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Chapter 7: Plane-Wave Propagation

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Chapter 7: Plane-Wave Propagation 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 8 ¢ H £ zˆ 30cos 10 t 0 5y (mA/m) ¨ 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. 8 £ (b) ω £ 10 rad/s, k 0 5 rad/m. ω 8 10 8 £ £ up £ 2 10 m/s k 0 5 £ £ (c) λ £ 2π k 2π 0 5 12 6 m. § § 2 8 2 c 3 10 £ £ ε £ (d) r 8 2 25. up 2 10 (e) From Eq. (7.39b), ¡ E £ ηkˆ H ¨ ¡ µ 120π 120π η £ £ £ £ Ω ¡ ε ε 251 33 ( ) ¡ r 1 5 j0 ¢ 5y 3 ˆ ¡ £ k £ yˆ and H zˆ30e 10 (A/m) Hence, ¢ j0 ¢ 5y 3 j0 5y £ ¡ E £ 251 33yˆ zˆ30e 10 xˆ7 54e (V/m) ¨ ¨ and ¡£¢ jωt 8 ¢¤£ ¢¤£ ¢ E y ¡ t Ee xˆ7 54cos 10 t 0 5y (V/m) ¨ ¨ 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 εr £ 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. 324 CHAPTER 7 £ ε £ Solution: For f £ 1 GHz, µr 1, and r 9, 9 £ ¡ ω £ 2π f 2π 10 rad/s π π π π 9 2 2 2 f 2 10 ε ε ¡ π £ £ £ £ £ ¡ ¡ k ¡ r r 8 9 20 rad/m λ λ0 c 3 10 9 ¢¦£ ¢ E y ¡ t xˆ6cos 2π 10 t 20πy φ0 (V/m) ¨ © £ £ At t £ 0 and y 2 cm, E 4 V/m: 2 ¢¤£ ¢ 4 £ 6cos 20π 2 10 φ0 6cos 0 4π φ0 ¨ © ¨ © Hence, § 1 4 £ φ π £ ¡ 0 0 4 cos 0 84 rad ¨ 6 which gives φ ¥ £ 0 £ 2 1 rad 120 19 and 9 ¥ ¢¦£ ¢ E y ¡ t xˆ 6cos 2π 10 t 20πy 120 19 (V/m) ¨ © Problem 7.3 The electric field phasor of a uniform plane wave is given by j0 ¢ 2z 8 E £ yˆ 10e (V/m). If the phase velocity of the wave is 1 5 10 m/s and the relative permeability of the medium is µr £ 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 ¢ H z ¡ t . Solution: j0 ¢ 2z £ (a) From E £ yˆ10e (V/m), we deduce that k 0 2 rad/m. Hence, 2π 2π £ £ £ λ £ 10π 31 42 m k 0 2 (b) 8 up 1 5 10 6 £ £ £ f £ λ 4 77 10 Hz 4 77 MHz 31 42 (c) From § § 2 2 c ε 1 c 1 3 £ ¡ £ £ £ up r 1 67 ε ¡ µr r µr up 2 4 1 5 CHAPTER 7 325 (d) ¡ ¡ ¡ µ µr 2 4 £ £ ¡ η £ π π Ω ε 120 ε 120 451 94 ( ) r 1 67 1 1 ¢ j0 ¢ 2z j0 2z ¢¤ £ ¢¤ £ ¡ H £ η zˆ E η zˆ yˆ10e xˆ 22 13e (mA/m) ¨ ¨ ¢¦£ ω ¢ ¡ H z ¡ t xˆ 22 13cos t 0 2z (mA/m) © ω π π 6 £ with £ 2 f 9 54 10 rad/s. Problem 7.4 The electric field of a plane wave propagating in a nonmagnetic material is given by 7 7 ¢ ¢¢¡ E £ yˆ 3sin π 10 t 0 2πx zˆ 4cos π 10 t 0 2πx (V/m) ¨ ¨ © Determine (a) the wavelength, (b) εr, and (c) H. Solution: π (a) Since k £ 0 2 , π π λ 2 2 £ £ £ 10 m k 0 2π (b) ω π 7 10 7 £ £ up £ 5 10 m/s k 0 