1. LECT 04© 2012 Raymond P. Jefferis III1 Satellite Communications Electromagnetic Wave Propagation Overview Electromagnetic Waves Propagation Polarization Antennas Antenna radiation patterns Propagation Losses Goldstone antenna at twilight, NASA

2. LECT 04© 2012 Raymond P. Jefferis III2 Reference Reference is specifically made to the following highly recommended source: Kraus, J. D. and Marhefka, R. J., Antennas For All Applications, Third Edition, McGraw-Hill, 2002 from which the antenna radiation equations used below were drawn.

3. LECT 04© 2012 Raymond P. Jefferis III3 Overview Satellite communication takes place through the propagation of focused and directed electromagnetic (EM) waves Since both received and transmitted waves are simultaneously present at very different power levels, in a satellite, both frequency separation and EM field polarization are used to decouple the channels

4. Maxwell’s Equations LECT 04© 2012 Raymond P. Jefferis III4 Maxwell’s equations in terms of free charge and current, WIKIPEDIA

5. Wave Equation LECT 04© 2012 Raymond P. Jefferis III5 For scalar variable, u (E & M Fields) Solutions are sinusoids in time and space (waves)

6. LECT 04© 2012 Raymond P. Jefferis III6 EM Wave Propagation Electromagnetic (EM) waves propagate energy, contained in their electric and magnetic fields, through space with velocity v, which is the speed of light under the conditions of propagation. Wikipedia

7. LECT 04© 2012 Raymond P. Jefferis III7 Transverse EM (Plane) Wave Properties Velocity of propagation (near light speed) Electric field is normal to the magnetic field Both electric and magnetic fields are normal to direction of propagation (plane wave) The relation of electric to magnetic fields is a constant for the medium (air, vacuum) Waves are polarized, as determined by the direction of the electric field orientation

8. LECT 04© 2012 Raymond P. Jefferis III8 Impedance The electric field strength E and magnetic field intensity H in a propagating wave are related by, where,  =magnetic permeability  Henry/meter]    -7  Henry/meter] in vacuum  =dielectric constant [Farads/meter]  0 = 1/36  *10 -8 [Farads/meter] in vacuum  =impedance of the medium (  0 =376.7 Ohms in free space)

9. LECT 04© 2012 Raymond P. Jefferis III9 Impedance Change At Boundaries At a boundary between two media of differing impedances (air and raindrops for instance), Z 1 and Z 2 [Ohms] –Part of the incident wave from Medium 1 is reflected –Part of the incident wave is transmitted into Medium 2

10. LECT 04© 2012 Raymond P. Jefferis III10 Wave Energy The electric and magnetic energy densities in a plane wave are equal. [J/m 2 ] The total energy is the sum of these energies. [J/m 2 ]

11. LECT 04© 2012 Raymond P. Jefferis III11 Wave Energy Density The energy density of a plane wave is the Poynting energy, S [Watts/m 2 ]

12. LECT 04© 2012 Raymond P. Jefferis III12 Vertical Polarization Behavior Radio frequency energy at frequency, f, propagates The wave propagates away from the observer (into the paper), along the z-axis Energy propagates with velocity, v, As a function of distance, z, and time, t, the vertical electric field is described by,

13. LECT 04© 2012 Raymond P. Jefferis III13 Horizontal Polarization Radio frequency energy at frequency, f, propagates The wave propagates away from the observer, along the z-axis Energy propagates with velocity, v, As a function of distance, z, and time, t, the horizontal E-field is described by,

14. Manipulated Variable Example Run mCos example: Vary the frequency and observe the results Pick a position (say z = 0.5), and change the z-variable to see how the wave propagates past the selected location LECT 04© 2012 Raymond P. Jefferis III Lect 00 - 14

15. Antennas Electromagnetic circuits comparable in size to the wavelength of an alternating current Have alternating electric and magnetic fields resulting in Electromagnetic (EM) radiation Have a polarization specified by the electric field direction (horizontal or vertical) Radiation pattern is affected by the shape of the current-carrying conductor(s) The EM radiation propagates in space LECT 04© 2012 Raymond P. Jefferis III Lect 00 - 15

