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1 Topic 3 - Basic EM Theory and Plane Waves EE 542 Antennas & Propagation for Wireless Communications.

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Presentation on theme: "1 Topic 3 - Basic EM Theory and Plane Waves EE 542 Antennas & Propagation for Wireless Communications."— Presentation transcript:

1 1 Topic 3 - Basic EM Theory and Plane Waves EE 542 Antennas & Propagation for Wireless Communications

2 O. Kilic EE542 2 Outline EM Theory Concepts Maxwell’s Equations –Notation –Differential Form –Integral Form –Phasor Form Wave Equation and Solution (lossless, unbounded, homogeneous medium) –Derivation of Wave Equation –Solution to the Wave Equation – Separation of Variables –Plane waves

3 O. Kilic EE542 3 EM Theory Concept The fundamental concept of em theory is that a current at a point in space is capable of inducing potential and hence currents at another point far away. J E, H

4 O. Kilic EE542 4 Introduction to EM Theory The existence of propagating em waves can be predicted as a direct consequence of Maxwell’s equations. These equations satisfy the relationship between the vector electric field, E and vector magnetic field, H in time and space in a given medium. Both E and H are vector functions of space and time; i.e. E (x,y,z;t), H (x,y,z;t.)

5 O. Kilic EE542 5 What is an Electromagnetic Field? The electric and magnetic fields were originally introduced by means of the force equation. In Coulomb’s experiments forces acting between localized charges were observed. There, it is found useful to introduce E as the force per unit charge. Similarly, in Ampere’s experiments the mutual forces of current carrying loops were studied. B is defined as force per unit current.

6 O. Kilic EE542 6 Why not use just force? Although E and B appear as convenient replacements for forces produced by distributions of charge and current, they have other important aspects. First, their introduction decouples conceptually the sources from the test bodies experiencing em forces. If the fields E and B from two source distributions are the same at a given point in space, the force acting on a test charge will be the same regardless of how different the sources are. This gives E and B meaning in their own right. Also, em fields can exist in regions of space where there are no sources.

7 O. Kilic EE542 7 Maxwell’s Equations Maxwell's equations give expressions for electric and magnetic fields everywhere in space provided that all charge and current sources are defined. They represent one of the most elegant and concise ways to state the fundamentals of electricity and magnetism. These set of equations describe the relationship between the electric and magnetic fields and sources in the medium. Because of their concise statement, they embody a high level of mathematical sophistication.

8 O. Kilic EE542 8 Notation: (Time and Position Dependent Field Vectors) E ( x,y,z;t ) Electric field intensity (Volts/m) H ( x,y,z;t ) Magnetic field intensity (Amperes/m) D ( x,y,z;t ) Electric flux density (Coulombs/m 2 ) B ( x,y,z;t ) Magnetic flux density (Webers/m 2, Tesla)

9 O. Kilic EE542 9 Notation: Sources and Medium J ( x,y,z;t ) Electric current density (Amperes/m 2 ) J d ( x,y,z;t ) Displacement current density (Amperes/m 2 ) ee Electric charge density (Coulombs/m 3 ) rr Permittivity of the medium (Farad/m) Relative permittivity (with respect to free space  o ) rr Permeability of the medium (Henry/m) Relative permittivity (with respect to free space  o )  Conductivity of the medium (Siemens/m)

10 O. Kilic EE Maxwell’s Equations – Physical Laws Faraday’s Law  Changes in magnetic field induce voltage. Ampere’s Law  Allows us to write all the possible ways that electric currents can make magnetic field. Magnetic field in space around an electric current is proportional to the current source. Gauss’ Law for Electricity  The electric flux out of any closed surface is proportional to the total charge enclosed within the surface. Gauss’ Law for Magnetism  The net magnetic flux out of any closed surface is zero.

11 O. Kilic EE Differential Form of Maxwell’s Equations Faraday’s Law: Ampere’s Law: Gauss’ Law: (1) (2) (3) (4)

12 O. Kilic EE Constitutive Relations Constitutive relations provide information about the environment in which electromagnetic fields occur; e.g. free space, water, etc. Free space values. (5) (6) permittivity permeability

13 O. Kilic EE Time Harmonic Representation - Phasor Form In a source free ( ) and lossless ( ) medium characterized by permeability  and permittivity , Maxwell’s equations can be written as:

14 O. Kilic EE Examples of del Operations The following examples will show how to take divergence and curl of vector functions

15 O. Kilic EE Example 1

16 O. Kilic EE Solution 1

17 O. Kilic EE Example 2 Calculate the magnetic field for the electric field given below. Is this electric field realizable?

