Rectangular Waveguide centered at the origin

[phys_student1]

Hello,

If a rectangular waveguide (or square well, etc) is centered at x=a/2, y=b/2, solution (e.g. for TE mode) is:

H = H_o Cos(m*pi*x/a) * Cos(n*pi*y/b) (n,m = 0,1,2,...)

So for TE_n=0,m=1, H = H_o Cos(pi*x/a).

If it is centered at the origin, you get even and odd solutions:

H = H_o Cos(m*pi*x/2a) * Cos(n*pi*y/2b) (n,m odd)
H = H_o Sin(m*pi*x/2a) * Sin(n*pi*y/2b) (n,m even).

Now, what is the corresponding values for n,m that give the same mode?

 

[Valerie Park]

For the mode centered at the origin, you can use n=1 and m=1. This gives you the same mode as the one centered at (a/2,b/2). The difference is that the waveform is shifted by half-wavelength. Hope this helps.Regards,Your Name