ECE 6310 Spring 2012 Exam 1 Solutions. Balanis The electric fields are given by. E r = ˆxe jβ 0 z

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1 ECE 6310 Spring 2012 Exam 1 Solutions Balanis 1.30 The electric fiels are given by E i ˆxe jβ 0 z E r ˆxe jβ 0 z The curl of the electric fiels are the usual cross prouct E i jβ 0 ẑ ˆxe jβ 0 z jβ 0 ŷe jβ 0 z E r jβ 0 ẑ ˆxe jβ 0 z ) jβ 0 ŷe jβ 0 z From Maxwell s curl of electric fiel equation E j H we fin the magnetic fiels H i H r 1 j E i β 0 ŷe jβ 0 z 1 η 0 ŷe jβ 0 z 1 j E r β 0 ŷe jβ 0 z 1 η 0 ŷe jβ 0 z where the wave amplitue has been simplifie as β 0 ω µ 0 1 η 0 Apply the bounary conition for transverse magnetic fiel ˆn H 2 H 1 ) J s which in this case is

2 ) + H r z 0) ) J s ẑ 0 H i z 0 Solve for the surface current. J s 2 η 0 ŷ ẑ Do the cross prouct. J s 2 η 0 ˆx 2 2 η A/m Balanis 2.9 See the solution to problem 2.6 in assignment one for general assumptions in parallel plate capacitor problems. a) The applie potential is equal to the line integral of the electric fiel across the capacitor E 1 + E 2 where is the thickness of each iniviual ielectric layer. Solving the constitutive relation for the electric fiel in each layer E 1 D E 2 D Insert the fiels into the potential equation. D + D Solve for the electric flux ensity ivie by the free space permittivity.

3 D Insert into the electric fiel equations above to obtain the esire electric fiel expression. E 1 E /m 750 /m b) Obtain an expression for the electric flux ensity from the constitutive relation. D E C/m 2 c) From Gauss s equation the surface charge on the plates is just equal to the electric flux ensity, an the total charge is the surface charge times the area. Q DA + A C ) The capacitance is the ratio of charge to potential. C Q A F Balanis 3.5 This problem is a bit saistic, but often times real problems have very complex analytic solutions, so it s useful to learn to maintain your focus an o this kin of thing. The general form of the electric fiel in cylinrical coorinates is E,φ,z) ˆE,φ,z) + ˆφE φ,φ,z) + ẑe,φ,z) Insert this form into the vector wave equation. E) E β 2 E

4 Start computing the ivergence of the graient using the forms of the ivergence an graient given in the appenix, II-19) an II-20). Here I have expane some but not all of the erivatives of proucts. E 1 E ) + 1 E φ φ + E z ψ ˆ ψ + ˆφ 1 ψ φ + ẑ ψ E) ˆ ) ˆ 1 E 1 E ) + 1 E φ φ + E z 2 E 2 ) 1 2 E ) ˆφ 1 1 φ E ) + 1 E φ φ + E z 2 E φ φ 1 E φ 2 φ + 2 E z + ˆφ φ E + ẑ 1 E ) E φ ) + 1 E φ φ 2 φ + E z E z φ + ẑ 1 2 E ) E φ φ + 2 E z 2 Now fin the curl of the curl, again some but not all of the erivatives of proucts. E ˆ 1 E z φ E φ E ˆ 1 1 φ E φ E ˆ 1 2 φ E φ + ˆφ E E z + ẑ 1 E φ ) 1 ) E ) 1 E φ E φ E E z + ˆφ 1 E z φ E φ 1 E φ 2 E + E z φ ˆφ 1 2 E z φ 2 E φ ) 1 E φ ) E φ E φ + ẑ 1 E E z 1 1 E z φ φ E φ ) E φ 1 E 2 φ + ẑ 2 E 2 E z E 1 E z E z φ E φ φ Now work out the raial component of the wave equation: expan proucts, cancel terms an finally substitute in the scalar Laplacian. ˆ E) ˆ E β 2 E 1 2 E 2 ) 1 2 E ) E φ φ 1 E φ 2 φ + 2 E z 1 2 φ E φ ) E + 2 E E z 2 φ 2 2 β 2 E 1 E + 1 E 1 E 1 E E φ 2 φ 1 E φ 2 φ + 2 E z 1 2 E φ φ 1 E φ 2 φ E 2 φ 2 1 E E 2 φ 2 2 E + E 2 E φ 2 2 φ β 2 E + 2 E 1 2 E 2 E φ 2 2 φ β 2 E + 2 E E z 2 β 2 E Now work out the angular component of the wave equation as above.

