Homework SourseGiven an airfilled circular waveguide Find the smallest diam
Solution
Waveguides are used to transfer electromagnetic power efficiently from one point in space to another. Some common guiding structures are shown in the figure below. These include the typical coaxial cable, the two-wire and mictrostrip transmission lines, hollow conducting waveguides, and optical fibers. In practice, the choice of structure is dictated by: (a) the desired operating frequency band, (b) the amount of power to be transferred, and (c) the amount of transmission losses that can be tolerated. Fig. 9.0.1 Typical waveguiding structures. Coaxial cables are widely used to connect RF components. Their operation is practical for frequencies below 3 GHz. Above that the losses are too excessive. For example, the attenuation might be 3 dB per 100 m at 100 MHz, but 10 dB/100 m at 1 GHz, and 50 dB/100 m at 10 GHz. Their power rating is typically of the order of one kilowatt at 100 MHz, but only 200 W at 2 GHz, being limited primarily because of the heating of the coaxial conductors and of the dielectric between the conductors (dielectric voltage breakdown is usually a secondary factor.) However, special short-length coaxial cables do exist that operate in the 40 GHz range. Another issue is the single-mode operation of the line. At higher frequencies, in order to prevent higher modes from being launched, the diameters of the coaxial conductors must be reduced, diminishing the amount of power that can be transmitted. Two-wire lines are not used at microwave frequencies because they are not shielded and can radiate. One typical use is for connecting indoor antennas to TV sets. Microstrip lines are used widely in microwave integrated circuits. 9.1. Longitudinal-Transverse Decompositions 363 Rectangular waveguides are used routinely to transfer large amounts of microwave power at frequencies greater than 3 GHz. For example at 5 GHz, the transmitted power might be one megawatt and the attenuation only 4 dB/100 m. Optical fibers operate at optical and infrared frequencies, allowing a very wide bandwidth. Their losses are very low, typically, 0.2 dB/km. The transmitted power is of the order of milliwatts. 9.1 Longitudinal-Transverse Decompositions In a waveguiding system, we are looking for solutions of Maxwell’s equations that are propagating along the guiding direction (the z direction) and are confined in the near vicinity of the guiding structure. Thus, the electric and magnetic fields are assumed to have the form: E(x, y, z, t)= E(x, y)ejtjz H(x, y, z, t)= H(x, y)ejtjz (9.1.1) where is the propagation wavenumber along the guide direction. The corresponding wavelength, called the guide wavelength, is denoted by g = 2/. The precise relationship between and depends on the type of waveguiding structure and the particular propagating mode. Because the fields are confined in the transverse directions (the x, y directions,) they cannot be uniform (except in very simple structures) and will have a non-trivial dependence on the transverse coordinates x and y. Next, we derive the equations for the phasor amplitudes E(x, y) and H(x, y). Because of the preferential role played by the guiding direction z, it proves convenient to decompose Maxwell’s equations into components that are longitudinal, that is, along the z-direction, and components that are transverse, along the x, y directions. Thus, we decompose: E(x, y)= xˆ Ex(x, y)+yˆ Ey(x, y) transverse + ˆ z Ez(x, y) longitudinal ET(x, y)+ˆz Ez(x, y) (9.1.2) In a similar fashion we may decompose the gradient operator: = xˆ x + yˆ y transverse + ˆ z z = T + ˆz z = T jˆz (9.1.3) where we made the replacement z j because of the assumed z-dependence. Introducing these decompositions into the source-free Maxwell’s equations we have: × E = jH × H = jE · E = 0 · H = 0 (T jˆz)×(ET + ˆz