Polarization and Related Antenna Parameters

Transcription

1 ANTENTOP , # 009 Polarization and Related Antenna Parameters Feel Yourself a Student! Dear friends, I would like to give to ou an interesting and reliable antenna theor. Hours searching in the web gave me lots theoretical information about antennas. Reall, at first I did not know what information to chose for ANTENTOP. Finall, I stopped on lectures Modern Antennas in Wireless Telecommunications written b Prof. Natalia K. Nikolova from McMaster Universit, Hamilton, Canada. You ask me: Wh? Well, I have read man tetbooks on Antennas, both, as in Russian as in English. So, I have the possibilit to compare different tetbook, and I think, that the lectures give knowledge in antenna field in great wa. Here first lecture Introduction into Antenna Stud is here. Net issues of ANTENTOP will contain some other lectures. So, feel ourself a student! Go to Antenna Studies! I.G. M Friends, the above placed Intro was given at ANTENTOP to Antennas ectures. Now I know, that the ecture is one of popular topics of ANTENTOP. Ever Antenna ecture was downloaded more than 1000 times! Now I want to present to ou one more ver interesting ecture - it is a ecture Polarization and Related Antenna Parameters. I believe, ou cannot find such info anwhere for free! Ver interesting and ver useful info for ever ham, for ever radio- engineer. So, feel ourself a student! Go to Antenna Studies! I.G. McMaster Universit Hall Prof. Natalia K. Nikolova Polarization and Related Antenna Parameters Polarization of EM fields revision. Polarization vector. Antenna polarization. Polarization loss factor and polarization efficienc. b Prof. Natalia K. Nikolova mirror: Page-5

2 ecture 5: Polarization and Related Antenna Parameters (Polarization of EM fields revision. Polarization vector. Antenna polarization. Polarization loss factor and polarization efficienc.) 1. Polarization of EM fields. The polarization of the EM field describes the time variations of the field vectors at a given point. In other words, it describes the wa the direction and magnitude the field vectors (usuall E ) change in time. The polarization is the figure traced b the etremit of the timevaring field vector at a given point. According to the shape of the trace, three tpes of polarization eist for harmonic fields: linear, circular and elliptical: E z ω z E z E (a) linear polarization (b) circular polarization (c) elliptical polarization An tpe of polarization can be represented b two orthogonal linear polarizations, ( E, E )or(, E H E V ), whose fields are out of phase b an angle of δ. If δ = 0 or nπ, then a linear polarization results. If δ = π / (90 ) and E = E, then a circular polarization results. In the most general case, elliptical polarization is defined. 6

3 It is also true that an tpe of polarization can be represented b a right-hand circular and a left-hand circular polarizations ( E, E ). We shall revise the above statements and definitions, while introducing the new concept of polarization vector.. Field polarization in terms of two orthogonal linearl polarized components. The polarization of an field can be represented b a suitable set of two orthogonal linearl polarized fields. Assume that locall a far field propagates along the z-ais, and the field vectors have onl transverse components. Then, the set of two orthogonal linearl polarized fields along the -ais and along the -ais, is sufficient to represent an TEM z field. We shall use this arrangement to demonstrate the idea of polarization vector. The field (time-dependent vector or phasor vector) is decomposed into two orthogonal components: e = e + e E = E + E, (5.1) e = Ecos( ωt βz) ˆ E = Eˆ (5.) jδ e = Ecos( ωt βz+ δ) ˆ E = Ee ˆ At a fied position (assume z = 0), equation (5.1) can be written as: et () = ˆ Ecosωt+ ˆ Ecos( ωt+ δ) (5.3) j E = ˆ E + ˆ E e δ R Case 1: inear polarization: δ = nπ, n= 0,1,, et ( ) = ˆ Ecos( ωt) + ˆ Ecos( ωt± nπ) E = ˆ E ± ˆ E (5.) 7

4 δ = kπ τ > 0 δ = (k + 1) π τ < 0 E τ E E E τ E E E τ =± arctan E (a) Case : Circular polarization: (b) π E = E = Em and δ =± + nπ, n= 0,1,, et () = ˆ Ecos( ωt) + ˆ Ecos[ ωt± ( π / + nπ)] E = E ( ˆ± ˆ j) m (5.5) 8

5 E = E ( ˆ+ jˆ) m π δ =+ + nπ 3π 5π 7π E = E ( ˆ jˆ) m π δ = nπ π 3π π ωt = π z 0 ωt = π z 0 3π π π If ( zˆ) is the direction of propagation: clockwise (CW) or right-hand polarization 5π 3π 7π If ( zˆ) is the direction of propagation: counterclockwise (CCW) or left-hand polarization 9