18 O. Kilic EE Solution

19 O. Kilic EE Solution continued

20 O. Kilic EE Solution continued To be realizable, the fields must satisfy Maxwell’s equations!

21 O. Kilic EE Solution Continued These fields are NOT realizable. They do not form em fields.

22 O. Kilic EE Time Harmonic Fields We will now assume time harmonic fields; i.e. fields at a single frequency. We will assume that all field vectors vary sinusoidally with time, at an angular frequency w; i.e.

23 O. Kilic EE Time Harmonics and Phasor Notation Using Euler’s identity The time harmonic fields can be written as Phasor notation

24 O. Kilic EE Note that the E and H vectors are now complex and are known as phasors Phasor Form Information on amplitude, direction and phase

25 O. Kilic EE Time Harmonic Fields in Maxwell’s Equations With the phasor notation, the time derivative in Maxwell’s equations becomes a factor of jw:

26 O. Kilic EE Maxwell’s Equations in Phasor Form (1)

27 O. Kilic EE Maxwell’s Equations in Phasor Form (2)

28 O. Kilic EE Phasor Form of Maxwell’s Equations (3) Maxwell’s equations can thus be written in phasor form as: Phasor form is dependent on position only. Time dependence is removed.

29 O. Kilic EE Examples on Phasor Form Determine the phasor form of the following sinusoidal functions: a)f(x,t)=(5x+3) cos(wt + 30) b)g(x,z,t) = (3x+z) sin(wt) c)h(y,z,t) = (2y+5)4z sin(wt + 45) d)V(t) = 0.5 cos(kz-wt)

30 O. Kilic EE Solutions a)

31 O. Kilic EE Solutions b)

32 O. Kilic EE Solution c)

33 O. Kilic EE Solution d)

34 O. Kilic EE Example Find the phasor notation of the following vector:

35 O. Kilic EE Solution

36 O. Kilic EE Example Show that the following electric field satisfies Maxwell’s equations.

37 O. Kilic EE Solution

38 O. Kilic EE The Wave Equation (1) If we take the curl of Maxwell’s first equation: Using the vector identity: And assuming a source free, i.e. and lossless; medium:i.e.

39 O. Kilic EE The Wave Equation (2) Define k, which will be known as wave number:

40 O. Kilic EE Wave Equation in Cartesian Coordinates where

41 O. Kilic EE Laplacian

42 O. Kilic EE Scalar Form of Maxwell’s Equations Let the electric field vary with x only. and consider only one component of the field; i.e. f(x).

43 O. Kilic EE Possible Solutions to the Scalar Wave Equation Standing wave solutions are appropriate for bounded propagation such as wave guides. When waves travel in unbounded medium, traveling wave solution is more appropriate. Energy is transported from one point to the other

44 O. Kilic EE The Traveling Wave The phasor form of the fields is a mathematical representation. The measurable fields are represented in the time domain. Then Let the solution to the  -component of the electric field be: Traveling in +x direction

45 O. Kilic EE Traveling Wave As time increases, the wave moves along +x direction

46 O. Kilic EE Standing Wave Then, in time domain:

47 O. Kilic EE Standing Wave Stationary nulls and peaks in space as time passes.

48 O. Kilic EE To summarize We have shown that Maxwell’s equations describe how electromagnetic energy travels in a medium The E and H fields satisfy the “wave equation”. The solution to the wave equation can be in various forms, depending on the medium characteristics

49 O. Kilic EE The Plane Wave Concept Plane waves constitute a special set of E and H field components such that E and H are always perpendicular to each other and to the direction of propagation. A special case of plane waves is uniform plane waves where E and H have a constant magnitude in the plane that contains them.

50 O. Kilic EE Plane Wave Characteristics amplitude Frequency (rad/sec) Wave number, depends on the medium characteristics phase Direction of propagation polarization amplitude phase

51 O. Kilic EE Plane Waves in Phasor Form Complex amplitude Position dependence polarization

52 O. Kilic EE Example 1 Assume that the E field lies along the x-axis (i.e. x- polarized) and is traveling along the z-direction. We derive the solution for the H field from the E field using Maxwell’s equation #1: Intrinsic impedance; 377  for free space wave number Note the I = V/R analogy in circuit theory. E H

53 O. Kilic EE Example 1 (2 of 4) direction of propagation x y z E, H plane E and H fields are not functions of x and y, because they lie on x-y plane

54 O. Kilic EE Example 1 (3 of 4) phase term *** The constant phase term  is the angle of the complex number E o In time domain:

55 O. Kilic EE Wavelength: period in space k = 2  Example 1 (4 of 4)

56 O. Kilic EE Velocity of Propagation (1/3) We observe that the fields progress with time. Imagine that we ride along with the wave. At what velocity shall we move in order to keep up with the wave???

