GUIDED MICROWAVES AND OPTICAL WAVES

Chapter 1 GUIDED MICROWAVES AND OPTICAL WAVES 1.1 Introduction In communication engineering, the carrier frequency has been steadily increasing for the obvious reason that a carrier wave with a higher frequency can accommodate more information. The frequency band currently used in communication is vastly wide, ranging from MHz (AM radio to 10 15 Hz (visible light. Electromagnetic waves with frequency higher than about 1GHz can be confined in a waveguide. (At lower frequencies, the size of waveguides would be impractically large. Exception is the natural, global waveguide formed by the ionospheric plasma and the earth, which can be used for low frequency communication, such as short wave radio. Microwaves (1 GHz ~300 GHz can be propagated in a hollow conductor tube, while optical waves in a dielectric fiber. In both cases, wave propagation mechanism can be qualitatively understood in terms of wave reflection at the waveguide wall. However, in contrast to electromagnetic waves in free space or in transmission lines, waves confined in waveguides cannot be purely TEM. In microwave waveguides, either TE (Transverse Electric or TM (Transverse Magnetic modes can exist, but not TEM modes. This important deviation from TEM propagation mode is due to the boundary conditions imposed on electromagnetic fields. In optical waveguides practically used, even pure TE or TM mode is not allowed, except for special propagation modes. Deviation from TEM nature not only complicates field analysis, but also causes inevitable wave dispersion, that is, the propagation velocity becomes dependent on the wave frequency. This in turn implies that the original waveform sent out is bound to deform. Of course, the merit of guided waves is in effi cient energy or signal transmission along a desired path. 1. Waveguides Microwave technology was greatly advanced during the World War II when radar detection became practical in military applications. In general, the resolution of radar detection improves with the 1

frequency, and the invention of high power, high frequency microwave tubes, such as the magnetron, played key roles in radar technology. Waveguides are used to transmit microwaves between various microwave devices. They play the role of wire conductors in low frequency electric circuits. At high frequency, an open transmission line will become an effi cient antenna, while closed transmission lines (such as coaxial cables are subject to strong dielectric losses. The only practical way to transmit high frequency electromagnetic waves is to confine them in a hollow conductor tube. It is desirable that the conductivity of the wall material is large to minimize wall losses. (Recall that the skin depth and consequent wall loss are inversely proportional to ωσ where σ is the conductivity. Waveguides most commonly used are either rectangular or circular, but the cross-section shape can be arbitrary as long as it does not change abruptly along the waveguide. In either case, there is a lower limit in the wave frequency allowed for propagation (cutoff frequency, similar to the case of wave propagation in a plasma. Waves having frequencies lower than the cutoff frequency cannot be propagated in a wave guide. The origin of the cutoff phenomena is in the boundary conditions at the conductor wall that should be satisfied by the electric and magnetic fields, and consequent deviation from the TEM nature. If a waveguide is filled with air, which is usually the case, the wave equation for the electromagnetic fields is still given by ( 1 c t ( 1c t E (r, t = 0 (1.1 H (r, t = 0 (1. These are formally identical to the wave equation in free space. However, solutions allowed for E and H as the electromagnetic fields in a waveguide will be distinctly different from the TEM mode we have been studying, because of the boundary conditions for the fields which impose rather stringent limitation for allowable solutions. Analysis of electromagnetic fields in a waveguide boils down to solving Eqs. (1.1 and (1. under appropriate boundary conditions. Each equation has three components. For example, in the cartesian coordinates, Eq. (1.1 can be decomposed into three scalar wave equations for E x, E y, and E z. In a cylindrical waveguide, the cylindrical coordinates (ρ, φ, z appear to be most convenient. Unlike the cartesian coordinates, the vector wave equation in the cylindrical coordinates cannot be separated into three scalar wave equations because of the complexity in the vector Laplacian. Fortunately, however, the axial (z component of the vector wave equation reduces to a scalar wave equation in the cylindrical coordinates as well which we know how to handle. This suggests the possibility that the entire electromagnetic fields in a waveguide may be described by two axial components, E z and H z, because both electric and magnetic field components are not entirely independent but constrained through the Maxwell s equations. Indeed, to describe electromagnetic fields in a waveguide, it is suffi cient to solve the following two scalar wave equations ( 1c t E z (r, t = 0 (1.3

( 1c t H z (r, t = 0 (1.4 as long as the waveguide cross-section does not change along the axis. The transverse components, E and H, can be fully described by the axial fields, E z and H z. Evidently, solving a scalar wave equation is much simpler than solving a vector wave equation. To see how this is done, let us go back to the Maxwell s equations from which the wave equations have been derived, H E = µ 0 t, H = ε E 0 t (1.5 Since the direction of wave (energy propagation is obviously along the waveguide (z direction, we may assume the phase function e j(ωt kzz where k z is the axial wavenumber in the z direction. (k z is called the phase constant in engineering and often denoted by β which, however, is confusing with the normalized velocity β = v/c in physics. As will be shown, the ratio ω/k z is not equal to c, as for TEM modes, but exceeds c. At this stage, k z is yet to be determined. For the assumed phase function, the gradient operator along the z axis and time derivative can be replaced by z jk z, t jω. Also, the electric and magnetic fields may be decomposed into axial and transverse components as E = E + E z, H = H + H z, (1.6 Then, Eq. (1.5 becomes E + E z e z jk z e z E = jωµ 0 (H + H z (1.7 H + H z e z jk z e z H = jωε 0 (E + E z (1.8 Note that e z = 0 because e z is a unidirectional constant vector. From the transverse components of Eq. (1.8, we obtain E = j ωε 0 ( H z e z jk z e z H (1.9 Substituting this into Eq. (1.7, we find the transverse magnetic field entirely in terms of the axial fields, j H = (ω/c (k z H z ωε 0 E z e z (1.10 kz 3

Similarly, the transverse electric field in terms of the axial fields is given by E = j (ω/c k z (k z E z + ωµ 0 H z e z (1.11 The results clearly indicate that if the axial fields, E z and H z are known, the transverse components of the electric and magnetic fields can be readily calculated. It is also apparent that there can be no TEM modes in a waveguide. TEM modes are characterized by E z = H z = 0. Then, all fields must vanish according to Eqs. (1.10 and (1.11. Although there are no TEM modes in a waveguide, modes with only transverse electric fields (TE and modes with only transverse magnetic (TM fields can exist independently. In fact, TE and TM modes constitute basic independent modes in general electrodynamics, and such classification is not limited to waveguide modes. TEM mode is a rather idealized mode of propagation and as long as the wave source is finite, there can be no pure TEM mode. For example, an oscillating electric dipole radiates TM modes and the radial (longitudinal component of the electric field does not completely vanishes. The radial component is responsible for the angular momentum flux which has been briefly discussed in Chapter 11. Similarly, a magnetic dipole radiates TE modes having a finite radial component of the magnetic field. Transverse electric (TE modes have no axial electric field, E z = 0, and the electric field of TE modes is determined by the axial magnetic field H z from E = jωµ 0 (ω/c H z e z (1.1 kz Similarly, TM modes are determined by the axial electric field E z from H = jωε 0 (ω/c E z e z (1.13 kz In the following Sections, TE and TM modes in rectangular and circular waveguides will be discussed. 1.3 Rectangular Waveguides We assume a rectangular waveguide having an inner cross-section a b with a > b, as shown in Fig. 1.1. (We do not lose generality by making this assumption. The wall material is assumed to have a suffi ciently large conductivity so that in the lowest order approximation we can regard the wall material as an ideal conductor. (Otherwise, analysis will be rather complicated. Such approximation is not bad, as long as the skin depth is small enough and the field penetration into the wall is negligible. Later, in evaluation of power loss due to the finite (non-infinite wall conductivity, we will remove this assumption, but for now, we assume that the wall is an ideal conductor. For TE modes, we need the solution for the axial magnetic field H z which obeys the following 4

y b a x z Figure 1-1: Rectangular waveguide. scalar wave equation, ( 1 c t H z = 0 or more explicitly, ( x + y + z 1 c t H z = 0 (1.14 The z and t dependence has already been assumed to be in the form e j(ωt kzz, that is, H z (x, y, z, t = H z (x, y e j(ωt kzz (1.15 Noting / z = jk z, / z = k z, / t = jω, / t = ω, we may rewrite Eq. (1.15 as ( x + y k z + c H z (x, y = 0 (1.16 This is a two dimensional Helmholtz equation and can be solved by the method of separation of variables as done for the Laplace equation. The boundary conditions for the electric and magnetic fields on the surface of an ideal conductor are E t = 0, H n = 0 (1.17 where "t" and "n" indicate the tangential and normal component, respectively. For the geometry assumed in Fig. 1.1, these boundary conditions can be translated into E x = 0 at y = 0, b (1.18 E y = 0 at x = 0, a (1.19 E z = 0 at x = 0, a and y = 0, b (1.0 H x = 0 at x = 0, a (1.1 H y = 0 at y = 0, b (1. 5

Let us assume that the solution for H(x, y is separable as H(x, y = H 0 X(xY (y (1.3 where X(x is a function of x only, and Y (y is a function of y only. Then, Eq. (1.17 reduces to 1 d X X dx + 1 Y d Y dy + k c z = 0 (1.4 Since (ω/c k z is a constant, the functions X and Y must be either sinusoidal or exponential function. However, exponential functions cannot satisfy the boundary conditions, Eq. (1.19. Therefore, solutions for X and Y must be sinusoidal. Since the axial magnetic field H z is tangential to the wall everywhere, the boundary condition for the magnetic field, Eq. (1.0, is not useful. However, the boundary condition for the electric field, Eq. (1.19, enables us to determine the magnetic field H z (x, y as follows. From the x component of Eq. (1.10 wherein E z = 0 for TE mode, we have E x = jωµ 0 H z (ω/c kz y (1.5 This should vanish at y = 0 and b. Since H z (x, y is sinusoidal, its derivative with respect to y is still sinusoidal. Therefore, the solution for E x should contain a function ( nπ sin b y (1.6 where n is an integer. Integration of Eq. (1.3 thus determines the function Y (y, ( nπ Y (y = cos b y (1.7 Similarly, from the boundary condition for the y component of the electric field, we find ( mπ X (x = cos a x (1.8 and the general solution for the axial magnetic field becomes ( mπ ( nπ H z (x, y, z, t = H 0 cos a x cos b y e j(ωt kzz (1.9 The integers m and n cannot be zero simultaneously. (If so, both E x and E y identically vanish. Substitution of Eq. (1.7 into the original wave equation, Eq. (1.15, yields the following relationship between the frequency ω and the axial wavenumber k z, c = k z + ( mπ ( nπ + a b (1.30 6

y 3 1 0 0 1 3 Figure 1-: Dispersion relation ω = (ck z + ω c. y axis: ω/ω c. x axis ck z /ω c. The straight line shows ω = ck (propagation in free space. x or ω = (ck z + ω c where ω c is given by (mπ ( nπ ω c = c + (1.31 a b In Hz, f c = c (m ( n + (1.3 a b Only waves having frequencies higher than ω c can exist in the waveguide. The frequency ω c is thus called the cutoff frequency. For a given generator frequency ω, the dispersion relation determines the value of k z uniquely. The lowest frequency allowed for wave propagation occurs at k z = 0. The ω k z relationship is shown in Fig. 1.. Note that the dispersion relation is identical to that of the electromagnetic waves in a plasma, if the plasma frequency ω p is replaced by the cutoff frequency. Although the dispersion relations are formally identical, the plasma mode is still TEM as we have seen in Chapter 10. Therefore, physics behind the waveguide mode and plasma mode is distinctly different. 1 + x For a given cross-section of a rectangular waveguide, the smallest cutoff frequency of TE mn modes occurs when m = 1, n = 0. (Recall the assumption a > b. The cutoff frequency of the TE 10 mode is given by f c10 = c a (1.33 The second smallest cutoff frequency is that of TE 0 or TE 01 mode because most rectangular waveguides have a ratio a/b. The cutoff frequency of these modes are given by f c0 = c a, f c01 = c b (1.34 7

