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Polarization y
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y |E| = Ex2 +Ey2
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y x z-axis out of page
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(a) Figure 5.2 (b) (c)
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Horizontal and vertical components of linear polarization.
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These would add vectorially, and the resultant would be a vector E (Fig. 5.2c) of amplitude 2E2 E2, at an angle to the horizontal given by x y a arctan Ey Ex (5.3)
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E varies sinusoidally in time in the same manner as the individual components. It is still linearly polarized but cannot be classified as simply horizontal or vertical. Arguing back from this, it is evident that E can be resolved into vertical and horizontal components, a fact which is of great importance in practical transmission systems. The power in the resultant wave is proportional to the voltage 2E2 E2 , x y squared, which is E2 E2 . In other words, the power in the resultant x y wave is the sum of the powers in the individual waves, which is to be expected. More formally, Ey and Ex are said to be orthogonal. The dictionary definition of orthogonal is at right angles, but a wider meaning will be attached to the word later. Consider now the situation where the two fields are equal in amplitude (denoted by E), but one leads the other by 90 in phase. The equations describing these are Ey Ex y E sinwt x E coswt (5.4a) (5.4b)
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Applying Eq. (5.3) in this case yields a wt. The tip of the resultant electric field vector traces out a circle, as shown in Fig. 5.3a, and the resultant wave is said to be circularly polarized. The amplitude of the resultant vector is E. The resultant field in this case does not go through zero. At wt 0, the y component is zero and the x component is E. At
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t = 90 E t = 180 t t=0
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t = 270
(a) RHC IEEE viewpoint Classical optics viewpoint
(b) LHC IEEE viewpoint Classical optics viewpoint
Figure 5.3 Circular polarization. (c)
wt 90 , the y component is E and the x component is zero. Compare this with the linear polarized case where at wt 0, both the x and y components are zero, and at wt 90 , both components are maximum at E. Because the resultant does not vary in time, the power must be found by adding the powers in the two linear polarized, sinusoidal waves. This gives a resultant proportional to 2E 2. The direction of circular polarization is defined by the sense of rotation of the electric vector, but this also requires that the way the vector is viewed must be specified. The Institute of Electrical and Electronics Engineers (IEEE) defines right-hand circular (RHC) polarization as a rotation in the clockwise direction when the wave is viewed along the direction of propagation, that is, when viewed from behind, as shown in Fig. 5.3b. Left-hand circular (LHC) polarization is when the rotation is in the counterclockwise direction when viewed along the direction of propagation, as shown in Fig. 5.3c. LHC and RHC polarizations are orthogonal. The direction of propagation is along the z axis.
Polarization
As a caution it should be noted that the classical optics definition of circular polarization is just the opposite of the IEEE definition. The IEEE definition will be used throughout this text. For a right-hand set of axes (Fig. 5.1) and with propagation along the z axis, then when viewed along the direction of propagation (from behind ) and with the y axis directed upward, the x axis will be directed toward the left. Consider now Eq. (5.4). At wt 0, Ey is 0 and Ex is a maximum at E along the x axis. At wt 90 , Ex is zero and Ey is a maximum at E along the y axis. In other words, the resultant field of amplitude E has rotated from the x axis to the y axis, which is a clockwise rotation when viewed along the direction of propagation. Equation (5.4) therefore represents RHC polarization. Given that Eq. (5.4) represents RHC polarization, it is left as an exercise to show that the following equations represent LHC polarization: Ey Ex y E sinwt x E coswt (5.5a) (5.5b)
In the more general case, a wave may be elliptically polarized. This occurs when the two linear components are Ey Ex y Ey sinwt x Ex sin(wt ) (5.6a) (5.6b)
Here, Ey and Ex are not equal in general, and d is a fixed phase angle. It is left as an exercise for the student to show that when Ey 1, Ex 1/3, and d 30 , the polarization ellipse is as shown in Fig. 5.4.
Ey( t)
1 0.4
0 Ex( t)
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