Plane Waves Chapter 13

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Chapter 13

∂ ∂ H =-e E ∂ z yx∂t Solutions of Maxwell’s Equations -- Uniform Plane Waves The sources of time varying electromagnetic fields are time varying charges and currents, whether man-made or naturally occurring. However, exami- nation of Maxwell’sr r equations shows that, even in empty space (with no sources), E and H fields cause each other. This means that, although these fields must originate in source regions, they can propagate through source-free regions! Because solutions in source regions are very hard to obtain, let us consider what kind of fields can exist in source-free regions. r r ∂ E —¥H = e eeee==n2 t or o ∂ r r ∂ H —¥E = -m ∂ t These equations are still very complicated (4 different partial derivatives). Let’s try to find simple solutions. We do this by making assumptions. After finding simplest solutions, we: (i) find what kind of source would generate this type of field (waves). (ii) show that these simple solutions are useful approximations to real electromagnetic waves. Assumption for Solution: Can we find a solution such that:

1.) Nor variationr exists in x and y directions. r 2.) H (\ B also) is oriented along one of the axes (we pick here H to be along y$ ). Ampere’s Law becomes:

$$$F ∂ ∂ I F ∂ ∂ I F ∂ ∂ I exzG H - He yyxJ +-G H He zzyJ +-G H H xJ H ∂ y ∂ z K H ∂ z ∂ x K H ∂ x ∂ y K ∂ E F ∂ Ex $$$y ∂ Ez I =++e G ex ey ez J , H ∂ t ∂ t ∂ t K ∂ ∂ E or, with our assumptions: -=e$$H e x e . xy∂ z ∂ t x ∂ ∂ Compare for transmission line: I =-C V ∂z ∂t ∂ ∂ H =-e E (1) ∂z yx∂t

In a similar manner, Faraday’s Law becomes: ∂ ∂ E =-m H (2) ∂ z xy∂t r r Thus, if H is only in the y$ direction, then E is only in the x$ . If we differentiate (2) with respect to z, we get: ∂ 2 ∂ ∂ ∂ F ∂ I 2 E xy=-m H =-m G H yJ ∂ z ∂ zt∂ ∂ tzH ∂ K

∂ ∂ Substitute from (2), H =-e E ∂z yx∂t ∂ 2 ∂ 2 E = me E ∂z 2 xx∂t 2 This is called the wave equation.

∂ 2V ∂ 2V Compare for transmission line: = LC (same). ∂z 2 ∂t 2

We will assume dielectric media; lossless: eeemm==or, o

Dimensional analysis: 11V V F I ∫ bgme F I 22H K oo H K m msm 1 m 2 ∫=(velocity)2 me s 2 Another way: Joules∫ Henry◊ Amp 2 aCoulombf2 Joules ∫ Farad 2 2 aCoulombf Henry◊af Amp ∫ Farad 2 F CoulombI Henry◊ Farad ∫ G J = s 2 H Amp K Henry Farad s 2 1 \∫meoo ◊ ∫= mmm 22()velocity In the units we are using: 1 me ∫ oo c e n me==r o c c ∂ 2 ∂ 2 E = me E ∂ z 2 xx∂t 2

By exactly the same method, we also get: ∂ 2 ∂ 2 H = me H ∂ z 2 yy∂t 2 n 1 c me = , or = c me n r r r r For example, the same equation is true for E and H . Thus, E and H must have the same type of functional dependence. This wave equation still has many possible solutions. In fact, it can be shown by direct substitution that z z Hft=-F I and ftF + I are solutions. y 1 H vK 2 H vK 1 c Here, v ∫=. me e It is important to note that f1 and f2 can be any function! r ∂ ∂ The E field we find directly from the equation: H =-e E ∂z yx∂t

Suppose: HHyo=-cosafwb tz ∂ bwbeHtzsin af-=-E (Differentiate with respect to z) ox∂ t b H E =-o sin awbtzdt - f (Integrate with respect to time) x e z b H =-o cosawbtzf weo

