1. 1 EE 543 Theory and Principles of Remote Sensing Reflection and Refraction from a Planar Interface

2. O. Kilic EE 543 2 Outline 1.Reflection and Transmission at a planar interface –Boundary conditions –Fresnel reflection coefficients –Special Cases: Total reflection (critical angle), Total transmission (Brewster angle) 2.Rough Surface 3.Scattering from objects

3. O. Kilic EE 543 3 Reflection and Transmission When we consider interaction of em waves with a target, we must account for the effects of boundaries between media. These boundary effects give rise to changes in the amplitude, phase and direction of propagation waves. These in turn either carry information about the target or cause clutter for the received signal.

4. O. Kilic EE 543 4 Boundary Conditions Medium 1 Medium 2 n i E, H * Boundary conditions are a direct consequence of Maxwell’s equations. * They can be derived from the integral form by assuming an infinitesimal closed path, or an infinitesimal volume across the boundary. t

5. O. Kilic EE 543 5 dA 2 dA 1 Continuity of the Normal Component Medium 1 Medium 2 n1n1 D1D1 D2D2  z  0 Medium 1 Medium 2 n2n2

6. O. Kilic EE 543 6 Continuity of the Tangential Component Medium 1 Medium 2 t1t1 D1D1 D2D2  z  0 Medium 1 Medium 2 t2t2 t

7. O. Kilic EE 543 7 General Form of Boundary Conditions Medium 1 Medium 2 n i E, H Components of E and H (Normal and Tangential): H n HnHn H tan n E EnEn E tan

8. O. Kilic EE 543 8 Special Cases – Dielectric Boundary For dielectric material or material with finite conductivity, J s = 0; i.e. surface current does not exist. Only volume current exists. Therefore:

9. O. Kilic EE 543 9 Special Cases – Perfect Conductors For perfect conductors, there can not be any voltage difference between any two points on the surface. Thus, E tan2 = 0 on the surface. Also, due to the same reasoning there can not be any fields inside a perfect conductor. Therefore:

10. O. Kilic EE 543 10 Example 1 A region contains a perfectly conducting half-space and air. We know that the surface current on the perfect conductor is. What is the tangential H field in air just above the conductor? y x JsJs H

11. O. Kilic EE 543 11 Plane of Incidence n i Defined by unit vectors i and n, where i is the direction of the incident fields, and n is the unit normal vector to the boundary between the two medium. In this example, x-z plane is the plane of incidence. x z

12. O. Kilic EE 543 12 Plane of Incidence and E, H Fields E H i Incident fields E i, H i lie on a plane (P) perpendicular to i. Therefore, E i, H i can be decomposed into two basis (orthogonal) vectors that describe (P). There are infinitely many possible such pair of components. For instance, if i=z then x, y or x+y, x-y would be possible orthogonal vectors. P

13. O. Kilic EE 543 13 E and H Decomposition We observe that E and H can be decomposed into infinitely many pairs of orthogonal vectors that lie on the plane P. One choice is to have t = i x n as one component, where i: incidence direction, and n: normal to the boundary. Since t is defined as the cross product of i with another vector, t is orthogonal to i, and thus can be a valid component of E or H.

14. O. Kilic EE 543 14 Polarization The selection of the orthogonal components for a reflection and transmission problem is usually done with respect to the plane of incidence. These define two orthogonally polarized components of E and H fields. These are called: –Perpendicular Polarization (TE, Horizontal) –Parallel Polarization (TM, Vertical) Defined with respect to the plane of incidence Defined for E field with respect to the interface Defined for E field

15. O. Kilic EE 543 15 Perpendicular (TE) Polarization (Horizontal) The electric field is perpendicular (transverse, TE) to the plane of incidence. n i E Plane formed by i and n REMARK: Since n is on the plane of incidence, this condition (TE) implies that E lies on the interface; i.e. is horizontal.

16. O. Kilic EE 543 16 Parallel Polarization (TM) (Vertical) The electric field does not have a component perpendicular to the plane of incidence; i.e. E lies on the plane of incidence. Hence, the magnetic field is perpendicular (transverse, TM) to the plane of incidence. n i H REMARK: Since n is on the plane of incidence, this condition (TM) implies that E has a component perpendicular to the interface i.e. is vertical.

17. O. Kilic EE 543 17 Decomposition into Two Polarizations Therefore, both E and H fields can be represented in the most general sense as a sum of these two orthogonal polarizations; i.e. TE (perpendicular) TM (parallel)

18. O. Kilic EE 543 18 Example: Planar Boundary with a Dielectric Interface x z i E, H n n: normal to the surface i: direction of propagation

19. O. Kilic EE 543 19 Reflection and Transmission – Perpendicular Polarization (TE) x i EiEi n z X HiHi ii o t rr rr  1,  1  2,  2 E is perpendicular to the plane of incidence. H lies on the plane of incidence. Plane of incidence: x-z

20. O. Kilic EE 543 20 Calculation of TE Coefficients Due to the law of reflection (from matching the boundary conditions) The incident, reflected and transmitted fields can be written in terms of the propagation direction, and reflection and transmission coefficients for TE waves.