Ez)= j(HT + ˆz Hz) (T jˆz)×(HT + ˆz Hz)= j(ET + ˆz Ez) (T jˆz)·(ET + ˆz Ez)= 0 (T jˆz)·(HT + ˆz Hz)= 0 (9.1.4) 364 9. Waveguides where , denote the permittivities of the medium in which the fields propagate, for example, the medium between the coaxial conductors in a coaxial cable, or the medium within the hollow rectangular waveguide. This medium is assumed to be lossless for now. We note that ˆz · ˆz = 1, ˆz × ˆz = 0, ˆz · ET = 0, ˆz · TEz = 0 and that ˆz × ET and ˆ z × TEz are transverse while T × ET is longitudinal. Indeed, we have: ˆ z × ET = ˆz × (xˆ Ex + yˆ Ey)= yˆ Ex xˆ Ey T × ET = (xˆ x + yˆ y)×(xˆ Ex + yˆ Ey)= ˆz(xEy yEx) Using these properties and equating longitudinal and transverse parts in the two sides of Eq. (9.1.4), we obtain the equivalent set of Maxwell equations: TEz × ˆz jˆz × ET = jHT THz × ˆz jˆz × HT = jET T × ET + jˆz Hz = 0 T × HT j ˆz Ez = 0 T · ET jEz = 0 T · HT jHz = 0 (9.1.5) Depending on whether both, one, or none of the longitudinal components are zero, we may classify the solutions as transverse electric and magnetic (TEM), transverse electric (TE), transverse magnetic (TM), or hybrid: Ez = 0, Hz = 0, TEM modes Ez = 0, Hz = 0, TE or H modes Ez = 0, Hz = 0, TM or E modes Ez = 0, Hz = 0, hybrid or HE or EH modes In the case of TEM modes, which are the dominant modes in two-conductor transmission lines such as the coaxial cable, the fields are purely transverse and the solution of Eq. (9.1.5) reduces to an equivalent two-dimensional electrostatic problem. We will discuss this case later on. In all other cases, at least one of the longitudinal fields Ez, Hz is non-zero. It is then possible to express the transverse field components ET, HT in terms of the longitudinal ones, Ez, Hz. Forming the cross-product of the second of equations (9.1.5) with ˆz and using the BAC-CAB vector identity, ˆz × (ˆz × HT)= ˆz(ˆz · HT)HT(ˆz · ˆz)= HT, and similarly, ˆ z × (THz × ˆz)= THz, we obtain: THz + jHT = j ˆz × ET Thus, the first two of (9.1.5) may be thought of as a linear system of two equations in the two unknowns ˆz × ET and HT, that is, ˆz × ET HT = jˆz × TEz ˆ z × ET HT = jTHz (9.1.6) 9.1. Longitudinal-Transverse Decompositions 365 The solution of this system is: ˆz × ET = j k2 c ˆz × TEz j k2 c THz HT = j k2 c ˆz × TEz j k2 c THz (9.1.7) where we defined the so-called cutoff wavenumber kc by: k2 c = 2 2 = 2 c2 2 = k2 2 (cutoff wavenumber) (9.1.8) The quantity k = /c = is the wavenumber a uniform plane wave would have in the propagation medium , . Although k2 c stands for the difference 2 2, it turns out that the boundary conditions for each waveguide type force k2 c to take on certain values, which can be positive, negative, or zero, and characterize the propagating modes. For example, in a dielectric waveguide k2 c is positive inside the guide and negative outside it; in a hollow conducting waveguide k2 c takes on certain quantized positive values; in a TEM line, k2 c is zero. Some related definitions are the cutoff frequency and the cutoff wavelength defined as follows: c = ckc , c = 2 kc (cutoff frequency and wavelength) (9.1.9) We can then express in terms of and c, or in terms of and c. Taking the positive square roots of Eq. (9.1.8), we have: = 1 c 2 2 c = c 1 2 c 2 and = 2 c + 2c2 (9.1.10) Often, Eq. (9.1.10) is expressed in terms of the wavelengths = 2/k = 2c/, c = 2/kc, and g = 2/. It follows from k2 = k2 c + 2 that 1 2 = 1 2 c + 1 2 g g = 1 2 2 c (9.1.11) Note that is related to the free-space wavelength 0 = 2c0/ = c0/f by the refractive index of the dielectric material = 0/n. It is convenient at this point to introduce the transverse impedances for the TE and TM modes by the definitions: TE = = c , TM = = c (TE and TM impedances) (9.1.12) 366 9. Waveguides where the medium impedance is = /, so that /c = and c = 1/. We note the properties: TETM = 2 , TE TM = 2 2c2 (9.1.13) Because c/ = 1 2 c/2, we can write also: TE = 1 2 c 2 , TM = 1 2 c 2 (9.1.14) With these definitions, we may rewrite Eq. (9.1.7) as follows: ˆz × ET = j k2 c ˆz × TEz + TE THz HT = j k2 c 1 TM ˆz × TEz + THz (9.1.15) Using the result ˆz × (ˆz × ET)= ET, we solve for ET and HT: ET = j k2 c TEz TE ˆz × THz HT = j k2 c THz + 1 TM ˆz × TEz (transverse fields) (9.1.16) An alternative and useful way of writing these equations is to form the following linear combinations, which are equivalent to Eq. (9.1.6): HT 1 TM ˆz × ET = j THz ET TE HT × ˆz = j TEz (9.1.17) So far we only used the first two of Maxwell’s equations (9.1.5) and expressed ET, HT in terms of Ez, Hz. Using (9.1.16), it is easily shown that the left-hand sides of the remaining four of Eqs. (9.1.5) take the forms: T × ET + jˆz Hz = j k2 c ˆz 2 THz + k2 cHz T × HT j ˆz Ez = j k2 c ˆz 2 TEz + k2 cEz T · ET jEz = j k2 c 2 TEz + k2 cEz T · HT jHz = j k2 c 2 THz + k2 cHz 9.1. Longitudinal-Transverse Decompositions 367 where 2T is the two-dimensional Laplacian operator: 2T = T · T = 2x + 2y (9.1.18) and we used the vectorial identities T × TEz = 0, T × (ˆz × THz)= ˆz 2THz, and T · (ˆz × THz)= 0. It follows that in order to satisfy all of the last four of Maxwell’s equations (9.1.5), it is necessary that the longitudinal fields Ez(x, y), Hz(x, y) satisfy the two-dimensional Helmholtz equations: 2TEz + k2cEz = 0 2THz + k2cHz = 0 (Helmholtz equations) (9.1.19) These equations are to be solved subject to the appropriate boundary conditions for each waveguide type. Once, the fields Ez, Hz are known, the transverse fields ET, HT are computed from Eq. (9.1.16), resulting in a complete solution of Maxwell’s equations for the guiding structure. To get the full x, y, z, t dependence of the propagating fields, the above solutions must be multiplied by the factor ejtjz. The cross-sections of practical waveguiding systems have either cartesian or cylindrical symmetry, such as the rectangular waveguide or the coaxial cable. Below, we summarize the form of the above solutions in the two types of coordinate systems. Cartesian Coordinates The cartesian component version of Eqs. (9.1.16) and (9.1.19) is straightforward. Using the identity ˆz × THz = yˆ xHz xˆ yHz, we obtain for the longitudinal components: (2x + 2y)Ez + k2cEz = 0 (2x + 2y)Hz + k2cHz = 0 (9.1.20) Eq. (9.1.16) becomes for the transverse components: Ex = jk2c xEz + TE yHz Ey = jk2c yEz TE xHz , Hx = jk2c xHz 1TM yEz Hy = jk2c yHz + 1TM xEz (9.1.21) Cylindrical Coordinates The relationship between cartesian and cylindrical coordinates is shown in Fig. 9.1.1. From the triangle in the figure, we have x = cos and y = sin . The transverse gradient and Laplace operator are in cylindrical coordinates: T = ˆ + ˆ 1 , 2T = 1 + 12 2 2 (9.1.22) 368 9. Waveguides Fig. 9.1.1 Cylindrical coordinates. The Helmholtz equations (9.1.19) now read: 1 Ez + 12 2Ez 2 + k2cEz = 0 1 Hz + 12 2Hz 2 + k2cHz = 0 (9.1.23) Noting that ˆz × ˆ = ˆ and ˆz × ˆ = ˆ, we obtain: ˆ z × THz = ( ˆ Hz)ˆ 1(Hz) The decomposition of a transverse vector is ET = Eˆ + E ˆ . The cylindrical coordinates version of (9.1.16) are: E = jk2c Ez TE 1Hz E = jk2c 1Ez + TEHz , H = jk2c Hz + 1 TMEz H = jk2c 1Hz 1TM Ez (9.1.24) For either coordinate system, the equations for HT may be obtained from those of ET by a so-called duality transformation, that is, making the substitutions: E H , H E , , (duality transformation) (9.1.25) These imply that 1 and TE 1 TM. Duality is discussed in greater detail in Sec. 18.2. 9.2 Power Transfer and Attenuation With the field solutions at hand, one can determine the amount of power transmitted along the guide, as well as the transmission losses. The total power carried by the fields along the guide direction is obtained by integrating the z-component of the Poynting vector over the cross-sectional area of the guide:
