Waveguides. 8.14 Problems. 8.14. Problems 361

8.4. Problems 36 improving liquid rystal displays, and other produts, suh as various optoeletroni omponents, osmetis, and hot and old mirrors for arhitetural and automotive windows. 8.4 Problems 9 Waveguides 8. Prove the refletane and transmittane formulas (8.4.6) in FTIR. 8. Computer Experiment FTIR. Reprodue the results and graphs of Figures 8.4.3 8.4.5. 8.3 Computer Experiment Surfae Plasmon Resonane. Reprodue the results and graphs of Figures 8.5.3 8.5.7. 8.4 Working with the eletri and magneti fields aross an negative-index slab given by Eqs. (8.6.) and (8.6.), derive the refletion and transmission responses of the slab given in (8.6.8). 8.5 Computer Experiment Perfet Lens. Study the sensitivity of the perfet lens property to the deviations from the ideal values of ɛ = ɛ and μ = μ, and to the presene of losses by reproduing the results and graphs of Figures 8.6.3 and 8.6.4. You will need to implement the omputational algorithm listed on page 39. 8.6 Computer Experiment Antirefletion Coatings. Reprodue the results and graphs of Figures 8.7. 8.7.3. 8.7 Computer Experiment Omnidiretional Dieletri Mirrors. Reprodue the results and graphs of Figures 8.8. 8.8.. 8.8 Derive the generalized Snel s laws given in Eq. (8..). Moreover, derive the Brewster angle expressions given in Eqs. (8..4) and (8..5). 8.9 Computer Experiment Brewster angles. Study the variety of possible Brewster angles and reprodue the results and graphs of Example 8... 8. Computer Experiment Multilayer Birefringent Strutures. Reprodue the results and graphs of Figures 8.3. 8.3.. Waveguides are used to transfer eletromagneti power effiiently from one point in spae to another. Some ommon guiding strutures are shown in the figure below. These inlude the typial oaxial able, the two-wire and mitrostrip transmission lines, hollow onduting waveguides, and optial fibers. In pratie, the hoie of struture is ditated by: (a) the desired operating frequeny band, (b) the amount of power to be transferred, and () the amount of transmission losses that an be tolerated. Fig. 9.. Typial waveguiding strutures. Coaxial ables are widely used to onnet RF omponents. Their operation is pratial for frequenies below 3 GHz. Above that the losses are too exessive. For example, the attenuation might be 3 db per m at MHz, but db/ m at GHz, and 5 db/ m at GHz. Their power rating is typially of the order of one kilowatt at MHz, but only W at GHz, being limited primarily beause of the heating of the oaxial ondutors and of the dieletri between the ondutors (dieletri voltage breakdown is usually a seondary fator.) However, speial short-length oaxial ables do exist that operate in the 4 GHz range. Another issue is the single-mode operation of the line. At higher frequenies, in order to prevent higher modes from being launhed, the diameters of the oaxial ondutors must be redued, diminishing the amount of power that an be transmitted. Two-wire lines are not used at mirowave frequenies beause they are not shielded and an radiate. One typial use is for onneting indoor antennas to TV sets. Mirostrip lines are used widely in mirowave integrated iruits.

9.. Longitudinal-Transverse Deompositions 363 364 9. Waveguides Retangular waveguides are used routinely to transfer large amounts of mirowave power at frequenies greater than 3 GHz. For example at 5 GHz, the transmitted power might be one megawatt and the attenuation only 4 db/ m. Optial fibers operate at optial and infrared frequenies, allowing a very wide bandwidth. Their losses are very low, typially,. db/km. The transmitted power is of the order of milliwatts. 9. Longitudinal-Transverse Deompositions In a waveguiding system, we are looking for solutions of Maxwell s equations that are propagating along the guiding diretion (the z diretion) and are onfined in the near viinity of the guiding struture. Thus, the eletri and magneti fields are assumed to have the form: E(x, y, z, t)= E(x, y)e jωt jβz H(x, y, z, t)= H(x, y)e jωt jβz (9..) where β is the propagation wavenumber along the guide diretion. The orresponding wavelength, alled the guide wavelength, is denoted by λ g = π/β. The preise relationship between ω and β depends on the type of waveguiding struture and the partiular propagating mode. Beause the fields are onfined in the transverse diretions (the x, y diretions,) they annot be uniform (exept in very simple strutures) and will have a non-trivial dependene on the transverse oordinates x and y. Next, we derive the equations for the phasor amplitudes E(x, y) and H(x, y). Beause of the preferential role played by the guiding diretion z, it proves onvenient to deompose Maxwell s equations into omponents that are longitudinal, that is, along the z-diretion, and omponents that are transverse, along the x, y diretions. Thus, we deompose: E(x, y)= ˆx E x (x, y)+ŷ E y (x, y) + ẑ E z (x, y) E } {{ } } {{ } T (x, y)+ẑ E z (x, y) (9..) transverse longitudinal In a similar fashion we may deompose the gradient operator: =ˆx x + ŷ y + ẑ z = T + ẑ z = T jβ ẑ (9..3) } {{ } transverse where we made the replaement z jβ beause of the assumed z-dependene. Introduing these deompositions into the soure-free Maxwell s equations we have: E = jωμh H = jωɛe E = H = ( T jβẑ) (E T + ẑ E z )= jωμ(h T + ẑ H z ) ( T jβẑ) (H T + ẑ H z )= jωɛ(e T + ẑ E z ) ( T jβẑ) (E T + ẑ E z )= ( T jβẑ) (H T + ẑ H z )= (9..4) where ɛ, μ denote the permittivities of the medium in whih the fields propagate, for example, the medium between the oaxial ondutors in a oaxial able, or the medium within the hollow retangular waveguide. This medium is assumed to be lossless for now. We note that ẑ ẑ =, ẑ ẑ =, ẑ E T =, ẑ T E z = and that ẑ E T and ẑ T E z are transverse while T E T is longitudinal. Indeed, we have: ẑ E T = ẑ (ˆx E x + ŷ E y )= ŷ E x ˆx E y T E T = (ˆx x + ŷ y ) (ˆx E x + ŷ E y )= ẑ( x E y y E x ) Using these properties and equating longitudinal and transverse parts in the two sides of Eq. (9..4), we obtain the equivalent set of Maxwell equations: T E z ẑ jβ ẑ E T = jωμh T T H z ẑ jβ ẑ H T = jωɛe T T E T + jωμ ẑ H z = T H T jωɛ ẑ E z = T E T jβe z = T H T jβh z = (9..5) Depending on whether both, one, or none of the longitudinal omponents are zero, we may lassify the solutions as transverse eletri and magneti (TEM), transverse eletri (TE), transverse magneti (TM), or hybrid: E z =, H z =, E z =, H z, E z, H z =, E z, H z, TEM modes TE or H modes TM or E modes hybrid or HE or EH modes In the ase of TEM modes, whih are the dominant modes in two-ondutor transmission lines suh as the oaxial able, the fields are purely transverse and the solution of Eq. (9..5) redues to an equivalent two-dimensional eletrostati problem. We will disuss this ase later on. In all other ases, at least one of the longitudinal fields E z,h z is non-zero. It is then possible to express the transverse field omponents E T, H T in terms of the longitudinal ones, E z, H z. Forming the ross-produt of the seond of equations (9..5) with ẑ and using the BAC-CAB vetor identity, ẑ (ẑ H T )= ẑ(ẑ H T ) H T (ẑ ẑ)= H T, and similarly, ẑ ( T H z ẑ)= T H z, we obtain: T H z + jβh T = jωɛ ẑ E T Thus, the first two of (9..5) may be thought of as a linear system of two equations in the two unknowns ẑ E T and H T, that is, β ẑ E T ωμh T = jẑ T E z ωɛ ẑ E T βh T = j T H z (9..6)

9.. Longitudinal-Transverse Deompositions 365 366 9. Waveguides The solution of this system is: ẑ E T = jβ k H T = jωɛ k ẑ T E z jωμ k T H z where we defined the so-alled utoff wavenumber k by: ẑ T E z jβ (9..7) k T H z k = ω ɛμ β = ω β = k β (utoff wavenumber) (9..8) The quantity k = ω/ = ω ɛμ is the wavenumber a uniform plane wave would have in the propagation medium ɛ, μ. Although k stands for the differene ω ɛμ β, it turns out that the boundary onditions for eah waveguide type fore k to take on ertain values, whih an be positive, negative, or zero, and haraterize the propagating modes. For example, in a dieletri waveguide k is positive inside the guide and negative outside it; in a hollow onduting waveguide k takes on ertain quantized positive values; in a TEM line, k is zero. Some related definitions are the utoff frequeny and the utoff wavelength defined as follows: ω = k, λ = π k (utoff frequeny and wavelength) (9..9) We an then express β in terms of ω and ω,orω in terms of β and ω. Taking the positive square roots of Eq. (9..8), we have: β = ω ω = ω ω ω and ω = ω + β (9..) Often, Eq. (9..) is expressed in terms of the wavelengths λ = π/k = π/ω, λ = π/k, and λ g = π/β. It follows from k = k + β that λ = λ + λ g λ λ g = λ λ (9..) Note that λ is related to the free-spae wavelength λ = π /ω = /f by the refrative index of the dieletri material λ = λ /n. It is onvenient at this point to introdue the transverse impedanes for the TE and TM modes by the definitions: η TE = ωμ β = η ω β, η TM = β ωɛ = η β ω (TE and TM impedanes) (9..) where the medium impedane is η = μ/ɛ, so that η/ = μ and η = /ɛ. We note the properties: η TE η TM = η, Beause β/ω = ω /ω, we an write also: η TE = η ω ω η TE = ω (9..3) η TM β, η TM = η With these definitions, we may rewrite Eq. (9..7) as follows: ẑ E T = jβ (ẑ ) T k E z + η TE T H z H T = jβ ( ) k ẑ T E z + T H z η TM Using the result ẑ (ẑ E T )= E T, we solve for E T and H T : E T = jβ T k E z η TE ẑ T H z H T = jβ ( T k H z + ) ẑ T E z η TM ω ω (9..4) (9..5) (transverse fields) (9..6) An alternative and useful way of writing these equations is to form the following linear ombinations, whih are equivalent to Eq. (9..6): H T η TM ẑ E T = j β T H z E T η TE H T ẑ = j β T E z (9..7) So far we only used the first two of Maxwell s equations (9..5) and expressed E T, H T in terms of E z,h z. Using (9..6), it is easily shown that the left-hand sides of the remaining four of Eqs. (9..5) take the forms: T E T + jωμ ẑ H z = jωμ k ẑ ( T H z + k H ) z T H T jωɛ ẑ E z = jωɛ ẑ ( T E z + k E ) z k T E T jβe z = jβ k T E z + k E z T H T jβh z = jβ k T H z + k H z

