ECE 6340 Intermediate EM Waves. Fall 2016 Prof. David R. Jackson Dept. of ECE. Notes 16

Transcription

1 C 6340 Intermediate M Waves Fall 2016 Prof. David R. Jackson Dept. of C Notes 16 1

2 Polarization of Waves Consider a plane wave with both and components (,, z) = ( ˆ + ˆ ) e jkz Assume j0 = ae = a = j be β (polar form of comple numbers) ( In general, β = phase of phase of.) 2

3 Polarization of Waves (cont.) Time Domain: At z = 0 (t) φ (t) (t) ( jωt ae ) acos( ωt) = Re = ( jβ jωt be e ) b cos( ωt β) = Re = + Depending on b and β, three different cases arise. 3

4 Linear Polarization β = 0 or π At z = 0: = acosωt = bcos ωt+ β =± bcosωt ( ) (+ for 0, - for π) = ( a ˆ ± b ˆ )cosωt b (shown for β = 0) a 4

5 Circular Polarization b = a AND β = ± π /2 At z = 0: = acosωt = bcos ωt+ β = acos( ωt± π / 2) = asinωt ( ) = + = a cos ωt+ a sin ωt = a 2 5

6 β = ± π /2 Circular Polarization (cont.) a φ(t) () t φ = = ωt = ωt Note: tan tan ( tan ) tan (tan ) so ω φ = ωt wave dφ = angular velocit of wave = = ω dt ( ) ωwave = ω 6

7 Circular Polarization (cont.) I convention β = π / 2 RHCP a φ(t) () t β = ± π /2 β = +π / 2 LHCP φ = ωt = a( ˆcos ωt ˆsin ωt) 7

8 Circular Polarization (cont.) Note: The rotation is space is opposite to that in time. Time and distance dependence: ( t kz) j e ω ( ωt ωt kz) ( ˆcos( ω ) ˆsin ( ω )) = a t kz + t kz RHCP ( ˆcos( ω ) ˆsin ( ω )) = a t + t z = 0 ( ˆcos( ) ˆsin ( )) = a kz kz t = 0 8

9 Circular Polarization (cont.) Snapshot" of wave RHCP z RHCP z 9

10 Circular Polarization (cont.) Animation of LHCP wave (Use ppt version in full-screen mode to see motion.) 10

11 Unit Rotation Vectors 1 rˆ = ( ˆ jˆ) 2 1 l = ( ˆ+ jˆ) 2 RHCP (β = - π / 2) LHCP (β = + π / 2) Add: Subtract: 1 ˆ = ( rˆ+ lˆ ) 2 j ˆ = ( rˆ lˆ ) 2 11

12 General Wave Representation = ˆ + ˆ 1 1 = ( ˆ ) ( ˆ r+ l + jr l) 2 2 = ra ˆ RH + lalh 1 1 ARH = ( + j) ALH = ( j) 2 2 Note: An polarization can be written as combination of RHCP and LHCP waves. This could be useful in dealing with CP antennas. 12

13 Circularl Polarized Antennas Method 1: The antenna is fundamentall CP. A helical antenna for satellite reception is shown here. The helical antenna is shown radiating above a circular ground plane. 13

14 Circularl Polarized Antennas (cont.) Method 2: Two perpendicular antennas are used, fed 90 o out of phase. V = 1 RHCP V = -j + - z A more complicated version using four antennas (omni CP) Two antennas are shown here being fed b a common feed point. Note: LHCP is radiated in the negative z direction. 14

15 Circularl Polarized Antennas (cont.) Method 3: A single antenna is used, but two perpendicular modes are ecited 90 o out of phase. (a) (b) (a) (b) (c) A square microstrip antenna with two perpendicular modes being ecited. A circular microstrip antenna using a single feed and slots. The feed is located at 45 o from the slot ais. A square patch with two corners chopped off, fed at 45 o from the edge. (c) 15

16 Circularl Polarized Antennas (cont.) Method 4: Sequential rotation of antennas is used (shown for N = 4). I = o LP elements CP elements I = o I = 1-90 o I = 1 0 o The elements ma be either LP or CP. This is an etension of method 2 when used for LP elements. Using multiple CP elements instead of one CP element often gives better CP in practice (due to smmetr). 16

17 Pros and Cons of CP Pros Alignment between transmit and receive antennas is not important. The reception does not depend on rotation of the electric field vector that might occur. When a RHCP bounces off an object, it mainl changes to a LHCP wave. Therefore, a RHCP receive antenna will not be as sensitive to multipath signals that bound off of other objects. Cons A CP sstem is usuall more complicated and epensive. Often, simple wire antennas are used for reception of signals in wireless communications, and hence the are alread directl compatible with linear polarization. (Using a CP transmitted signal will result in 50% of the transmitted power being wasted, or a 3 db drop in the received signal). Satellite transmission often uses CP because of Farada rotation in the ionosphere. 17

