Waves. Daniel S. Weile. ELEG 648 Waves. Department of Electrical and Computer Engineering University of Delaware. Plane Waves Reflection of Waves

Transcription

1 Waves Daniel S. Weile Department of Electrical and Computer Engineering University of Delaware ELEG 648 Waves

2 Outline

3 Outline

4 Introduction Let s start by introducing simple solutions to Maxwell s equations without sources. These are Source-Free Curl Equations E = ẑh H = ŷe Therefore, E = ẑ H = ẑŷe Define the Wavenumber k = ẑŷ

5 Source-Free General Wave Equations The magnetic field solves the same equation. Thus Vector Wave Equations E k 2 E = 0 H k 2 H = 0 Taking the divergence of these equations, or appealing directly to the source-free Maxwell Equations demonstrates the Divergence-Free Conditions E = 0 H = 0

6 Simplified Vector Wave Equation Usually, these equations are simplified by defining the Vector Laplacian 2 A = ( A) A The reason for this definition is that, in Cartesian coordinates 2 (A x u x + A y u y + A z u z ) = u x 2 A x + u y 2 A y + u z 2 A z This is not true in other coordinate systems though. Using the vector Laplacian definition with the divergenceless nature of the fields gives

7 Simplified Vector Wave Equation Usually, these equations are simplified by defining the Vector Laplacian 2 A = ( A) A The reason for this definition is that, in Cartesian coordinates 2 (A x u x + A y u y + A z u z ) = u x 2 A x + u y 2 A y + u z 2 A z This is not true in other coordinate systems though. Using the vector Laplacian definition with the divergenceless nature of the fields gives The Wave Equation 2 E + k 2 E = 0 2 H + k 2 H = 0

8 Simplifications of the Wave Equation Note that the new wave equation does not enforce the divergence-free nature of allowable solutions. Thus, any solution we find must be tested to ensure it is divergence free. With this caveat, our goal is to solve the equation 2 ψ + k 2 ψ = 0 for one of the components of the (electric) field. Let us assume

9 Simplifications of the Wave Equation Note that the new wave equation does not enforce the divergence-free nature of allowable solutions. Thus, any solution we find must be tested to ensure it is divergence free. With this caveat, our goal is to solve the equation 2 ψ + k 2 ψ = 0 for one of the components of the (electric) field. Let us assume 1 Lossless medium so ẑ = jωµ, ŷ = jωɛ, and k = ω µɛ

10 Simplifications of the Wave Equation Note that the new wave equation does not enforce the divergence-free nature of allowable solutions. Thus, any solution we find must be tested to ensure it is divergence free. With this caveat, our goal is to solve the equation 2 ψ + k 2 ψ = 0 for one of the components of the (electric) field. Let us assume 1 Lossless medium so ẑ = jωµ, ŷ = jωɛ, and k = ω µɛ 2 An x-directed field so E = E x u x

11 Simplifications of the Wave Equation Note that the new wave equation does not enforce the divergence-free nature of allowable solutions. Thus, any solution we find must be tested to ensure it is divergence free. With this caveat, our goal is to solve the equation 2 ψ + k 2 ψ = 0 for one of the components of the (electric) field. Let us assume 1 Lossless medium so ẑ = jωµ, ŷ = jωɛ, and k = ω µɛ 2 An x-directed field so E = E x u x 3 Only z variation, so = u z d dz 2 = d2 dz 2

12 Simplifications of the Wave Equation Note that the new wave equation does not enforce the divergence-free nature of allowable solutions. Thus, any solution we find must be tested to ensure it is divergence free. With this caveat, our goal is to solve the equation 2 ψ + k 2 ψ = 0 for one of the components of the (electric) field. Let us assume 1 Lossless medium so ẑ = jωµ, ŷ = jωɛ, and k = ω µɛ 2 An x-directed field so E = E x u x 3 Only z variation, so = u z d dz 2 = d2 dz 2 Then d 2 E x dz 2 + k 2 E x = 0

13 Solutions of the Simplified Wave Equation The solutions of the above equation are 1 E x (z) = E 0 e jkz

14 Solutions of the Simplified Wave Equation The solutions of the above equation are 1 E x (z) = E 0 e jkz 2 E x (z) = E 0 e jkz

15 Solutions of the Simplified Wave Equation The solutions of the above equation are 1 E x (z) = E 0 e jkz 2 E x (z) = E 0 e jkz 3 E x (z) = E 0 cos(kz)

16 Solutions of the Simplified Wave Equation The solutions of the above equation are 1 E x (z) = E 0 e jkz 2 E x (z) = E 0 e jkz 3 E x (z) = E 0 cos(kz) 4 E x (z) = E 0 sin(kz) We will examine the first of these, though the sum of any two is the most general solution.