9.. Longitudinal-Transverse Deompositions 367 368 9. Waveguides where T is the two-dimensional Laplaian operator: T = T T = x + y (9..8) and we used the vetorial identities T T E z =, T (ẑ T H z )= ẑ T H z, and T (ẑ T H z )=. It follows that in order to satisfy all of the last four of Maxwell s equations (9..5), it is neessary that the longitudinal fields E z (x, y), H z (x, y) satisfy the two-dimensional Helmholtz equations: Fig. 9.. Cylindrial oordinates. T E z + k E z = T H z + k H z = (Helmholtz equations) (9..9) These equations are to be solved subjet to the appropriate boundary onditions for eah waveguide type. One, the fields E z,h z are known, the transverse fields E T, H T are omputed from Eq. (9..6), resulting in a omplete solution of Maxwell s equations for the guiding struture. To get the full x, y, z, t dependene of the propagating fields, the above solutions must be multiplied by the fator e jωt jβz. The ross-setions of pratial waveguiding systems have either artesian or ylindrial symmetry, suh as the retangular waveguide or the oaxial able. Below, we summarize the form of the above solutions in the two types of oordinate systems. The Helmholtz equations (9..9) now read: ( ρ E ) z + E z ρ ρ ρ ρ φ + k E z = ( ρ H ) z + H z ρ ρ ρ ρ φ + k H z = Noting that ẑ ˆρ = ˆφ and ẑ ˆφ = ˆρ, we obtain: ẑ T H z = ˆφ( ρ H z ) ˆρ ρ ( φh z ) (9..3) Cartesian Coordinates The artesian omponent version of Eqs. (9..6) and (9..9) is straightforward. Using the identity ẑ T H z = ŷ x H z ˆx y H z, we obtain for the longitudinal omponents: ( x + y)e z + k E z = ( x + y )H z + k H z = Eq. (9..6) beomes for the transverse omponents: E x = jβ x k E z + η TE y H z E y = jβ, y k E z η TE x H z Cylindrial Coordinates H x = jβ ( x k H z ) y E z η TM H y = jβ ( y k H z + ) x E z η TM (9..) (9..) The relationship between artesian and ylindrial oordinates is shown in Fig. 9... From the triangle in the figure, we have x = ρ os φ and y = ρ sin φ. The transverse gradient and Laplae operator are in ylindrial oordinates: T = ˆρ ρ + ˆφ ρ φ, ( T = ρ ) + (9..) ρ ρ ρ ρ φ The deomposition of a transverse vetor is E T = ˆρE ρ + ˆφE φ. The ylindrial oordinates version of (9..6) are: E ρ = jβ ( ρ k E z η TE ρ ) φh z E φ = jβ ( k ρ ) φe z + η TE ρ H z, H ρ = jβ ( ρ k H z + η TM ρ ) φe z H φ = jβ ( k ρ φh z ) ρ E z η TM (9..4) For either oordinate system, the equations for H T may be obtained from those of E T by a so-alled duality transformation, that is, making the substitutions: E H, H E, ɛ μ, μ ɛ (duality transformation) (9..5) These imply that η η and η TE η TM. Duality is disussed in greater detail in Se. 8.. 9. Power Transfer and Attenuation With the field solutions at hand, one an determine the amount of power transmitted along the guide, as well as the transmission losses. The total power arried by the fields along the guide diretion is obtained by integrating the z-omponent of the Poynting vetor over the ross-setional area of the guide:

9.. Power Transfer and Attenuation 369 37 9. Waveguides P T = P z ds, S where P z = Re(E H ) ẑ (9..) It is easily verified that only the transverse omponents of the fields ontribute to the power flow, that is, P z an be written in the form: P z = Re(E T H T ) ẑ (9..) For waveguides with onduting walls, the transmission losses are due primarily to ohmi losses in (a) the ondutors and (b) the dieletri medium filling the spae between the ondutors and in whih the fields propagate. In dieletri waveguides, the losses are due to absorption and sattering by imperfetions. The transmission losses an be quantified by replaing the propagation wavenumber β by its omplex-valued version β = β jα, where α is the attenuation onstant. The z-dependene of all the field omponents is replaed by: e jβz e jβz = e (α+jβ)z = e αz e jβz (9..3) The quantity α is the sum of the attenuation onstants arising from the various loss mehanisms. For example, if α d and α are the attenuations due to the ohmi losses in the dieletri and in the onduting walls, then Seond, the magneti fields on the ondutor surfaes are determined and the orresponding indued surfae urrents are alulated by J s = ˆn H, where ˆn is the outward normal to the ondutor. Third, the ohmi losses per unit ondutor area are alulated by Eq. (.8.7). Figure 9.. shows suh an infinitesimal ondutor area da = dl dz, where dl is along the ross-setional periphery of the ondutor. Applying Eq. (.8.7) to this area, we have: dp loss da = dp loss dldz = R s J s (9..6) where R s is the surfae resistane of the ondutor given by Eq. (.8.4), ωμ ωɛ R s = σ = η σ = σδ, δ = = skin depth (9..7) ωμσ Integrating Eq. (9..6) around the periphery of the ondutor gives the power loss per unit z-length due to that ondutor. Adding similar terms for all the other ondutors gives the total power loss per unit z-length: P loss = dp loss dz = C a R s J s dl + C b R s J s dl (9..8) α = α d + α (9..4) The ohmi losses in the dieletri an be haraterized either by its loss tangent, say tan δ, or by its ondutivity σ d the two being related by σ d = ωɛ tan δ. More generally, the effetive dieletri onstant of the medium may have a negative imaginary part ɛ I that inludes both ondutive and polarization losses, ɛ(ω)= ɛ jɛ I, with ɛ I = ɛ tan δ. Then, the orresponding omplex-valued wavenumber β is obtained by the replaement: β = ω μɛ k β = ω μɛ(ω) k Fig. 9.. Condutor surfae absorbs power from the propagating fields. For weakly lossy dieletris (ɛ I ɛ), we may make the approximation: β = ω μ(ɛ jɛ I ) k = β jω μɛ I = β j ω μɛ I β j ω μɛ I β β Resulting in the attenuation onstant, after setting μɛ = / and β/ω = ω /ω, α d = ω μɛ I β = ω μɛ β tan δ = ω tan δ (dieletri losses) (9..5) ω /ω The ondutor losses are more ompliated to alulate. In pratie, the following approximate proedure is adequate. First, the fields are determined on the assumption that the ondutors are perfet. where C a and C b indiate the peripheries of the ondutors. Finally, the orresponding attenuation oeffiient is alulated from Eq. (.6.): α = P loss P T (ondutor losses) (9..9) Equations (9..) (9..9) provide a systemati methodology by whih to alulate the transmitted power and attenuation losses in waveguides. We will apply it to several examples later on. Eq. (9..9) applies also to the dieletri losses so that in general P loss arises from two parts, one due to the dieletri and one due to the onduting walls, α = P loss P T = P diel + P ond P T = α d + α (attenuation onstant) (9..)

9.3. TEM, TE, and TM modes 37 37 9. Waveguides Eq. (9..5) for α d an also be derived diretly from Eq. (9..) by applying it separately to the TE and TM modes. We reall from Eq. (.9.6) that the losses per unit volume in a dieletri medium, arising from both a ondution and polarization urrent, J tot = J + jωd, are given by, dp loss dv = Re[ J tot E ] = ωɛ I E E Integrating over the ross-setional area of the guide gives the dieletri loss per unit waveguide length (i.e., z-length), P diel = ωɛ I E ds S Applying this to the TE ase, we find, P diel = ωɛ I E ds = S ωɛ I P T = S Re(E T H T ) ẑ ds = η TE α d = P diel = ω μɛ I P T β E T ds S E T ds = β S ωμ E T ds S The TM ase is a bit more involved. Using Eq. (9.3.) from Problem 9., we find, after using the result, β + k = ω μɛ, P diel = ωɛ I E ds = S ωɛ [ I Ez + E T ] ds S = [ ] ωɛ I E z + β S k 4 T E z ds = ωɛ I + β k E z ds S P T = E T ds = ωɛ β η TM S β S k 4 T E z ds = ωɛβ k E z ds S α d = P diel P T = ωɛ I + β k ωɛβ k 9.3 TEM, TE, and TM modes = ω μɛ I β The general solution desribed by Eqs. (9..6) and (9..9) is a hybrid solution with nonzero E z and H z omponents. Here, we look at the speialized forms of these equations in the ases of TEM, TE, and TM modes. One ommon property of all three types of modes is that the transverse fields E T, H T are related to eah other in the same way as in the ase of uniform plane waves propagating in the z-diretion, that is, they are perpendiular to eah other, their ross-produt points in the z-diretion, and they satisfy: H T = η T ẑ E T (9.3.) where η T is the transverse impedane of the partiular mode type, that is, η, η TE,η TM in the TEM, TE, and TM ases. Beause of Eq. (9.3.), the power flow per unit ross-setional area desribed by the Poynting vetor P z of Eq. (9..) takes the simple form in all three ases: TEM modes P z = Re(E T H T ) ẑ = η T E T = η T H T (9.3.) In TEM modes, both E z and H z vanish, and the fields are fully transverse. One an set E z = H z = in Maxwell equations (9..5), or equivalently in (9..6), or in (9..7). From any point view, one obtains the ondition k =, or ω = β. For example, if the right-hand sides of Eq. (9..7) vanish, the onsisteny of the system requires that η TE = η TM, whih by virtue of Eq. (9..3) implies ω = β. It also implies that η TE,η TM must both be equal to the medium impedane η. Thus, the eletri and magneti fields satisfy: H T = η ẑ E T (9.3.3) These are the same as in the ase of a uniform plane wave, exept here the fields are not uniform and may have a non-trivial x, y dependene. The eletri field E T is determined from the rest of Maxwell s equations (9..5), whih read: T E T = T E T = (9.3.4) These are reognized as the field equations of an equivalent two-dimensional eletrostati problem. One this eletrostati solution is found, E T (x, y), the magneti field is onstruted from Eq. (9.3.3). The time-varying propagating fields will be given by Eq. (9..), with ω = β. (For bakward moving fields, replae β by β.) We explore this eletrostati point of view further in Se.. and disuss the ases of the oaxial, two-wire, and strip lines. Beause of the relationship between E T and H T, the Poynting vetor P z of Eq. (9..) will be: P z = Re(E T H T ) ẑ = η E T = η H T (9.3.5)