18 Pros and Cons of CP (cont.) ffects of alignment errors α = alignment angle error TX-RX LP-LP CP-LP LP-CP Gain Loss cos 2 α : The received signal can var from zero db to - db -3 db, no matter what the alignment -3 db, no matter what the alignment CP-CP 0 db (polarizations same) or - db (polarizations opposite), no matter what the alignment 18

19 amples of Polarization AM Radio ( MHz): linearl polarized (vertical) (Note 1) FM Radio ( MHz): linearl polarized (horizontal) TV (VHF, MHz; UHF, MHz: linearl polarized (horizontal) (some transmitters are CP) Cell phone antenna (about 2 GHz, depending on sstem): linearl polarized (direction arbitrar) Cell phone base-station antennas (about 2 GHz, depending on sstem): often linearl polarized (dual-linear slant 45 o ) (Note 2) DBS Satellite TV ( GHz): transmits both LHCP and RHCP (frequenc reuse) (Note 3) GPS (1.574 GHz): RHCP (Note 3) Notes: 1) Low-frequenc waves travel better along the earth when the are polarized verticall instead of horizontall. 2) Slant linear is used to switch between whichever polarization is stronger. 3) Satellite transmission often uses CP because of Farada rotation in the ionosphere. 19

20 lliptical Polarization Includes all other cases β 0 or π (not linear) β ±π / 2 (not CP) β = ± π / 2 and b a (not CP) () t φ(t) llipse 20

21 lliptical Polarization (cont.) Note: The rotation is not at a constant speed. RHP φ(t) LHP llipse 0 < β < π π < β < 0 LHP RHP 21

22 Proof of llipse Propert = bcos( ωt+ β) = acosωt = bcosωtcos β bsinωtsin β so = bcos β bsin β 1 a a 2 2 b 2 b 2 cos β sin β b = a a b cos 2 b β cos β sin β b b + = a a a 22

23 Proof of llipse Propert (cont.) b cos 2 b β cos β sin β b b + = a a a 2 2 b ( 2 2 ) 2 b 2 2 cos β + sin β + 2 cos β = b sin β a a 2 2 b b cos β + = b sin a a β Consider the following quadratic form: A + B + C = D

24 Proof of llipse Propert (cont.) Discriminant: so B 2 = 4AC b b = 4 cos β 4 a a 2 b 2 4 cos 1 = β a 2 b 2 4 sin β 0 = < a From analtic geometr: > 0, hperbola = 0, line < 0, ellipse Hence, this is an ellipse. If β = 0 or π we have = 0, and this is linear polarization. 24

25 Rotation Propert We now prove the rotation propert: 0 < β < π π < β < 0 LHP RHP RHP LHP φ(t) ellipse Recall that = acosωt = bcos( ωt+ β) = bcosωtcos β bsinωtsin β 25

26 Rotation Propert (cont.) Hence b tanφ = = cos β tanω sin β a [ t ] Take the derivative: so dφ b φ = ( ωt)( ω) β dt a 2 2 sec sec sin dφ b = sin cos sec dt a ( t)( ) 2 2 β φ ω ω ( a) 0< β < π dφ < 0 dt LHP ( b) π < β < 0 dφ > 0 dt RHP (proof complete) 26

27 Phasor Picture ( a) 0< β < π LHP () t Rule: The electric field vector rotates in time from the leading ais to the lagging ais. Im β leads LHP Re A phase angle 0 < φ < π is a leading phase angle (leading with respect to zero degrees). A phase angle -π < φ < 0 is a lagging phase angle (lagging with respect to zero degrees). 27

28 Phasor Picture (cont.) ( b) π < β < 0 RHP Rule: The electric field vector rotates in time from the leading ais to the lagging ais. Im lags () t β RHP Re A phase angle 0 < φ < π is a leading phase angle (leading with respect to zero degrees). A phase angle -π < φ < 0 is a lagging phase angle (lagging with respect to zero degrees). 28

29 ample = zˆ (1 + j) + ˆ (2 j) e [ ] jk z z What is this wave s polarization? 29

30 ample (cont.) = zˆ (1 + j) + ˆ (2 j) e [ ] jk Im z z leads Re Therefore, in time, the wave rotates from the z ais to the ais. 30

31 ample (cont.) = zˆ(1 + j) + ˆ (2 j) e [ ] jk z z Note: LHP or LHCP π z and β ± (so this is not LHCP) 2 31

32 ample (cont.) = zˆ (1 + j) + ˆ (2 j) e [ ] jk z z LHP 32

33 Aial Ratio (AR) and Tilt Angle (τ ) C () t τ B A D AR major ais = AB 1 minor ais CD db 10 ( ) AR = 20 log AR 33

34 Aial Ratio (AR) and Tilt Angle (τ ) j0 = ae = a = j be β We first calculate γ : b γ = tan a 1 Tilt Angle tan 2τ = tan 2γ cos β (-90 o < τ < 90 o ) Note: The tilt angle τ is ambiguous b the addition of ± 90 o. (We cannot tell the difference between the major and minor aes.) sin 2ξ = sin 2γ sin β ( o o 45 ξ + 45 ) AR = ξ > ξ < 0 0 Aial Ratio cot ( ξ ) implies LHP implies RHP 34