21. O. Kilic EE 543 21 TE Geometry Wave vectors: Let the unit vectors along the direction of propagation be: incident reflected transmitted 11 z o 11 k 1i tt k 1r k2k2 i t

22. O. Kilic EE 543 22 Electric Field Expressions, TE Unknowns:

23. O. Kilic EE 543 23 Magnetic Field Expressions, TE The corresponding magnetic fields are obtained from Maxwell’s equations where,

24. O. Kilic EE 543 24 Application of Boundary Conditions Boundary conditions on the surface (dielectric medium): x i EiEi X HiHi o t  1,  1  2,  2 z X ErEr HrHr X HtHt EtEt

25. O. Kilic EE 543 25 BC for Electric Field

26. O. Kilic EE 543 26 BC for Magnetic Field

27. O. Kilic EE 543 27 Equations for TE In summary, the following has to be satisfied for all points across the boundary; i.e. for all x values. This implies that exponential terms have to be equal; i.e. Snell’s Law Phase matching condition

28. O. Kilic EE 543 28 Solution for TE Thus, the boundary conditions result in: We obtain

29. O. Kilic EE 543 29 Reflection and Transmission – Parallel Polarization (TM) x i EiEi n X HiHi ii o t tt rr  1,  1  2,  2 H is perpendicular to the plane of incidence. E lies on the plane of incidence; i.e. is parallel to it.

30. O. Kilic EE 543 30 Calculation of TM Coefficients Due to duality principle, the reflection and transmission coefficients for TM can be obtained from the TE case by letting

31. O. Kilic EE 543 31 Solution for TM HW Problem: Carry out the derivation for TM mode following similar steps as in TE case, and prove the equation above.

32. O. Kilic EE 543 32 Power Coefficients The R, T coefficients describe field (E,H) behavior. The power reflection coefficient (reflectivity)  and transmission coefficient (transmissivity)  are defined with respect to the normal components of the time average Poynting vector.

33. O. Kilic EE 543 33 Power Conservation S av_z, inc S av_z, ref S av_z, tr S av_x, inc S av_x, ref S av_x, tr S av_x, inc S av_x, ref S av_x, tr S av_z, inc S av_z, tr S av_z, ref = +

34. O. Kilic EE 543 34 Reflectivity and Transmissivity where Time average Poynting Vector

35. O. Kilic EE 543 35 Reflectivity and Transmissivity, TE Note that the field coefficients satisfy: While the power coefficients satisfy energy conservation: Due to b.c. Due to power conservation

36. O. Kilic EE 543 36 Reflectivity and Transmissivity, TM Note that cos(  2 ) can be a complex number for certain incidence angles, such as in total reflection phenomenon.

37. O. Kilic EE 543 37 Total Reflection (Critical Angle) If the wave is incident from a dense to a less dense medium (i.e. k1>k2), it is possible to have no transmission into the second medium. Recall Snell’s Law: Can be > 1 for certain values of  1 >1 will result in pure imaginary  2

38. O. Kilic EE 543 38 Transmission Angle for Incidence Beyond Critical Angle Pure imaginary

39. O. Kilic EE 543 39 Critical Angle The smallest incidence angle at which no real transmission angle exists is called the critical angle. Thus all waves that arrive at incidence angles greater than the critical angle suffer total reflection.

40. O. Kilic EE 543 40 Total Reflection for Incidence Beyond Critical Angle Therefore, for power considerations and

41. O. Kilic EE 543 41 Example – Light source under water For an isotropic (i.e. equal radiation in all directions) light source, only light rays within a cone of  c can be transmitted to air. The permittivity of water at optical frequencies is 1.77  o. Calculate the value of the critical angle. k1k1 k2k2

42. O. Kilic EE 543 42 Total Transmission (Brewster Angle) There exists an angle at which total transmission occurs for TM (parallel = vertically polarized) wave.

43. O. Kilic EE 543 43 Proof

44. O. Kilic EE 543 44 Total Transmission Note that total transmission phenomenon only occurs for TM (parallel) polarization. As a result, randomly polarized waves become polarized on reflection. (TM portion is not reflected back) Note that: – for incidence angles smaller that  B, R // term is less than zero. –for incidence angles larger than  B, R // term is greater than zero.