9.3. TEM, TE, and TM modes 373 374 9. Waveguides TE modes TE modes are haraterized by the onditions E z = and H z. It follows from the seond of Eqs. (9..7) that E T is ompletely determined from H T, that is, E T = η TE H T ẑ. The field H T is determined from the seond of (9..6). Thus, all field omponents for TE modes are obtained from the equations: T H z + k H z = H T = jβ k T H z E T = η TE H T ẑ (TE modes) (9.3.6) The relationship of E T and H T is idential to that of uniform plane waves propagating in the z-diretion, exept the wave impedane is replaed by η TE. The Poynting vetor of Eq. (9..) then takes the form: P z = Re(E T H T ) ẑ = E T = η TE η TE H T = η β TE k 4 T H z (9.3.7) The artesian oordinate version of Eq. (9.3.6) is: T E z + k E z = E T = jβ k T E z H T = η TM ẑ E T (TM modes) (9.3.) Again, the relationship of E T and H T is idential to that of uniform plane waves propagating in the z-diretion, but the wave impedane is now η TM. The Poynting vetor takes the form: P z = Re(E T H T ) ẑ = E T = β η TM η TM k 4 T E z (9.3.) 9.4 Retangular Waveguides Next, we disuss in detail the ase of a retangular hollow waveguide with onduting walls, as shown in Fig. 9.4.. Without loss of generality, we may assume that the lengths a, b of the inner sides satisfy b a. The guide is typially filled with air, but any other dieletri material ɛ, μ may be assumed. ( x + y )H z + k H z = H x = jβ x H z, k E x = η TE H y, And, the ylindrial oordinate version: H y = jβ k y H z E y = η TE H x ( ρ H ) z + H z ρ ρ ρ ρ φ + k H z = H ρ = jβ H z k, H φ = jβ H z ρ k ρ φ E ρ = η TE H φ, E φ = η TE H ρ where we used H T ẑ = (ˆρH ρ + ˆφH φ ) ẑ = ˆφH ρ + ˆρH φ. TM modes (9.3.8) (9.3.9) Fig. 9.4. Retangular waveguide. The simplest and dominant propagation mode is the so-alled TE mode and depends only on the x-oordinate (of the longest side.) Therefore, we begin by looking for solutions of Eq. (9.3.8) that depend only on x. In this ase, the Helmholtz equation redues to: x H z(x)+k H z(x)= The most general solution is a linear ombination of os k x and sin k x. However, only the former will satisfy the boundary onditions. Therefore, the solution is: TM modes have H z = and E z. It follows from the first of Eqs. (9..7) that H T is ompletely determined from E T, that is, H T = η TMẑ E T. The field E T is determined from the first of (9..6), so that all field omponents for TM modes are obtained from the following equations, whih are dual to the TE equations (9.3.6): H z (x)= H os k x (9.4.) where H is a (omplex-valued) onstant. Beause there is no y-dependene, it follows from Eq. (9.3.8) that y H z =, and hene H y = and E x =. It also follows that: H x (x)= jβ k x H z = jβ k ( k )H sin k x = jβ H sin k x H sin k x k

9.4. Retangular Waveguides 375 376 9. Waveguides Then, the orresponding eletri field will be: where we defined the onstants: E y (x)= η TE H x (x)= η TE jβ k H sin k x E sin k x H = jβ k H E = η TE H = η TE jβ k H = jη ω ω H where we used η TE = ηω/β. In summary, the non-zero field omponents are: (9.4.) Fig. 9.4. 9.5 Higher TE and TM modes Eletri field inside a retangular waveguide. To onstrut higher modes, we looor solutions of the Helmholtz equation that are fatorable in their x and y dependene: H z (x)= H os k x H x (x)= H sin k x E y (x)= E sin k x H z (x, y, z, t)= H os k xe jωt jβz H x (x, y, z, t)= H sin k xe jωt jβz E y (x, y, z, t)= E sin k xe jωt jβz (9.4.3) Assuming perfetly onduting walls, the boundary onditions require that there be no tangential eletri field at any of the wall sides. Beause the eletri field is in the y-diretion, it is normal to the top and bottom sides. But, it is parallel to the left and right sides. On the left side, x =, E y (x) vanishes beause sin k x does. On the right side, x = a, the boundary ondition requires: E y (a)= E sin k a = sin k a = This requires that k a be an integral multiple of π: k a = nπ k = nπ (9.4.4) a These are the so-alled TE n modes. The orresponding utoff frequeny ω = k, f = ω /π, and wavelength λ = π/k = /f are: ω = nπ a, f = n a, λ = a (TE n modes) (9.4.5) n The dominant mode is the one with the lowest utoff frequeny or the longest utoff wavelength, that is, the mode TE having n =. It has: k = π a, ω = π a, f = a, λ = a (TE mode) (9.4.6) Fig. 9.4. depits the eletri field E y (x)= E sin k x = E sin(πx/a) of this mode as a funtion of x. Then, Eq. (9.3.8) beomes: H z (x, y)= F(x)G(y) F (x)g(y)+f(x)g (y)+k F(x)G(y)= F (x) F(x) + G (y) G(y) + k = (9.5.) Beause these must be valid for all x, y (inside the guide), the F- and G-terms must be onstants, independent of x and y. Thus, we write: F (x) F(x) = k x, G (y) G(y) = k y F (x)+k x F(x)=, G (y)+k yg(y)= (9.5.) where the onstants k x and k y are onstrained from Eq. (9.5.) to satisfy: k = k x + k y (9.5.3) The most general solutions of (9.5.) that will satisfy the TE boundary onditions are os k x x and os k y y. Thus, the longitudinal magneti field will be: H z (x, y)= H os k x x os k y y (TE nm modes) (9.5.4) It then follows from the rest of the equations (9.3.8) that: H x (x, y) = H sin k x x os k y y H y (x, y) = H os k x x sin k y y where we defined the onstants: H = jβk x k H, H = jβk y k H or E x (x, y) = E os k x x sin k y y E y (x, y) = E sin k x x os k y y (9.5.5) E = η TE H = jη ωk y ω k H, E = η TE H = jη ωk x ω k H

9.5. Higher TE and TM modes 377 The boundary onditions are that E y vanish on the right wall, x = a, and that E x vanish on the top wall, y = b, that is, E y (a, y)= E y sin k x a os k y y =, E x (x, b)= E x os k x x sin k y b = The onditions require that k x a and k y b be integral multiples of π: k x a = nπ, k y b = mπ k x = nπ a, k y = mπ (9.5.6) b These orrespond to the TE nm modes. Thus, the utoff wavenumbers of these modes k = k x + k y take on the quantized values: nπ mπ k = + (TE nm modes) (9.5.7) a b The utoff frequenies f nm = ω /π = k /π and wavelengths λ nm = /f nm are: f nm = n + a m, λ nm = b ( n a ) + (9.5.8) m b The TE m modes are similar to the TE n modes, but with x and a replaed by y and b. The family of TM modes an also be onstruted in a similar fashion from Eq. (9.3.). Assuming E z (x, y)= F(x)G(y), we obtain the same equations (9.5.). Beause E z is parallel to all walls, we must now hoose the solutions sin k x and sin k y y. Thus, the longitudinal eletri fields is: E z (x, y)= E sin k x x sin k y y (TM nm modes) (9.5.9) The rest of the field omponents an be worked out from Eq. (9.3.) and one finds that they are given by the same expressions as (9.5.5), exept now the onstants are determined in terms of E : E = jβk x k E, E = jβk y k H = η TM E = jωk y ω k η E, H = η TM E = jωk x ω k η H where we used η TM = ηβ/ω. The boundary onditions on E x,e y are the same as before, and in addition, we must require that E z vanish on all walls. These onditions imply that k x,k y will be given by Eq. (9.5.6), exept both n and m must be non-zero (otherwise E z would vanish identially.) Thus, the utoff frequenies and wavelengths are the same as in Eq. (9.5.8). Waveguide modes an be exited by inserting small probes at the beginning of the waveguide. The probes are hosen to generate an eletri field that resembles the field of the desired mode. E 378 9. Waveguides 9.6 Operating Bandwidth All waveguiding systems are operated in a frequeny range that ensures that only the lowest mode an propagate. If several modes an propagate simultaneously, one has no ontrol over whih modes will atually be arrying the transmitted signal. This may ause undue amounts of dispersion, distortion, and errati operation. A mode with utoff frequeny ω will propagate only if its frequeny is ω ω, or λ<λ.ifω<ω, the wave will attenuate exponentially along the guide diretion. This follows from the ω, β relationship (9..): ω = ω + β β = ω ω If ω ω, the wavenumber β is real-valued and the wave will propagate. But if ω<ω, β beomes imaginary, say, β = jα, and the wave will attenuate in the z- diretion, with a penetration depth δ = /α: e jβz = e αz If the frequeny ω is greater than the utoff frequenies of several modes, then all of these modes an propagate. Conversely, if ω is less than all utoff frequenies, then none of the modes an propagate. If we arrange the utoff frequenies in inreasing order, ω <ω <ω 3 <, then, to ensure single-mode operation, the frequeny must be restrited to the interval ω <ω<ω, so that only the lowest mode will propagate. This interval defines the operating bandwidth of the guide. These remarks apply to all waveguiding systems, not just hollow onduting waveguides. For example, in oaxial ables the lowest mode is the TEM mode having no utoff frequeny, ω =. However, TE and TM modes with non-zero utoff frequenies do exist and plae an upper limit on the usable bandwidth of the TEM mode. Similarly, in optial fibers, the lowest mode has no utoff, and the single-mode bandwidth is determined by the next utoff frequeny. In retangular waveguides, the smallest utoff frequenies are f = /a, f = /a = f, and f = /b. Beause we assumed that b a, it follows that always f f.ifb a/, then /a /b and therefore, f f, so that the two lowest utoff frequenies are f and f. On the other hand, if a/ b a, then f f and the two smallest frequenies are f and f (exept when b = a, in whih ase f = f and the smallest frequenies are f and f.) The two ases b a/ and b a/ are depited in Fig. 9.6.. It is evident from this figure that in order to ahieve the widest possible usable bandwidth for the TE mode, the guide dimensions must satisfy b a/ so that the bandwidth is the interval [f, f ], where f = f = /a. In terms of the wavelength λ = /f, the operating bandwidth beomes:.5 a/λ, or, a λ a. We will see later that the total amount of transmitted power in this mode is proportional to the ross-setional area of the guide, ab. Thus, if in addition to having the Murphy s law for waveguides states that if a mode an propagate, it will.