17 Solutions of the Simplified Wave Equation The solutions of the above equation are 1 E x (z) = E 0 e jkz 2 E x (z) = E 0 e jkz 3 E x (z) = E 0 cos(kz) 4 E x (z) = E 0 sin(kz) We will examine the first of these, though the sum of any two is the most general solution. Note that any of these satisfy the divergence free condition. They would not if the z-component were chosen.

18 More About the FIrst Solution Associated with the first solution is a magnetic field given by H = 1 jωµ E = u k y ωµ E 0e jkz E 0 = u y η e jkz Note that the ratio between the electric field and magnetic field at any point is given by the Characteristic Impedance In free space, η = η 0 = η = E x H y = µ ɛ = ẑ ŷ µ0 ɛ 0 377Ω 120πΩ

19 Interpretation of the solution In the time domain, the electromagnetic field is given by Time Domain Solution E x (z, t) = E 0 cos(ωt kz) H y (z, t) = E 0 η cos(ωt kz) This is a wave. Consider what happens for successive fixed values of t: the same pattern is plotted with origin (ωt) moving right.

20 Plot of the solution k = 1, ωt = 0π 8

21 Plot of the solution k = 1, ωt = 1π 8

22 Plot of the solution k = 1, ωt = 2π 8

23 Plot of the solution k = 1, ωt = 3π 8

24 Plot of the solution k = 1, ωt = 4π 8

25 Plot of the solution k = 1, ωt = 5π 8

26 Plot of the solution k = 1, ωt = 6π 8

27 Plot of the solution k = 1, ωt = 7π 8

28 Plot of the solution k = 1, ωt = 8π 8

29 Wave Terminology We call these plane waves because they have constant phase over a set of planes. These are the equiphase surfaces of the wave. We call these plane waves uniform because the amplitude is constant in the equiphase plane. The wave is travelling in the +z direction as shown in the last slides. The wave is linearly polarized in the x-direction since at any point in space the tip of the E-field vector traces out a line segment along the x axis.

30 Phase Velocity The formula for an equiphase plane is ωt kz = constant

31 Phase Velocity The formula for an equiphase plane is ωt kz = constant The velocity with which the equiphase plane moves is the phase velocity. Differentiating the above gives ω k dz dt = 0

32 Phase Velocity The formula for an equiphase plane is ωt kz = constant The velocity with which the equiphase plane moves is the phase velocity. Differentiating the above gives ω k dz dt = 0 This gives the formula for phase velocity Phase Velocity v p = dz dt = ω k = 1 µɛ

33 More About Waves In free space, the phase velocity is The Speed of Light in Vacuum c = 1 µ0 ɛ 0 = 299, 792, 458 m/s m/s The period T is the amount of time it takes for the wave to repeat. Clearly T = 2π ω. The frequency f is the number of times the wave repeats in a given amount of time. f = 1 T = ω 2π. The wavelength is the spatial period λ = 2π k.

34 More About Waves In free space, the phase velocity is The Speed of Light in Vacuum c = 1 µ0 ɛ 0 = 299, 792, 458 m/s m/s The period T is the amount of time it takes for the wave to repeat. Clearly T = 2π ω. The frequency f is the number of times the wave repeats in a given amount of time. f = 1 T = ω 2π. The wavelength is the spatial period λ = 2π k. An Important Formula v p = ω k = 2πf 2π λ = f λ

35 Power and Energy In the time domain, our wave is E x (z, t) = E 0 cos(ωt kz) H y (z, t) = E 0 η cos(ωt kz) Corresponding with this, we have the expressions for Power and Energy Density w e = 1 2 ɛe 2 = 1 2 ɛe 2 0 cos 2 (ωt kz) w m = 1 2 µh2 = 1 2 ɛe 2 0 cos 2 (ωt kz) S = E H = u z E 2 0 η cos2 (ωt kz)

36 Energy Velocity We may think of the wave as packets of energy moving at each point in space with an energy velocity. The flux of power must then be the energy velocity times the energy density. We thus have Energy Velocity v e = S = 1 w e + w h ɛη = 1 µɛ Note that energy velocity is not always equal to phase velocity. Phase velocity can be greater than c, but not energy velocity. Finally Frequency Domain Poynting Vector S = 1 2 E H = 1 2 u E 0 2 z η

37 General z-directed Lossless Wave Travel In a lossless dielectric, the most general formula for wave travel is E + x = Ae jkz E + y = Be jkz E x = Ce +jkz H + y = A η e jkz H + x = B η e jkz H y = C η e+jkz Ey = De +jkz Hx = D η e+jkz Any linear combination of these is a solution to our original ordinary differential equation. Indeed, the three new solutions are the same as the original solution under a rotation of coordinates.