5 ˆφ E) ˆφ E β 2 E φ φ E ) E φ 2 φ E φ + 1 E 2 φ E φ 2 φ 2 1 E φ E φ 2 φ E z φ 1 2 E z E z 2 E φ + E φ + 2 E 2 2 φ β 2 E φ φ + 2 E φ 2 φ 1 2 E z φ + 2 E φ E φ 1 2 E 2 φ + 2 E 2 φ β 2 E φ E 2 φ + 1 ) 1 2 E φ ) 1 E φ E φ + 1 E 2 φ β 2 E φ E φ 1 E φ 1 E 1 2 E 2 φ φ E φ β 2 E φ An finally the z component. ẑ E) ẑ E β 2 E z ) E 2 E φ φ + 2 E z 2 2 E + 1 E E φ φ + 2 E z 2 2 E z E z E z 2 φ + 2 E z β 2 E 2 2 z E z E z 2 φ 2 2 E z β 2 E z + 2 E z 2 2 E + 2 E z 1 E E z E z 2 φ 2 2 E + 2 E z 1 E E z E z 2 φ 2 β 2 E z 1 2 E φ φ β 2 E z 1 2 E φ φ β 2 E z It s interesting how for the z-component equation, the raial erivative term of the scalar Laplacian operator came in pieces, one each from the graient of the ivergence term an the ouble curl term. Balanis 4.27 The ecomposition of a wave traveling in the positive z-irection into two waves traveling in the positive z-irection looks like this. j ωt βz E ˆxE x + ŷe y )e ) j ωt βz ˆxE Rx + ŷe Ry )e ) j ωt βz + ˆxE Lx + ŷe Ly )e ) We can solve a set of four complex equations in four unknowns: the first two equations inicate the ecomposition an the secon two equations specify that that the two waves are right an left circularly polarize. E x E Rx + E Lx E y E Ry + E Ly E Ry je Rx E Ly je Lx

6 At this point we nee not constrain the original wave to be linearly polarize. Since the pair of right an left circularly polarize waves form a complete polarization basis, a wave of arbitrary polarization can be so ecompose. Solving the four equations for the complex coefficients of the right an left circularly polarize waves. E Rx 1 2 E + je x y ) E Ry 1 2 E je y x ) E Lx 1 2 E je x y ) E Ly 1 2 E + je y x ) You can see by inspection that these coefficients satisfy the require constraints: x- an y- components have the same magnitue, the proper phase relationship an the two waves a up to the original wave. If we constrain the original wave to be linearly polarize E x E x0 e jφ E y E y0 e jφ where the zero subscripte coefficients are real-value magnitues. The coefficients become E Rx 1 2 E + je x0 y0 )e jφ E Ry 1 2 E je y0 x0 )e jφ E Lx 1 2 E je x0 y0 )e jφ E Ly 1 2 E + je y0 x0 )e jφ

7 Balanis 5.27 The incient angle for total transmission is the Brewster angle. For parallel polarization in nonmagnetic meia the correct expression is equation 5-33) ε θ i θ B sin 1 2 ε 1 + ε 2 1 sin o For total reflection the incient angle must be greater than the critical angle. For non-magnetic meia the correct expression is equation 5-36) ε θ i θ c sin sin 1 2 π 6 30o ε 1