This leads to a new quantity that relates the electric and magnetic fields: b me w m w ===h b ∫ we ew e v

Rewriting: r $ EeH=-xohwbcosaf t z Units E volt / m of h ∫∫ =W H amp / m

mo ho ∫==impedance of free space 377 W eo m m 1377 h ∫===impedance of a lossless dielectric oo W e e o eerr

r r Note that E and H are in time phase and space quadrature. The directions of the vectors are such that:

rr 2 $$Eo 2 EH¥= ez =eHzoh z > 0 h E 2 =-e$$o =-eH2 h z < 0 z h zo rr This vector, EH¥ , always points in the direction of propagation. These types of waves are called Uniform Plane Electromagnetic Waves. “Plane” refers to the fact that, at any instant in time, the surfaces of constant phase are planes (here, z = constant). “Uniform” means no variation in transverse direction. The wave derived here is one oscillating at a single frequency w . The frequency is determined by the source. If the source has multiple frequencies, so will the wave. In communications, modulating a carrier wave results in multiple frequencies. But before we can study such fields, we must thor- oughly understand single frequency sinusoidal waves.

Phasor Plane Waves The following discussion represents another slightly more general derivation of plane waves. This derivation also uses phasor forms. rr rr —¥EH = - j wm —¥HE = jwe rro r

—¥di —¥EH = -jjjwmoo —¥ = - wm◊ we E B B rr r 22 -—EE +—di —◊ = wmeo E r But —=◊ E 0 rr 22 \—EE +wmeo =0 r If we work with H field, we get, in exactly the same manner: rr 22 —+HHwmeo =0 w 2 We note that wme2 ==b 2 . o v 2 We can write this wave equation for each cartesian component of each field:

L E xO M P This is called the Helmholz M Ey P M P equation. It is the complex 22E z mr—+b M P = 0 equivalent of the wave M H P x equation. M H P M yP NHzQ r Note: — 2 A , the Laplacian of a vector, is only defined properly in rectangular coordinates. Its meaning is: Take Laplacian operator and operate on each rectangular component of the vector.

Up to this point, the derivationT wasT general, with no assumptions made as to the components of E and H .

r 2 r 22 F 2 ∂ I ch—+b E =0 G —- 2 J E = 0 H ∂ t K or the time dependent equivalent forms r 2 r 22 F ∂ I c—+b h H =0 2 G —- 2 J H = 0 H ∂ t K have many, many different solutions, depending on the sources (wherever they may be) and on the boundary conditions. Plane waves are the simplest solutions. But plane waves are not just idealized, mathematical solutions. Many real fields can be approximated by uniform plane waves (so called).

Uniform Plane Wave Assumption: There is no variation of the fields in planes perpendicular to the direction of propagation. Previously, we have chosen the direction of propagation to be z$ . We can just as easily chose x$ or y$ . Other directions, not parallel to the coordinate axes are mathemati- cally a little bit more difficult. We shall do such arbitrary direction later. What about the possibility of a field component in the direction of propagation? Let’s consider that possibility and again chose z as the direction of propagation. Then,

2 ∂ ∂ 2 d , Æ 0 —Æ ∂ xy∂ dz2 d 2 rr EE+=b 2 0 dz2 r What components of E exist? r —=◊ E 0

∂E ∂Ey ∂E dE x ++=z 0 \=z 0 ∂ xyz∂ ∂ dz

\=Ez constant or zero. It must be zero.

The uniform plane wave assumption also means that the fields have no component along the direction of propagation.

So we have: 2 2 L d 2 O L d 2 O M 2 + b P E x = 0 M 2 + b P E y = 0 N dz Q N dz Q

The solution for each is Ae± jzb .