2π But c up £ ε ¡ r Hence, § § 2 8 2 c 3 10 ε £ £ £ r 7 36 up 5 10 (c) 1 ˆ 1 7 7 £ ¢ ¢ £ π π π π H ηk E ηxˆ yˆ3sin 10 t 0 2 x zˆ4cos 10 t 0 2 x ¡ ¨ © ¨ 3 π 7 π 4 π 7 π ¢ ¢ ¡ £ zˆ η sin 10 t 0 2 x yˆ η cos 10 t 0 2 x (A/m) ¨ ¨ ¨ 326 CHAPTER 7 with η π η 0 120 π Ω £ £ £ ε 20 62 83 ( ) ¡ r 6 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. λ £ λ ε Solution: From 0 ¡ r, § § 2 2 λ0 60 £ £ ε £ 9 r λ 20 Problem 7.6 The electric field of a plane wave propagating in a lossless, nonmagnetic, dielectric material with εr £ 2 56 is given by 9 ¢ E £ yˆ 20cos 6π 10 t kz (V/m) ¨ Determine: (a) f , up, λ, k, and η, and (b) the magnetic field H. Solution: (a) 9 £ ¡ ω £ 2π f 6π 10 rad/s 9 £ ¡ f £ 3 10 Hz 3 GHz 8 c 3 10 8 £ £ ¡ up £ ε 1 875 10 m/s ¡ ¡ r 2 56 8 λ up 1 875 10 £ £ ¡ £ 9 3 12 cm f 6 10 2π 2π £ £ ¡ k £ 2 201 4 rad/m λ 3 12 10 η0 377 377 £ £ £ η £ Ω ε 235 62 ¡ ¡ r 2 56 1 6 CHAPTER 7 327 (b) 20 π 9 ¢ H £ xˆ η cos 6 10 t kz ¨ ¨ 20 9 π ¢ £ xˆ cos 6 10 t 201 4z ¨ ¨ 235 62 2 9 π ¢ £ xˆ 8 49 10 cos 6 10 t 201 4z (A/m) ¨ ¨ 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 ωt=0 z x ωt=π/2 Figure P7.7: Locus of E versus time. Solution: For an RHC wave traveling in zˆ, let us try the following: ¨ ω ω ¢ ¢ E £ xˆ acos t kz yˆasin t kz © © © 2 2 ¡ ¡ £ £ £ Modulus E a a a 2 2 (V/m). Hence, © 2 ¡ £ a £ 2 ¡ 2 328 CHAPTER 7 Next, we need to check the sign of the yˆ-component relative to that of the xˆ-component. We do this by examining the locus of E versus t at z £ 0: Since the wave is traveling along zˆ, when the thumb of the right hand is along zˆ (into ¨ the page), the other four fing¨ ers point in the direction shown (clockwise as seen from above). Hence, we should reverse the sign of the yˆ-component: ¡ ¡ ω ω ¢ ¢ E £ xˆ 2cos t kz yˆ 2sin t kz (V/m) © © ¨ with 2π 2π £ £ ¡ k £ 2 104 72 (rad/m) λ 6 10 and π ω 2 8 π 10 £ £ £ kc λ 3 10 10 (rad/s) Problem 7.8 For a wave characterized by the electric field ¢¦£ ¢ ¢ ¡ E z ¡ t xˆax cos ωt kz yˆay cos ωt kz δ ¨ ¨ © © γ χ ¢ identify the polarization state, determine the polarization angles ¡ , and sketch the ¢ locus of E 0 ¡ t for each of the following cases: £ δ £ (a) ax £ 3 V/m, ay 4 V/m, and 0, δ ¥ £ £ (b) ax £ 3 V/m, ay 4 V/m, and 180 , δ ¥ £ £ (c) ax £ 3 V/m, ay 3 V/m, and 45 , £ £ ¥ (d) ax £ 3 V/m, ay 4 V/m, and δ 135 .
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