16. LECT 04© 2012 Raymond P. Jefferis III16 Vertically Polarized Antenna Total antenna length typically  /2 Electric field shown normal to the plane of the earth (vertical) Oscillating electric fields produce accelerating and decelerating conduction electrons, with consequent radiation of EM-energy A magnetic field surrounds the current- carrying wire The phases of the electric and magnetic fields differ by 90 degrees

17. LECT 04© 2012 Raymond P. Jefferis III17 Horizontally Polarized Antenna Total antenna length typically  /2 where λ = c/f Electric field shown parallel to the plane of the earth (horizontal) Oscillating electric fields produce accelerating and decelerating conduction electrons, with consequent radiation of EM-energy A magnetic field surrounds the current- carrying wire The phases of the electric and magnetic fields differ by 90 degrees

18. LECT 04© 2012 Raymond P. Jefferis III18 Polarization Match Angles A match angle,  M, is defined as the angular polarization difference between a transmitting and a receiving antenna Smaller match angles result in greater coupling between transmitting and receiving antennas If the antennas are at opposite polarizations (vertical - horizontal) the received power will be zero, theoretically.

19. LECT 04© 2012 Raymond P. Jefferis III19 Circular Polarization Radio frequency energy at frequency, f, propagates as an EM wave, away from the observer, along the z-axis (into the paper) The energy propagates with velocity, v The electric and magnetic fields rotate in time (space) according to,

20. LECT 04© 2012 Raymond P. Jefferis III20 Circularly Polarized Antenna Circular Polarization, Wikipedia Note the spiral net electric field resolves into time-varying E x and E y components. Conductor (black); E x => Green; E y => Red

21. The Isotropic (Ideal) Antenna The gains of antennas can be stated relative to an isotropic ideal antenna as G [dBi], where G > 0. This antenna is a (theoretical) point source of EM energy It radiates uniformly in all directions A sphere centered on this antenna would exhibit constant energy per unit area over its surface The gain of an isotropic antenna is 0 dBi Lect 05© 2012 Raymond P. Jefferis III Lect 00 - 21

22. LECT 04© 2012 Raymond P. Jefferis III22 Radiation Patterns of Antennas Electric field intensity is a function of the radial distance and the angle from the antenna A radiation pattern can be plotted to show field strength (shown as a radial distance) vs angle The angle between half-power points (denoted as HPBW) is a measure of the focusing (Gain) of the antenna. [Note: Half-power = 3 dB] Note: Antenna Gain is with respect to an ideal isotropic antenna (Gain = 1.0 or 0.0 dBi)

23. LECT 04© 2012 Raymond P. Jefferis III23 Antenna Gain Calculation G = P A /P I where, P I is the power per unit area radiated by an isotropic antenna, and P A is the antenna power per unit area radiated by a non-isotropic antenna, G is the amount by which the isotropic power would be multiplied to give the same power per unit area as the gain antenna exhibits in the chosen direction

24. LECT 04© 2012 Raymond P. Jefferis III24 Antenna Gain Calculation P r = radiated power per unit area W = total applied power R r = antenna radiation resistance I m = maximum value of antenna current

25. LECT 04© 2012 Raymond P. Jefferis III25 Antenna Gain and Aperture Calculations G = antenna gain A e = effective aperture area  = carrier wavelength η = aperture efficiency A = aperture area (  r 2 )

26. LECT 04© 2012 Raymond P. Jefferis III26 Half-Wave Dipole Power θ is the angle normal to the antenna

27. LECT 04© 2012 Raymond P. Jefferis III27 Dipole Radiation Patterns Two dipole lengths shown:  L =  /2 (half wave dipole) HPBW = 78˚ Gain = 2.15 dBi  L =  (full wave dipole) HPBW = 47˚ Gain = 3.8 dBi The longer antenna focuses the energy into a more narrow beam and thus has higher Gain. Electric field intensity, half-wave dipole

28. LECT 04© 2012 Raymond P. Jefferis III28 Half-Wave Dipole Radiation The radiated field and power of a half-wave dipole antenna are expressed by: Radiated power pattern, half-wave dipole

29. LECT 04© 2012 Raymond P. Jefferis III29 Half-Wave Dipole Radiation Pattern zro = 0.000001; e0 = 1.0; e1 = Cos[p/2*Cos[theta]]/Sin[theta]; e2 = e1^2; PolarPlot[{e2}, {theta, zro, Pi}, PlotStyle -> {Directive[Thick, Black]}, PlotRange -> Automatic]

30. LECT 04© 2012 Raymond P. Jefferis III30 Half-Power Beam Width The Half-Power Beam Width (HPBW) is defined as the included angle between the half-power points on the radiation pattern. The power is down by 3 dB at these points. For a half-wave dipole antenna this is calculated as shown on the Mathematica ® notebook output that continues below.