The reason for the particular ratio a/b is to avoid degeneracy between modes, that is, for a given generator frequency, only a single mode can be excited in a waveguide. In practice, the TE 10 mode is most frequently used, and we will study more about this particular mode in the following Section. 1.4 Field Profiles of TE 10 Mode In a waveguide, the direction of energy propagation is evidently along the waveguide, that is, in the z direction. The Poynting vector is therefore expected to be directed in the z direction. In the cartesian coordinates, the z component of the Poynting vector is given by S z = (E H z = E x H y E y H x, W/m (1.35 For the TE 10 mode, E x = 0 (because n = 0, and the Poynting vector reduces to If the axial magnetic fields of the TE 10 mode is assumed to be S z = E y H x (1.36 ( π H z (x, z, t = H 0 cos a x e j(ωt kzz (1.37 the y component of the electric field can be found from Eq. (1.11 E y (y, z, t = j ωµ ( 0 π π/a H 0 sin a x e j(ωt kzz (1.38 and the x component of the magnetic field from either Eq. (1.10 or more directly from H = 0, H y = jk ( z π π/a H 0 sin a x e j(ωt kzz (1.39 Other components are zero. If we introduce a complex amplitude for the electric field through the field components can be rewritten as E 0 = ωµ 0 π/a H 0 (1.40 ( π E y = E 0 sin a x e j(ωt kzz (1.41 H x = k ( z π E 0 sin ωµ 0 a x e j(ωt kzz (1.4 H z = j π/a ( π E 0 cos ωµ 0 a x e j(ωt kzz (1.43 8

The real parts of these fields are ( π Re E y = E 0 sin a x cos (ωt k z z (1.44 Re H z = k ( z π E 0 sin ωµ 0 a x cos (ωt k z z (1.45 Re H z = π/a ( π E 0 cos ωµ 0 a x sin (ωt k z z (1.46 which allow us to plot the field profiles (snap-shot at a given instant, say t = 0. This is shown in Fig. 1.3 over one axial wavelength, λ z = π/k z. The profiles shown propagate with the phase velocity ω/k z in the z direction. Observe that the boundary conditions for the fields E and H are satisfied on the waveguide wall. The electric field is normal to the wall and the magnetic field is tangential. The induced surface charge σ (C/m and the electric field at the wall surface are related through E n = σ ε 0 (1.47 as we learned in Chapter 3. Since the induced charge is varying with time, a surface current must flow according to the charge conservation σ t + J s = 0 (1.48 where J s (A/m is the surface (sheet current density on the wall surface. The surface current and the tangential component of the magnetic field at the wall are in turn related through n H t = J s (1.49 where n is the normal unit vector directed on the inner wall surface. The surface current profiles are shown in Fig. 1.5. Note that the surface current at the midway, x = a/, flows along the z axis. A thin slit cut along the waveguide at this location should not very much disturb the current profile and thus the whole electromagnetic field profile. This fact is exploited in microwave standing wave meters. A probe can be inserted through the slit to measure the electric field intensity along the waveguide. 1.5 Power Associated with TE 10 Mode The concrete expressions for the relevant fields of TE 10 mode in the preceding Section allow us to calculate the amount of power carried by the TE 10 mode. The power going through the cross-section 9

E y b E y E y a λ z ( π Figure 1-3: Electric field profile of the TE 10 mode E y (x, z = E 0 sin a x cos (k z z 1.5 z1 0.5 00 0. 0.4 0.6 0.8 x 1 Figure 1-4: Magnetic field profile of the TE 10 mode in a rectangular waveguide. of the waveguide can be evaluated by integrating the Poynting flux, P = where the z component of the Poynting flux is a 0 dx b 0 S z (1.50 S z = E y Hx = k ( z E0 sin π ωµ 0 a x (1.51 10

1.5 z1 0.5 00 0.0.40.60.8 x 1 Figure 1-5: Surface current density pattern on the upper inner surface. Then P = k z ωµ 0 E 0 = k z ωµ 0 E 0 a 0 ab sin ( π a x dx b (W (1.5 Noting we can rewrite k z ε0 = ωµ 0 µ 0 P rms = ε0 µ 0 1 1 ( fc f ( fc f E0 ab 4 (1.53 Example: A rectangular waveguide having an inner cross-section 1 cm (a = cm, b = 1 cm is excited by a klystron at a frequency 9 GHz and an RMS power of W. Estimate the peak electric field in the waveguide. What is the phase velocity? Group velocity? First, let us calculate the phase and group velocities. since the dispersion relation of waveguide modes and that of the plasma mode are identical, we can use the results in Chapter 10, v p = c 1 (fc /f = 5.43 108 m/sec, v g = 1 (f c /f c = 1.66 10 8 m/sec, where the cutoff frequency is f c = c/a = 7.5 10 9 Hz. The peak electric field, which occurs 11