38 z-directed, x-polarized Waves To simplify our study, we concentrate on the most general waves polarized in x, traveling in the ±z direction. E x = Ae jkz + Ce jkz H y = 1 η [Ae jkz Ce jkz] For the moment, we examine the A = C = E 2 case, i.e., the Pure Standing Wave E x = E cos(kz) H y = je η sin(kz)

39 Pure Standing Waves In the time domain, this becomes Time Domain Standing Wave E x = Re { E cos(kz)e jωt} = E cos(kz) cos(ωt) { } H y = Re je η sin(kz)ejωt = E sin(kz) sin(ωt) η

40 Pure Standing Waves In the time domain, this becomes Time Domain Standing Wave E x = Re { E cos(kz)e jωt} = E cos(kz) cos(ωt) { } H y = Re je η sin(kz)ejωt = E sin(kz) sin(ωt) η 1 At each point in space, there is a sinusoidal oscillation in time.

41 Pure Standing Waves In the time domain, this becomes Time Domain Standing Wave E x = Re { E cos(kz)e jωt} = E cos(kz) cos(ωt) { } H y = Re je η sin(kz)ejωt = E sin(kz) sin(ωt) η 1 At each point in space, there is a sinusoidal oscillation in time. 2 The amplitude of these oscillations is sinusoidally distributed in space.

42 Pure Standing Waves In the time domain, this becomes Time Domain Standing Wave E x = Re { E cos(kz)e jωt} = E cos(kz) cos(ωt) { } H y = Re je η sin(kz)ejωt = E sin(kz) sin(ωt) η 1 At each point in space, there is a sinusoidal oscillation in time. 2 The amplitude of these oscillations is sinusoidally distributed in space. 3 There are nodes (where the amplitude is zero) and antinodes (where it is maximum).

43 Pure Standing Waves In the time domain, this becomes Time Domain Standing Wave E x = Re { E cos(kz)e jωt} = E cos(kz) cos(ωt) { } H y = Re je η sin(kz)ejωt = E sin(kz) sin(ωt) η 1 At each point in space, there is a sinusoidal oscillation in time. 2 The amplitude of these oscillations is sinusoidally distributed in space. 3 There are nodes (where the amplitude is zero) and antinodes (where it is maximum). 4 The nodes and antinodes are spaced λ 2 apart.

44 Pure Standing Waves In the time domain, this becomes Time Domain Standing Wave E x = Re { E cos(kz)e jωt} = E cos(kz) cos(ωt) { } H y = Re je η sin(kz)ejωt = E sin(kz) sin(ωt) η 1 At each point in space, there is a sinusoidal oscillation in time. 2 The amplitude of these oscillations is sinusoidally distributed in space. 3 There are nodes (where the amplitude is zero) and antinodes (where it is maximum). 4 The nodes and antinodes are spaced λ 2 apart. 5 The nodes of the electric field are λ from the magnetic field nodes. 4

45 Pure Standing Waves In the time domain, this becomes Time Domain Standing Wave E x = Re { E cos(kz)e jωt} = E cos(kz) cos(ωt) { } H y = Re je η sin(kz)ejωt = E sin(kz) sin(ωt) η 1 At each point in space, there is a sinusoidal oscillation in time. 2 The amplitude of these oscillations is sinusoidally distributed in space. 3 There are nodes (where the amplitude is zero) and antinodes (where it is maximum). 4 The nodes and antinodes are spaced λ 2 apart. 5 The nodes of the electric field are λ from the magnetic field nodes. 4 6 It is still a uniform plane wave.

46 Power and Energy in Pure Standing Waves For this wave we have w e = ɛ 2 E 2 cos 2 (kz) cos 2 (ωt) w m = ɛ 2 E 2 sin 2 (kz) sin 2 (ωt) S = E 2 u z sin(2kz) sin(2ωt) 2η S = je 2 u z 4η sin(2kz)

47 Power and Energy in Pure Standing Waves For this wave we have w e = ɛ 2 E 2 cos 2 (kz) cos 2 (ωt) w m = ɛ 2 E 2 sin 2 (kz) sin 2 (ωt) S = E 2 u z sin(2kz) sin(2ωt) 2η S = je 2 u z 4η sin(2kz) Why is S purely imaginary?

48 Power and Energy in Pure Standing Waves For this wave we have w e = ɛ 2 E 2 cos 2 (kz) cos 2 (ωt) w m = ɛ 2 E 2 sin 2 (kz) sin 2 (ωt) S = E 2 u z sin(2kz) sin(2ωt) 2η S = je 2 u z 4η sin(2kz) Why is S purely imaginary? Why do the expressions for energy depend on 2k instead of k?

49 Power and Energy in Pure Standing Waves For this wave we have w e = ɛ 2 E 2 cos 2 (kz) cos 2 (ωt) w m = ɛ 2 E 2 sin 2 (kz) sin 2 (ωt) S = E 2 u z sin(2kz) sin(2ωt) 2η S = je 2 u z 4η sin(2kz) Why is S purely imaginary? Why do the expressions for energy depend on 2k instead of k? Where could we put PECs without altering the field?