45. O. Kilic EE 543 45 Brewster Angle Impact on Polarization Impact on Polarization Polarization changes direction when R // changes sign. Consider circular polarization:

46. O. Kilic EE 543 46 HW Problem Plot the magnitude of reflection and transmission coefficients for a plane wave incident on a flat ground for incidence angles [0-90] degrees. Assume a lossless ground, i.e. zero conductivity and  r2 = 4.0. Plot both TE and TM cases on the same chart.

47. O. Kilic EE 543 47 Brewster vs Critical Angle Note that Brewster angle is always less than critical angle. cc BB

48. O. Kilic EE 543 48 Example A perpendicularly polarized em wave impinges from medium 1 to medium 2 as shown below. Calculate: 60 o EiEi  2,  2  1,  1 x z a)the critical angle b)k x, k z in terms of w,  ,   c)k t z in terms of w,  ,   d)Reflection and transmission coefficients

49. O. Kilic EE 543 49 About Lossy Media The Brewster angle loses its meaning if one of the media is lossy, in which case the Brewster angle θ B will be complex-valued.

50. O. Kilic EE 543 50 Solution (1/3)

51. O. Kilic EE 543 51 Solution (2/3)

52. O. Kilic EE 543 52 Solution (3/3) Plane of incidence: x-z TE polarization because E is along y-direction

53. O. Kilic EE 543 53 Reflection and Transmission for Perfect Conductor Boundary

54. O. Kilic EE 543 54 Examples 1.Normal incidence of a plane wave on a perfect conductor 2.Oblique incidence of a plane wave on a perfect conductor

55. O. Kilic EE 543 55 Normal Incidence on a Perfect Conductor Note that normal incidence is a special case of TE. E lies on the interface, so it is horizontally polarized (TE).

56. O. Kilic EE 543 56 Total Fields (Perfect Conductor Boundary)

57. O. Kilic EE 543 57 Oblique Incidence on a Perfect Conductor

58. O. Kilic EE 543 58 Reflection and Transmission for Lossy Media (Conducting Medium) When a plane wave is incident on a finitely conducting medium, the laws of Fresnel and Snell are still valid in a purely formal way. Since the medium permittivity is complex, the Fresnel coefficients are now complex. This implies that the incident wave will be modified in both amplitude and phase upon reflection. The problem of using Snell’s law to find the angle of transmission is more involved. The complex value of the angle results in attenuating transmitted waves.

59. O. Kilic EE 543 59 Plane Wave Parameters for Conducting Material >>

60. O. Kilic EE 543 60 Incidence on Lossy Medium

61. O. Kilic EE 543 61 Lossy medium

62. O. Kilic EE 543 62 Examples of Conducting Medium

63. O. Kilic EE 543 63 f=100 MHz  = 0.001  r = 7 BB

64. O. Kilic EE 543 64 f=100 MHz  = 0.02  r = 30

65. O. Kilic EE 543 65 Effect of Conductivity on Brewster Angle Labels reversed!!!!!!!

66. O. Kilic EE 543 66 Layered Media Examples: soil, vegetation, ionosphere The problem of reflection and transmission at a plane boundary can be generalized to a multilayer case by evaluating the fields existing within a layer and then using a matrix technique to sum the effects of all layers.

67. O. Kilic EE 543 67 Layered Media Waves in each layer should satisfy: Maxwell’s equations Boundary conditions The wave in each layer is the sum of transmitted and reflected waves from all layers. z oo * Will be discussed in depth later.

68. O. Kilic EE 543 68 Rough Surface

69. O. Kilic EE 543 69 What is rough?

70. O. Kilic EE 543 70 Rayleigh Criterion

71. O. Kilic EE 543 71 Scattering from Finite Size Objects D E i, H i E s, H s E = E inc + E s H = H inc + H s Use boundary conditions on the surface S (1)

72. O. Kilic EE 543 72 Radiation Condition i o r Generic form of scattered field (2) Function of: object characteristics, direction of incidence direction of scattered field

73. O. Kilic EE 543 73 Object Size Depending on object size (D), different approximations can be applied.  >> DLow Frequency Approximation (quasi-static, Rayleigh)  ~ D resonant region; modal solution  << Doptical region; ray optics

74. O. Kilic EE 543 74 Scattering from a Cylinder Assume infinitely long and circularly symmetric z x y EiEi

75. O. Kilic EE 543 75 Scattering from a Cylinder Approach would be to write Maxwell’s equations. The scattered fields and internal fields (i.e. fields transmitted inside the cylinder) should both satisfy Maxwell’s equations. Analytic solution is an infinite series summation. Assumption of infinite length helps reduce the problem, as no variation with z is expected due to symmetry along that axis.

76. O. Kilic EE 543 76 Scattering from Cylinders Outward going wave

77. O. Kilic EE 543 77 References Applied electromagnetism, L. C. Shen, J. A. Kong, PWS Microwave Remote Sensing, F. T. Ulaby, et.al. Addison-Wesley