14.1 Rectangular Waveguides TM modes TE modes Cylindrical waveguides TM modes TE modes...

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1 14 TM and TE modes Contents 14.1 Rectangular Waveguides TM modes TE modes Cylindrical waveguides TM modes TE modes Keywords: Rectangular and cylindrical waveguides, TM and TE modes, Bessel functions, Zeros of Bessel functions Ref: J. D. Jackson: Classical Electrodynamics; D. J. Griffith: Introduction to Electrodynamics Rectangular Waveguides In this section we shall consider a rectangular metallic waveguide of height described in the previous chapter. The height (which is along the x axis of the waveguide is a and the width is b (along the y axis TM modes The TM modes are the transverse magnetic modes where the component of the magnetic field vanishes along the waveguide. For our case this amounts to

2 2 14 TM and TE modes B Z = 0 everywhere inside the guide. Hence the equation (13.32 is satisfied trivially. We are left the equation(13.31, which has to be solved subjected to the boundary condition E = 0 (Dirichlet boundary condition as discussed in the previous chapter. On the boundary one of the parallel components is always E z. The whole boundary condition now reads as E z (0,y = E z (a,y = E z (x,0 = E z (x,b = 0, (14.1 E x (x,0 = E x (x,b = 0, (14.2 E y (0,y = E y (a,y = 0. (14.3 One can write down the solution of equation (13.31 as, E z (x,y = E 0 sin(mπx/asin(nπy/b, (14.4 satisfying the boundary condition (14.1. m,n = 1,2,3 and a solution like (14.4 this is called a TM mn mode. For a TM mn mode, k 2 = (ω/c 2 (mπ/a 2 (nπ/b 2, (14.5 k 2 = k 2 0 (mπ/a 2 (nπ/b 2, (14.6 (2π/λ 2 = (2π/λ 0 2 (2π/λ c 2, ( λ = 1 2 λ 2 1, ( λ 2 c where λ c defined as, 1 = 1 (m 2 ( n 2, + (14.9 λ c 2 a b

3 14.1 Rectangular Waveguides 3 is the cut-off wavelength. λ and λ 0 are the effective wavelength inside the guide and the free space wavelength respectively. It is essential to have λ c > λ 0 for any propagation of electromagnetic wave through the guide. If the case is opposite then the square of effective wavelength will be negative and one will have a complex wavenumber which in turn attenuate the wave inside the guide as explained in the previous chapter. Now E x and E y are calculated from the expressions (13.26 and (13.24 respectively, as E x (x,y = E y (x,y = ike ( 0 mπ (ω/c 2 k 2 a ike 0 (ω/c 2 k 2 ( nπ b cos(mπx/a sin(nπy/b, (14.10 sin(mπx/a cos(nπy/b. (14.11 One can verify easily that solutions (14.10 and (14.11 satisfy respective boundary conditions (14.2 and (14.3. The components B x and B y are obtained from the expressions (13.25 and (13.27 respectively, giving B x (x,y = iωe ( 0 nπ sin(mπx/a cos(nπy/b, (14.12 ω 2 c 2 k 2 b B y (x,y = iωe ( 0 mπ cos(mπx/a sin(nπy/b. (14.13 ω 2 c 2 k 2 a We see that B x (0,y = B x (a,y = 0 and B y (x,0 = B y (x,b = 0. These are synonymous to the normal components of the magnetic field vanishing on the boundary of the guide as required. The figures 14.1 and 14.2 show the field lines for TM 11 and TM 21 respectively. The wave is propagating into the paper.

4 4 14 TM and TE modes Fig. 14.1: Electric and magnetic field lines of the TM 11 mode Fig. 14.2: Electric and magnetic field lines of the TM 21 mode Problem 1: Show that TEM modes cannot propagate through hollow waveguides TE modes For transverse electric or TE modes one has E z = 0 everywhere inside the guide. In this case the equation (13.31 is trivially satisfied. We have to solve the equation (13.32 with N B z = 0 (normal derivative of B z on the boundary. Solution for this can be written as, B z (x,y = B 0 cos(mπx/acos(nπy/b,. (14.14

5 14.1 Rectangular Waveguides 5 This solution will denoted by TE m,n. In this case either of m and n can be be zero as well, but not both. So we will have modes like TE 10, TE 01, TE 11, TE 02 etc.. The cut-off wavelength is given by the same formula -(14.9. On the boundary x = 0&a the normal direction is x and on the boundary y = 0&bthenormaldirectionisy so,wefindthatthesolution(14.14satisfies, x B z (0,y = x B z (a,y = 0 and y B z (x,0 = y B z (x,b = 0, (14.15 the required boundary conditions. We find B x and B y using (13.25 and (13.27 respectively B x (x,y = ikb ( 0 mπ (ω/c 2 k 2 a B y (x,y = ikb 0 (ω/c 2 k 2 ( nπ b sin(mπx/a cos(nπy/b, (14.16 cos(mπx/a sin(nπy/b. (14.17 The above expressions are satisfying B x (0,y = B x (a,y = 0 and B y (x,0 = B y (x,b = 0, i.e. the B = 0 on the boundary. Problem 2: Obtain the expressions for E x and E y for the transverse electric case and show that they satisfy the condition E = 0 on the boundary. Problem 3: A rectangular metallic waveguide has dimensions, a = 3.0 cm and b = 4.0 cm. Find out the modes (TE and TM in which a free space wavelength of 4.5 cm could propagate through it.