6 A picture of the field vector (at a particular moment of time) along the direction of propagation (left-hand circularl polarized wave): β z β z 10

7 Case 3: Elliptical polarization The field vector at a given point traces an ellipse as a function of time. This is the most general tpe of polarization of time-harmonic fields, obtained for an phase difference δ and an ratio ( E/ E ). Mathematicall, the linear and the circular polarizations are special cases of the elliptical polarization. In practice, however, the term elliptical polarization is used to indicate polarizations other than linear or circular. et ( ) = ˆ Ecosωt+ ˆ Ecos( ωt+ δ) j E = ˆ E + ˆ E e δ Show that the trace of the time-dependent vector is an ellipse: e () t = E (cosωt cosδ sinωt sin δ ) where: e cos () t E and e sin 1 ( t) ωt = E (5.6) () () e() t e() t e t e t sin δ = cosδ E E + E E or 1 = () t () t ()cos t δ + () t, (5.7) e () t cos t t () = ω Esinδ = sinδ ; e () t cos( ωt + δ ) t () = = E sinδ sinδ Equation (5.7) is the parametric equation of an ellipse centered in the plane. It describes the motion of a point of coordinates e( t) and e( t) along an ellipse with a frequenc ω. The elliptical polarization can also be right-hand and left-hand polarization, depending on the relation between the direction of propagation and the direction of rotation. 11

8 e( t) E major ais ( OA) E ω minor ais ( OB) τ E e () t The parameters of the polarization ellipse are given below. Their derivation is given in Appendi I. a) major ais ( OA ) OA = 1 E + E + E + E + EE cos( δ ) (5.8) b) minor ais ( OB) OB = 1 E + E E + E + EE cos( δ ) (5.9) c) tilt angle τ 1 EE τ = arctan cos δ E E (5.10) d) aial ratio AR = major ais OA minor ais OB (5.11) Note: The linear and circular polarizations can be considered as special cases of the elliptical polarization. π If δ =± + nπ and E = E,thenOA= OB= E = E; the ellipse becomes a circle. 1

9 E If δ = nπ,thenob= 0and τ =± arctan ; the ellipse collapses into E a line. 3. Field polarization in terms of two circularl polarized components The representation of a comple vector field in terms of circularl polarized components is somewhat less eas to perceive but it is actuall more useful in the calculation of the polarization ellipse parameters. E = E R( ˆ+ jˆ) + E ( ˆ jˆ) (5.1) Assuming a relative phase difference of δc = ϕ ϕr, one can write (5.1) as: jδc E = ER( ˆ + jˆ) + Ee ( ˆ jˆ) (5.13) The relation between the linear-component and the circular-component representations of the field polarization is easil found as: E = ˆ( E R + E ) + ˆ j( E R E ) (5.1) E E. Polarization vector and polarization ratio The polarization vector is the normalized phasor of the electric field vector. It is a comple-number vector of unit magnitude and direction coinciding with the direction of the electric field vector. E E E ˆ ρ = = ˆ + ˆ e, E = E + E jδ m Em Em Em (5.15) The polarization vector takes the following forms in some special cases: Case 1: inear polarization ˆ ρ = E E ˆ ˆ, Em E E E ± m E = + (5.16) m Case : Circular polarization 13

10 1 ˆ ρ = ( ˆ± ˆ j), Em = E = E (5.17) The polarization ratio is the ratio of the phasors of the two orthogonal polarization components. It is a comple number. j E δ E e E δ r = re = = or r = E E E V H (5.18) Point of interest: In the case of circular polarization, the polarization ratio is defined as: jδ E C R r C = rce = (5.19) E The circular polarization ratio r C is of particular interest since the aial ratio of the polarization ellipse AR can be epressed as: rc + 1 AR = (5.0) r 1 Besides, its tilt angle is: δ τ = C (5.1) Comparing (5.10) and (5.1) readil shows the relation between the phase difference of the circular-polarization representation and the linear polarization j ratio r = re δ : r δc = arctan cosδ (5.) 1 r One can calculate r C from the linear polarization ratio r making use of (5.11) and (5.0): C δ δ rc r + 1+ r + r cos( ) = rc 1 1+ r 1+ r + r cos( ) (5.3) 1

