TE mode in circular wave-guide

If the frequency is very close to the cutoff frequency, the field amplitude is not accurate because \(\sqrt{1-(\frac{f_c}{f})^2}\) tends to be zero.
constants:c mu0


Parameters

Azimuth mode number
m \(\)
Radial mode number
n \(\)
Radius of the wave guide
a \(mm\)
Frequency
f \(GHz\)
Power
p \(W\)

Output

\(\chi_{mn}'\) 

\(J'_m(\chi)=0\)

cutoff wave number 

\(k_c=\chi_{mn}/a\)

cutoff frequency 

\(k_c=k=\frac{\omega}{c}=\frac{2 \pi f_c}{c}\)

cutoff wave length 

\(c=\lambda_c k_c\)

wave number for f0 

\(k=\frac{\omega}{c}=\frac{2 \pi f_0}{c}\)

axial wave number 

\(k^2=k_c^2+k_z^2\)

waveguide wavelength 

\(\lambda_g k_z=2\pi\)

wave impendance  

\(Z_{mn}=\frac{\omega \mu}{k_z}=\frac{120\pi}{\sqrt{1-(f_c/f)^2}} \)

Hz0 

\(P=H_{z0}^2 \frac{Z_0}{2}(\frac{f}{f_c})^2\sqrt{1-(\frac{f_c}{f})^2} \pi \frac{1}{2} a^2[J_m(\chi_{mn}')^2-J_{m-1}(\chi_{mn}')J_{m+1}(\chi_{mn}')]\)

\(E_{\phi0}\) 

\(E_{\phi}=j(\frac{\omega\mu_0 H_{z0}}{k_c})J'_m(k_c r) cos(m\phi) e^{-j k_z z}\)

\(E_{r0}\) 

\(E_{r}=j(\frac{\omega \mu_0 m H_{z0}}{k_c^2}) \frac{J_m(k_c r)}{ r} sin(m\phi)e^{-j k_z z}\)

\(C_{mn}\) 

\( \frac{1}{\sqrt{\pi(\chi_{mn}^{'2}-m^2)J_m^2(\chi_{mn}')}}\)