ECE 6130 Rectangular Waveguides Text Sections: 3.3 Chapter 3, Problem 3 (See Appendix I) and Derive the TM modes of a rectangular waveguide following the methods described here for TE modes. Rectangular Waveguides Recall: Method of solution: 1. Solve Helmholtz equation for either Hz (TE) or Ez (TM). This can be done analytically or numerically. In the analytical case, you guess the form of the solution, which will have several unknown constants (like magnitude, phase, number of cycles) 2. Use 3.19 to 3.23 to find transverse components from Ez or Hz. 3. Solve for the constants from the boundary conditions. In metal boundaries, these are that tangential E and normal H = 0 on the boundary. Now you have Ez or Hz. 4. Use Maxwell’s equation to find the other E or H components. TE Solution 1. Solve Helmholtz wave equation: a) Use Method of Separation of Variables: hx(x,y) = X(x) Y(y) b) Substitute into wave equation (Helmholtz equation): where kc2 = kx2 + ky2 c) Separate the Variables: d) “Guess” the form of the solution: 0),(22222=⎟⎟⎠⎞⎜⎜⎝⎛+∂∂+∂∂yxhkyxzc022222=++ XYkdyYdXdxXdYc00222222=⎥⎦⎤⎢⎣⎡+=⎥⎦⎤⎢⎣⎡+YkdyYdXXkdxXdYyxhz(x,y) = (A coskxx + B sin kxx) (C cos kyy + D sin kyy) The unknown constants are ABCD. Also kx and ky. 2. Solve for the unknown constants from boundary conditions. a) Define the boundary conditions Tangential E fields =0 on the metal surfaces (walls of the waveguide) ex = 0 at y=0,b ey = 0 at x=0,a b) Obtain appropriate expressions for the boundary condition fields From equations 3.19: Ex =(-jωμ / kc2) ∂Hz / ∂y Ey =(jωμ / kc2) ∂Hz / ∂x So: ex(x,y) =(-jωμ / kc2) ky (A coskxx + B sin kxx) (- Csin kyy + D cos kyy) ey(x,y) = (jωμ / kc2) kx (-A sinkxx + B cos kxx) (C cos kyy + D sin kyy) c) Use boundary conditions to solve for ABCD: Substitute ex = 0 at y=0,b into equations above. ex(x,0) =(-jωμ / kc2) ky (A coskxx + B sin kxx) (- Csin ky0 + D cos ky0)=0 So, D = 0 Note that a “trivial solution” also exists if ky =0 ex(x,b) =(-jωμ / kc2) ky (A coskxx + B sin kxx) (- Csin kyb + 0 cos kyb)=0 when ky = nπ/b and kx = mπ/a Substitute ey =0 at x=0,a into equations above: ey(0,y) = (jωμ / kc2) kx (-A sinkx0 + B cos kx0) (C cos kyy + D sin kyy) =0 So, B=0 ey(x,a) = (jωμ / kc2) kx (-A sinkxa + B cos kxa) (C cos kyy + D sin kyy)=0 So, kx = mπ/a Now we can simplify the form: ex(x,b) =(-jωμ / kc2) (nπ/b) (-AC) cos( mπx/a ) sin (nπy/b ) ex(x,b) =Amn cos( mπx/a ) sin (nπy/b ) d) Apply constants to H: hz(x,y) = (A coskxx + 0 sin kxx) (C cos kyy + 0 sin kyy) Hz(x,y) = AC cos (mπx/a) cos (nπy/b) = Amn cos (mπx/a) cos (nπy/b) e-jβz e) Ex, Ey, Hx,Hy components are found from equations 3.19 again. TEmn modes: Modes are numbered m-n, indicating how many cosine wave cycles are in the waveguide.Cutoff frequency for each mode is different: fc = kc / (2π√με) = “Lowest Order Mode” Assuming b>a TE10 mode has the lowest cutoff frequency, so will be the first mode in the waveguide.