11 Using (5.) and (5.3) allows eas switching between the representation of the wave polarization in terms of linear and circular components. 5. Antenna polarization The polarization of a radiated wave (polarization of a radiating antenna) at a specific point in the far zone is the polarization of the locall plane wave. The polarization of a received wave (polarization of a receiving antenna) is the polarization of a plane wave, incident from a given direction, and having given power flu densit, which results in maimum available power at the antenna terminals. 6. Polarization loss factor and polarization efficienc Generall, the polarization of the receiving antenna is not the same as the polarization of the incident wave. This is called polarization mismatch. The polarization loss factor (PF) characterizes the loss of EM power because of polarization mismatch. PF ˆ ˆ The above definition is based on the representation of the incident field and the antenna polarization b their polarization vectors. If the incident field is i i E = Em ˆ ρi, then the field of the same magnitude that would produce maimum received power at the antenna terminals is i E = E ˆ ρ. = ρi ρa (5.) a m a 15

12 If the antenna is polarization matched, then PF = 1, and there is no polarization power loss. If PF = 0, then the antenna is incapable of receiving the signal. 0 PF 1 (5.5) The polarization efficienc has the same phsical meaning as the PF. Eamples Eample 5.1. The electric field of a linearl polarized EM wave is i j z E = ˆ Em(, ) e β It is incident upon a linearl polarized antenna whose polarization is: Ea = ( ˆ+ ˆ) E( r, θ, ϕ) Find the PF. 16

13 1 1 PF = ˆ ( ˆ+ ˆ) = PF = 10log 0.5 = 3 [db] 10 Eample 5.. A transmitting antenna produces a far-zone field, which is right-circularl polarized. This field impinges upon a receiving antenna, whose polarization (in transmitting mode) is also right-circular. Determine the PF. Both antennas (the transmitting one and the receiving one) are rightcircularl polarized in transmitting mode. et s assume that a transmitting antenna is located at the center of a spherical coordinate sstem. The farzone field it would produce is described as: far E = E ˆ m θ cosωt+ ˆ ϕ cos ( ωt π / ) It is a right-circularl polarized field with respect to the outward radial direction. Its polarization vector is: ˆ θ j ˆ ϕ ˆ ρ = According to the definitions in Section, this is eactl the polarization vector of a transmitting antenna. 17

14 z E ϕ ϕ θ r E θ far This same field E is incident upon a receiving antenna, which has the ˆ θa j ˆ ϕa polarization vector ˆ ρa = in its own coordinate sstem ( ra, θa, ϕ a). However, this field propagates along r a in the ( ra, θa, ϕa) coordinate sstem, and, therefore, its polarization vector becomes: ˆ θa + j ˆ ϕa ˆ ρi = The PF is calculated as: ˆ ˆ ( θa + jˆ ϕa)( θa jˆ ϕa) PF = ˆ ρi ˆ ρa = = 1, PF[dB] = 10log101 = 0 There are no polarization losses. 18

15 Eercise: Show that an antenna of right-circular polarization (in transmitting mode) cannot receive left-circularl polarized incident wave (or a wave emitted b a left-circularl polarized antenna). Appendi I Find the tilt angle τ, the length of the major ais OA, and the length of the minor ais OB of the ellipse described b the equation: e () () ( ) ( ) t e t e t e t sin δ = cosδ E E + (A-1) E E e( t) E major ais ( OA) E ω minor ais ( OB) τ E e( t) Equation (A-1) can be written as: a b + c = 1, (A-) where: = e( t) and = e( t) are the coordinates of a point of the ellipse centeredinthe plane; 1 a E sin δ cosδ b = ; EE sin δ 19

16 1 c = E sin δ. After dividing both sides of (A-) b ( ), one obtains 1 a b+ c = () Introducing ξ = e t = e() t, one obtains that 1 = cξ bξ + a (1 + ξ ) ρ ( ξ) = + = (1 + ξ ) = (A-) cξ bξ + a Here, ρ is the distance of the ellipse point from the center of the coordinate sstem. We want to know at what ξ values the maimum and the minimum of ρ occur. This will produce the tilt angle τ. We also want to know what are the values of ρma and ρ min. Then, we have to solve d( ρ ) = 0,or dξ ( a c) ξ ξ 1 = 0 (A-5) b (A-5) is better solved for the angle α, such that π ξ = tan α = ; α = τ (A-6) Substituting (A-6) in (A-5) ields: sinα sinα C 1= 0 cos α cosα cosα a c E E where C = =. b E E cosδ The solution of (A-7) is: (A-3) (A-7) 0

17 1 EE cos 1 arctan δ π α = ; α = α1+ E E τ = α 1, 1, Substituting α 1 and α back in ρ ields the epressions for OA and OB. (A-8) 1