at x = a/, can be estimated from Eq. (1.50 by equating the RMS power to W. The result is E 0 = 5. 10 3 V/m. Often the characteristic impedance of the TE 10 mode is defined by Z T E = E y µ0 = H x 1 ε 0 Note that the negative sign here is related to that of the Poynting vector in Eq. (1.34. In terms of the impedance, the power may be rewritten as It should be cautioned that the impedance defined in Eq. (1.51 is meaningful only for TE 1 ( fc f modes. For TM modes, the impedance takes a rather different form Z T M = 1.6 Circular Waveguides µ0 ε 0 1 ( fc f Electromagnetic waves in a conducting cylinder with circular cross-section can also be divided into TE and TM modes, as in rectangular wave guides. The axial components E z and H z satisfy the scalar wave equation ( 1 c t E z = 0, where the Laplacian in the cylindrical coordinates is ( 1c t H z = 0 (1.54 = ρ + 1 ρ ρ + 1 ρ φ + z (1.55 We consider a cylindrical waveguide with inner wall radius a as shown in Fig. 1.6. The boundary conditions for the electric and magnetic fields are E φ = E z = 0 at ρ = 0 H ρ = 0 at ρ = a For TE modes, solutions of the wave equation for H z (ρ, φ, z, t are required. Separating the z and t dependence as we can reduce the original wave equation to H z (ρ, φ, z, t = H z (ρ, φe j(ωt kzz (1.56 ( ρ + 1 ρ ρ + 1 ρ φ k z + c H z (ρ, φ = 0 (1.57 1

φ ρ a x z Figure 1-6: Circular waveguide with radius a and the cylyndrical coordinates (ρ, φ, z. We have encountered this type of equation in Chapter 3 on electrostatic boundary value problems. Since H z (ρ, φ should be a periodic function of φ, the φ dependence may be assumed either sin(nφ or cos(nφ where n is an integer. Then, noting ( φ sin (nφ cos (nφ = n ( sin (nφ cos (nφ we can further reduce Eq. (1.73 to an ordinary differential equation with respect to ρ, (1.58 ( d dρ + 1 d ρ dρ n k z + c ρ R (ρ = 0 (1.59 where H z (ρ, φ = R(ρF (φ with F (φ being either sin(nφ or cos(nφ. Introducing k defined by k = k c z (1.60 we finally reduce Eq. (1.75 to the standard form of the Bessel s equation ( d dρ + 1 d ρ dρ + k n ρ R (ρ = 0 (1.61 whose solution is the n-th order Bessel function J n (kρ. J 0 (x and J 1 (x are shown in Fig. 1.5. J 0 (x, J 1 (x y 1.0 0.8 0.6 0.4 0. 0.0 0. 0.4 4 6 8 10 1 14 16 18 x Bessel functions J 0 (x (solid and J 1 (x (dashed. J 0 (0 = 1.0, J 1 (0 = 0. 13

(The Bessel function of the second kind N n (kρ should be discarded because it diverges at ρ = 0. Therefore, the general solution for the axial magnetic field of circular TE modes may be written as H z (ρ, φ, z, t = H 0 J n (kρ e jnφ e j(ωt kzz (1.6 where sin(nφ and cos(nφ have been replaced by their equivalents, e ±jnφ. At this stage, the axial wavenumber k z is yet to be determined. It can be uniquely determined from the boundary conditions as follows. In TE modes we are considering, the axial electric field is evidently zero, and E z = 0 at ρ = a is automatically satisfied. The useful boundary condition for TE modes is therefore E φ = 0 at ρ = a. Recalling Eq. (1.11, ( E H z e z = e ρ ρ + e 1 φ H z e z ρ φ 1 H z = e ρ ρ φ e H z φ ρ (1.63 we see that the φ component of the electric field is generated by the radial derivative of the axial magnetic field. Therefore, the boundary condition E φ = 0 at ρ = 0 requires that dj n (kρ dρ = 0 at ρ = a (1.64 This is the basic condition to determine the dispersion relation of TE modes in a circular waveguide. Introducing a dimensionless variable x = kρ, we seek roots of the following equation dj n (x dx = 0 (1.65 Since the Bessel function J n (x oscillates, there are in fact infinitely many roots satisfying Eq. (1.81. In the Table, some low order solutions are listed. x mn means the m-th root of dj n (x/dx = 0. n = 0 n = 1 n = m = 1 3.83 1.84 3.05 m = 7.0 5.33 8.54 m = 3 10.17 8.54 9.97 Table 1.1 x mn (m-th root of dj n (x/dx = 0 The solution for k is therefore given by k mn = x mn a (1.66 14