57 O. Kilic EE Velocity of Propagation (2/3) Constant phase points E field as a function of different times kz

58 O. Kilic EE Velocity of Propagation (3/3) In free space: Note that the velocity is independent of the frequency of the wave, but a function of the medium properties.

59 O. Kilic EE Example 2 A uniform em wave is traveling at an angle  with respect to the z-axis as shown below. The E field is in the y-direction. What is the direction of the H field? x z y k  E

60 O. Kilic EE Solution: Example 2 x z y k  The E field is along y The direction of propagation is the unit vector Because E, H and the direction of propagation are perpendicular to each other, H lies on x-z plane. It should be in the direction parallel to: E

61 O. Kilic EE Example 3 Write the expression for an x-polarized electric field that propagates in +z direction at a frequency of 3 GHz in free space with unit amplitude and 60 o phase. =1 W = 2  f = 2  *3* o x + z-direction

62 O. Kilic EE Solution 3 =1 W = 2  f = 2  *3* o x + z-direction

63 O. Kilic EE Example 4 If the electric field intensity of a uniform plane wave in a dielectric medium where  =  o  r and  =  o is given by: Determine: The direction of propagation and frequency The velocity The dielectric constant (i.e. permittivity) The wavelength

64 O. Kilic EE Solution: Example 4 (1/2) 1.+y direction; w = 2  f = Velocity: 3.Permittivity:

65 O. Kilic EE Solution: Example 4 (2/2) 4. Wavelength: m

66 O. Kilic EE Example 5 Assume that a plane wave propagates along +z-direction in a boundless and a source free, dielectric medium. If the electric field is given by: Calculate the magnetic field, H.

67 O. Kilic EE Example 5 - observations Note that the phasor form is being used in the notation; i.e. time dependence is suppressed. We observe that the direction of propagation is along +z-axis.

68 O. Kilic EE Solution: Example 5 (1/2) Intrinsic impedance, I = V/R E k

69 O. Kilic EE Solution: Example 5 (2/2) E, H and the direction of propagation are orthogonal to each other. Amplitudes of E and H are related to each other through the intrinsic impedance of the medium. Note that the free space intrinsic impedance is 377 

70 O. Kilic EE Example 6 Sketch the motion of the tip of the vector A(t) as a function of time.

71 O. Kilic EE Solution: Example 6 (1/2)

72 O. Kilic EE Solution: Example 6 (2/2) wt = 0 x y wt = 90 o wt = 180 o wt = 270 o The vector A(t) rotates clockwise wrt z-axis. The tip traces a circle of radius equal to unity with angular frequency w.

73 O. Kilic EE Polarization The alignment of the electric field vector of a plane wave relative to the direction of propagation defines the polarization. Three types: –Linear –Circular –Elliptical (most general form) Polarization is the locus of the tip of the electric field at a given point as a function of time.

74 O. Kilic EE Linear Polarization Electric field oscillates along a straight line as a function of time Example: wire antennas y x x y E E

75 O. Kilic EE Example 7 For z = 0 (any position value is fine) x t = 0 t =  - E o EoEo y Linear Polarization: The tip of the E field always stays on x- axis. It oscillates between ±E o

76 O. Kilic EE Example 8 Let z = 0 (any position is fine) x y t =  /2 t = Linear Polarization E xo =1 E yo =2

77 O. Kilic EE Circular Polarization Electric field traces a circle as a function of time. Generated by two linear components that are 90 o out of phase. Most satellite antennas are circularly polarized. y x y x RHCP LHCP

78 O. Kilic EE Example 7 E xo =1 E yo =1 Let z= 0 x RHCP y t=0 t=  /2w t=  t=3  /2w

79 O. Kilic EE Elliptical Polarization This is the most general form Linear and circular cases are special forms of elliptical polarization Example: log spiral antennas y x LH y x RH

80 O. Kilic EE Example 8 ExEx EyEy Linear when Circular when Elliptical if no special condition is met.

81 O. Kilic EE Example 9 Determine the polarization of this wave.

82 O. Kilic EE Solution: Example 9 (1/2) Note that the field is given in phasor form. We would like to see the trace of the tip of the E field as a function of time. Therefore we need to convert the phasor form to time domain.