50 Combinations of Standing and Travelling Waves In general we have E x = Ae jkz + Ce +jkz. The amplitude at of the field oscillation at any z is E x (z). Now max E x (z) = max Ae jkz 1 + C z z A e2jkz = max A 1 + C z A e2jkz [ ] = A 1 + = A + C C A By the same token min z E x (z) = A C.

51 Standing Wave Ratio The ratio of the maximum field amplitude to the minimum field amplitude is Standing Wave Ratio SWR = A + C 1 + A C = 1 C A C A

52 Standing Wave Ratio The ratio of the maximum field amplitude to the minimum field amplitude is Standing Wave Ratio SWR = A + C 1 + A C = 1 C A C A What is the SWR for a pure travelling wave?

53 Standing Wave Ratio The ratio of the maximum field amplitude to the minimum field amplitude is Standing Wave Ratio SWR = A + C 1 + A C = 1 C A C A What is the SWR for a pure travelling wave? What is the SWR for a pure standing wave?

54 Standing Wave Ratio The ratio of the maximum field amplitude to the minimum field amplitude is Standing Wave Ratio SWR = A + C 1 + A C = 1 C A C A What is the SWR for a pure travelling wave? What is the SWR for a pure standing wave? What physical situation would give rise to the above cases? To a case in between?

55 Outline

56 Let us now consider the most general case of wave travel in the positive z-direction: E = (Au x + Bu y ) e jkz H = ( Bu x + Au y ) e jkz The polarization of this wave is the figure traced by the tip of the electric field vector at a fixed location in space. Without loss of generality we will assume A is real and positive; A = A. Also, B = B e jβ.

57 Let us now consider the most general case of wave travel in the positive z-direction: E = (Au x + Bu y ) e jkz H = ( Bu x + Au y ) e jkz The polarization of this wave is the figure traced by the tip of the electric field vector at a fixed location in space. Without loss of generality we will assume A is real and positive; A = A. Also, B = B e jβ. Why is there no loss of generality in this assumption?

58 Time Domain Fields At z = 0 the time domain expression for these fields is easily seen to be E x = A cos(ωt) E y = B cos(ωt + β) There are three cases of interest: 1 Linear 2 Circular 3 Elliptic We examine each in turn.

59 Linear The wave is linearly polarized if β = 0 or β = π. In this case E x = A cos(ωt) E y = ± B cos(ωt) The angle with the x-axis is given by tan 1 ( Ey E x ) = tan 1 ( ± B A independent of time. The vector oscillates back and forth maintaining the same angle at all points in space and all times. )

60 Circular The wave is circularly polarized if A = B and β = ± π 2. In this case E x = A cos(ωt) E y = A sin(ωt) In either case E 2 x + E 2 y = A 2 so the length of the vector is independent of time.

61 Circular The wave is circularly polarized if A = B and β = ± π 2. In this case E x = A cos(ωt) E y = A sin(ωt) In either case E 2 x + E 2 y = A 2 so the length of the vector is independent of time. If β = π 2, the wave is right-hand circularly polarized (RHCP), otherwise, left-hand circularly polarized (LHCP).

62 Circular The wave is circularly polarized if A = B and β = ± π 2. In this case E x = A cos(ωt) E y = A sin(ωt) In either case E 2 x + E 2 y = A 2 so the length of the vector is independent of time. If β = π 2, the wave is right-hand circularly polarized (RHCP), otherwise, left-hand circularly polarized (LHCP). The name is determined by curling the fingers of the appropriate hand with the thumb pointing in the direction of travel.

63 Elliptical In general the vector traces the curve x = E x = A cos(ωt) y = E y = B cos(β) cos(ωt) B sin(β) sin(ωt)

64 Elliptical In general the vector traces the curve x = E x = A cos(ωt) y = E y = B cos(β) cos(ωt) B sin(β) sin(ωt) What sort of curve is this? Well...

65 Elliptical In general the vector traces the curve x = E x = A cos(ωt) y = E y = B cos(β) cos(ωt) B sin(β) sin(ωt) What sort of curve is this? Well... cos(ωt) = x A sin(ωt) = 1 ( ) x 2 A

66 Elliptical Therefore, y = ( ) x 2 B sin β 1 = A [ ( ) ] x 2 B 2 sin 2 β 1 = A B 2 sin 2 β = ( ) B cos β x 2 x B sin β 1 A A B cos β x y A [ ] B cos β 2 x y A [ B cos β 2 ( ) x 2 x y] + sin 2 β A A

67 Elliptical Therefore, y = ( ) x 2 B sin β 1 = A [ ( ) ] x 2 B 2 sin 2 β 1 = A B 2 sin 2 β = ( ) B cos β x 2 x B sin β 1 A A B cos β x y A [ ] B cos β 2 x y A [ B cos β 2 ( ) x 2 x y] + sin 2 β A A What kind of curve is this? How do you know?

