2.4 Circular Waveguide y a x Figure 2.5: A circular waveguide of radius a. For a circular waveguide of radius a (Fig. 2.5), we can perform the same sequence of steps in cylindrical coordinates as we did in rectangular coordinates to find the transverse field components in terms of the longitudinal (i.e. Ez, Hz) components. In cylindrical coordinates, the transverse field is ^ ^ ET =½E ^ ½ + ÁEÁ HT =½H ^ ½ + ÁHÁ (2.66) Using this in Maxwell’s equations (where the curl is applied in cylindrical coordinates) leads to µ ¶ µ ¶ j !² @Ez @Hz ¡j @Ez !¹ @Hz H½ = 2 ¡ ¯ (2.67) E½ = 2 ¯ ¡ (2.69) kc ½ @Á @½ kc @½ ½ @Á µ ¶ µ ¶ ¡j @Ez ¯ @Hz ¡j ¯ @Ez @Hz HÁ = 2 !² ¡ (2.68) EÁ = 2 ¡ !¹ (2.70) kc @½ ½ @Á kc ½ @Á @½ 2 2 2 where kc = k ¡ ¯ as before. Please note that here (as well as in rectangular waveguide derivation), we have assumed e¡j¯z propagation. For e+j¯z propagation, we replace ¯ with ¡¯. 2.4.1 TE Modes We don’t need to prove that the wave travels as e§j¯z again since the differentiation in z for the Laplacian is the same in cylindrical coordinates as it is in rectangular coordinates (@2=@z2). However, the ½ and Á derivatives of the Laplacian are different than the x and y derivatives. The wave equation for Hz is 2 2 (r + k )Hz = 0 (2.71) µ ¶ @2 1 @ 1 @2 @2 + + + + k2 H (½; Á; z) = 0 (2.72) @½2 ½ @½ ½2 @Á2 @z2 z ECEn 462 17 September 12, 2003 ¡j¯z Using the separation of variables approach, we let Hz(½; Á; z) = R(½)P (Á)e , and obtain 2 3 6 1 1 7 4R00P + R0P + RP 00 + (k2 ¡ ¯2) RP 5 e¡j¯z = 0 (2.73) ½ ½2 | {z } 2 kc Multiplying by a common factor leads to 00 0 00 2 R R 2 2 P ½ + ½ + ½ kc + = 0 (2.74) | R {zR } |{z}P function of ½ function of Á Because the terms in this equation sum to a constant, yet each depends only on a single coordinate, each term must be constant: P 00 = ¡k2 ! P 00 + k2P = 0 (2.75) P Á Á so that P (Á) = A0 sin(kÁÁ) + B0 cos(kÁÁ) (2.76) Using this result in (2.74) leads to R00 R0 ½2 + ½ + (½2k2 ¡ k2) = 0 (2.77) R R c Á or 2 00 0 2 2 2 ½ R + ½R + (½ kc ¡ kÁ)R = 0 (2.78) This is known as Bessel’s Differential Equation. Now, we could use the Method of Frobenius to solve this equation, but we would just be repeating a well- known solution. The series you obtain from such a solution has very special properties (a lot like sine and cosine: you may recall that sin(x) and cos(x) are really just shorthand for power series that have special properties). The solution is R(½) = C0JkÁ (kc½) + D0NkÁ (kc½) (2.79) where Jº(x) is the Bessel function of the first kind of order º and Nº(x) is the Bessel function of the second kind of order º. 1. First, let’s examine kÁ. £ ¤ ¡j¯z Hz(½; Á; z) = C0JkÁ (kc½) + D0NkÁ (kc½) [A0 sin(kÁÁ) + B0 cos(kÁÁ)] e (2.80) Clearly, Hz(½; Á; z) = Hz(½; Á + 2¼`; z) where ` is an integer. This can only be true if kÁ = º , where º = integer. ¡j¯z Hz(½; Á; z) = [C0Jº(kc½) + D0Nº(kc½)] [A0 sin(ºÁ) + B0 cos(ºÁ)] e (2.81) ECEn 462 18 September 12, 2003 2. It turns out that Nº(kc½) ! ¡1 as ½ ! 