No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]
Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jordan Professor of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, Urbana, Illinois, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India 3.6 Polarization of Sinusoidally Time-Varying Fields 3.6-3 Polarization is the characteristic which describes how the position of the tip of the vector varies with time Linear Polarization: Tip of the vector describes a line Circular Polarization: Tip of the vector describes a circle 3.6-4 Elliptical Polarization: Tip of the vector describes an ellipse (i) Linear Polarization F1 F1 cos (t ) a x Magnitude varies sinusoidally with time Direction remains along the x axis Linearly polarized in the x direction 3.6-5 F2 F2 cos (t ) a y Magnitude varies sinusoidally with time Direction remains along the y axis Linearly polarized in the y direction If two (or more) component linearly polarized vectors are in phase, (or in phase opposition), then their sum vector is also linearly polarized Ex: F F1 cos (t ) a x F2 cos (t ) a y 3.6-6 y F2 F F1 x F2 cos (t ) –1 tan F1 cos (t ) F2 –1 tan F1 constant (ii) Circular Polarization If two component linearly polarized vectors are (a) equal to amplitude (b) differ in direction by 90˚ (c) differ in phase by 90˚, then their sum vector is circularly polarized 3.6-7 Example: F F1 cos t ax F1 sin t a y F F1 cos t F1 sin t 2 F1 , constant tan tan 1 1 F1 sin t F1 cos t tan t t y F2 F F1 x 3.6-8 (iii) Elliptical Polarization In the general case in which either of (i) or (ii) is not satisfied, then the sum of the two component linearly polarized vectors is an elliptically polarized vector Ex: F F1 cos t a x F2 sin t a y y F2 F F1 x 3.6-9 Example: F F0 cos t a x F0 cos (t 4)a y y F0 F2 F /4 –F0 –F0 F1 F x 3.6-10 D3.17 F1 F0 cos 2 108t 2 z ax F2 F0 cos 2 108t 3 z a y F1 and F2 are equal in amplitude (= F0) and differ in direction by 90˚ The phase difference (say ) depends on z in the manner –2z – (–3z) = z (a) At (3, 4, 0), = (0) = F1 F2 is linearly polarized (b) At (3, –2, 0.5), = (0.5) = 0.5 F1 F2 is circularly polarized 3.6-11 (c) At (–2, 1, 1), = (1) = F1 F2 is linearly polarized (d) At (–1, –3, 0.2) = = (0.2) = 0.2 F1 F2 is elliptically polarized