68 Outline

69 Wave Constants In general, wave constants are complex to include loss. This gives rise to the expressions General Wave Constants ẑ = jω (µ jµ ) = ωµ + jωµ ŷ = σ + jω (ɛ jɛ ) = ( σ + ωɛ ) + jωɛ

70 Wave Constants In general, wave constants are complex to include loss. This gives rise to the expressions General Wave Constants ẑ = jω (µ jµ ) = ωµ + jωµ ŷ = σ + jω (ɛ jɛ ) = ( σ + ωɛ ) + jωɛ Why do you think the imaginary parts of ɛ and µ are taken as negative?

71 Wave Constants When the wave is not in a lossless dielectric, the general expressions for wavenumber and impedance are Wave Constants k = ẑŷ η = ẑ ŷ Sometimes the inverted expressions are useful: Wave Constants ẑ = jkη ŷ = jk η

72 The Propagation Constant We can express the propagation constant in terms of its real and imaginary parts, and the impedance in polar form: k = k jk η = η e jζ Frequency Domain Fields E x = E 0 e jkz = E 0 e k z e jk z H y = E 0 η e jkz = E 0 η e k z e j(k z ζ)

73 The Propagation Constant Time Domain Fields E x = E 0 e k z cos(ωt k z) H y = E 0 η e k z cos(ωt k z ζ)

74 The Propagation Constant Time Domain Fields E x = E 0 e k z cos(ωt k z) H y = E 0 η e k z cos(ωt k z ζ) What is the function of k? Of k? Of ζ?

75 The Propagation Constant Time Domain Fields E x = E 0 e k z cos(ωt k z) H y = E 0 η e k z cos(ωt k z ζ) What is the function of k? Of k? Of ζ? Is k usually positive or negative? How about k?

76 The Propagation Constant Time Domain Fields E x = E 0 e k z cos(ωt k z) H y = E 0 η e k z cos(ωt k z ζ) What is the function of k? Of k? Of ζ? Is k usually positive or negative? How about k? Is either of these conditions an absolute limitation on the value of k? Why?

77 The Propagation Constant Time Domain Fields E x = E 0 e k z cos(ωt k z) H y = E 0 η e k z cos(ωt k z ζ) What is the function of k? Of k? Of ζ? Is k usually positive or negative? How about k? Is either of these conditions an absolute limitation on the value of k? Why? How does this change if the energy is travelling in the z direction?

78 Good Dielectrics In a good dielectric, we assume σ = µ = 0 and ɛ ɛ. We also need the Taylor Approximation for 1 + x 1 + x = 1 + Given this we compute d 1 + y x + O(x dy y=0 2 ) = 1 + x 2 + O(x 2 ) k = ( jωµ) (ωɛ + jωɛ ) ( ) = ω 2 µɛ 1 j ɛ ɛ ω ( ) µɛ 1 j ɛ 2ɛ

79 Good Dielectrics By the same token, because in a good dielectric we have Since ẑ = jωµ = j ẑ, we find Thus, since ŷ ωɛ η = ω µɛ ωɛ η = ẑ jk = ẑk jkk = j k ẑ ẑ ŷ η = k ŷ ) ) (1 + j ɛ µ 2ɛ = (1 ɛ + j ɛ 2ɛ

80 Good Dielectric Summary Summary k = ω µɛ k = ωɛ µ 2 ɛ ) µ η = (1 ɛ + j ɛ 2ɛ ζ = tan 1 ɛ 2ɛ There is attenuation, but it is not very pronounced. E and H are out of phase, but not by much. The attenuation is frequency dependent, but only because we used a model with ɛ instead of σ.

81 Good Conductor In a good conductor, we assume σ >> ωɛ, so ŷ σ. In this case, k = σ(jωµ) = ωµσ j Of course, Therefore, ( j) 1 2 = ( e j π ) = e j π 4 = 1 (1 j) 2 ωµσ k = (1 j) 2

82 Good Conductor Impedance is easily computed: η = k ŷ = ωµ (1 + j) = 2σ ωµ π σ ej 4 Summary k = k = η = ωµσ 2 ωµσ 2 ωµ σ ζ = π 4

83 More About Good Conductors Conductors are often characterized by their Skin Depth δ = 1 2 k = ωµσ = 1 πf µσ i.e. the distance in which the field decays to 1 e important is the of its value. Also Surface Resistance R = ωµσ 2 = 1 σδ which is just the real part of the impedance.

84 Power Dissipated by Good Conductors Assuming that the power enters the conductor at a normal to its surface (we will see it does), the power density entering the surface is Power Density Absorbed by a Conductor Re{S} = 1 2 Re{E H } = 1 2 R H 0 2 u z where H 0 is the magnetic field on the surface.

85 Power Dissipated by Good Conductors Assuming that the power enters the conductor at a normal to its surface (we will see it does), the power density entering the surface is Power Density Absorbed by a Conductor Re{S} = 1 2 Re{E H } = 1 2 R H 0 2 u z where H 0 is the magnetic field on the surface. How else might we compute the power absorbed by the conductor?