31. LECT 04© 2012 Raymond P. Jefferis III31 Half-Wave Dipole HPBW Calculation r1 = FindRoot[e1^2 - 0.5 == 0.0, {theta, 60.0 Degree}]; Print[r1] w1 = theta /. r1 Print[w1/Degree] r2 = FindRoot[e1^2 - 0.5 == 0.0, {theta, 120.0 Degree}]; Print[r2] w2 = theta /. R2 Print[w2/Degree] Print[(w2 - w1)/Degree]

32. LECT 04© 2012 Raymond P. Jefferis III32 HPBW for Half-Wave Dipole From the foregoing notebook, the Half- Power Beam Width is found to be: HPBW = 78.0777 degrees At the outer edges of the beam (HPBW), the power will be 70.7% of the maximum power value.

33. LECT 04© 2012 Raymond P. Jefferis III33 Full-Wave Dipole Radiation Power pattern, full-wave dipole The radiated field and power of a full-wave dipole antenna are expressed, as a function of angle, by:

34. LECT 04© 2012 Raymond P. Jefferis III34 Full-Wave Dipole Radiation Pattern zro = 0.000001; e0 = 1.0; en = 2.0; e1 = (Cos[p*Cos[theta]] + 1)/(Sin[theta]*en); e2 = e1^2; PolarPlot[{e2}, {theta, zro, p}, PlotStyle -> {Directive[Thick, Black]}]

35. LECT 04© 2012 Raymond P. Jefferis III35 Half-Power Beam Width The Half-Power Beam Width (HPBW) is defined as the included angle between half- power points on the radiation pattern. The power is down by 3 dB at these points. For a full-wave dipole antenna this is calculated as shown on the Mathematica ® notebook output that continues below.

36. LECT 04© 2012 Raymond P. Jefferis III36 Full-Wave HPBW Calculation r1 = FindRoot[e1^2 - 0.5 == 0.0, {theta, 60.0 Degree}]; Print[r1] w1 = theta /. r1 Print[w1/Degree] r2 = FindRoot[e1^2 - 0.5 == 0.0, {theta, 120.0 Degree}]; Print[r2] w2 = theta /. r2 Print[w2/Degree] Print[(w2 - w1)/Degree]

37. LECT 04© 2012 Raymond P. Jefferis III37 HPBW for Full-Wave Dipole From the foregoing notebook, the Half- Power Beam Width is found to be: HPBW = 47.8351 degrees At the outer edges of the beam (HPBW), the power will be 70.7% of the full value.

38. LECT 04© 2012 Raymond P. Jefferis III38 Circular Aperture Antenna The electric field of a circular aperture antenna can be calculated from: where, D/  gives the aperture diameter in wavelengths and ϕ is the angle relative to the normal to the plane of the aperture.

39. LECT 04© 2012 Raymond P. Jefferis III39 Radiated E-Field of Aperture Antenna The Mathematica ® notebook follows, for D/  = 10: E-field for aperture with D/ = 10

40. LECT 04© 2012 Raymond P. Jefferis III40 Radiation Pattern of Aperture Antenna Dlam = 10; e2 = (2.0/p*Dlam)*(BesselJ[1, p*Dlam*Sin[theta]])/Sin[theta]; PolarPlot[Abs[e2]/100, {theta, -p/6, p/6}, PlotStyle -> {Directive[Thick, Black]}]

41. LECT 04© 2012 Raymond P. Jefferis III41 Radiated Power from an Aperture The normalized radiated power can be found from E 2 [  ] as shown below: Normalized radiated power for aperture with D/  = 10