9.7. Power Transfer, Energy Density, and Group Veloity 379 38 9. Waveguides Fig. 9.6. Operating bandwidth in retangular waveguides. where H = jβ H, E = η TE H = jη ω H (9.7.) k ω The Poynting vetor is obtained from the general result of Eq. (9.3.7): P z = η TE E T = η TE E y (x) = η TE E sin k x The transmitted power is obtained by integrating P z over the ross-setional area of the guide: widest bandwidth, we also require to have the maximum power transmitted, the dimension b must be hosen to be as large as possible, that is, b = a/. Most pratial guides follow these side proportions. If there is a anonial guide, it will have b = a/ and be operated at a frequeny that lies in the middle of the operating band [f, f ], that is, f =.5f =.75 (9.6.) a Table 9.6. lists some standard air-filled retangular waveguides with their naming designations, inner side dimensions a, b in inhes, utoff frequenies in GHz, minimum and maximum reommended operating frequenies in GHz, power ratings, and attenuations in db/m (the power ratings and attenuations are representative over eah operating band.) We have hosen one example from eah mirowave band. name a b f f min f max band P α WR-5 5..55.6.45. L 9 MW.7 WR-84.84.34.8.6 3.95 S.7 MW.9 WR-59.59.795 3.7 4.64 7.5 C.9 MW.43 WR-9.9.4 6.56 8..5 X 5 kw. WR-6.6.3 9.49.9 8. Ku 4 kw.76 WR-4.4.7 4.5 7.6 6.7 K 5 kw.37 WR-8.8.4.8 6.4 4. Ka 7 kw.583 WR-5.48.74 39.87 49.8 75.8 V 7.5 kw.5 WR-..5 59. 73.8. W 3.5 kw.74 Table 9.6. Charateristis of some standard air-filled retangular waveguides. Noting the definite integral, P T = a b η TE E sin k x dxdy a a sin k xdx= sin ( πx ) a dx = a and using η TE = ηω/β = η/ ω /ω, we obtain: (9.7.3) P T = E ab = 4η TE 4η E ab ω ω (transmitted power) (9.7.4) We may also alulate the distribution of eletromagneti energy along the guide, as measured by the time-averaged energy density. The energy densities of the eletri and magneti fields are: w e = Re( ɛe E ) = 4 ɛ E y w m = Re( μh H ) = 4 μ( H x + H z ) Inserting the expressions for the fields, we find: w e = 4 ɛ E sin k x, w m = 4 μ( H sin k x + H os k x ) 9.7 Power Transfer, Energy Density, and Group Veloity Next, we alulate the time-averaged power transmitted in the TE mode. We also alulate the energy density of the fields and determine the veloity by whih eletromagneti energy flows down the guide and show that it is equal to the group veloity. We reall that the non-zero field omponents are: H z (x)= H os k x, H x (x)= H sin k x, E y (x)= E sin k x (9.7.) Beause these quantities represent the energy per unit volume, if we integrate them over the ross-setional area of the guide, we will obtain the energy distributions per unit z-length. Using the integral (9.7.3) and an idential one for the osine ase, we find: W e = a W m = a b b w e (x, y) dxdy = a b 4 ɛ E sin k x dxdy = 8 ɛ E ab 4 μ( H sin k x + H os k x ) dxdy = 8 μ( H + H ) ab

9.8. Power Attenuation 38 38 9. Waveguides Although these expressions look different, they are atually equal, W e = W m. Indeed, using the property β /k + = (β + k )/k = k /k = ω /ω and the relationships between the onstants in (9.7.), we find: μ ( H + H ) = μ ( H β k + H ) = μ H ω ω = μ η E = ɛ E The equality of the eletri and magneti energies is a general property of waveguiding systems. We also enountered it in Se..3 for uniform plane waves. The total energy density per unit length will be: The field expressions (9.4.3) were derived assuming the boundary onditions for perfetly onduting wall surfaes. The indued surfae urrents on the inner walls of the waveguide are given by J s = ˆn H, where the unit vetor ˆn is ±ˆx and ±ŷ on the left/right and bottom/top walls, respetively. The surfae urrents and tangential magneti fields are shown in Fig. 9.8.. In partiular, on the bottom and top walls, we have: W = W e + W m = W e = 4 ɛ E ab (9.7.5) Aording to the general relationship between flux, density, and transport veloity given in Eq. (.6.), the energy transport veloity will be the ratio v en = P T /W. Using Eqs. (9.7.4) and (9.7.5) and noting that /ηɛ = / μɛ =, we find: v en = P T W = ω ω (energy transport veloity) (9.7.6) This is equal to the group veloity of the propagating mode. For any dispersion relationship between ω and β, the group and phase veloities are defined by v gr = dω dβ, v ph = ω β (group and phase veloities) (9.7.7) For uniform plane waves and TEM transmission lines, we have ω = β, so that v gr = v ph =. For a retangular waveguide, we have ω = ω + β. Taking differentials of both sides, we find ωdω = βdβ, whih gives: v gr = dω dβ = β ω = ω (9.7.8) ω where we used Eq. (9..). Thus, the energy transport veloity is equal to the group veloity, v en = v gr. We note that v gr = β /ω = /v ph,or v gr v ph = (9.7.9) The energy or group veloity satisfies v gr, whereas v ph. Information transmission down the guide is by the group veloity and, onsistent with the theory of relativity, it is less than. 9.8 Power Attenuation In this setion, we alulate the attenuation oeffiient due to the ohmi losses of the onduting walls following the proedure outlined in Se. 9.. The losses due to the filling dieletri an be determined from Eq. (9..5). Fig. 9.8. Currents on waveguide walls. J s =±ŷ H =±ŷ (ˆx H x + ẑh z )=±( ẑ H x + ˆx H z )=±( ẑ H sin k x + ˆx H os k x) Similarly, on the left and right walls: J s =±ˆx H =±ˆx (ˆx H x + ẑh z )= ŷ H z = ŷ H os k x At x = and x = a, this gives J s = ŷ(±h )= ŷ H. Thus, the magnitudes of the surfae urrents are on the four walls: { J s H, (left and right walls) = H os k x + H sin k x, (top and bottom walls) The power loss per unit z-length is obtained from Eq. (9..8) by integrating J s around the four walls, that is, P loss = a R s J s dx + b R s J s dy a ( = R s H os k x + H sin k x ) b dx + R s H dy a ( = R s H + H ) + R s b H = R sa ( H + H + b a H ) Using H + H = E /η from Se. 9.7, and H = ( E /η )ω/ω, whih follows from Eq. (9.4.), we obtain: ( P loss = R sa E η + b a ) ω ω The attenuation onstant is omputed from Eqs. (9..9) and (9.7.4):

9.8. Power Attenuation 383 384 9. Waveguides whih gives: α = R s ηb α = P loss P T = ( + b a ) ω ω ω ω ( R s a E + b η a 4η E ab ) ω ω ω ω (attenuation of TE mode) (9.8.) This is in units of nepers/m. Its value in db/m is obtained by α db = 8.686α. For a given ratio a/b, α inreases with dereasing b, thus the smaller the guide dimensions, the larger the attenuation. This trend is noted in Table 9.6.. The main tradeoffs in a waveguiding system are that as the operating frequeny f inreases, the dimensions of the guide must derease in order to maintain the operating band f f f, but then the attenuation inreases and the transmitted power dereases as it is proportional to the guide s area. Example 9.8.: Design a retangular air-filled waveguide to be operated at 5 GHz, then, redesign it to be operated at GHz. The operating frequeny must lie in the middle of the operating band. Calulate the guide dimensions, the attenuation onstant in db/m, and the maximum transmitted power assuming the maximum eletri field is one-half of the dieletri strength of air. Assume opper walls with ondutivity σ = 5.8 7 S/m. Solution: If f is in the middle of the operating band, f f f, where f = /a, then f =.5f =.75/a. Solving for a, wefind a =.75 f.75 3 GHz m = = 4.5 m 5 For maximum power transfer, we require b = a/ =.5 m. Beause ω =.5ω,we have ω /ω = /3. Then, Eq. (9.8.) gives α =.37 db/m. The dieletri strength of air is 3 MV/m. Thus, the maximum allowed eletri field in the guide is E =.5 MV/m. Then, Eq. (9.7.4) gives P T =. MW. At GHz, beause f is doubled, the guide dimensions are halved, a =.5 and b =.5 m. Beause R s depends on f like f /, it will inrease by a fator of. Then, the fator R s /b will inrease by a fator of. Thus, the attenuation will inrease to the value α =.37 =.4 db/m. Beause the area ab is redued by a fator of four, so will the power, P T =./4 =.8 MW = 8 kw. The results of these two ases are onsistent with the values quoted in Table 9.6. for the C-band and X-band waveguides, WR-59 and WR-9. Example 9.8.: WR-59 Waveguide. Consider the C-band WR-59 air-filled waveguide whose harateristis were listed in Table 9.6.. Its inner dimensions are a =.59 and b = a/ =.795 inhes, or, equivalently, a = 4.386 and b =.93 m. α (db/m)..8.6.4. The utoff frequeny of the TE mode is f = /a = 3.7 GHz. The maximum operating bandwidth is the interval [f, f ]= [3.7, 7.4] GHz, and the reommended interval is [4.64, 7.5] GHz. Assuming opper walls with ondutivity σ = 5.8 7 S/m, the alulated attenuation onstant α from Eq. (9.8.) is plotted in db/m versus frequeny in Fig. 9.8.. Attenuation Coeffiient bandwidth 3 4 5 6 7 8 9 f (GHz) Fig. 9.8. P T (MW).5.5 Power Transmitted bandwidth 3 4 5 6 7 8 9 f (GHz) Attenuation onstant and transmitted power in a WR-59 waveguide. The power transmitted P T is alulated from Eq. (9.7.4) assuming a maximum breakdown voltage of E =.5 MV/m, whih gives a safety fator of two over the dieletri breakdown of air of 3 MV/m. The power in megawatt sales is plotted in Fig. 9.8.. Beause of the fator ω /ω in the denominator of α and the numerator of P T, the attenuation onstant beomes very large near the utoff frequeny, while the power is almost zero. A physial explanation of this behavior is given in the next setion. 9.9 Refletion Model of Waveguide Propagation An intuitive model for the TE mode an be derived by onsidering a TE-polarized uniform plane wave propagating in the z-diretion by obliquely bouning bak and forth between the left and right walls of the waveguide, as shown in Fig. 9.9.. If θ is the angle of inidene, then the inident and refleted (from the right wall) wavevetors will be: k = ˆx k x + ẑ k z = ˆx k os θ + ẑ k sin θ k = ˆx k x + ẑ k z = ˆx k os θ + ẑ k sin θ The eletri and magneti fields will be the sum of an inident and a refleted omponent of the form: E = ŷ E e jk r + ŷ E e jk r = ŷ E e jkxx e jkzz + ŷ E e jkxx e jkzz = E + E H = η ˆk E + η ˆk E