86 Outline Normal Incidence Oblique Wave Travel Oblique Incidence

87 Normal Incidence Normal Incidence Oblique Wave Travel Oblique Incidence Reflected Wave Incident Wave Transmitted Wave In region 1: E (1) x = E 0 ( e jk 1z + Γe jk 1z ) H (1) y = E 0 η 1 ( e jk 1z Γe jk 1z ) In region 2: E (2) x = TE 0 e jk 2z H (2) y = TE 0 η 2 e jk 2z

88 Normal Incidence Normal Incidence Oblique Wave Travel Oblique Incidence Enforcing continuity of E x and H y at z = 0 gives 1 + Γ = T 1 η 1 (1 Γ) = T η 2 Solving these equations results in formulas for Reflection and Transmission Coefficients Γ = η 2 η 1 η 2 + η 1 2η 2 T = η 2 + η 1

89 Normal Incidence Oblique Wave Travel Oblique Incidence Power and SWR for Normal Incidence Since E x = E 0 e jk 1z (1 + Γe 2jk 1z ) Standing Wave Ratio SWR = 1 + Γ 1 Γ Power can be computed directly: In region 1 S = 1 2 Re{E H } = 1 2 Re { E 0 ( e jk 1z + Γe jk 1z ) u x E 0 η 1 ( e jk 1z Γ e jk 1z ) u y } = E 0 2 2η 1 u z Re {1 Γ 2 + Γe 2jk 1z Γ e 2jk 1z }

90 Normal Incidence Oblique Wave Travel Oblique Incidence Power and SWR for Normal Incidence Because a a = 2jIm {a}, we get In region 2, we find S = E 0 2 2η 1 u z ( 1 Γ 2) = S inc ( 1 Γ 2) S = 1 2 Re{E H } { = 1 2 Re TE 0 e jk2z T E0 } e jk 2z η 2 = T 2 E 0 2 2η 2 = η 1 T 2 S inc η 2

91 Normal Incidence Oblique Wave Travel Oblique Incidence Power and SWR for Normal Incidence Note that the power transmitted is not T 2 S inc. This is because the characteristic impedance of the second medium is not the same as in the first medium. In fact Theorem 1 Γ 2 = η 1 η 2 T 2 This is a consequence of energy conservation (i.e. the Poynting theorem) but this remains to be proved.

92 Proof of Power Conservation Normal Incidence Oblique Wave Travel Oblique Incidence Proof. η 1 η 2 T 2 = η 1 = η 2 2η 2 η 2 + η 1 4η 1 η 2 (η 1 + η 2 ) 2 = (η 2 + η 1 ) 2 (η 2 η 1 ) 2 (η 1 + η 2 ) 2 = 1 Γ 2 2

93 Outline Normal Incidence Oblique Wave Travel Oblique Incidence

94 Coordinate Rotation Normal Incidence Oblique Wave Travel Oblique Incidence Y Y y ξ z Z To study wave propagation in other directions, the coordinate system is rotated. From the figure we can see that z = z cos ξ + y sin ξ. ξ z }{{}}{{} z cos ξ y sin ξ Z Similarly, it is not hard to show that u y = u y cos ξ u z sin ξ.

95 Obliquely Traveling Wave Normal Incidence Oblique Wave Travel Oblique Incidence Electric Field Magnetic Field E = u x E 0 e jkz jk(y sin ξ+z cos ξ) = u x E 0 e H = u E 0 y η e jkz = (u y cos ξ u z sin ξ) E 0 η e jk(y sin ξ+z cos ξ) Note that there is no new physics here; only a rotation of the coordinate system. We can however define a z-directed impedance Z TE z = E x H y = η cos ξ

96 Wavevector Normal Incidence Oblique Wave Travel Oblique Incidence It is also useful to define a Wavevector k = k(u y sin ξ + u z cos ξ) = k y u y + k z u z Also, let the position vector be r = xu x + yu y + zu z, and the polarization vector p = u x. Then Electric Field Magnetic Field E = pe 0 e jk r H = u k p E 0 η e jk r

97 Observations Normal Incidence Oblique Wave Travel Oblique Incidence Because of physics E, H, and k form a right handed triplet. In particular, u E u H = u k u k u E = u H u H u k = u E

98 Observations Normal Incidence Oblique Wave Travel Oblique Incidence Because of physics E, H, and k form a right handed triplet. In particular, u E u H = u k u k u E = u H u H u k = u E Also E = η H and k = k = ω µɛ

99 Observations Normal Incidence Oblique Wave Travel Oblique Incidence Because of physics E, H, and k form a right handed triplet. In particular, Also u E u H = u k u k u E = u H u H u k = u E E = η H and k = k = ω µɛ Appealing to our physical understanding of the phenomenon makes this computation far easier than solving differential equations!!!!!

