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17 Pages

### 325_reflection_transmission

Course: MER 325, Fall 2009
School: University of Texas
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Word Count: 1775

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wave plane reflections Maxwell, time harmonic, transverse-to-z collecting all the terms, assuming time harmonic solutions using Ohms law assuming there is no component of either E or H in the z direction Maxwells equations reduce to E y z = j H x H y z = ( + j ) Ex E y x Ex =0 y Ex = j H y z things to notice H x = ( + j ) E y z H y x i D = v i B = 0 H x =0 y J =E D= E Ey is...

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wave plane reflections Maxwell, time harmonic, transverse-to-z collecting all the terms, assuming time harmonic solutions using Ohms law assuming there is no component of either E or H in the z direction Maxwells equations reduce to E y z = j H x H y z = ( + j ) Ex E y x Ex =0 y Ex = j H y z things to notice H x = ( + j ) E y z H y x i D = v i B = 0 H x =0 y J =E D= E Ey is connected to Hx via d/dz and Ex is connected to Hy via d/dz and Hy is connected to Ex via d/dz and Hx is connected to Ey via d/dz and Ey and Ex are connected via d/dx and d/dy Hy and Hx are connected via d/dx and d/dy 1 r o B= H r o EE 325, Dept. of ECE, Univ. of Texas at Austin Dean P. Neikirk 2004, last update April 27, 2004 plane wave reflections Uniform plane wave solution to Maxwells equations the complete, time harmonic solution is E plane = Exo e jt z x wave H plane = ( Exo ) e jt z y wave = j ( + j ) = + j 2 = j ( ) d d = vg = d d Ps ,ave = 1 is called the complex propagation constant = vp = direction of propagation Eo H o k = j EH EH k = j ( + j ) 2 1 Re ES H S 2 ( ) x E plane = Exo e wave j ( t k z ) Dean P. Neikirk 2004, last update April 27, 2004 EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Plane waves and boundaries we have found the TEM traveling wave solution to Maxwells equations that would work in a world that is made on one and only one material what would happen if there were a simple dielectric interface, i.e., half the world is filled with 1, and half is 2 ? we must still satisfy boundary conditions at the interface between the materials medium 1 in anticipation of what we x need to solve this problem, r1 lets assume now that as r1 a result of the incident wave there will be a transmitted 1 medium 2 wave and a reflected wave r2 r2 incident = 1 = j1 ( 1 + j1 ) Eincident 2 kin kincident = j 1 z Eincident = Ex+1o e 1z x H incident y z H incident = Ex+1o 1 e 1z y Dean P. Neikirk 2004, last update April 27, 2004 ( ) 1 = j (1 1 ) 3 EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Plane waves and boundaries as a result of the incident wave there will be a transmitted wave and a reflected wave incident = 1 = j1 ( 1 + j1 ) transmit = 2 = j2 ( 2 + j 2 ) Etransmit = Ex+2 o e 2 z x + H tramsit = Ex 2o 2 e 2 z y Eincident = Ex+1o e 1z x H incident = Ex+1o 1 e 1z y ( ) ( ) Eincident kin H incident x Etransmit ktransmit reflect = 1 = j1 ( 1 + j1 ) medium 1 r1 r1 1 y H transmit Ereflect = Ex1o e + 1z x H reflect = Ex1o 1 e+ 1z y Dean P. Neikirk 2004, last update April 27, 2004 H reflect kreflect 4 ( ) Ereflect medium 2 r2 r2 2 z EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Fields at the interface the total fields at the interface between the two materials (i.e., at z=0) are infinitesimally to the left of the interface + Eleft = Ex1o + Ex1o H left = Ex+1o 1 + Ex1o 1 H right = Ex+2 o 2 ( ) ( ) ) infinitesimally to the right of the interface Eright = Ex+2 o ( since the fields are tangential to the interface, and were assuming there is no surface current in this problem, the fields must be CONTINUOUS across the interface Eleft = Eright + Ex+1o + Ex1o = Ex 2 o H left = H right (E + x1o 1 ) + ( Ex1o 1 ) = ( Ex+2o 2 ) so we have two unknowns, Ereflect (Ex1o-) and Etransmit (Ex2o+) Dean P. Neikirk 2004, last update April 27, 2004 5 EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Transmitted and reflected fields two unknowns, two equations + Ex+1o + Ex1o = Ex 2 o medium 1 medium 2 Ex+1o Ex1o 2 x1o (E + x1o + x1o 1 ) + ( Ex1o 1 ) = ( Ex+2o 2 ) x1o 1 + x1o x1o Ex+2 o ( E ) + ( E ) = ( E + E ) (E ) (E ) = (E ) + (E 1 + x1o 1 + x1o 2 x1o 2 1 ) + (1 1 ) (1 2 ) Ex1o = (1 2 ) + (1 1 ) Ex1o E x1o (1 1 ) (1 2 ) E + = 2 1 E + = x1o 1 2 ) + (1 1 ) 2 + 1 x1o ( j ( + j ) = 6 = j = j Dean P. Neikirk 2004, last update April 27, 2004 j ( + j ) EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Reflection coefficient E x1o (1 1 ) (1 2 ) E + = 2 1 E + = x1o 1 2 ) + (1 1 ) 2 + 1 x1o ( medium 1 medium 2 we now have simple relation that gives the ratio of the reflected electric field to the incident electric field the reflection coefficient is Ex+1o Ex1o Ex+2 o E x1o = E E + x1o Ex1o 2 1 E = + = Ex1o 2 + 1 = j ( + j ) for our assumed coordinate system the sign of will tells us which way the reflected electric field points Eincident pointed in the +x^ direction if is positive, then Ereflect also points in the +x^ direction if is negative, then Ereflect points in the -x^ direction Dean P. Neikirk 2004, last update April 27, 2004 7 EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Transmission coefficient recall we had the equation from continuity of total tangential electric field at the interface, and we also have the reflection coefficient, so + Ex+1o + Ex1o = Ex 2 o Ex1o = E Ex+1o E + x 2o medium 1 medium 2 E + x1o + (E E + x1o )=E + x 2o = (1 + E ) E Ex+1o + x1o Ex1o Ex+2 o we define the transmission coefficient to be the ratio of the transmitted electric field to the incident electric field Ex+2 o 2 2 = E = 1+ E = 1+ 2 1 = + 2 + 1 2 + 1 Ex1o = Dean P. Neikirk 2004, update last April 27, 2004 8 j ( + j ) EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Magnetic fields recall we had the simple relation between the electric and magnetic fields + H o ,incident = H y1o y = ( Ex+1o 1 ) y H o ,reflect = H y1o y = Ex1o 1 y ( ) + + H tramsit = H y 2 o y = ( H y1o + H y1o ) y so we can also define a magnetic field reflection coefficient H = H y1o H + y1o ( = (E Ex1o 1 + x1o 1 )=E E ) )) x1o + x1o = E and the transmitted magnetic field is + + + + H y 2 o = H y1o + H y1o = H y1o 1 + H y1o H y1o ( ( H + + = H y1o (1 + H ) = H y1o (1 E ) H = + H y 2o + y1o = 1+ H = 1 E 9 EE 325, Dept. of ECE, Univ. of Texas at Austin Dean P. Neikirk 2004, last update April 27, 2004 plane wave reflections Power flow: reflected power Poynting vectors at the interface (z = 0) 2 1 * 1 1 1 + + Ex1o 1 y = Ex+1o Re Pin ,ave = Re Ein H in = Re Ex1o x z 2 2 2 1 * * 1 1 Ex1o 1 y Preflect ,ave = Re Erefl H refl = Re Ex1o x 2 2 ( ) ( ) ( ) ( ) ( ) 1 1 2 1 1 + 2 = Ex1o Re ( z ) = E Ex1o Re (z) 2 2 1 * 1 * 1 1 2 + 2 = E Ex1o Re (z ) 2 1 * so the ratio of the incident power to the reflected power is Preflect ,ave Pin ,ave Dean P. Neikirk 2004, last update April 27, 2004 = E 10 2 EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Power flow: transmitted power Poynting vectors at the interface (z = 0) Pin ,ave = 1 + 2 1 Ex1o Re z 2 1 * Ptrans ,ave = * 1 1 + Re Etrans H trans = Re Ex 2 o x Ex+2 o 2 y 2 2 1 1 + 2 1 1 + 2 = Ex 2 o Re ( z ) = E Ex1o Re ( z) 2 2 2 * 2 * ( ) ( ) ( ) 1 1 2 + 2 = E Ex1o Re (z) 2 2 * so the ratio of transmitted power to incident power is Ptrans ,ave Pin ,ave Dean P. Neikirk 2004, last update April 27, 2004 = E 2 Re (1 2 *) Re (1 1 *) 11 EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Example: two lossless dielectrics assume the two materials are lossless, = 0, = o so here medium 1 medium 2 =r2o = j ( j ) = j o r o =r1o Ex+1o Ex1o = o r o Ex1o E = + = Ex1o limits: r1 << r2 r1 >> r2 ~ -1, ~ 0 ~ +1, ~ 2 Ex+2 o o r 2 o o r1 o 1 r 2 1 r1 r1 r 2 = = o r 2 o + o r1 o r1 + r 2 1 r 2 + 1 r1 r1 r 2 2 r1 E = 1+ = r1 + r 2 r1 + r 2 1 < E < 1 0 <E < 2 12 EE 325, Dept. of ECE, Univ. of Texas at Austin Dean P. Neikirk 2004, last update April 27, 2004 plane wave reflections Example: two lossless dielectrics assume the two materials are lossless, = 0, = o lets assume that r1 < r2 1 = o r1 o > o r 2 o = 1 E = 1 r 2 1 r1 1 r 2 + 1 r1 r1 r 2 = <0 r1 + r 2 E = Ex+1o H + y1o 2 r1 medium 1 =r1o medium 2 =r2o r1 + r 2 H y1o >0 H y1o Ex+2 o = Ex+1o H + y 2o H = H + y1o = E > 0 + + Ex1o = Ex1o = Ex1o H = + H y 2o H + y1o = 1+ H > 0 Dean P. Neikirk 2004, last update April 27, 2004 13 EE 325, Dept. of ECE, Univ. of Texas at Austin plane wave reflections Example: two lossless dielectrics, power flow assume two lossless materials, = 0, r1 , r2 , = o E = r1 r 2 r1 + r 2 E = 2 r1 r1 + r 2 >0 1 = o r1 o 2 2 ...

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