9.9. Refletion Model of Waveguide Propagation 385 386 9. Waveguides The boundary ondition on the right wall requires sin k x a =, whih gives rise to the same ondition as (9.4.4), that is, k a = nπ. This model larifies also the meaning of the group veloity. The plane wave is bouning left and right with the speed of light. However, the omponent of this veloity in the z-diretion will be v z = sin θ. This is equal to the group veloity. Indeed, it follows from Eq. (9.9.3) that: v z = sin θ = ω ω = v gr (9.9.5) Fig. 9.9. Refletion model of TE mode. where the eletri field was taken to be polarized in the y diretion. These field expressions beome omponent-wise: E y = ( E e jkxx + E e jkxx) e jkzz H x = η sin θ( E e jkxx + E e jkxx) e jkzz H z = η os θ( E e jkxx E e jkxx) e jkzz (9.9.) The boundary ondition on the left wall, x =, requires that E + E =. We may write therefore, E = E = je /. Then, the above expressions simplify into: E y = E sin k x xe jkzz H x = η sin θe sin k x xe jkzz H z = j η os θe os k x xe jkzz (9.9.) These are idential to Eq. (9.4.3) provided we identify β with k z and k with k x,as shown in Fig. 9.9.. It follows from the wavevetor triangle in the figure that the angle of inidene θ will be given by os θ = k x /k = k /k, or, Eq. (9.9.3) implies also that at ω = ω, we have sin θ =, or θ =, that is, the wave is bouning left and right at normal inidene, reating a standing wave, and does not propagate towards the z-diretion. Thus, the transmitted power is zero and this also implies, through Eq. (9..9), that α will be infinite. On the other hand, for very large frequenies, ω ω, the angle θ will tend to 9 o, ausing the wave to zoom through guide almost at the speed of light. The phase veloity an also be understood geometrially. Indeed, we observe in the rightmost illustration of the above figure that the planes of onstant phase are moving obliquely with the speed of light. From the indiated triangle at points,,3, we see that the effetive speed in the z-diretion of the ommon-phase points will be v ph = / sin θ so that v ph v gr =. Higher TE and TM modes an also be given similar geometri interpretations in terms of plane waves propagating by bouning off the waveguide walls [886]. 9. Resonant Cavities Cavity resonators are metalli enlosures that an trap eletromagneti fields. The boundary onditions on the avity walls fore the fields to exist only at ertain quantized resonant frequenies. For highly onduting walls, the resonanes are extremely sharp, having a very high Q of the order of,. Beause of their high Q, avities an be used not only to effiiently store eletromagneti energy at mirowave frequenies, but also to at as preise osillators and to perform preise frequeny measurements. Fig. 9.. shows a retangular avity with z-length equal to l formed by replaing the sending and reeiving ends of a waveguide by metalli walls. A forward-moving wave will boune bak and forth from these walls, resulting in a standing-wave pattern along the z-diretion. os θ = ω ω, sin θ = ω ω (9.9.3) The ratio of the transverse omponents, E y /H x, is the transverse impedane, whih is reognized to be η TE. Indeed, we have: η TE = E y = η H x sin θ = η ω ω (9.9.4) Fig. 9.. Retangular avity resonator (and indued wall urrents for the TE np mode.)

9.. Resonant Cavities 387 388 9. Waveguides Beause the tangential omponents of the eletri field must vanish at the end-walls, these walls must oinide with zero rossings of the standing wave, or put differently, an integral multiple of half-wavelengths must fit along the z-diretion, that is, l = pλ g / = pπ/β, orβ = pπ/l, where p is a non-zero integer. For the same reason, the standingwave patterns along the transverse diretions require a = nλ x / and b = mλ y /, or k x = nπ/a and k y = mπ/b. Thus, all three artesian omponents of the wave vetor are quantized, and therefore, so is the frequeny of the wave ω = kx + k y + β : nπ mπ pπ ω nmp = + + (resonant frequenies) (9..) a b l Suh modes are designated as TE nmp or TM nmp. For simpliity, we onsider the ase TE np. Eqs. (9.3.6) also desribe bakward-moving waves if one replaes β by β, whih also hanges the sign of η TE = ηω/β. Starting with a linear ombination of forward and bakward waves in the TE n mode, we obtain the field omponents: H z (x, z) = H os k x ( Ae jβz + Be jβz), H x (x, z) = jh sin k x ( Ae jβz Be jβz), H = β k H E y (x, z) = je sin k x ( Ae jβz + Be jβz), E = ω ω ηh (9..) where ω = k. By requiring that E y (x, z) have z-dependene of the form sin βz, the oeffiients A, B must be hosen as A = B = j/. Then, Eq. (9..) speializes into: H z (x, z) = H os k x sin βz, where we used the following definite integrals (valid beause k = nπ/a, β = pπ/l): a a sin k xdx= os k xdx= a l l, sin βz dz = os βz dz = l (9..5) The ohmi losses are alulated from Eq. (9..6), integrated over all six avity sides. The surfae urrents indued on the walls are related to the tangential magneti fields by J s = ˆn H tan. The diretions of these urrents are shown in Fig. 9... Speifially, we find for the urrents on the six sides: H sin βz (left & right) J s = H os k x sin βz + H sin k x os βz (top & bottom) H sin k x (front & bak) The power loss an be omputed by integrating the loss per unit ondutor area, Eq. (9..6), over the six wall sides, or doubling the answer for the left, top, and front sides. Using the integrals (9..5), we find: P loss = R s J s da = R s [H bl walls + (H + H) al ] 4 + ab H [ ] = (9..6) 4 R sh l(b + a)+ β k a(b + l) where we substituted H = Hβ /k. It follows that the Q-fator will be: Q = ω W = ωμ (k + β )(abl) P loss R s k l(b + a)+β a(b + l) H x (x, z) = H sin k x os βz, H = β k H (9..3) For the TE np mode we have β = pπ/l and k = nπ/a. Using Eq. (9..7) to replae R s in terms of the skin depth δ, we find: E y (x, z) = je sin k x sin βz, E = ω ω ηh As expeted, the vanishing of E y (x, z) on the front/bak walls, z = and z = l, and on the left/right walls, x = and x = a, requires the quantization onditions: β = pπ/l and k = nπ/a. The Q of the resonator an be alulated from its definition: Q = ω W (9..4) P loss where W is the total time-averaged energy stored within the avity volume and P loss is the total power loss due to the wall ohmi losses (plus other losses, suh as dieletri losses, if present.) The ratio Δω = P loss /W is usually identified as the 3-dB width of the resonane entered at frequeny ω. Therefore, we may write Q = ω/δω. It is easily verified that the eletri and magneti energies are equal, therefore, W may be alulated by integrating the eletri energy density over the avity volume: W = W e = ɛ E y (x, z) dx dy dz = a b l 4 vol ɛ E sin k x os βz dx dy dz [ ] = 8 ɛ E (abl)= 8 μ H ω ω (abl)= k 8 μ H + β k (abl) n Q = a + p l δ n ( a a + ) + p b l ( l + ) (9..7) b The lowest resonant frequeny orresponds to n = p =. For a ubi avity, a = b = l, the Q and the lowest resonant frequeny are: Q = a 3δ, ω = π, f = ω a π = a (9..8) For an air-filled ubi avity with a = 3 m, we find f = 7.7 GHz, δ = 7.86 5 m, and Q = 74. As in waveguides, avities an be exited by inserting small probes that generate fields resembling a partiular mode. 9. Dieletri Slab Waveguides A dieletri slab waveguide is a planar dieletri sheet or thin film of some thikness, say a, as shown in Fig. 9... Wave propagation in the z-diretion is by total internal

9.. Dieletri Slab Waveguides 389 39 9. Waveguides x, and exist effetively within a skin depth distane /α from the slab. Setting k = k and k = jα, Eqs. (9..) beome in this new notation: k = k n β α = k n β k = k n β α = β k n (9..3) Similarly, Eqs. (9..) read: xh z (x)+k H z (x)= for x a x H z(x) α H z(x)= for x a (9..4) Fig. 9.. Dieletri slab waveguide. refletion from the left and right walls of the slab. Suh waveguides provide simple models for the onfining mehanism of waves propagating in optial fibers. The propagating fields are onfined primarily inside the slab, however, they also exist as evanesent waves outside it, deaying exponentially with distane from the slab. Fig. 9.. shows a typial eletri field pattern as a funtion of x. For simpliity, we assume that the media to the left and right of the slab are the same. To guarantee total internal refletion, the dieletri onstants inside and outside the slab must satisfy ɛ >ɛ, and similarly for the refrative indies, n >n. We only onsider TE modes and looor solutions that depend only on the x oordinate. The utoff wavenumber k appearing in the Helmholtz equation for H z (x) depends on the dieletri onstant of the propagation medium, k = ω ɛμ β. Therefore, k takes different values inside and outside the guide: k = ω ɛ μ β = ω ɛ μ n β = k n β k = ω ɛ μ β = ω ɛ μ n β = k n β (inside) (outside) (9..) where k = ω/ is the free-spae wavenumber. We note that ω, β are the same inside and outside the guide. This follows from mathing the tangential fields at all times t and all points z along the slab walls. The orresponding Helmholtz equations in the regions inside and outside the guide are: x H z(x)+k H z (x)= for x a xh z (x)+k H z (x)= for x a (9..) Inside the slab, the solutions are sin k x and os k x, and outside, sin k x and os k x, or equivalently, e ±jkx. In order for the waves to remain onfined in the near viinity of the slab, the quantity k must be imaginary, for if it is real, the fields would propagate at large x distanes from the slab (they would orrespond to the rays refrated from the inside into the outside.) If we set k = jα, the solutions outside will be e ±αx.ifα is positive, then only the solution e αx is physially aeptable to the right of the slab, x a, and only e αx to the left, x a. Thus, the fields attenuate exponentially with the transverse distane The two solutions sin k x and os k x inside the guide give rise to the so-alled even and odd TE modes (referring to the evenness or oddness of the resulting eletri field.) For the even modes, the solutions of Eqs. (9..4) have the form: H sin k x, if a x a H z (x)= H e αx, if x a H 3 e αx, if x a (9..5) The orresponding x-omponents H x are obtained by applying Eq. (9.3.8) using the appropriate value for k, that is, k = α outside and k = k inside: H x (x)= jβ k x H z (x)= jβ H os k x, if a x a k jβ α x H z (x)= jβ H e αx, if x a α (9..6) jβ α x H z (x)= jβ H 3 e αx, if x a α The eletri fields are E y (x)= η TE H x (x), where η TE = ωμ /β is the same inside and outside the slab. Thus, the eletri field has the form: E os k x, if a x a E y (x)= E e αx, if x a (even TE modes) (9..7) E 3 e αx, if x a where we defined the onstants: E = jβ η TE H, E = jβ η TE H, E 3 = jβ η TE H 3 (9..8) k α α The boundary onditions state that the tangential omponents of the magneti and eletri fields, that is, H z,e y, are ontinuous aross the dieletri interfaes at x = a and x = a. Similarly, the normal omponents of the magneti field B x = μ H x and therefore also H x must be ontinuous. Beause E y = η TE H x and η TE is the same in both media, the ontinuity of E y follows from the ontinuity of H x. The ontinuity of H z at x = a and x = a implies that:

9.. Dieletri Slab Waveguides 39 39 9. Waveguides H sin k a = H e αa and H sin k a = H 3 e αa (9..9) Similarly, the ontinuity of H x implies (after aneling a fator of jβ): We note that the eletri fields E y (x) given by Eqs. (9..7) and (9..5) are even or odd funtions of x for the two families of modes. Expressing E and E 3 in terms of E, we summarize the forms of the eletri fields in the two ases: H os k a = H e αa and k α Eqs. (9..9) and (9..) imply: k H os k a = α H 3 e αa (9..) E os k x, if a x a E y (x)= E os k ae α(x a), if x a E os k ae α(x+a), if x a (even TE modes) (9..9) H = H 3 = H e αa sin k a = H e αa α k os k a (9..) Similarly, we find for the eletri field onstants: E sin k x, if a x a E y (x)= E sin k ae α(x a), if x a E sin k ae α(x+a), if x a (odd TE modes) (9..) E = E 3 = E e αa os k a = E e αa k α sin k a (9..) The onsisteny of the last equations in (9..) or (9..) requires that: os k a = k sin k a α = k tan k a (9..3) α For the odd TE modes, we have for the solutions of Eq. (9..4): The resulting eletri field is: H os k x, if a x a H z (x)= H e αx, if x a H 3 e αx, if x a E sin k x, if a x a E y (x)= E e αx, if x a E 3 e αx, if x a The boundary onditions imply in this ase: and, for the eletri field onstants: (9..4) (odd TE modes) (9..5) H = H 3 = H e αa os k a = H e αa α k sin k a (9..6) E = E 3 = E e αa sin k a = E e αa k α os k a (9..7) The onsisteny of the last equation requires: α = k ot k a (9..8) Given the operating frequeny ω, Eqs. (9..3) and (9..3) or (9..8) provide three equations in the three unknowns k,α,β. To solve them, we add the two equations (9..3) to eliminate β: α + k = k (n n )= ω (n n ) (9..) Next, we disuss the numerial solutions of these equations. Defining the dimensionless quantities u = k a and v = α a, we may rewrite Eqs. (9..3), (9..8), and (9..) in the equivalent forms: v = u tan u v + u = R (even modes), where R is the normalized frequeny variable: R = k an A = ωa v = u ot u v + u = R (odd modes) (9..) N A = πfa N A = πa λ N A (9..3) where N A = n n is the numerial aperture of the slab and λ = /f, the free-spae wavelength. Beause the funtions tan u and ot u have many branhes, there may be several possible solution pairs u, v for eah value of R. These solutions are obtained at the intersetions of the urves v = u tan u and v = u ot u with the irle of radius R, that is, v + u = R. Fig. 9.. shows the solutions for various values of the radius R orresponding to various values of ω. It is evident from the figure that for small enough R, that is, R<π/, there is only one solution and it is even. For π/ R<π, there are two solutions, one even and one odd. For π R<3π/, there are three solutions, two even and one odd, and for an optial fiber, the single-mode ondition reads πan A /λ <.45, where a is the ore radius.

9.. Dieletri Slab Waveguides 393 394 9. Waveguides tan(u mπ/)= sin u os(mπ/) os u sin(mπ/) os u os(mπ/)+ sin u sin(mπ/) Therefore, to find the mth mode, whether even or odd, we must find the unique solution of the following system in the u-range R m u<r m+ : Fig. 9.. Even and odd TE modes at different frequenies. v = u tan(u R m ) v + u = R (mth mode) (9..9) If one had an approximate solution u, v for the mth mode, one ould refine it by using Newton s method, whih onverges very fast provided it is lose to the true solution. Just suh an approximate solution, aurate to within one perent of the true solution, was given by Lotspeih [96]. Without going into the detailed justifiation of this method, the approximation is as follows: so on. In general, there will be M + solutions, alternating between even and odd, if R falls in the interval: Mπ (M + )π R< (9..4) Given a value of R, we determine M as that integer satisfying Eq. (9..4), or, M R/π<M+, that is, the largest integer less than R/π: R M = floor (maximum mode number) (9..5) π Then, there will be M+ solutions indexed by m =,,...,M, whih will orrespond to even modes if m is even and to odd modes if m is odd. The M + branhes of tan u and ot u being interseted by the R-irle are those ontained in the u-ranges: where R m u<r m+, m =,,...,M (9..6) R m = mπ, m =,,..., M (9..7) If m is even, the u-range (9..6) defines a branh of tan u, and if m is odd, a branh of ot u. We an ombine the even and odd ases of Eq. (9..) into a single ase by noting the identity: tan u, if m is even tan(u R m )= (9..8) ot u, if m is odd This follows from the trigonometri identity: u = R m + w (m)u (m)+w (m)u (m), m =,,...,M (9..3) where u (m), u (m) are approximate solutions near and far from the utoff R m, and w (m), w (m) are weighting fators: + R(R Rm ) u (m)=, u (m)= π R R m R + w (m)= exp ( (R R m ) /Vm), w (m)= w (m) π/4 + Rm V m = ln.5 os(π/4) R m (9..3) This solution serves as the starting point to Newton s iteration for solving the equation F(u)=, where F(u) is defined by Newton s iteration is: F(u)= u tan(u R m ) v = u tan(u R m ) R u (9..3) for i =,...,N it do: u = u F(u) G(u) where G(u) is the derivative F (u), orret to order O(F): (9..33) G(u)= v u + u v + R (9..34) u The solution steps defined in Eqs. (9..9) (9..34) have been implemented in the MATLAB funtion dslab.m, with usage: [u,v,err] = dslab(r,nit); % TE-mode utoff wavenumbers in a dieletri slab

9.. Dieletri Slab Waveguides 395 396 9. Waveguides where N it is the desired number of Newton iterations (9..33), err is the value of F(u) at the end of the iterations, and u, v are the (M + )-dimensional vetors of solutions. The number of iterations is typially very small, N it = 3. The related MATLAB funtion dguide.m uses dslab to alulate the solution parameters β, k,α, given the frequeny f, the half-length a, and the refrative indies n,n of the slab. It has usage: [be,k,a,f,err] = dguide(f,a,n,n,nit); % dieletri slab guide where f is in GHz, a in m, and β, k,α in m. The quantity f is the vetor of the M + utoff frequenies defined by the branh edges R m = mπ/, that is, R m = ω m an A / = πf m an A / = mπ/, or, f m = m, m =,,...,M (9..35) 4aN A The meaning of f m is that there are m + propagating modes for eah f in the interval f m f<f m+. Example 9..: Dieletri Slab Waveguide. Determine the propagating TE modes of a dieletri slab of half-length a =.5 m at frequeny f = 3 GHz. The refrative indies of the slab and the surrounding dieletri are n = and n =. Solution: The solution is obtained by the MATLAB all: f = 3; a =.5; n = ; n = ; Nit = 3; [be,k,a,f,err] = dguide(f,a,n,n,nit) The frequeny radius is R = 5.444, whih gives R/π = 3.464, and therefore, M = 3. The resulting solutions, depited in Fig. 9..3, are as follows: v 7 6 5 4 3 TE Modes for R = 5.44 3 4 5 6 7 u Fig. 9..3 3 E y (x)/e Eletri Fields 3 3 x/a TE modes and orresponding E-field patterns. m u v β k α f m.348 5.777.838.6497.5553..6359 4.763.47 5.78 9.57 8.663 3.95 3.7837 9.8359 7.8 7.5675 7.35 3 5.793.959 7.397.585 3.937 5.988 3 The utoff frequenies f m are in GHz. We note that as the mode number m inreases, the quantity α dereases and the effetive skin depth /α inreases, ausing the fields outside the slab to be less onfined. The eletri field patterns are also shown in the figure as funtions of x. The approximation error, err, is found to be 4.885 5 using only three Newton iterations. Using two, one, and no (the Lotspeih approximation) iterations would result in the errors.38 8,4.9 4, and.58. The lowest non-zero utoff frequeny is f = 8.663 GHz, implying that there will be a single solution if f is in the interval f<f. For example, if f = 5 GHz, the solution is β =.5649 rad/m, k =.39 rad/m, and α =.69 nepers/m. The frequeny range over whih there are only four solutions is [5.988, 34.64] GHz, where the upper limit is 4f. We note that the funtion dguide assumes internally that = 3 GHz m, and therefore, the alulated values for k,α would be slightly different if a more preise value of is used, suh as 9.979458 of Appendix A. Problem 9.3 studies the sensitivity of the solutions to small hanges of the parameters f, a,,n,n. In terms of the ray piture of the propagating wave, the angles of total internal refletion are quantized aording to the values of the propagation wavenumber β for the various modes. If we denote by k = k n the wavenumber within the slab, then the wavenumbers β, k are the z- and x-omponents k z,k x of k with an angle of inidene θ. (The vetorial relationships are the same as those in Fig. 9.9..) Thus, we have: β = k sin θ = k n sin θ (9..36) k = k os θ = k n os θ The value of β for eah mode will generate a orresponding value for θ. The attenuation wavenumber α outside the slab an also be expressed in terms of the total internal refletion angles: α = β k n = k n sin θ n Sine the ritial angle is sin θ = n /n, we may also express α as: α = k n sin θ sin θ (9..37) Example 9..: For the Example 9.., we alulate k = 6.83 and k =.5664 rad/m. The ritial and total internal refletion angles of the four modes are found to be: θ = asin ( n n ) = 3 o β θ = asin k ={77.875 o, 65.96 o, 5.5 o, 36.69 o } As required, all θs are greater than θ.

9.. Asymmetri Dieletri Slab 397 9. Asymmetri Dieletri Slab The three-layer asymmetri dieletri slab waveguide shown in Fig. 9.. is a typial omponent in integrated optis appliations [98 99]. A thin dieletri film n f of thikness a is deposited on a dieletri substrate n s. Above the film is a dieletri over or ladding n, suh as air. To ahieve propagation by total internal refletion within the film, we assume that the refrative indies satisfy n f > n s n. The ase of the symmetri dieletri slab of the previous setion is obtained when n = n s. 398 9. Waveguides whih ombine to define the allowed range of β for the guided modes: n n s β k n f (9..3) where the lower limit β = k n s defines the utoff frequenies, see Eq. (9..3). TE modes We onsider the TE modes first. Assuming only x-dependene for the H z omponent, it must satisfy the Helmholtz equations in the three regions: Fig. 9.. Three-layer asymmetri dieletri slab waveguide. ( x + )H z(x)=, x a ( x α )H z(x)=, x a ( x α s )H z(x)=, x a The solutions, deaying exponentially in the substrate and over, an be written in the following form, whih automatially satisfies the ontinuity onditions at the two boundaries x =±a: H sin( x + φ), x a H z (x)= H sin( a + φ)e α(x a), x a (9..4) H sin( a φ)e αs(x+a), x a In this setion, we briefly disuss the properties of the TE and TM propagation modes. Let k = ω μ = ω/ = πf/ = π/λ be the free-spae wavenumber at the operating frequeny ω or f in Hz. The t, z dependene of the fields is assumed to be the usual e jωt jβt. If we orient the oordinate axes as shown in the above figure, then the deay onstants α s and α within the substrate and ladding must be positive so that the fields attenuate exponentially with x within both the substrate and ladding, hene, the orresponding transverse wavenumbers will be jα s and jα. On the other hand, the transverse wavenumber within the film will be real-valued. These quantities satisfy the relations (we assume μ = μ in all three media): k f = k n f β k f + α s = k (n f n s) where φ is a parameter to be determined. The remaining two omponents, H x and E y, are obtained by applying Eq. (9.3.8), that is, This gives in the three regions: H x = jβ k x H z, E y = η TE H x η TE = ωμ f β j β H os( x + φ), H x (x)= j β H sin( a + φ)e α(x a), α x a x a (9..5) α s = β k n s k f + α = k (n f n s)( + δ)= k (n f n ) (9..) j β H sin( a φ)e αs(x+a), α s x a α = β k n where we defined the asymmetry parameter δ: α α s = k (n f n s )δ = k (n s n ) Sine we assumed that μ = μ in all three regions, the ontinuity of E y aross the boundaries x =±a implies the same for the H x omponents, resulting in the two onditions: δ = n s n n f n s (9..) Note that δ sine we assumed n f >n s n. Beause,α s,α are assumed to be real, it follows that β must satisfy the inequalities, β k n f, β k n s, and β k n, os( a + φ) = α sin( a + φ) os( a φ) = α s sin( a φ) tan( a + φ) = α tan( a φ) = α s (9..6)