100 Outline Normal Incidence Oblique Wave Travel Oblique Incidence

101 Oblique Incidence Normal Incidence Oblique Wave Travel Oblique Incidence Reflected Wave y Transmitted Wave θ i z θ r θ t Incident Wave k i = k 1 (u y sin θ i + u z cos θ i ) = k i yu y + k i zu z k r = k 1 (u y sin θ r u z cos θ r ) = k r yu y k r zu z k t = k 2 (u y sin θ t + u z cos θ t ) = k t yu y + k t zu z

102 Transverse Electric Reflection Normal Incidence Oblique Wave Travel Oblique Incidence Assuming that E is in the x-direction, we can write In region 1: ( ) E (1) x = E 0 e j(k y i y+k zz) i + Γe j(k y r y kzz) r In region 2: H (1) y = E 0 η 1 ( cos θ i e j(k i y y+k i z z) Γ cos θ r e j(k r y y k r z z)) E (2) x = TE 0 e j(k t y y+k t zz) H (2) y = TE 0 η 2 cos θ t e j(k t y y+k t z z) These components must be continuous at z = 0.

103 Matching Boundary Conditions Normal Incidence Oblique Wave Travel Oblique Incidence Setting z = 0 gives the following equations: e jk i y y + Γe jk r y y = Te jk t y y 1 ( ) cos θ i e jk y i y Γ cos θ r e jk y r y = T cos θ t e jk y t y η 1 η 2

104 Matching Boundary Conditions Normal Incidence Oblique Wave Travel Oblique Incidence Setting z = 0 gives the following equations: e jk i y y + Γe jk r y y = Te jk t y y 1 ( ) cos θ i e jk y i y Γ cos θ r e jk y r y = T cos θ t e jk y t y η 1 η 2 From this we can conclude k i y = k r y = k t y WHY???

105 Reflection and Refraction Normal Incidence Oblique Wave Travel Oblique Incidence k i y = k r y k 1 sin θ i = k 1 sin θ r

106 Reflection and Refraction Normal Incidence Oblique Wave Travel Oblique Incidence k i y = k r y k 1 sin θ i = k 1 sin θ r The Law of Reflection θ i = θ r

107 Reflection and Refraction Normal Incidence Oblique Wave Travel Oblique Incidence k i y = k r y k 1 sin θ i = k 1 sin θ r The Law of Reflection θ i = θ r k i y = k t y

108 Reflection and Refraction Normal Incidence Oblique Wave Travel Oblique Incidence k i y = k r y k 1 sin θ i = k 1 sin θ r The Law of Reflection θ i = θ r k i y = k t y The Law of Refraction or Snell s Law k 1 sin θ i = k 2 sin θ t n 1 sin θ i = n 2 sin θ t

109 TE Reflection Normal Incidence Oblique Wave Travel Oblique Incidence With Snell s Law and the law of reflection in tow, we find (letting θ 1 = θ i and θ 2 = θ r ) 1 + Γ = T cos θ 1 (1 Γ) = cos θ 2 T η 1 Recalling the z-directed wave impedance, we can define η 2 Z TE z1 = η 1 sec θ 1 Z TE z2 = η 2 sec θ 2 Now these equations have a familiar form...

110 Normal Incidence Oblique Wave Travel Oblique Incidence Oblique Reflection and Transmission Coefficients TE Reflection and Transmission Coefficients Γ TE = Z z2 TE T TE = Z TE z2 Z TE z2 Zz1 TE + Z z1 TE 2Z TE z2 + Z TE z1

111 Normal Incidence Oblique Wave Travel Oblique Incidence Oblique Reflection and Transmission Coefficients TE Reflection and Transmission Coefficients Γ TE = Z z2 TE T TE = Z TE z2 Z TE z2 Zz1 TE + Z z1 TE 2Z TE z2 + Z TE z1 TM Reflection and Transmission Coefficients Γ TM = Z z2 TM T TM = Z TM z2 Z TM z2 Zz1 TM + Z z1 TM 2Z TM z2 + Z TM z1

112 Normal Incidence Oblique Wave Travel Oblique Incidence Oblique Reflection and Transmission Coefficients TE Reflection and Transmission Coefficients Γ TE = Z z2 TE T TE = Z TE z2 Z TE z2 Zz1 TE + Z z1 TE 2Z TE z2 + Z TE z1 TM Reflection and Transmission Coefficients Γ TM = Z z2 TM T TM = What do these (TM) equations mean?? Z TM z2 Z TM z2 Zz1 TM + Z z1 TM 2Z TM z2 + Z TM z1

113 Brewster Angle Normal Incidence Oblique Wave Travel Oblique Incidence If Zz2 TM = Z z1 TM, there will be no reflection of TM polarized waves. Let us work out the conditions under which this happens: η 2 1 η1 2 cos2 θ 1 = η2 2 cos2 θ ) 2 ) (1 sin 2 θ 1 = η2 (1 2 sin 2 θ 2 ( ) 1 sin 2 θ 1 = η2 2 η1 2 1 k 1 2 k2 2 sin 2 θ 1 ) sin 2 θ 1 (η 2 2 k 2 1 η 2 1 k (ɛ sin θ 1 ɛ ) = η2 2 η = µ 2ɛ 1 µ 1 ɛ 2 1

