Homework Assignment #1 Solutions

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1 Physics 56 Winter 8 Textook prolems: h. 8: 8., 8.4 Homework Assignment # Solutions 8. A trnsmission line consisting of two concentric circulr cylinders of metl with conductivity σ nd skin depth δ, s shown, is filled with uniform lossless dielectric (µ, ɛ). A TEM mode is propgted long this line. Section 8. pplies. ) Show tht the time-verged power flow long the line is ( ) µ P = ɛ π H ln where H is the pek vlue of the zimuthl mgnetic field t the surfce of the inner conductor. A TEM mode is essentilly two-dimensionl electrosttic prolem. Thus we strt y finding the electric field etween the two cylinders. By elementry mens, it should e cler tht E t = A ρ ˆρ where A is constnt tht will e determined shortly. Assuming wve propgtion in the +z direction, we use B t = µɛ ẑ E t to otin the mgnetic field ɛ A H t = µ ρ ˆφ This indictes tht the mgnitude of the mgnetic field t the inner conductor is H() = ɛ/µ(a/). Defining this s H gives µ E t = ɛ H ρ ˆρ, Ht = H ρ ˆφ () The (hrmonic) Poynting vector is then S = E H = µ ɛ H ρ ẑ so the power flow is P = ẑ S d = A µ ɛ H µ πρ dρ = π ρ ɛ H ln ( ) () ) Show tht the trnsmitted power is ttenuted long the line s P (z) = P e γz

2 where γ = ɛ σδ µ ( + ) ln ( ) We compute the ttenution coefficient ccording to γ = P dp dz (3) The power P ws clculted in prt. For the power loss per unit length of the wveguide, we use dz = ˆn H σδ dl = σδ H ρ dl Note tht there re two oundries, one t ρ = (with circumference π) nd the other t ρ = (with circumference π). This gives dz = σδ H [π + (/) π] = π σδ H ( + ) (4) Inserting this power loss expression nd the power () into (3) yields γ = ɛ + σδ µ ln(/) = ɛ σδ µ ( + ) ln ( ) c) The chrcteristic impednce Z of the line is defined s the rtio of the voltge etween the cylinders to the xil current flowing in one of them t ny position z. Show tht for this line Z = ( ) µ π ɛ ln Since Z = V /I, we need to compute the voltge difference etween the cylinders s well s the current. For the voltge difference, we hve V = E d µ ( ) l = ɛ H µ ρ dρ = ɛ H ln where we hve used () for the electric field. In ddition, the current is given y integrting the surfce current density. For the inside conductor, we hve K = ˆn H = ˆρ ( ) H ρ ˆφ = H ẑ ρ=

3 Hence I = Tking the rtio Z = V /I results in K dl = πh Z = ( ) µ π ɛ ln d) Show tht the series resistnce nd inductnce per unit length of the line re R = ( πσδ + ) { ( ) µ L = π ln + µ cδ 4π ( + )} where µ c is the permeility of the conductor. The correction to the inductnce comes from the penetrtion of the flux into the conductors y distnce of order δ. We my otin the series resistnce from the power loss I R = dz where R denotes the resistnce per unit length. Using dp/dz from (4) s well s the current computed ove, we find R = ( I ) = + dz πσδ For the inductnce per unit length, we compute the energy per unit length stored in the mgnetic field. Inside the volume of the wveguide, we hve U vol = A µ 4 H d = µ 4 H ρ πρ dρ = µ H π ln ( ) In ddition, since some of the mgnetic field penetrtes the conducting wlls, we use the pproximtion H(ζ) = H e ζ/δ e iζ/δ where ζ is the distnce into the conductor. Assuming the skin depth is much less thn the thickness of the conductor s well s the rdius of curvture, we pproximte U wll = µ c 4 H(ξ) dξ = µ c 4 H e ξ/δ dξ = µ c 8 δ H

4 where is the circumference of the wll. On the inside wll, we hve = π nd H = H, while on the outside wll, we hve = π nd H = H (/). Hence U wlls = µ c 8 δ H [π + π(/) ] = µ c 4 πδ H ( + ) Using we end up with 4 L I = U vol + U wlls L = µ ( ) π ln + µ cδ + 4π 8.4 Trnsverse electric nd mgnetic wves re propgted long hollow, right circulr cylinder with inner rdius R nd conductivity σ. ) Find the cutoff frequencies of the vrious TE nd TM modes. Determine numericlly the lowest cutoff frequency (the dominnt mode) in terms of the tue rdius nd the rtio of cutoff frequencies of the next four higher modes to tht of the dominnt mode. For this prt ssume tht the conductivity of the cylinder is infinite. The eigenvlue eqution for either TE or TM modes is [ t + γ ]ψ(ρ, φ) = where ψ(r, φ) = for TM modes or dψ(ρ, φ)/dρ ρ=r = for TE modes. Writing ψ(ρ, φ) = ψ(ρ)e ±imφ, the rdil eqution (in cylindricl coordintes) ecomes ( ρ ρ ρ ) ρ + γ m ρ ψ(ρ) = which is solved y Bessel functions. Avoiding the Neumnn function which lows up t ρ =, we hve ψ(ρ, φ) J m (γρ)e ±imφ The oundry conditions then plce conditions on γ. For TM modes (Dirichlet conditions), we demnd J m (γr) =. Hence (TM) γ mn = x mn R or ω mn = x mn µɛr where x mn is the n-th zero of J m. For TE modes (Neumnn conditions), on the other hnd, we demnd J m(γr) =. Hence (TE) γ mn = x mn R or ω mn = x mn µɛr

