ac 1 The average of cos t of . Hence, over a period T is equal to zero, regardless of the value e 40b cos 36 85 E 2 dt Pav P t dt P t dt 1 T T (b) 2 2 e cos 4 80y 36 85 cos 36 85 40b 109 t 32ac cos 4 109 t Pt 36 85 cos 36 85 St 64 40y e cos 4 109 t 2 S t Ay 0 S t Ay b 80y 36 85 Using the identity cos cos 1 2 cos cos , cos 36 85 346 CHAPTER 7 2 (b) av 0 (c) av Problem 7.34 A 60-MHz plane wave traveling in the x-direction in dry soil with relative permittivity r 4 has an electric field polarized along the z-direction. Assuming dry soil to be approximately lossless, and given that the magnetic field has a peak value of 10 (mA/m) and that its value was measured to be 7 (mA/m) at t 0 and x 0 75 m, develop complete expressions for the wave's electric and magnetic fields. c 6 107 3 108 ^ Given that E points along z and wave travel is along ^ x, we can write where E0 and 0 are unknown constants at this time. The intrinsic impedance of the medium is 0 120 60 () r 2 Ext ^ z E0 cos 2 60 106 t 0 8x 0 (V/m) k r 2 4 0 8 Solution: For f 60 MHz 6 107 Hz, r 4, r 1, (rad/m) wm 2 E0 42 2 E0 4 2 E0 4 we wm 1 2T 1 2T 2 E0 42 T 0 T H 2 dt 2 E0 cos2 t 2 H ^ y E0 cos t kz kz dt av av we With T , 2 E0 2 cos2 t kz dt 4 0 2 E0 2 cos2 t kz d t 4 0 2 E0 4 CHAPTER 7 ^ ^ With E along z and k along ^ x, (7.39) gives 347 or Hence, Also, Problem 7.35 At 2 GHz, the conductivity of meat is on the order of 1 (S/m). When a material is placed inside a microwave oven and the field is activated, the presence of the electromagnetic fields in the conducting material causes energy dissipation in the material in the form of heat. (a) Develop an expression for the time-average power per mm 3 dissipated in a material of conductivity if the peak electric field in the material is E 0 . Solution: (a) Let us consider a small volume of the material in the shape of a box of length d and cross sectional area A. Let us assume the microwave oven creates a wave traveling along the z direction with E along y, as shown. (b) Evaluate the result for meat with E 0 4 104 (V/m). Hxt ^ y 10 cos 1 2 10 t 8 0 8x 153 6 (mA/m) Ext ^ z 0 6 cos 1 2 which leads to 0 Hence, 153 6 . 108 t 0 8x 153 6 (V/m) H 0 75 m 0 7 10 3 10 cos 0 8 0 75 E0 E0 10 10 (mA/m) 60 10 3 0 6 (V/m) 0 Hxt ^ y E0 cos 1 2 108 t H 1 ^ k E 0 8x 0 (A/m) 10 3 348 z CHAPTER 7 A y E d k ^ x E Along y, the E field will create a voltage difference across the length of the box V , where V Ed Conduction current through the cross sectional area A is Hence, the instantaneous power is where V Ad is the small volume under consideration. The power per mm 3 is obtained by setting V 10 3 3 , 0 1 2 E 2 0 10 9 (W/mm3 ) Pav As a time harmonic signal, E E0 cos t. The time average dissipated power is 1 T T 2 E0 cos2 t dt 10 9 P P 10 9 E 2 10 9 E 2 V P IV E 2 Ad (W/mm3 ) I JA EA CHAPTER 7 (b) 349 Problem 7.36 A team of scientists is designing a radar as a probe for measuring the depth of the ice layer over the antarctic land mass. In order to measure a detectable echo due to the reflection by the ice-rock boundary, the thickness of the ice sheet should not exceed three skin depths. If r 3 and r 10 2 for ice and if the maximum anticipated ice thickness in the area under exploration is 1.2 km, what frequency range is useable with the radar? Solution: Hence, Since increases with increasing frequency, the useable frequency range is f 41 6 MHz For 25 10 3 6f 10 11 , f 41 6 MHz 2 Since 1, we can use (7.75a) for : 2 f r 0 0 2 r 0 f r c r f 10 2 3 108 3 6f 10 11 1 s 1 400 25 s 400 m 3s 1 2 km 1200 m 10 3 (Np/m) Np/m Pav 1 1 2 4 104 2 10 9 08 (W/mm3 ) ...
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This note was uploaded on 09/15/2008 for the course EE 330 taught by Professor Tofighimohammad- during the Spring '08 term at Penn State.

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