These are two independent equations. We can have only Ex , only Ey , or both. What we have depends on the nature of the source. We take the e- jzb solution and restore e jtw dependence. r $ jtafwb- z E = eEexo r E jtafwb- z H = e$ o e y h These correspond to the solutions previously obtained for z = 0 . Another “Pure” Phasor Approach r This time, let’s assume that both x$ and y$ components of E exist. But let us still assume only solutions travelling in the +z direction. - jzb jtw For example,r we take e . Restoring e dependence, we get the complex E field components:

jtafwb- z jtafwb- z E = Ee E = Ee r xxo yyor r H also can have two independent components. E and H , however, are related by the two Maxwell curl equations.r rUsing the complex form, it is easy to get a relationship between H and E . $$ $ eexy e z ∂ 00 =-jewm $ HH + e ∂ z ox x y y

EExy0 We note that for e jtw dependence differentiation with respect to t becomes multiplication by jw . For e- jzb dependence, differentiation with respect to z becomes multiplication by - jb . Notice here the advantage of using complex form. The function does not change, unlike real sinusoids. We get: $$ $$ --ejxyyxafbbwmEE +- ej af =- j oxxyych e HH + e $ $ Equating the corresponding ex and ey parts: b b HExy=- HEyx= wmo wmo r r r r e$ ¥ E b ¥ E H = b = h wmo T This is a general result for plane waves, no matter what the direction of b vector is. r r r r rr rrr ABC¥¥ = BACdidi◊◊ - CAB rr rr1 r r EE◊ r r EH¥= E ¥ee$$ ¥ E = +0 $ dibb , since eb ◊ E = 0 . Remember, E and hhoo r rr H have no component in the direction of propagation. The EH¥ vector points in the direction of propagation. This gives an easy way of finding the ee$ , $ $ direction of one field vector from the other. We see that HE and eb are three mutually orthogonal unit vectors.

Example: An electromagneticr wave is propagating in a vacuum in the -y$ direction. The phasor E field has amplitude 754 and a phase angle of 45 degrees.r The electric field vector is oriented as shown below. Find the phasor H field and write out the complete fields in real form. z y

r 3 1 E eee$$$=+ 754–∞ 45 vm/ Exz2 2 30∞ x

$ $ $ eH mustr be perpendicular to e and to - ey . Graphically, itrr is easy to see that H must be as shown by the dashed line. This gives EH¥ in the - y$ direction.

3 1 Thus, eee$$$=- Hzx2 2

$$L 3 $1 $O L 3 $1 $O Math check: eeEH¥=M e x + e zP ¥-M e z e xP N 2 2 Q N 2 2 Q 3 1 =-eee$$$ - =- 4 yyy4 r $$$754–∞ 45 F 3 1 I H = eeeH =-2 G zxJ –∞45 ho H 2 2 K The real fields are (by inspection): r F 3 $$1 I Eztaf, =+G exz eJ 754cos awb t++∞ y 45 f H 2 2 K r F 3 $$1 I Hztaf, =-G ezx eJ 245cos awb t++∞ y f H 2 2 K

p t = 4w t = 0

2p b

p ty=++∞045: cosa b f b Pictures of the field lines for this wave: 2p F lpI F=b y =G - J =- =-45 ∞ l H 84K ﬁ-+cosafwt 45 45 ﬁ peak! y l In the plane y =- at t = 0 etc. 8 r E = 754 x etc. Strength is indicated by number and by density of lines.

p same plane, t = y 4w etc. r E = 653

etc.

p r t = : E = 0 2w y r p t = : E = 754 w EEC 130A Plane Waves Fall, ‘97

Return now briefly to the derivation.r We derived E r = h H rr

from —¥EH = - jwmo . rr

We can also equally well derive it from —¥HE = jweo . rr We get jjbweHE= . Therefore, r E b r = . H we

This must again equal h . It does.

b wme m ==o o we we e b wm =ﬁo wme22 = b we b o

Or, in real form: rr r EEeEe=+ejxx yycosbwb tza -- o g y

r r H E y r E

r x

z This is linear polarization. Now we consider a more general case.

f130a13_planewaves_intro.fm 13 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

EExx=+-cosafwb taz rr EEyy=+-cosafwb tbz p Suppose we let a = 0 and b = . Then: 2 rr EE=-cosafwb tz rrxx EEyy=-sin afwb tz - rrr

What can we see about Eztaf, =+ Exxee E yy ?