# WR 284 Waveguide (Homework 1-323)

WR284 waveguide can be purchased. It’s rectangular waveguide with inner dimensions 34.04 mm by 72.14 mm.

$a=34.04*10^{-3}$ m and $b=72.14*10^{-3}$ m

a) What is the cutoff frequency of the lowest mode?

b) What is the cutoff frequency of the next-lowest mode?

a)

The rule here is (this is not in the book)

for TE modes   $m? 0$ or $n? 0$;    for TM modes  $m? 0$ and $n? 0$
Lowest TE mode is 10 or 01 (the mode 00 does not exist). The cutoff frequency is (see question 1):

$ω_{m n}^2=c^2 π^2 [(m/a)^2+(n/b)^2]$

And the lowest frequency is

$ω_{01}=cπ*\sqrt{1/b^2}=cπ\sqrt{1/(72.14*10^{-3})^2}=1.306*10^{10} (rad/s )$

or $ν=2.08*10^9 Hz=2.08 GHz$

The next lowest mode is 10

$ω_{10}=cπ/a=2.77*10^{10} (rad/s)$   or $ν=4.407 GHz$

b)

$λ_0=100 mm$    means $ω_0=2πc/λ_0 =1.885*10^{10} (rad/sec) (=ω)$
From question 1)a) red equation  (or question 1b first equation) one has

$k=1/c \sqrt(ω^2-ω_{m n}^2 )$

The wave velocity is by definition (it is found greater than c also in the book):

$v=ω/k=c/\sqrt{1-(ω_{m n}/ω)^2 }=(3*10^8)/\sqrt{1-((1.3*10^{10})/(1.885*10^{10}))^2}=$

$=4.14*10^8 (m/s)$
The group velocity is (see question 1b or equation 9.192)

$v_g=(dω/d k)=c\sqrt{(1-(ω_m n/ω)^2}=3*10^8 \sqrt{(1-(1.3/1.885)^2)}=2.17*10^8 (m/s)$