35 Aial Ratio (AR) and Tilt Angle (τ) (cont.) AR = ( ) cot ξ () t B C τ ξ D A Phsical interpretation of the angle ξ Note : 0 < ξ < 45 LP o CP 35

36 Special Case The tilt angle is zero or 90 o if: β = ± π /2 Tilt Angle: tan 2τ = tan 2γ cos β = 0 τ = 0 or π /2 = acosωt = bcos( ωt± π / 2) = bsinωt () t a = b 36

37 LHCP/RHCP Ratio Sometimes it is useful to know the ratio of the LHCP and RHCP wave amplitudes. j RH j LH ra ˆ la rˆ A e φ φ = + = + l A e RH LH RH LH () t 1 1 rˆ = ( ˆ jˆ) l = ( ˆ+ jˆ) 2 2 A Point A: () t = ARH + ALH B Point B: () t = ARH ALH Hence AR A + A A + A = = A A A A RH LH RH LH RH LH RH LH 37

38 LHCP/RHCP Ratio (Cont.) Hence 1 + ALH / AR = 1 A / LH A A RH RH or AR 1 + ALH / A RH = ± 1 ALH / ARH (Use whichever sign gives AR > 0.) From this we can also solve for the ratio of the CP components, if we know the aial ratio: A A LH RH = AR 1 AR + 1 or AR + 1 AR 1 (We can t tell which one is correct from knowing onl the AR.) 38

39 ample = zˆ (1 + j) + ˆ (2 j) e [ ] jk LHP z Find the aial ratio and tilt angle. Find the ratio of the CP amplitudes. z Re-label the coordinate sstem: z z 39

40 ample (cont.) = ˆ(2 j) + ˆ(1 + j) e jkz z LHP z 1+ j = = j0.6 = e = j j1.249 o b/ a = , β = o b γ = = ( ) = a 1 1 tan tan radians 40

41 ample (cont.) Formulas tan 2τ = tan 2γ cos β sin 2ξ = sin 2γ sin β + o o 45 ξ 45 AR = cot ξ τ = or Results o o o o ξ = AR = ξ > 0: LHP LHP ξ < 0: RHP 41

42 ample (cont.) () t AR = o τ = τ Normalized field (a = 1): = cosωt o ( ) ωt = cos( ) Note: Plotting the ellipse as a function of time will help determine which value is correct for the tilt angle τ. 42

43 ample (cont.) AR = () t A A LH RH = AR 1 AR + 1 or AR + 1 AR 1 τ Hence A A LH RH = or We know that the polarization is LHP, so the LHCP amplitude must dominate. Hence, we have A A LH RH =

44 A ξ > ξ < AR 0 0 C = ξ ( ) cot ξ () t τ implies LHP D implies RHP Poincaré Sphere B latitude = 2ξ 2ξ 2τ Unit sphere longitude = 2τ z LHCP RHCP θ = π /2 2ξ φ = 2τ 44

45 Poincaré Sphere (cont.) latitude = 2ξ longitude = 2τ LHCP (north pole) ξ = 45 o LHP linear (equator) ξ = 0 o φ = 90 o (45 o tilt) RHP RHCP (south pole) ξ = -45 o φ = 0 (no tilt) 45

46 Poincaré Sphere (cont.) View in full-screen mode to watch the world spinning. 46

47 Poincaré Sphere (cont.) Two diametricall opposite points on the Poincaré sphere have the propert that the electric field vectors (in the phasor domain) are orthogonal in the comple sense. (A proof is given in the Appendi.) z amples A B A = ˆ A: B = ˆ B: ( ) ( ) equator on ais equator on - ais A = ˆ+ ˆ + j (LHCP) A: B = ˆ+ ˆ j (RHCP) B: north pole south pole A B* = ( A B* ) 0 H zˆ = 0 ( ) 1 H zˆ = zˆ zˆ η A B* A B* ( ) ( ) 0 0 A B* B* A ( ( ) ( )) 1 = z ˆ z ˆ z ˆ η (orthogonal in the comple power sense) Note : A ( B C) = B( AC ) C( AB ) 47

48 Appendi Proof of orthogonal propert z A B* = 0 A B From eamining the polarization equations: A B: ( θφ, ) ( π θφ, + π) ; ( ξτ, ) ( ξτ, + π/2 ); ( βγ) ( π+ βπ γ),, /2 ; b/ a a/ b Hence j ( β + jβ ) ( ) a ˆ + ˆ be b ˆ ˆ ae ( jβ ( ) ) + jβ ( ) ( ) * a ˆ + ˆ be b ˆ ˆ ae = ab ab = 0 48