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Plane Waves, Polarization and the Poynting Vector

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Plane Waves, Polarization and the Poynting Vector Plane Waves, Polarization and the Poynting Vector EE H =-xx+- y hh Uniform Plane Wave in Free Space We have previously established the following properties of plane waves in free space: • Electric and magnetic field components in propagation direction are zero. rr rr • Electric and magnetic fields (EBxy, or EByx, ) are related. Each is the “source” of the other. • A set of three second order differential equations apply, one for each field component in rectangular coordinates. For example: rr ∂ 2 E me∂2 E xoox= , ∂ z 2 ∂t 2 or for harmonic fields, in phasor notation: r ∂ 2E r x =-wme2 E ∂ z 2 oo x A general solution to the Helmholz equation is written: r rrr - jzbboo jz ECx =+12ee C (C i can be complex), where bwmewooo==/c, c = speed of light. Propagation of Magnetic Field Suppose we haverr an electric field wave travelling in the positive z direction. Recall that curl EB=-jw , or for our travelling wave, r rr r - jzb ∂ E EE= e o x xxo+ =-jw B y r ∂ z - jzb o r ∂ diE + xo e =-jw B ∂ z y rrrb -=-fi=jebwEBBE- jzbbo jo e- jzo oxo+ yyw + xo w r = E e - jzb o cw + xo r • Electric and magnetic (B ) field are orthogonal (perpendicular, Right Hand Rule), in-phase, and the ratio of the field magnitudes is the imped- ance. It isr very useful to expressr this ratio of electric and magnetic fields in terms of H , rather than B . r r E r mmeH ==x E , oy c xoo r Er x m o m o or, after rearranging: ==m o c = H y emoo e o Henry/ m Joule/ amp 2 volt Dimensional analysis: ===Ohm!!! Farad/ m Joule/ volt 2 amp or: r Electric fields are volts/meter and H fields are amps/meter. r Er x m o =∫ho Ohms, hpo ª=120 377 Ohmsr (impedance of freespace) H y e o $ Er x For a wave travelling in the - ez direction, =-ho . This just means that r H r $ y E x is oriented along the - ex direction (or equivalently with H y ). Summary: r r •E and H are perpendicular to each other and propagation direction. Right Hand Rule gives direction. r r • Ratio of H to E is the intrinsic wave impedance, ho . The wavelength is the distance that the wave travels so that the phase changes by 2p radi- ans. 222p p pcc blo = 2 p, or l == = = bo w/c w f •Picture: • Phase velocity, vp To understand the phase velocity, we must return to the real representation of the field with both space and time dependence; that is, EE=-+ xocosbgwb t o z Consider an observer moving along in z at the same point, say the peak, of the oscillating field. This means mathematically that: wbtz-=o 0 or any constant. Thus, wbdt= o dz , dz w or =∫vp , the phase velocity. dt bo Polarization of Place Waves Consider the propagation characteristics of a plane wave in which the elec- tric field has components in both the x and y directions: rr r $$- jzb o , EE=+dixxeee E yy where the field components may be complex. That is, rr rr ja jb EExx= e and EEyy= e In phase ab= , rr r jza $$--bgb o EE=+ejxxeee E yy or in real form: rr r EEeEe=+e xx yyj cosbgwb t -- o za y r Ey r E r x Ex z This is linear polarization. Now we consider a more general case. Elliptical Polarization In this case we allow arbitrary phase relationships a and b: rr r ja jb - jzb o EE=+ejeexxeeeyy E It is easier to see what this means if we write each component out in real form: rr EE=+-cosafwb taz rrxx EEyy=+-cosafwb tbz Suppose we let a = 0 and b = p /2. Then: rr EExx=-cosafwb tz rr EEyy=-sin afwb tz - rrr What can we say about Eztaf, =+ Exxee E yy? rr1 Make a plot in the x-y plane with t as a parameter and EE==1, . xy2 r r wt E x Ey 0 1 1 0 5 6 7 .707p /4 2 −./707 2 1 p/2 3 0 − 12/ 4 − 2 p 4 1 0 3 54p/ 5 −.707 ./707 2 32p / 6 0 12/ 74.707p/ 7 ./707 2 This clockwise rotation describes an ellipse, with major axes parallel to the x axis. As the wave propagates along z, the ellipse spreads out to anrr elliptical helix. This is called elliptical polarization. For the special case of EExy= , we have circular polarization. If we had chosen b =-p /2, we would have found that an identical ellipse would be formed except that the rotation would be counter-clockwise. Poynting Theorem We know that energy is propagated by waves, in general, and electromag- netic waves, in particular. We need to quantify the associated power flow. We easily obtain a hint of how to calculate power flow by recalling our circuit theory, where PVI= * , or by a dimensional analysis of the fields. Remember that the units of H are Amp/meter and the units of E are V/m. Their product Amp Volts/m2 has units of Watts/area, which is a power den- sity, just what we want. The product must clearly have a direction associ- ated with it, and it ought to somehow point in a reasonable direction. A reasonable direction for power flow in lossless, homogeneous, linear media, such as free space, wouldrr be in the direction of propagation. We might therefore guess that EH¥ would be a reasonable definition of power den- sity. We will consider the complex Poynting vector for time harmonic plane elec- tromagnetic waves in phasor notation. This requires some thought because rr of the nonlinear nature of EH¥ . r r jtw $ Let Eztaf, = Re E af ze ex and r r jt w $ Hztaf, = Re H af ze ey We will write out the real and imaginary parts of E and H. r rr jtwwjt $ Ezta, f=+Re EEr af ze ji af ze ex and r rr jtwwjt $ Hztaf, =+Re HHr af ze ji af ze ey Now expand the complex exponential and take the real part: rr r $ Ezta,cossin f=- Erix af zww t Ez af te rr r $ Hzta,cossin f=- Hriy af zww t Hz af te rrr The Poynting vector is: Pztaf, =¥ E H = rr rr rr rr 22 $ EHrrcoswwwwww t+- EH ii sin t EH ri cos t sin t - EH ir sin t cos te z The time average Poynting vector is: rr rrrr 11T 1 Pzav af==+Pztdt a, f EHrr EH ii T z0 2 2 The very same expression can be obtained directly from the phasors by the following rule: r rr r r r r 1 1 $$ Pztaf,*=¥=+¥-RediEH Re d Erixriy j E idi e H j H e 2 2 rr rr 1 $ =+diEHrr EH iiz e 2 Example What is the time average Poynting vector for a plane wave propagating in free space with the following (phasor) fields: r r E+0 - jzb o - jzb o $ H afz = ee$ E afzEee= +0 x and y ? h o r 1 rr Then Pztav af, =¥=RediEH* 2 1 F E I 1 E 2 - jzbboo$$+00- jz + $, ReG Ee+0 ex ¥ eey J = Re ez 2 H hho K 2 o or: r E 2 +0 $ Pztav af, = ez 2ho.
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