Cylindrical Waves Guided Waves Cylindrical Waves Daniel S. Weile Department of Electrical and Computer Engineering University of Delaware ELEG 648—Waves in Cylindrical Coordinates D. S. Weile Cylindrical Waves 2 Guided Waves Cylindrical Waveguides Radial Waveguides Cavities Cylindrical Waves Guided Waves Outline 1 Cylindrical Waves Separation of Variables Bessel Functions TEz and TMz Modes D. S. Weile Cylindrical Waves Cylindrical Waves Guided Waves Outline 1 Cylindrical Waves Separation of Variables Bessel Functions TEz and TMz Modes 2 Guided Waves Cylindrical Waveguides Radial Waveguides Cavities D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Outline 1 Cylindrical Waves Separation of Variables Bessel Functions TEz and TMz Modes 2 Guided Waves Cylindrical Waveguides Radial Waveguides Cavities D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes The Scalar Helmholtz Equation Just as in Cartesian coordinates, Maxwell’s equations in cylindrical coordinates will give rise to a scalar Helmholtz Equation. We study it first. r2 + k 2 = 0 In cylindrical coordinates, this becomes 1 @ @ 1 @2 @2 ρ + + + k 2 = 0 ρ @ρ @ρ ρ2 @φ2 @z2 We will solve this by separating variables: = R(ρ)Φ(φ)Z(z) D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Separation of Variables Substituting and dividing by , we find 1 d dR 1 d2Φ 1 d2Z ρ + + + k 2 = 0 ρR dρ dρ ρ2Φ dφ2 Z dz2 The third term is independent of φ and ρ, so it must be constant: 1 d2Z = −k 2 Z dz2 z This leaves 1 d dR 1 d2Φ ρ + + k 2 − k 2 = 0 ρR dρ dρ ρ2Φ dφ2 z D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Separation of Variables Now define the Radial Wavenumber 2 2 2 kρ = k − kz and multiply the resulting equation by ρ2 to find ρ d dR 1 d2Φ ρ + + k 2ρ2 = 0 R dρ dρ Φ dφ2 ρ The second term is independent of ρ and z, so we let 1 d2Φ = −n2 Φ dφ2 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Separation of Variables This process leaves an ordinary differential equation in ρ alone. We thus have: The Cylindrical Helmholtz Equation, Separated d dR h i ρ ρ + (k ρ)2 − n2 R = 0 dρ dρ ρ d2Φ + n2Φ = 0 dφ2 d2Z + k 2Z = 0 dz2 z 2 2 2 kρ + kz = k The first of these equations is called Bessel’s Equation; the others are familiar. D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes The Harmonic Equations We have already seen equations like those in the z and φ directions; the solutions are trigonometric, or exponential. The only novelty is that φ is periodic or finite; it therefore is always expanded in a series and not an integral. If there is no limit in the φ direction we find The Periodic Boundary Condition Φ(φ) = Φ(φ + 2π) This implies that n 2 Z if the entire range is included. D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Outline 1 Cylindrical Waves Separation of Variables Bessel Functions TEz and TMz Modes 2 Guided Waves Cylindrical Waveguides Radial Waveguides Cavities D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel’s Equation For Statics The remaining equation to be solved is the radial equation, i.e. Bessel’s Equation. Note that the problem simplifies considerably if kρ = 0 (which would be the case if ρ = 0. In this case, we have Bessel’s Equation for Statics d dR ρ ρ − n2R = 0 dρ dρ To solve it, let ρ = ex so dρ = ex = ρ. dx D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel’s Equation For Statics This implies that d dρ d d = = ρ dx dx dρ dρ Our equation therefore becomes d2R − n2R = 0 dx2 The solutions to this are A + Bx n = 0 R(x) = Aenx + Be−nx n 6= 0 and thus the solutions really are Static Solutions of Bessel’s Equation A + B ln ρ n = 0 R(ρ) = Aρn + Bρ−n n 6= 0 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel Functions We are generally more interested in the dynamic case in which we must solve the full Bessel Equation: d dR h i ξ ξ + ξ2 − n2 R = 0 dξ dξ (We normalize kρ = 1, and rewrite the equation in terms of ρ instead of ξ) To solve this equation, we suppose 1 α X m R(ξ) = ξ cmξ m=0 Now 1 dR X = (α + m)c ξα+m−1 dξ m m=0 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel Functions Thus 1 dR X ξ = (α + m)c ξα+m dξ m m=0 and 1 d dR X ξ = (α + m)2c ξα+m−1 dξ dξ m m=0 so finally 1 d dR X ξ ξ = (α + m)2c ξα+m dξ dξ m m=0 Now, we can plug in... 1 1 X 2 α+m h 2 2i X α+m (α + m) cmξ + ξ − n cmξ = 0 m=0 m=0 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel Functions We now have 1 1 X h 2 2i α+m X α+m+2 (α + m) − n cmξ + cmξ = 0 m=0 m=0 or 1 1 X h 2 2i α+m X α+m (α + m) − n cmξ + cm−2ξ = 0 m=0 m=2 We can proceed by forcing the coefficients of each term to vanish. We fix c0 6= 0 because of the homogeneity of the equation. D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel Functions For ξα: (α2 − n2) = 0 since c0 6= 0 by assumption, α = ±n: For ξα+1: h 2 2i (α + 1) − n c1 = 0 Thus c1 = 0 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel Functions Finally, for all other ξα+m: h 2 2i (α + m) − n cm + cm−2 = 0 Assuming α = n, −1 c = c m m(m + 2n) m−2 Thus, immediately, for p 2 Z c2p+1 = 0 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel Functions Given that the odd coefficients vanish, we let m = 2p and let ap = c2p So... −1 a = c = c = a p 2p 4p(p + n) 2p−2 p−1 and −1 a = a 1 4(n + 1) 0 −1 −1 a = a 2 4(n + 2)(2) 4(n + 1) 0 −1 −1 −1 a = a 3 4(n + 3)(3) 4(n + 2)(2) 4(n + 1) 0 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Bessel Functions Thus, in general, (−1)pn! a = a p 4pp!(n + p)! 0 −n Thus, if we choose 2 n!a0 = 1, and recall 1 1 α X m X 2p+n R(ξ) = ξ cmξ = apξ m=0 p=0 we can (finally!) define the Bessel Function of Order n 1 2p+n X (−1)p ξ J (ξ) = n p!(n + p)! 2 p=0 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes Observations This function is entire; it exists and is differentiable for all ξ. This is only one solution of the equation: We will get to the other shortly. The other solution is not regular at the origin since the coefficient of the second order derivative vanishes there. Note that the solution looks like the corresponding static (ρn) solution at the origin. Note also that fractional orders are possible, but do not arise as commonly in applications (Why not?) D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes The Other Solution Our original equation (normalized) was d du h i ξ ξ + ξ2 − n2 u = 0 dξ dξ The other solution, v must solve d dv h i ξ ξ + ξ2 − n2 v = 0 dξ dξ Multiply the first equation by v and the second by u, subtract, and divide by ξ: d du d dv v ξ − u ξ = 0 dξ dξ dξ dξ D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes The Other Solution Expanding this out ξ(u00v − uv 00) + u0v − uv 0 = 0; or d ξ(u0v − uv 0) = 0 dξ It therefore stands to reason that ξ(u0v − uv 0) = C or u0v − uv 0 C = 2 v 2 ξv 2 D. S. Weile Cylindrical Waves Separation of Variables Cylindrical Waves Bessel Functions Guided Waves TEz and TMz Modes The Other Solution This of course implies d u C = 2 dξ v ξv 2 This can be integrated to give u Z dξ = C + C v 1 2 ξv 2 or Z dξ u(ξ) = C v(ξ) + C v(ξ) 1 2 ξv(ξ)2 D.