42. LECT 04© 2012 Raymond P. Jefferis III42 Radiated Power Calculation Dlam = 10; e2 = (2.0/p*Dlam)*(BesselJ[1, p*Dlam*Sin[theta]])/Sin[theta]; PolarPlot[Abs[e2/100], {theta, -p/6, p/6}, PlotStyle -> {Directive[Thick, Black]}, PlotRange -> {{0, 1}, {-0.04, 0.04}}]

43. LECT 04© 2012 Raymond P. Jefferis III43 Half Power Beam Width The HPBW of an aperture having D/  = 10 is calculated to be: 5.89831 Degrees The Mathematica ® notebook for this calculation follows:

44. LECT 04© 2012 Raymond P. Jefferis III44 Aperture HPBW Calculation p20 =((2.0/p*Dlam)* (BesselJ[1,p*Dlam*Sin[0.00001]])/Sin[0.00001])^2 p2 = ((2.0/p*Dlam)* (BesselJ[1,p*Dlam*Sin[theta]])/Sin[theta])^2/p20; r1 = FindRoot[p2 - 0.5 == 0.0, {theta,1 Degree}]; w1 = theta /. r1; Print[w1/Degree] r2 = FindRoot[p2 - 0.5 == 0.0,{theta,-1 Degree}]; w2 = theta /. r2 Print[w2/Degree] Print[Abs[(w2 - w1)]/Degree]

45. Workshop 04 - Antenna HPBW A circular aperture antenna has D/  = 20. Plot the radiation pattern of this antenna and calculate its Half Power Beam Width. What can you say about the aiming requirements for such an antenna mounted on a satellite? LECT 04© 2012 Raymond P. Jefferis III Lect 00 - 45

46. LECT 04© 2012 Raymond P. Jefferis III46 Transmission Losses Transmitted electromagnetic energy from a satellite is lost on its way to the receiving station due to a number of factors, including: – Antenna efficiency– Rain/Cloud loss – Antenna aperture gain– Atmospheric loss – Path loss– Diffraction loss

47. LECT 04© 2012 Raymond P. Jefferis III47 Antenna Gain and Link Losses P t = transmitted power P r = received power A t = transmit antenna aperture A r = receive antenna aperture L p = path loss L a = atmospheric attenuation loss L d = diffraction losses Antenna Gain (t or r): G t/r = 4  A e t/r /  2 Combined Antenna Gain (t + r): G = G t G r

48. LECT 04© 2012 Raymond P. Jefferis III48 Antenna Gain A e = effective antenna aperture G = 4  A e /  2 (Antenna Gain) d = antenna diameter λ = wavelength  = aperture efficiency

49. Compensating for Link Losses Increase antenna gain Increase power input to antenna Net effect: increase EIRP (Equivalent Isotropically Radiated Power) - Make sure tracking of beam is accurate (target on beam axis). LECT 04© 2012 Raymond P. Jefferis III Lect 00 - 49

50. EIRP Equivalent Isotropic Radiated Power – the equivalent power input that would be needed for an isotropic antenna to radiate the same power over the angles of interest LECT 04© 2012 Raymond P. Jefferis III Lect 00 - 50

51. LECT 04© 2012 Raymond P. Jefferis III51 Path Loss Calculation Effective Aperture (transmit or receive): A e =  A Effective Radiated Power: EIRP = P t G t = P t  t A t Path Loss (for path length R): L p = (4  R/  2 Received Power: P r = EIRP*G r /L p where, G t = 4  A et /  2 G r = 4  A er /  2

52. LECT 04© 2012 Raymond P. Jefferis III52 Decibel (dB) Scale Definition P dB = 10 log 10 P t /P r Logarithmic scale changes division and multiplication into subtraction and addition dBW refers to power with respect to 1 Watt. Received power (Pratt & Bostian, Eq. 4.11): P r = EIRP + G r - L p [dBW]

53. LECT 04© 2012 Raymond P. Jefferis III53 Received Power - dB Model (Pratt & Bostian, Eq. 4.11) P r = EIRP + G r - L p - L a - L t - L r [dBW] –EIRP => Effective radiated power –G r => Receiving antenna gain –L p => Path loss –L a => Atmospheric attenuation loss –L t => Transmitting antenna losses –L r => Receiving antenna losses

54. LECT 04© 2012 Raymond P. Jefferis III54 End