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Lecture 5: Polarization and Related Antenna Parameters (Polarization of EM Fields – Revision

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Lecture 5: Polarization and Related Antenna Parameters (Polarization of EM Fields – Revision Lecture 5: Polarization and Related Antenna Parameters (Polarization of EM fields – revision. Polarization vector. Antenna polarization. Polarization loss factor and polarization efficiency.) 1. Introduction and Definition The polarization of the EM field describes the orientation of its vectors at a given point and how it varies with time. In other words, it describes the way the direction and magnitude of the field vectors (usually E) change in time. Polarization is associated with TEM time-harmonic waves where the H vector relates to the E vector simply by H= rˆ E / . This is why it suffices to know the polarization of the E vector only. In antenna theory, we are concerned with the polarization of the field in the plane orthogonal to the direction of propagation (the polarization plane)—this is the plane defined by the far-zone vectors E and H. Remember that the far field is a quasi-TEM field. The polarization is the locus traced by the extremity of the time-varying field vector at a fixed observation point. According to the shape of the trace, three types of polarization exist for harmonic fields: linear, circular and elliptical. Any polarization can be represented by two orthogonal linearly polarized fields, (xyˆˆEExy, ) or (EEHV, ), which, in general, may have different magnitudes and phases. What matters is their phase difference L=− y x . y y y E E z x z x z E x (a) linear polarization (b) circular polarization (c) elliptical polarization Nikolova 2019 1 • If L = n (n is integer), the field is linearly polarized, i.e., the locus traced by the tip of the E vector as time flows is a line since E(t )= ( xˆˆ mxy y m )cos( t ) . (5.1) 22 The magnitude of the total field is Em=+ m x m y . t = 0 t = 0 Etx ( ) only Etx () Ety () t = t = Ety () Etx ( ) only Etx () Animation: Linear Polarization, L = 0, mmxy= Nikolova 2019 2 • If L = / 2 (90 ) and mmxy= , then the field is circularly polarized. This is evident from the fact that the magnitude of the E-field vector below remains the same at all times: E(t )= m xˆ cos( t ) + y ˆ cos( t / 2) = m x ˆ cos( t ) y ˆ sin( t ) (5.2) indicating that the tip of the E-field vector traces a circle as time flows. t1 tt21=+ /2 Animation: Clockwise Circular Rotation • In the most general case, the polarization is elliptical. t = 0 t = /2 Animation: Counter-clockwise Elliptical Rotation Alternatively, any type of polarization can be represented by a right-hand circular and a left-hand circular polarizations (ELH , ERH ). [Animation] Next, we review the above statements and definitions, and introduce the new concept of polarization vector. Nikolova 2019 3 2. Field Polarization in Terms of Two Orthogonal Linearly Polarized Components The polarization of any field can be represented by the superposition of two orthogonal linearly polarized fields. Assume that locally a far-field wave propagates along the z-axis. The far-zone field vectors have only transverse components. Then, the set of two orthogonal linearly polarized fields along the x-axis and along the y-axis, is sufficient to represent any TEMz field. We use this arrangement to introduce the concept of polarization vector. The field (time-dependent or phasor vector) is decomposed into two orthogonal components: e()()()t= ex t + e y t E = E x + E y , (5.3) − jz ex(t )= E x cos( t − z) xˆˆ E x = E x e x (5.4) − j z j ey(t )= E y cos( t − z + L) yˆˆ E y = E y e e L y . At a fixed position (assume z = 0), equation (5.3) can be written as e(t )= xˆˆ Ex cos t + y E y cos( t + L ) (5.5) j E = xˆˆ Exy + y E e L The x-component is always assumed to have a zero phase, setting the reference for the phase of the y-component. As a result, the x-component is always real positive whereas the y-component may be complex. Case 1: Linear polarization: L ==nn, 0,1,2, e(t )= xˆˆ Exy cos( t ) + y E cos( t n ) (5.6) E = xˆˆ EExy y Note that the phasor vector of a linearly polarized wave is real-valued. y y L = 2k L =+(2k 1) 0 0 E y E E E y Ex x x Ex Ey =arctan Ex (a) (b) Nikolova 2019 4 Case 2: Circular polarization: Ex= E y = E mand L = + n , n = 0,1,2, 2 e(t )= xˆˆ Exy cos( t ) + y E cos[ t ( / 2 + n )] (5.7) E =Ejm () xˆˆ y Note that the phasor of a circularly polarized wave is complex with the x- component being real-positive whereas the y-component is purely imaginary. E=+Ejm () xˆˆ y E=−Ejm () xˆˆ y = + + 2n = − − 2n L 2 L 2 3 t = y t = y 2 2 5 7 3 t = t = t = t = 4 4 4 4 z t = 0 t = 0 t = t = z x x 7 t = 5 t = 3 t = 4 t = 4 3 4 t = 4 t = 2 2 If +zˆ is the direction of propagation: If +zˆ is the direction of propagation: counterclockwise (CCW) or left- clockwise (CW) or right-hand (RH) hand (LH) polarization polarization The sense of rotation, clockwise or counterclockwise, depends on the direction into which the observer looks. To avoid this ambiguity, in antenna engineering, the direction into which the observer looks must coincide with the direction of wave propagation. In the plot on the left, where E()()z=+ Em xˆˆ j y , if the wave propagates along −zˆ (into the plot), we have a right-hand (RH or CW) wave. In the plot on the right, where E()()z=− Em xˆˆ j y , if the wave propagates along (into the plot), we have a left-hand (LH or CCW) wave. Nikolova 2019 5 Vice versa, if the wave propagates along +zˆ , then the left plot shows a LH (or CCW) wave, whereas the right plot shows a RH (or CW) wave. The terms RH and LH are associated with the use of the right hand or the left hand to relate the direction of propagation to the direction of rotation of the E-field vector. The thumb of the hand has to be aligned with the direction of propagation. With this alignment, one has to determine whether it is the fingers of the right hand or those of the left hand that show properly the direction of rotation of the E-field. There is a simple rule of thumb that can be used to tell whether a wave has RH or LH sense of rotation. If the sign in the propagation factor of the wave matches the sign of the y-component in the field pahsor, then the wave is RHCP. If the signs do not match, the wave is LHCP. Assume, for example, a wave travelling along +z . The propagation factor is then e− jz . The phasor − jz E()()z= Em xˆˆ + j y e (5.8) represents an LHCP wave. This is the case in the left plot on the previous page. On the contrary, the phasor − jz E()()z= Em xˆˆ − j y e (5.9) represents an RHCP wave. This is the case in the right plot. For a wave travelling along −z, the propagation factor is e+ jz and so + jz E()()z= Em xˆˆ + j y e (5.10) is an RHCP wave whereas + jz E()()z= Em xˆˆ − j y e (5.11) is LHCP. A snapshot of the field vector along the axis of propagation is given below for an RHCP wave. Pick an observing position along the axis of propagation (see the plane defined by the x and y axes in the plot below) and imagine that the whole helical trajectory of the tip of the field vector moves along the wave vector k . Are you going to see the vector rotating clockwise or counter- clockwise as you look along ? (Ans.: Clockwise, which is equivalent to RH sense of rotation.) Nikolova 2019 6 [Hayt, Buck, Engineering Electromagnetics, 8th ed., p. 399] Case 3: Elliptic polarization The field vector at a given point traces an ellipse as a function of time. This is the most general type of polarization, obtained for any phase difference and any ratio (/)EExy. Mathematically, 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. e(t )= xˆˆ Ex cos( t ) + y E y cos( t + L ) (5.12) j E = xˆˆExy + y E e L Show that the trace of the time-dependent vector is an ellipse: ey( t )= E y (cos t cos L − sin t sin L ) 2 etx () etx () cost = and sint =− 1 Ex Ex 2 2 2 exx()() t e t eyy()() t e t sinLL= − 2 cos + EEEEx x y y 2 or (dividing both sides by sin L ), 22 1=x ()2()()cos t − x t y tL + y () t , (5.13) where e( t ) cos( t ) et() cos(t + ) xt()==x and yt()==y L . Exsin L sin L Eysin L sin L Nikolova 2019 7 Equation (5.13) is the equation of an ellipse centered in the xy plane. It describes the trajectory of a point of coordinates x(t) and y(t), i.e., normalized etx () and ety () values, along an ellipse where the point moves with an angular frequency .
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