83 O. Kilic EE Solution: Example 9 (2/2) Let z=0 Elliptical polarization

84 O. Kilic EE Example 10 Find the polarization of the following fields: a) b) c)

85 O. Kilic EE Solution: Example 10 (1/4) a) x y z Let kz=0 t=0 t=  /2w t=  t=3  /2w RHCP Observe that orthogonal components have same amplitude but 90 o phase difference. Circular Polarization

86 O. Kilic EE Solution: Example 10 (2/4) b) Observe that orthogonal components have same amplitude but 90 o phase difference. y z Let kx=0 t=-  /4w t=+  /4w t=3  t=5  /4w RHCP x Circular Polarization

87 O. Kilic EE Solution: Example 10 (3/4) c) Observe that orthogonal components have different amplitudes and are out of phase. Elliptical Polarization z Let ky=0 t=-  /w t=+  /w x y Left Hand

88 O. Kilic EE Solution: Example 10 (4/4) d) Observe that orthogonal components are in phase. Linear Polarization x y z

89 O. Kilic EE Coherence and Polarization In the definition of linear, circular and elliptical polarization, we considered only completely polarized plane waves. Natural radiation received by an anatenna operating at a frequency w, with a narrow bandwidth,  w would be quasi-monochromatic plane wave. The received signal can be treated as a single frequency plane wave whose amplitude and phase are slowly varying functions of time.

90 O. Kilic EE Quasi-Monochromatic Waves amplitude and phase are slowly varying functions of time

91 O. Kilic EE Degree of Coherence where denotes the time average.

92 O. Kilic EE Degree of Coherence – Plane Waves

93 O. Kilic EE Unpolarized Waves An em wave can be unpolarized. For example sunlight or lamp light. Other terminology: randomly polarized, incoherent. A wave containing many linearly polarized waves with the polarization randomly oriented in space. A wave can also be partially polarized; such as sky light or light reflected from the surface of an object; i.e. glare.

94 O. Kilic EE Poynting Vector As we have seen, a uniform plane wave carries em power. The power density is obtained from the Poynting vector. The direction of the Poynting vector is in the direction of wave propagation.

95 O. Kilic EE Poynting Vector

96 O. Kilic EE Example 11 Calculate the time average power density for the em wave if the electric field is given by:

97 O. Kilic EE Solution: Example 11 (1/2)

98 O. Kilic EE Solution: Example 11 (2/2)

99 O. Kilic EE Plane Waves in Lossy Media Finite conductivity,  results in loss Ohm’s Law applies: Conductivity, Siemens/m Conduction current

100 O. Kilic EE Complex Permittivity From Ampere’s Law in phasor form:

101 O. Kilic EE Wave Equation for Lossy Media Attenuation constant Phase constant Wave number: Loss tangent, 

102 O. Kilic EE Example 12 (1/2) Plane wave propagation in lossy media: complex number

103 O. Kilic EE Example 12 (2/2) Plane wave is traveling along +z-direction and dissipating as it moves. attenuation propagation

104 O. Kilic EE Field Attenuation in Lossy Medium

105 O. Kilic EE Attenuation and Skin Depth Attenuation coefficient, , depends on the conductivity, permittivity and frequency. Skin depth,  is a measure of how far em wave can penetrate a lossy medium

106 O. Kilic EE Lossy Media

107 O. Kilic EE Example 13 Calculate the attenuation rate and skin depth of earth for a uniform plane wave of 10 MHz. Assume the following properties for earth:  =  o  = 4  o  = 10 -4

108 O. Kilic EE Solution: Example 13 First we check if we can use approximate relations. Slightly conducting

109 O. Kilic EE References HYS/Class/waves/u10l1b.htmlhttp://www.glenbrook.k12.il.us/GBSSCI/P HYS/Class/waves/u10l1b.html Applied Electromagnetism, Liang Chi Shen, Jin Au Kong, PWS

110 O. Kilic EE Homework Assignments Due 9/25/08

111 O. Kilic EE Homework 3.1 The magnetic field of a uniform plane wave traveling in free space is given by 1.What is the direction of propagation? 2.What is the wave number, k in terms of permittivity,  o and permeability,  o ? 3.Determine the electric field, E.

112 O. Kilic EE Homework 3.2 Find the polarization state of the following plane wave:

113 O. Kilic EE Homework 3.3 How far must a plane wave of frequency 60 GHz propagate in order for the phase of the wave to be retarded by 180 o in a lossless medium with  r =1 and  r = 3.5?

114 O. Kilic EE What is the direction of propagation? Ans: -z 2.What is the wave number, k in terms of permittivity,  o and permeability,  o ? Ans: free space  3.Determine the electric field, E. Solution Homework 3.1 k H E

115 O. Kilic EE Solution: Homework 3.2 Observe that orthogonal components are in phase. Linear Polarization x y z

116 O. Kilic EE Solution 3.3 (1/2) Wavelength: period in space k = 2 

117 O. Kilic EE Solution 3.3 (2/2)


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