6 6 14 TM and TE modes 14.2 Cylindrical waveguides The basic strategy of solving this problem is the same as the above. Since the geometry is different the problem is solved in the cylindrical co-ordinate system (ρ,φ,z TM modes The equation (13.31 in cylindrical co-ordinates read as 2 E z ρ + 1 E z 2 ρ ρ E z ρ 2 φ 2 +(ω2 /c 2 k 2 E z = 0. (14.18 Assume E z = ξ(ρφ(φ and substitute in ( Dividing the equation by E z and multiplying by ρ 2 we obtain the following after some re-arrangements, ρ 2 ξ 2 ξ ρ + ρ ξ 2 ξ ρ ++ρ2 (ω 2 /c 2 k 2 = 1 2 Φ (14.19 Φ φ 2 Inthiswaythevariablesareseparatedandtherighthandsideoftheequation is only dependent of φ, whereas the left hand side is dependent on ρ. The angular part of the solution can be solved easily giving two independent solution, Φ(φ = A cos(pφ or B sin(pφ, (14.20 where p 2 is the separation constant in the equation ( Since electric and magnetic fields are measurable quantities one would like to have Φ(φ single-valued and this would demand integer ps. The radial equation now

7 14.2 Cylindrical waveguides 7 reduces to, ρ 2 2 ξ ρ 2 +ρ ξ ρ ++(ρ2 (ω 2 /c 2 k 2 p 2 ξ = 0. (14.21 The above equation is identified as cylindrical Bessel differential equation. The solution of this is known as cylindrical Bessel function, ξ(ρ = J p ( (ω/c 2 k 2 ρ (14.22 Bessel functions of first few orders are shown in Fig. (14.3. One can write J J 1 J2 J 3 J 4 J Fig. 14.3: Cylindrical Bessel functions two degenerate solutions for each non zero integer p as E z (ρ,φ = AJ p ( (ω/c 2 k 2 ρcos(pφ, (14.23 and E z (ρ,φ = AJ p ( (ω/c 2 k 2 ρsin(pφ. (14.24

8 8 14 TM and TE modes For p = 0 there is only one solution, E z (ρ,φ = AJ 0 ( (ω/c 2 k 2 ρ. (14.25 Now imposing the boundary condition, E z (R,φ = 0 (tangential component, we have (ω/c 2 k 2 R = ρ p,n, where ρ p,n is the n th zero of the p th order Bessel function J p. First few zeros of Bessel functions of different orders are the following. ρ 0,1 = , ρ 1,1 = , ρ 2,1 = , ρ 0,2 = , ρ 3,1 = , ρ 1,2 = etc.. Now where the cut-off λ c defined as, k 2 = (ω/c 2 (ρ p,n /R 2, (14.26 k 2 = k 2 0 (ρ p,n /R 2, (14.27 (2π/λ 2 = (2π/λ 0 2 (2π/λ c 2, ( λ = 1 2 λ 2 1, ( λ 2 c λ c = 2πR ρ p,n, (14.30 The modes are designated by TM p,n, like TM 01, TM 12 etc.. Problem 4: Obtain the expressions for E x, E y, B x and B y for the transverse magnetic case of cylindrical waveguides and show that they satisfy proper boundary conditions.

9 14.2 Cylindrical waveguides TE modes For TE modes of cylindrical waveguide analysis is similar to that of the TM case. In this case we work with the field B z instead of E z and the the boundary condition is Neumann, i.e. N B Z (R,φ = r B z (R,φ = 0. Like the TM case we write the solutions of (13.32 as, or for non-zero integer p and B z (ρ,φ = AJ p ( (ω/c 2 k 2 ρcos(pφ (14.31 B z (ρ,φ = AJ p ( (ω/c 2 k 2 ρsin(pφ (14.32 B z (ρ,φ = AJ 0 ( (ω/c 2 k 2 ρ. (14.33 for p = 0. Imposing the boundary condition we obtain, (ω/c 2 k 2 R = ρ p,n, where ρ p,n is the n th zero of the derivative of p th order Bessel function J p, i.e. ρ J p (ρ p,n = 0. FirstfewzerosofderivativeofBesselfunctionsofdifferent orders are the following. ρ 1,1 = , ρ 2,1 = , ρ 0,1 = , ρ 3,1 = , ρ 4,1 = , ρ 1,2 = etc.. So the cut-off λ c defined as, λ c = 2πR, (14.34 ρ p,n The modes are designated by TE p,n, like TE 11, TE 02 etc.. Problem 5: Obtain other components of magnetic and electric fields for the transverse electric case of cylindrical waveguides and show that they satisfy

10 10 14 TM and TE modes proper boundary condition. Problem 6: A cylindrical metallic waveguide has radius, R = 1.0 cm. Find out the modes (TE and TM in which a free space wavelength of 1.5 cm could propagate through it. Problem 7: Plot field lines of TE 11 mode of cylindrical waveguide.