5 where x mn is the n-th zero of J m. Sorting through the zeros of J m nd J m, the lowest five modes re given y mode µɛrωmn ω mn /ω dominnt TE.84 TM TE TE nd TM Note tht the TE nd TM modes re degenerte. This is specil cse where the Bessel identity J (ζ) = J (ζ) demonstrtes tht x,n+ = x n. ) lculte the ttenution constnts of the wveguide s function of frequency for the lowest two distinct modes nd plot them s function of frequency. The computtion of the ttenution coefficients involves computing oth power P nd power loss dp/dz. We first consider TM modes. The power is given y P = Using ψ = J m (γρ)e ±imφ gives A ψ d = π R ( ) ( ) / ɛ ω µ ω mn ω ψ d (5) A J m (x mn ρ/r) ρ dρ = π[ R J m+ (x mn ) ] = πr J m+ (x mn ) where the expression in the squre rckets comes from the Bessel function orthogonlity reltion J ν (x νm ρ/)j ν (x νn ρ/)ρ dρ = J ν+ (x νm ) δ mn Hence P = ( ) ( ) / ɛ ω µ ω mn ω πr J m+(x mn) (6) For TM mode, the power loss is given y dz = ( ) ω σδ ω mn ψ µ ωmn n dl In this cse ψ n = ψ ρ = γ mn J m(x mn )e ±imφ ρ=r Using γ mn = µɛω mn, we otin dz = ɛ σδ µ (πr)j m(x mn )

6 We my now hve some fun with Bessel functions. Using the recursion reltion J m+ (ζ) = m ζ J m(ζ) J m(ζ) s setting ζ = x mn to e zero of J m, we otin J m+ (x mn ) = J m(x mn ) This llows us to rewrite the power loss s dz = ɛ σδ µ (πr)j m+(x mn ) (7) Given (6) nd (7), the TM mn ttenution coefficient is otined y setting β mn = P dp dz = ( ) / ɛ πr σδ µ ω πr = ( ) / ɛ σδ µ ω R Note tht /R = /(A) were = πr nd A = πr re the circumference nd re of the cylindricl wveguide. Since δ = δ mn ωmn /ω (where δ mn is the skin depth t the cutoff frequency ω mn ), we get the stndrd TM expression with the geometric fctor ξ mn =. For the TE mode, the power loss clcultion is somewht lengthier, s it involves oth H z nd H t. We egin with the power, which is given y similr expression s (5), however with fctor of µ/ɛ insted. The Bessel normliztion integrl is now which gives P = R µ ɛ J m (x mnρ/r) ρ dρ = R ( m /x mn)j m (x mn) ( ω ω mn This time, the power loss expression is ) ( ) / ) ω πr ( m x mn J m (x mn) (8) dz = ( ) [ ( ) ] ω σδ ω mn γmn ω ˆn t ψ + ω mn ω ψ dl There re two terms to evlute. The simple one is ψ dl = (πr)j m (x mn)

7 For the grdient term, we note tht ˆn = ˆρ on the inside of the cylinder. And t = ˆρ ρ + (/ρ) ˆφ φ. Hence ˆn t ψ dl = (πr) ψ ρ φ = (πr) m R J m(x mn) omining these two terms yields dz = ( ) [ ( ) ] ω m (πr) σδ ω mn x mn ω + ω mn ω J m (x mn) Using this for the power loss nd (8) for the power itself gives n ttenution coefficient β mn = dp P dz = ɛ σδ µ = ɛ σδ µ ( ω ( ω ) / [ πr m ) / R πr [ x mn ( ω m x mn m + ω mn ω ] ) ] ] + ω mn [ ω m x mn This demonstrtes tht the TE geometric fctors re ξ mn = m /(x mn m ) nd η mn =. The ttenution constnts re plotted s follows β / β 4 3 TM, TE, where β = ɛ σδ mn µ R ω/ ω dominnt

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