and the desired dispersion relation by with the cutoff frequency defined by ( ω = (ck z x + c mn = (ck z + ω c (1.67 a ω c = c x mn a (1.68 The lowest order TE mode (having the smallest cutoff frequency corresponds to the smallest root of x mn, which occurs when m = 1, n = 1, x 11 1.84. As an example, consider a circular waveguide having a radius of 5 mm. The cutoff frequency of the TE 11 mode in the waveguide is f c = c 1.84 17.6GHz. πa Note that the indices m and n of the circular waveguide modes have entirely different meanings from those in rectangular modes. Let us assume that the axial magnetic field of the TE 11 mode is of the form H z (ρ, φ, z, t = H 0 J 1 (k 11 ρ cos φe j(ωt kzz (1.69 We have chosen cos φ function because sin φ function corresponds to the rotation by an angle π/ in φ direction. In so doing, we do not lose generality. The transverse electric and magnetic fields can be found by referring to the general formulae Eqs. (1.10 and (1.11, E = jωµ 0 (ω/c H z e z kz = jωµ 0 k H 0 ( 1 ρ J 1 (kρ sin φe ρ + dj 1 (kρ cos φe φ dρ (1.70 H = = jωµ 0 (ω/c H z kz ( jk z H 0 dj1 (kρ (ω/c H 0 cos φe ρ 1 kz dρ ρ J 1 (kρ sin φe φ (1.71 The rectangular TE 10 mode and circular TE 11 mode are in fact in the same family and mutually convertible when the sizes of both waveguides are not vastly different. This is illustrated in Fig. 1. 7. Conversion can be achieved by gradual tapering. (Abrupt change in the cross section shape causes large wave reflection and should be avoided. Such mutual conversion between rectangular and circular modes is often required because many microwave components are based on circular waveguide modes. Typical examples are attenuators and isolators. (See "Introduction to Microwave Technology". 15

Taper circular TE 11 mode rectangular TE 10 mode Figure 1-7: Circular TE 11 mode and rectangular TE 10 mode are topologically in the same family. One mode can be converted to another through a taper with gradual change in the cross-section. 1.7 TM Mode in Circular Slow Waveguide Waveguides used in linear electron accelerators must accommodate TM modes having a phase velocity close to c. Waveguides with smooth inner walls can only accommodate modes having phase velocities larger than c and thus cannot be used for this purpose. Modes must be TM because TE modes have no electric field in the axial direction needed to accelerate electrons. Figure 1-8: Circular slow waveguide of radius a with diaphragms of radius b. The spacing between diaphragms is much smaller than the axial wavelength λ z. Slow waveguides have conductor diaphragms placed periodically along the axis as shown in Fig. 1-8. The purpose of the diaphragms is to increase the capacitance per unit length of the waveguide which contributes to slowing down the phase velocity of electromagnetic waves. We consider modes symmetric about the axis, m = 0. The waveguide has a radius a and diaphragms have holes of a b Figure 1-9: In the region b < ρ < a, the eelctric field lines are straight, E z / z = 0. In the region ρ < b, the field lines are curved and E z / z = jk z E z. 16

radius b. In the region ρ < b, the Helmholtz equation for the axial electric field is ( ρ + 1 ρ ρ + c k z E z (ρ = 0, ρ < b. (1.7 Since we are interested in modes having an axial phase velocity slightly smaller than c, that is, ω k z c, (1.73 the quantity (ω/c k z must be negative, and thus solution for E z (ρ may be assumed to be E z (ρ = AI 0 (kρ, ρ < b (1.74 where k = kz > 0. (1.75 c Y 0 (x 1.0 0.5 0.0 1 3 4 5 6 7 8 9 10 x 0.5 1.0 J 0 (x and Y 0 (x (dashed In the diaphragm region b < ρ < a, the electric fields lines are essentially straight provided the axial period of the diaphragms is suffi ciently smaller than the axial wavelength. See Fig. 1-9. We assume that this condition is met. Then, the wave equation in the region b < ρ < a may be approximated by ( ρ + 1 ρ ρ + c General solutions are E z (ρ = 0, b < ρ < a. (1.76 E z (ρ = BJ 0 c ρ + CN 0 c ρ, b < ρ < a. (1.77 The boundary conditions are: and E z (ρ = a = 0, (1.78 E z and H φ be continuous at ρ = b. (1.79 17

