About TE and TM mode cutoffs or something

So listen, I have the formula below

$$fc_{m,n} = \frac{v}{2} \sqrt{(\frac{m}{a})^2 + (\frac{n}{b})^2}$$

Sometimes they don't give you $v$ by itself. They make you work for it (bitches)

$$v= \frac{1}{\sqrt{\mu \varepsilon}}$$

Listen here monkey. It is $\frac{v}{2}$, not the opposite. $v$ is on the top, the numerator

$$v= \frac{1}{\sqrt{\mu_o \mu_r \varepsilon_o \varepsilon_r}}$$

Ok so far so good. IDK how $v$ is that, but it is so it is.

Group Velocity

WTF is group velocity? I have no fucking idea. They say it's this:

$$v_g = v \sqrt{1 - (\frac{f_c}{f})^2}$$

Here, I guess $f_c$ is the cutoff frequency? But what is $f$?

Looks like $f$ is the signal frequency. Ok, whatevs.

Phase Velocity

Why are there so many fucking velocities?

$$v_p = \frac{v}{\sqrt{1-(\frac{f_c}{f})^2}}$$

There looks to be some relation between $v$, $v_g$ and $v_p$. It's this:

$$v_g = \frac{v^2}{v_p}$$

Ok thanks.

I think this is enough for exam. :)

Adding things

So okay, something cool I've found. When question says, find "dominant mode", just find $f_c$ for $TE_{1,0}$ for rectangular and $TE_{1,1}$ for cylindrical.

MAKE SURE $a$ is greater than $b$