114 Brewster Angle Normal Incidence Oblique Wave Travel Oblique Incidence Solving this gives sin θ 1 = ɛ 2 /ɛ 1 µ 2 /µ 1 ɛ 2 /ɛ 1 ɛ 1 /ɛ 2 For nonmagnetic media we have ɛ2 sin θ 1 = ɛ 2 + ɛ 1 TM Nonmagnetic Brewster Angle tan θ B = ɛ2 ɛ 1

115 TE Brewster Angle? Normal Incidence Oblique Wave Travel Oblique Incidence For TE incidence, the angle at which the reflection coefficient vanishes is µ 2 /µ 1 ɛ 2 /ɛ 1 sin θ 1 = µ 2 /µ 1 µ 1 /µ 2

116 TE Brewster Angle? Normal Incidence Oblique Wave Travel Oblique Incidence For TE incidence, the angle at which the reflection coefficient vanishes is µ 2 /µ 1 ɛ 2 /ɛ 1 sin θ 1 = µ 2 /µ 1 µ 1 /µ 2 Note that TE waves incident on nonmagnetic media are never fully transmitted.

117 TE Brewster Angle? Normal Incidence Oblique Wave Travel Oblique Incidence For TE incidence, the angle at which the reflection coefficient vanishes is µ 2 /µ 1 ɛ 2 /ɛ 1 sin θ 1 = µ 2 /µ 1 µ 1 /µ 2 Note that TE waves incident on nonmagnetic media are never fully transmitted. How can this be used to make a polarizer?

118 Total Reflection Normal Incidence Oblique Wave Travel Oblique Incidence n 1 sin θ 1 = n 2 sin θ 2 If n 1 > n 2 then for some θ 1 there is no solution for θ 2.

119 Total Reflection Normal Incidence Oblique Wave Travel Oblique Incidence n 1 sin θ 1 = n 2 sin θ 2 If n 1 > n 2 then for some θ 1 there is no solution for θ 2. What happens in this case, when the incident angle is greater than the Critical Angle θ c = sin 1 n 2 = sin 1 ɛ2 n 1 ɛ 1 Well, we came to this by forcing k i y = k t y. We can still force this, but now the the value of k t z must change.

120 Total Reflection Normal Incidence Oblique Wave Travel Oblique Incidence We can always take sin θ 1 = n 1 n 2 sin θ 2 What kind of angle has a sine greater than one?

121 Total Reflection Normal Incidence Oblique Wave Travel Oblique Incidence We can always take sin θ 1 = n 1 n 2 sin θ 2 What kind of angle has a sine greater than one? An imaginary angle! After all, sin ψ = ejψ e jψ 2j

122 Total Reflection Normal Incidence Oblique Wave Travel Oblique Incidence We can always take sin θ 1 = n 1 n 2 sin θ 2 What kind of angle has a sine greater than one? An imaginary angle! After all, sin ψ = ejψ e jψ 2j In any case, we can define β = k t y k t z = k 2 cos θ 2 = k 2 1 sin 2 θ 2 = ±jα

123 Evanescent Waves Normal Incidence Oblique Wave Travel Oblique Incidence Our original formula for the transmitted wave was (ignoring subscripts) Plugging in jk(y sin ξ+z cos ξ) E = u x TE 0 e H = (u y cos ξ u z sin ξ) TE 0 sin ξ+z cos ξ) e jk(y η E = u x TE 0 e jβy e αz H = (u y jα u z β) TE 0 kη e jβy e αz This is called an evanescent wave.

124 Normal Incidence Oblique Wave Travel Oblique Incidence Observations About Evanescent Waves The wave in the second medium decays exponentially in the z-direction. The ratio between E x and H y is pure imaginary Z z = jα kη The z-component of the Poynting vector vanishes S = Re { 1 2 E H } = 0 The reflection coefficient has unit magnitude. (This does not mean that T = 0!)

125 Normal Incidence Oblique Wave Travel Oblique Incidence More Observations About Planar Interfaces All of the above does not change in other situations such as lossy media. It is often easier to work directly with k y and k z then the angles. To be a solution to Maxwell s Equations, we need k 2 y + k 2 z = k 2. (This is the same as cos 2 θ + sin 2 θ = 1.) This avoids confusion with trigonometry and complex numbers. For incidence on a good conductor (σ 2 >> jωɛ 2 ) we have sin θ 2 = k 1 = ω µ0 ɛ 1 jωɛ 1 = 0 sin θ 1 k 2 jωµ0 σ 2 σ 2 Since sin θ 2 0, fields enter good conductors on the normal.