These boundary conditions yield BJ 0 c a + CN 0 c a = 0, (1.80 AI 0 (kb = BJ 0 c b + CN 0 c b, (1.81 A k I 1(kb = c ] [BJ 1 ω c b + CN 1 c b, (1.8 where J 0 (x = J 1(x, N 0 (x = N 1(x, I 0 (x = I 1(x are noted. Eqs. (1.80 through (1.8 give the following dispersion relation ck ω c a N 0 c b J 0 c b N 0 c a I 0 (kb J 0 I 1 (kb =. (1.83 J 0 c a N 1 c b J 1 c b N 0 c a For the purpose of accelerating highly relativistic electrons, the axial phase velocity ω/k z must be close to c or k 0. Then I 0 (kb 1, I 1 (kb kb/, and the LHS of Eq. (1.83 reduces to c/ωb. For a given rf frequency ω and the size of the waveguide a, the dispersion relation c a N 0 c b J 0 c b N 0 c a c J 0 ωb, (1.84 J 0 c a N 1 c b J 1 c b N 0 c a can be solved numerically to determine the aspect ratio a/b of a slow wave circular waveguide. Fig.1-10 shows the function ( ( b b f(x = a J 0 (x N 0 xb a x J 0 a x N 0 (x ωa ( (, x = b b J 1 a x N 0 (x J 0 (x N 1 a x c, (1.85 when a/b =.5. The first root occurs at x 3.89 and for a given rf frequency ω, the outer radius a can thus be determined. 1.8 Dielectric Waveguides An optical fiber can confine light waves because of total reflection at the surface. In contrast to conductor waveguides, light waves in optical waveguides cannot be pure TE or TM modes. This is because electromagnetic fields outside, as well as inside, the optical fiber must be considered simultaneously. Although the outer fields are evanescent (otherwise waves cannot be confined, the fields near the surface do affect those inside. 18

1 0.5 03.5 3.6 3.7 x 3.8 3.9 4 0.5 1 Figure 1-10: Root of f(x = 0 when a/b =.5. We first consider a simple case of step change in the index of refraction, n(ρ = { n, ρ < a, 1, ρ > a. Such an optical fiber is of no practical interest, for fibers used in optical communication all have graded index of refraction with a gradual change with the radius ρ. The axial electric field E z (r, t satisfies the wave equations in both regions, ( ρ + 1 ρ ρ + 1 ρ φ + z + µ 0εω E z < (r = 0, ρ < a, (1.86 ( ρ + 1 ρ ρ + 1 ρ φ + z + µ 0ε 0 ω E z > (r = 0, ρ > a. (1.87 The azimuthal dependence may be assumed to be e imφ and the axial dependence e ikzz, E(r = E(ρe imφ+ikzz. Then, ( d dρ + 1 d ρ dρ m ρ + µ 0εω kz ( d dρ + 1 d ρ dρ m ρ + µ 0ε 0 ω kz which admit the following bounded solutions, E < z (ρ = 0, ρ < a, (1.88 E > z (ρ = 0, ρ > a, (1.89 E < z (ρ = AJ m (k 1 ρ, ρ < a, (1.90 E > z (ρ = BK m (k ρ, ρ > a. (1.91 19

Here k 1 = µ 0 εω k z, k = k z µ 0 ε 0 ω. (1.9 Note that the outer field should be evanescent for the waveguide to confine light waves. Similarly, the axial magnetic field H z (r may be assumed to be H < z (ρ = CJ m (k 1 ρ, ρ < a, (1.93 H > z (ρ = DK m (k ρ, ρ > a. (1.94 The transverse fields E and H can then be calculated by referring to Eqs. (1.9 and (1.10. The azimuthal components of the fields are and E < φ (ρ = i k 1 E φ > (ρ = i ( im k k z H < φ (ρ = i k 1 H > φ (ρ = i k 1 The continuity of E z, E φ, H z and H φ yields ( im k z ρ AJ m(k 1 ρ ωµ 0 k 1 CJ m(k 1 ρ, (1.95 ρ BJ m(k ρ ωµ 0 k DK m(k ρ ( k z im ρ CJ m(k 1 ρ + ωεk 1 AJ m(k 1 ρ, (1.96, (1.97 ( im k z ρ DK m(k ρ + ωε 0 k BK m(k 1 ρ. (1.98 AJ m (k 1 a = BK m (k a, (1.99 CJ m (k 1 a = DK m (k a, (1.100 1 ( im k1 k z a AJ m(k 1 a ωµ 0 k 1 CJ m(k 1 a = 1 ( im k k z a BK m(k a ωµ 0 k DK m(k a, ( (1.101 1 im k z a CJ m(k 1 a + ωεk 1 AJ m(k 1 a = 1 ( im k k z a DK m(k a + ωε 0 k BK m(k 1 a. (1.10 k 1 Then the determinantal dispersion relation is ( 1 k 1 + 1 k ( mkz = a ( ω ( J m (k 1 a k 1 c 1 J m (k 1 a ( + ω 1 k 1 k c 1 + 1 c + k c ( K m (k a K m (k a J m (k 1 ak m(k a J m (k 1 ak m (k a. (1.103 Figure 1-11 shows the dispersion relation, namely, axial wavenumber k z normalized by k 0 = ω/c as a function of the normalized frequency k 0 a = ωa/c when m = 1 and n = 1.1. It can be seen that a cutoff occurs at ωa/c 1.386. Fig.1-1 shows the case when m =. The cutoff frequency increases to ωa/c 5.43. 0

1.1 1.08 1.06 1.04 1.0 10 5 10 15 x 0 5 30 Figure 1-11: ck z /ω vs. ωa/c when n = 1.1, m = 1. The cutoff frequency is ω c a/c 1.386. 1.1 1.08 1.06 1.04 1.0 10 5 10 15 x 0 5 30 Figure 1-1: ck z /ω vs. ωa/c when n = 1.1, m =. The cutoff frequency is ω c a/c 5.43. 1.9 Graded Index Fibers In optical fibers used in practical communication, the index of refraction is designed to have gradual, rather than step, variation with the radius. Quadratic variation is commonly employed, n(ρ = n 0 [ 1 α ρ ], α = constant (1.104 because the electromagnetic fields are then well confined with a Gaussian profile e a ρ. corresponding permittivity is The ε(ρ = ε 0 n [ 0 1 α ρ ]. (1.105 Since [ε(re] = ε(r E + E ε = 0, (1.106 1

the wave equation E = ω µ 0 ε(re, (1.107 reduces to ( E ε E + ω µ 0 ε(re + = 0. (1.108 ε If the change in the permittivity is small, the last term can be ignored in the lowest order approximation, and we obtain a simple wave equation with an inhomogeneous permittivity, E + ω µ 0 ε(re 0. (1.109 In the cylindrical geometry, a cartesian component of the transverse electric field satisfies For weak variation of ε(ρ, ( ρ + 1 ρ ρ + 1 ρ φ + z + ω µ 0 ε(ρ E i = 0. (1.110 ε(ρ = ε 0 n 0(1 α ρ ε 0 n 0 ( 1 α ρ, and axially symmetric mode / φ = 0, Eq. (1.110 reduces to ( d dρ + 1 d ρ dρ k z + k0 ( 1 α ρ E(ρ = 0, (1.111 where Assuming we find k 0 = ω ε 0 µ 0 n 0. (1.11 E(ρ = E 0 e a ρ, a = 1 k 0 α, (1.113 k z = k 0 4 k 0 α. (1.114 The electric field is confined with a Gaussian profile in the radial direction. The e-folding radial distance is w = 1 a = 4 k0 α, (1.115 which is called beam radius. A constant beam radius is maintained only for appropriate injection of light wave at the input end. If not, the beam radius varies with the axial distance accompanied by periodic focusing and defocusing as intuitively expected from the picture of