9.. Asymmetri Dieletri Slab 399 4 9. Waveguides Sine the argument of the tangent is unique up to an integral multiple of π, we may invert the two tangents as follows without loss of generality: α a + φ = artan + mπ αs a φ = artan whih result in the harateristi equation of the slab and the solution for φ: a = mπ + αs artan + α artan (9..7) φ = mπ + α artan αs artan (9..8) where the integer m =,,,..., orresponds to the mth mode. Eq. (9..7) and the three equations (9..) provide four equations in the four unknowns {β,,α s,α }. Using the trig identities tan(θ +θ )= (tan θ +tan θ )/( tan θ tan θ ) and tan(θ)= tan(θ + mπ), Eqs. (9..7) and (9..8) may also be written in the following forms: tan( a)= (α + α s ) k f α, tan(φ)= (α α s ) α s k f + α (9..9) α s The form of Eq. (9..7) is preferred for the numerial solution. To this end, we introdue the dimensionless variables: R = k a n f n s = πfa u = a, v = α s a, w= α a n f n s = π a λ n f n s (9..) For a given operating frequeny f, the value of R is fixed. All allowed propagating modes must satisfy R m R, or, mπ + artan( δ ) R m R artan( δ ) π This fixes the maximum mode index M to be: M = floor ( ) R artan δ π (maximum TE mode index) (9..4) Thus, there are (M + ) modes labeled by m =,,...,M. In the symmetri ase, δ =, and (9..4) redues to Eq. (9..5) of the previous setion. The orresponding utoff frequenies are obtained by setting: R m = πf ma n f n s f m = mπ + artan( δ ) πa n f n s (9..5) whih an be written more simply as f m = fr m /R, m =,,...,M, where f = /λ. For eah of the M+ propagating modes one an alulate the orresponding angle of total internal refletion of the equivalent ray model by identifying with the transverse propagation wavenumber, that is, = k n f os θ, as shown in Fig. 9... Then, Eqs. (9..7) and (9..) an be written in the normalized forms: u = mπ + artan ( v u ) + w artan u u + v = R w v = R δ (9..) One these are solved for the three unknowns u, v, w, or,α s,α, the propagation onstant β, or equivalently, the effetive index n β = β/k an be obtained from: β = k n f n β = β = n f k k = n f u k (9..) a To determine the number of propagating modes and the range of the mode index m, we set v = in the harateristi equation (9..) to find the radius R m of the mth mode. Then, u = R m and w = R m δ, and we obtain: R m = mπ + artan( δ ), m =,,,... (9..3) Fig. 9.. Ray propagation model. The harateristi equation (9..7) an be given a nie interpretation in terms of the ray model [9]. The field of the upgoing ray at a point A at (x, z) is proportional, up to a onstant amplitude, to e j x e jβz Similarly, the field of the upgoing ray at the point B at (x, z + l) should be e j x e jβ(z+l) (9..6) But if we follow the ray starting at A along the zig-zag path AC CS SB, the ray will have traveled a total vertial roundtrip distane of 4a and will have suffered two total internal refletion phase shifts at points C and S, denoted by ψ and ψ s. We

9.. Asymmetri Dieletri Slab 4 4 9. Waveguides reall that the refletion oeffiients have the form ρ = e jψ for total internal refletion, as given for example by Eq. (7.8.3). Thus, the field at point B would be e j (x+4a) e jψs e jψ e jβ(z+l) This must math (9..6) and therefore the extra aumulated phase 4 a ψ s ψ must be equal to a multiple of π, that is, 4 a ψ s ψ = mπ a = mπ + ψ s + ψ As seen from Eq. (7.8.3), the phase terms are exatly those appearing in Eq. (9..7): tan ψ = α, tan ψ s = α s α αs ψ = artan, ψ s = artan A similar interpretation an be given for the TM modes. It is ommon in the literature to represent the harateristi equation (9..) by means of a universal mode urve [93] defined in terms of the following saled variable: b = v R = β k n s k (n f n s) (9..7) whih ranges over the standardized interval b, so that u = R b, v = R b, w= R b + δ (9..8) Then, Eq. (9..) takes the universal form in terms of the variables b, R: R b = mπ + artan b + artan b + δ (9..9) b b b.9.8.7.6.5.4.3 Universal mode urves for TE modes m =. δ =. δ = δ = 3 4 5 6 7 8 9 R Fig. 9..3 3 Universal mode urves. where φ is a parameter to be determined. Then, the E x omponent is: j β E os( x + φ), E x (x)= j β E sin( a + φ)e α(x a), α 4 x a x a (9..) It is depited in Fig. 9..3 with one branh for eah value of m =,,,..., and for the three asymmetry parameter values δ =,,. A vertial line drawn at eah value of R determines the values of b for the propagating modes. Similar urves an be developed for TM modes. See Example 9.. for a onrete example that inludes both TE and TM modes. TM modes The TM modes are obtained by solving Eqs. (9.3.) in eah region and applying the boundary onditions. Assuming x-dependene only, we must solve in eah region: ( x + k f )E z =, E x = jβ k x E z, H y = E x, η TM = β f η TM ωɛ The solution for E z (x) is given by a similar expression as Eq. (9..4): E sin( x + φ), x a E z (x)= E sin( a + φ)e α(x a), x a E sin( a φ)e αs(x+a), x a (9..) j β E sin( a φ)e αs(x+a), α s x a The boundary onditions require the ontinuity of the normal omponent of displaement field D x = ɛe x aross the interfaes at x =±a, whih is equivalent to the ontinuity of the tangential field H y beause H y = E x /η TM = ɛe x ω/β = D x ω/β. Thus, the boundary onditions at x =±a require: ɛ f os( a + φ) = ɛ sin( a + φ) α ɛ f os( a φ) = ɛ s sin( a φ) α s where we defined the ratios: p = ɛ f = n f, p s = ɛ f = n f ɛ ɛ s n tan( a + φ) = p α tan( a φ) = p s α s n s (9..) (9..3) R is usually denoted by the variable V.

9.. Asymmetri Dieletri Slab 43 44 9. Waveguides Inverting the tangents we obtain: α a + φ = artan p + mπ α s a φ = artan p s These give the harateristi equation of the slab and φ: a = mπ + artan α s p s + k artan α p f φ = mπ + artan α p k artan α s p s f and, as in Eq. (9..9), we an write: (9..4) (9..5) tan( a)= (p s α s + p α s ) k f p sp α s α, tan(φ)= (p α p s α s ) k f + p sp α s α (9..6) In terms of the normalized variables u, v, w, R, we have: u = mπ + artan v p s + u artan w p u u + v = R w v = R δ (9..7) The number of propagating modes and the range of the mode index m, are again determined by setting v =, u = R m, and w = R m δ: R m = mπ + artan( p δ ), m =,,,... The allowed propagating modes must satisfy R m R, or, mπ + artan( p δ ) R m R artan( ) p δ π whih fixes the maximum mode index M to be: M = floor ( ( ) ) R artan p δ π (maximum TM mode index) (9..8) The (M + ) modes are again labeled by m =,,...,M. The orresponding utoff frequenies are obtained by setting: R m = πf ma n f n s f m = mπ + artan( p δ ) πa n f n s (9..9) whih an be written more simply as f m = fr m /R, m =,,...,M, where f = /λ. The orresponding angles of total internal refletion in the equivalent ray model are obtained by solving = k n f os θ. Beause p >, we observe that the maximum mode index M and the utoff frequenies f m will satisfy the following inequalities for the TE and TM ases: Numerial Solutions M TM M TE, f m,te f m,tm (9..3) Next we look at the numerial solutions of Eqs. (9..7). The TE ase is also inluded by setting p s = p =. A simple and effetive iterative method for solving suh harateristi equations was given in Ref. [959]. By replaing v, w in terms of u, let F(u) denote the right-hand side of Eq. (9..7): F(u)= mπ + artan v p s + u artan w p u The problem then beomes that of finding the fixed-point solutions u = F(u). The method suggested in Ref. [959] is to use the iteration: u n+ = F(u n ), n=,,,... initialized at u = R. This simple iteration does onverges in many ases, but fails in others. We have found that a simple modifiation that involves the introdution of a relaxation parameter r suh that r, enables the onvergene of even the most diffiult ases. The modified iteration has the form: u n+ = rf(u n )+( r)u n Expliitly, the iteration starts with the initial values: u = R, v =, w = R δ (9..3) and proeeds iteratively, for n =,,,..., until two suessive u n values beome loser to eah other than some speified error tolerane, say tol, suh as tol = : [ u n+ = r mπ + ( artan v n p s u n ) + artan ( if u n+ u n < tol, then exit, else ontinue v n+ = R u n+ w n+ = R δ vn+ p w n u n )] + ( r)u n The MATLAB funtion dguide3.m implements the method and has usage: [be,kf,as,a,fm,nit] = dguide3(a,ns,nf,n,mode,r,tol); (9..3)