0. Clearly, ½ = 0 is in the domain of the waveguide. Physically, however, we can’t have infinite field intensity at this point. This leads us to conclude that D0 = 0 . We now have ¡j¯z Hz(½; Á; z) = [A sin(ºÁ) + B cos(ºÁ)] Jº(kc½)e (2.82) 3. The relative values of A and B have to do with the absolute coordinate frame we use to define the waveguide. For example, let A = F cos(ºÁ0) and B = ¡F sin(ºÁ0) (you can find a value of F and Á0 to make this work). Then A sin(ºÁ) + B cos(ºÁ) = F sin [º(Á ¡ Á0)] (2.83) The value of Á0 that makes this work can be thought of as the coordinate reference for measuring Á. So, we really are left with finding F , which is simply the mode amplitude and is therefore determined by the excitation. 4. We still need to determine kc. The boundary condition that we can apply is EÁ(a; Á; z) = 0, where ½ = a represents the waveguide boundary. Since j!¹ @Hz EÁ(½; Á; z) = 2 (2.84) kc @½ j!¹ 0 ¡j¯z = 2 [A sin(ºÁ) + B cos(ºÁ)] kcJº(kc½)e (2.85) kc where d J 0 (x) = J (x); (2.86) º dx º 0 our boundary condition indicates that Jº(kca) = 0. So p0 k a = p0 ! k = ºn (2.87) c ºn c a 0 0 0 where pºn is the nth zero of Jº(x). Below is a table of a few of the zeros of Jº(x): 0 Jº(kca) = 0 n = 1 n = 2 n = 3 º = 0 0.0000 3.8317 7.0156 º = 1 1.8412 5.3314 8.5363 º = 2 3.0542 6.7061 9.9695 2 2 2 5. We have already defined kc = k ¡ ¯ , so µ ¶ p0 2 ¯2 = k2 ¡ ºn (2.88) a Note that there is no “Á” term here. However, the Á variation of the fields in the waveguide does influence ¯. (How?) 6. Cutoff frequency (¯ = 0): Since k = kc = 2¼fc;ºn=c at the mode cutoff frequency, c p0 f = ºn (2.89) c;ºn 2¼ a ECEn 462 19 September 12, 2003 0 7. Dominant Mode: We don’t count the º = 0, n = 1 mode (TE01) since p01 = 0 resulting in zero 0 fields. The dominant TE mode is therefore the mode with the smallest non-zero value of pºn, which is the TE11 mode. 8. The expressions for wavelength and phase velocity derived for the rectangular waveguide apply here as well. However, you must use the proper value for the cutoff frequency in these expressions. 2.4.2 TM Modes The derivation is the same except that we are solving for Ez. We can therefore write ¡j¯z Ez(½; Á; z) = [A sin(ºÁ) + B cos(ºÁ)] Jº(kc½)e (2.90) Our boundary condition in this case is Ez(a; Á; z) = 0 or Jº(kca) = 0. This leads to p k a = p ! k = ºn (2.91) c ºn c a where pºn is the nth zero of Jº(x). Jº(kca) = 0 n = 1 n = 2 n = 3 º = 0 2.4048 5.5201 8.6537 º = 1 3.8317 7.0156 10.1735 º = 2 5.1356 8.4172 11.6198 In this case, we have ³p ´2 ¯2 = k2 ¡ ºn (2.92) a c p f = ºn (2.93) c;ºn 2¼ a It becomes clear the the TE11 mode is the dominant overall mode of the waveguide. ECEn 462 20 September 12, 2003 2.4.3 Bessel Functions Here are some of the basic properties of Bessel functions: X1 (¡1)n(x=2)2n+º J (x) = (2.94) º n!(n + º)! n=0 Jp(x) cos (p¼) ¡ J¡p(x) Nº(x) = lim (2.95) p!º sin (p¼) º Jº(¡x) = (¡1) Jº(x); º = integer (2.96) r 2 J (x) ' cos (x ¡ ¼=4 ¡ º¼=2); x ! 1 (2.97) º ¼x r 2 N (x) ' sin (x ¡ ¼=4 ¡ º¼=2); x ! 1 (2.98) º ¼x d Z (x) = Z (x) ¡ ºZ (x)=x (2.99) dx º º¡1 º where Z is any Bessel function. Figures 2.6 and 2.7 show Bessel functions of the first and second kinds of orders 0, 1, 2, 3. 1 J (x) 0 J (x) 1 J (x) 2 J (x) 3 0.5 0 −0.5 0 5 10 15 20 x Figure 2.6: Bessel functions of the first kind. ECEn 462 21 September 12, 2003 1 N (x) 0 N (x) 1 N (x) 2 N (x) 3 0.5 0 −0.5 −1 −1.5 0 5 10 15 20 x Figure 2.7: Bessel functions of the second kind. ECEn 462 22 September 12, 2003.