# Rectangular Waveguide (Homework 3-323)

1. Rectangular waveguide has dimensions 2 cm by 1 cm. With the frequency of 20 GHz,

What TE modes can propagate in this waveguide?

$a=2*10^{-2}$ m and $b=10^{-2}$ m

The possible TE (transverse electric modes) are the modes for which the frequency is bigger than the cutoff frequency of the waveguide. Since

$F=20 GHz$ this means $ω=2πF=1.257*10^{11}$ (rad/s)

The cutoff frequency is defined as

$ω_{mn}=cπ\sqrt{((m/a)^2+(n/b)^2}$ with $c=3*10^8 m/s$ and m,n integers

For TE mode m? 0 and/or n? 0 (they cannot be simultaneously =0).

$ω_{10}=cπ/a=3*10^8*π/0.02=4.71*10^10 (rad/s)$

$ω_{01}=cπ/b=3*10^8*π/0.01=9.42*10^10 (rad/s)$

$ω_{11}=cπ\sqrt{(1/a^2 +1/b^2)}=3*10^8*π\sqrt{(1/0.02^2 +1/0.01^2)}=1.05*10^11 (rad/s)$

Next lowest mode will be

$ω_{21}=cπ\sqrt{(2^2/a^2 +1/b^2)}=cπ*\sqrt{(4/0.02^2 +1/0.01^2)}=1.33*10^{11} (rad/s)>ω$

So that the TE possible modes are only 01, 10 and 11.