9.. Asymmetri Dieletri Slab 45 46 9. Waveguides where the inputs and outputs have the following meanings: a = half-width of slab in units of the free-spae wavelength λ n s,n f,n = refrative indies of substrate, film, and ladding (n f >n s >= n ) mode = TE or TM r = relaxation parameter (default r =.5) tol = error tolerane (default tol = ) β = propagation wavenumbers in units of k = π/λ = transverse wavenumbers inside slab in units of k α s,α = deay wavenumbers in substrate and ladding in units of k f m = utoff frequenies in units of f = /λ N it = number of iterations it takes to onverge to within tol Internally, the funtion determines M from Eq. (9..4) or (9..8) and alulates β,,α s,α,f m as (M + )-dimensional olumn vetors representing the M + modes. To larify the omputations, the essential part of the ode is listed below: k = *pi; R = k*a * sqrt(nf^-ns^); d = (ns^-n^)/(nf^-ns^); % la = *pi/k = in the assumed units % (u,v) irle radius, note k*a = *pi*(a/la) % asymmetry parameter if strmpi(mode, TE ) % mode an also be entered in lower ase ps = ; p = ; else ps = nf^/ns^; p = nf^/n^; end M = floor((*r - atan(p*sqrt(d)))/pi); m = (:M) ; u = R*ones(M+,); v = zeros(m+,); w = R*sqrt(d)*ones(M+,); % highest mode index % vetor of mode indies % initialize iteration variables u,v,w % u,v,w are (M+)x vetors Nit = ; % number of iterations % while loop repeats till onvergene while unew = r*(m*pi/ + atan(ps*v./u)/ + atan(p*w./u)/) + (-r)*u; if norm(unew-u) <= tol, break; end Nit = Nit + ; u = unew; v = sqrt(r^ - u.^); w = sqrt(r^*d + v.^); if Nit>, break; end % safeguard against possible divergene end kf = u/(k*a); as = v/(k*a); a = w/(k*a); be = sqrt(nf^ - kf.*kf); Rm = m*pi/ + atan(p*sqrt(d))/; fm = Rm/R; thm = aos(kf/nf); % kf in units of k, i.e., kf/k = u/(k*a) % beta in units of k, i.e., beta/k % utoff radius for m-th mode % utoff frequenies in units of f = /la % angles of total internal refletion Example 9..: For omparison purposes, we onsider the same benhmark example disussed in [959] onsisting of a silion film of thikness of μm with n f = 3.5, an oxide substrate with n s =.45, and air over, with operating wavelength λ =.55 μm. The following MATLAB ode generates both the TE and TM modes, with the numerial outputs listed in the tables below. nf=3.5; ns=.45; n=; la =.55; a =.5; a = a/la; r=.3; tol=e-; % oxide substrate silion film air over % units of mirons % half-thikness in units of la % default value r=.5 fails to onverge for the TE modes [be,kf,as,a,fm,nit] = dguide3(a,ns,nf,n, te,r,tol); [be,kf,as,a,fm,nit] = dguide3(a,ns,nf,n, tm,r,tol); m β/k /k α s /k α /k f m /f θm o 3.434746.677 3.37 3.86.47 78.9 3.3789.343.8894 3.74.679 67.47.873..4794.696.5 55.5 3.35.6364.788.735.7545 4.3 4.4597 3.846.756.57.9978 4.5 m β/k /k α s /k α /k f m /f θm o 3.4657.7599 3.935 3.669.8 77.46 3.549.569.8.995.346 64.3.66893.64.47.4745.5894 49.69 3.86544.966.733.5745.837 3. 4.76 % TE modes % TM modes (TE modes) (TM modes) The β/k olumn is the effetive phase index of the modes. The default value of the relaxation parameter r =.5 did not work in this ase and aused the TE iteration to diverge and the smaller value r =.3 was hosen. The number of iterations were N it = 57 for TE and N it = 66 for TM. The TIR angles were omputed by the following ommand: thm = aos(kf/nf)*8/pi; % degrees We note that all TIR angles are greater than the ritial angles omputed by: ns θ s = arsin = 4.47 o n, θ = arsin = 6.6 o n f n f There are five TE modes and four TM ones. The fifth TE mode is very weakly bound to the substrate side beause its deay parameter α s is very small, its utoff frequeny is very near the operating frequeny f = /λ, and its TIR angle, very lose to the ritial angle. With referene to the inequality (9..3), it so happened that in this example f falls in the range f 4,TE <f<f 4,TM, and therefore, the fifth TM mode f 4,TM is not exited, but f 4,TE is. The onvergene an be verified for all modes at one by omputing the vetor error norm of the harateristi equations, that is,

9.. Asymmetri Dieletri Slab 47 M = length(be)-; m = (:M) ; Err = norm(kf**pi*a - m*pi/ - atan(ps*as./kf)/ - atan(p*a./kf)/); This error is of the order of the assumed tolerane, indeed, we have Err =.94 for TE, and Err =.53 for TM. We note that the quantity kf**pi*a represents the variable u in our units, indeed, u = a = ( /k )k a = ( /k )π(a/λ ). Finally, Fig. 9..4 displays the TE and TM solutions on the universal mode urves, see e.g. Eq. (9..9). b.9.8.7.6.5.4.3.. m = 3 4 5 6 7 8 9 R 3 4 48 9. Waveguides 9.3 Problems 9. An air-filled.5 m 3 m waveguide is operated at a frequeny that lies in the middle of its TE mode band. Determine this operating frequeny in GHz and alulate the maximum power in Watts that an be transmitted without ausing dieletri breakdown of air. The dieletri strength of air is 3 MV/m. 9. It is desired to design an air-filled retangular waveguide suh that (a) it operates only in the TE mode with the widest possible bandwidth, (b) it an transmit the maximum possible power, and () the operating frequeny is GHz and it lies in the middle of the operating band. What are the dimensions of the guide in m? 9.3 An air-filled retangular waveguide is used to transfer power to a radar antenna. The guide must meet the following speifiations: The two lowest modes are TE and TE. The operating frequeny is 3 GHz and must lie exatly halfway between the utoff frequenies of these two modes. The maximum eletri field within the guide may not exeed, by a safety margin of 3, the breakdown field of air 3 MV/m. a. Determine the smallest dimensions a, b for suh a waveguide, if the transmitted power is required to be MW. b. What are the dimensions a, b if the transmitted power is required to be maximum? What is that maximum power in MW? 9.4 It is desired to design an air-filled retangular waveguide operating at 5 GHz, whose group veloity is.8. What are the dimensions a, b of the guide (in m) if it is also required to arry maximum power and have the widest possible bandwidth? What is the utoff frequeny of the guide in GHz and the operating bandwidth? 9.5 Show the following relationship between guide wavelength and group veloity in an arbitrary air-filled waveguide: v g λ g = λ, where λ g = π/β and λ is the free-spae wavelength. Moreover, show that the λ and λ g are related to the utoff wavelength λ by: Fig. 9..4 Universal mode urves. TE (solid lines/filled irles), TM (dashed lines/open irles). λ = λ + g λ Example 9..: A seond, more diffiult, example from [959] has the parameters λ =.55 μm, a =.5 μm, n f = 3.3, n s = 3.56, n =. The same MATLAB ode applies here, but we used the default value r =.5, whih onverges in 8 and iterations respetively for the TE and TM modes. Only one (M = ) TE and one TM mode are supported with parameters given in the table below. The ritial TIR angles are in this example: ns θ s = arsin = 8.63 o n, θ = arsin = 7.64 o n f n f mode β/k /k α s /k α /k f m /f θm o TE 3.65996.475.553 3.9.647 8.77 TM 3.63384.49.94 3.64.74 8.46 The omputational errors in the harateristi equations were Err =.63 for TE, and Err =.5 for TM. 9.6 Determine the four lowest modes that an propagate in a WR-59 and a WR-9 waveguide. Calulate the utoff frequenies (in GHz) and utoff wavelengths (in m) of these modes. 9.7 An air-filled WR-9 waveguide is operated at 9 GHz. Calulate the maximum power that an be transmitted without ausing dieletri breakdown of air. Calulate the attenuation onstant in db/m due to wall ohmi losses. Assume opper walls. 9.8 A retangular waveguide has sides a, b suh that b a/. Determine the utoff wavelength λ of this guide. Show that the operating wavelength band of the lowest mode is.5λ λ λ. Moreover, show that the allowed range of the guide wavelength is λ g λ / 3. 9.9 The TE mode operating bandwidth of an air-filled waveguide is required to be 4 7 GHz. What are the dimensions of the guide? 9. Computer Experiment: WR-59 Waveguide. Reprodue the two graphs of Fig. 9.8.. 9. A TM mode is propagated along a hollow metalli waveguide of arbitrary but uniform ross setion. Assume perfetly onduting walls. a. Show that the E z (x, y) omponent satisfies: T E z ds = k S E z ds, (k = utoff wavenumber) (9.3.) S

9.3. Problems 49 4 9. Waveguides b. Using the above result, show that the energy veloity is equal to the group veloity. Hint: Use the identity: T (A T B)= T A T B + A TB, for salar A, B. 9. Computer Experiment: Dieletri Slab Waveguide. Using the MATLAB funtions dslab and dguide, write a program that reprodues all the results and graphs of Examples 9.. and 9... 9.3 Show that if the speed of light is slightly hanged to = + Δ (e.g. representing a more exat value), then the solutions of Eq. (9..9) for k,α hange into: k Δ k + Δk = k + α a ( α + Δα = α α + k a ) Δ + α a For Example 9.., alulate the orreted values when = 3 and = 9.979458 GHz m. Compare with the values obtained if is replaed by inside the funtion dguide. More generally, onsider the sensitivity of the solutions of Eq. (9..9) to any of the parameters ω,a,,n,n, whih affet the solution through the value of R = aω n n. A small hange in one or all of the parameters will indue a small hange R R + ΔR. Show that the solutions are hanged to u ΔR u + Δu = u + + v R ( v + Δv = v + v + u + v ) ( ΔR R In partiular, for simultaneous hanges in all of the parameters, show that ΔR R = Δa a + Δω ω Δ + n Δn n Δn n n From these results, show that the hanges due to a hange a a + Δa of the slab thikness are given by, k + Δk = k k α + α a Δa α + Δα = α + k + α a Δa 9.4 For the dieletri slab waveguide shown in Fig. 9.., demonstrate that the energy transport veloity is equal to the group veloity. Speifially, onsider the ase of even TE modes defined by Eqs. (9..3) (9..3), and show that v en = v gr, where v en = P T W, v gr = dω dβ ) (9.3.) a. Show that P T is given as the sum of the following two terms, where the first one represents the power flowing within the slab, and the seond, the power flowing outside the slab: P T = H βωμ [ ak + sin(ak )os(ak ) ] k 3 + H βωμ sin (ak ) α 3 where H is the amplitude defined in Eq. (9..5). Without loss of generality, from now on set, H =. b. Show that the eletri and magneti energy densities are given as follows, where again, the first terms represent the energy ontained within the slab, and the seond, the energy outside the slab: W e = μ (β + k ) [ ak + sin(ak )os(ak ) ] 4k 3 W m = μ (β + k )ak + μ (β k )sin(ak )os(ak ) 4k 3 + μ (β α )sin (ak ) 4α 3. Using the above expressions and Eq. (9..3), show the equality W e = W m Thus, the total energy density is W = W e + W m = W e. d. From parts (a,b,), show that P T = ωβμ ( + aα ) α k W = μ [ (β + k )aα + β ] α k + μ (β + α )sin (ak ) 4α 3 e. By differentiating Eqs. (9..3) and (9..3) with respet to ω, show that ωβ dβ dω = (β + k )aα + β + aα f. Combining the results of parts (e,f), show finally that v en = v gr = ωβ( + aα ) (β + k )aα + β = ωβ( + aα ) ω ɛ μ n aα + β 9.5 Computer Experiment. Asymmetri Slab Waveguide. Reprodue all the results and Fig. 9..4 of Example 9... Moreover, make a separate graph of Fig. 9..4 that zooms into the neighborhood of the fifth TE mode to make sure that it is indeed below utoff. where P T is the time-averaged power transmitted in the z-diretion through the rosssetional area defined by y and <x<, and W is the energy ontained in the volume defined by the above area and unit-z-length, i.e., P T = E y (x) dx, W = [ ɛ Ey (x) + H x (x) + μ H z (x) ] dx η TE 4 Beause of the substantial amount of algebra involved, break the alulation as follows: