TM modes theory (Homework 3-323)

2. Work out the theory of TM modes for a rectangular waveguide. Find the maximum frequencies and the phase and group velocities.

For TM modes the component of the magnetic field is only transverse. This means

$B_z (x,y)=0$   and $E_z (x,y)? 0$

E z(x,y) needs to satisfy the equation

$[∂^2/(∂x^2 )+∂^2/(∂y^2 )+(ω_0/c)^2-k_g^2 ] E_z=0$

$ω_0$  is the frequency of the wave propagating in free space

and $k_g$   is the wave vector inside the waveguide

The boundary condition is

$E^{(||)}=0$

We separate the variables

$E_z (x.y)=X(x)Y(y)$

so that $Y (∂^2 X)/(∂x^2 )+X (∂^2 Y)/(∂y^2 )+(ω_0/c)^2 X Y-k_g^2 X Y=0$

divide by XY  and get   $1/X*(∂^2 X)/(∂x^2 )+1/Y*(∂^2 Y)/(∂y^2 )+(ω_0/c)^2-k_g^2=0$
which is true only when

$1/X*(∂^2 X)/(∂x^2 )=-k_x^2$   and  $1/Y*(∂^2 Y)/(∂y^2 )=-k_y^2$    with

$-k_x^2-k_y^2+(ω_0/c)^2-k_g^2=0$
The solutions to the above two equations are

$X(x)=A*\sin⁡(k_x x)+B*\cos⁡(k_x x)$

From the boundary condition one gets $Ex=0$  at $x=0$ and $x=a$

$E_x=i/((ω/c)^2-k^2 ) (k*(∂E_z)/∂x+ω (∂B_z)/∂y)=i/((ω/c)^2-k^2 )*(k (∂E_z)/∂x)$

it means also  $(∂E_z)/∂x→0$  at $x=0$ and  $x=a$

and therefore

$A=0$  and $k_x=mπ/a$

The same is true for $Y(y)$ so that

$k_y=nπ/b$

and thus

$E_z (x,y)=X(x)Y(y)=E_0*\cos⁡(mπ/a)*cos⁡(nπ/b)$

Back into the brown equation with the values of $k_x=mπ/a$   and $k_y=nπ/b$  we obtain

$(ω_0/c)^2=k_g^2+π^2 [(m/a)^2+(n/b)^2 ]$   or

$k_g=\sqrt{((ω_0/c)^2-π^2 [(m/a)^2+(n/b)^2 ] )}$

$k_g=\sqrt{((ω_0/c)^2-(ω_{m n}/c)^2 )}$     with
$ω_{m n}=cπ\sqrt([(m/a)^2+(n/b)^2 ] )$

Thus

$ck_g=\sqrt(ω_0^2-ω_{m n}^2)$

The wave velocity is

$v=ω_0/k_g =c/\sqrt{(1-(ω_{m n}/ω)^2)}$

The group velocity is

$v_g=(dω_0)/(d k_g )=c\sqrt{(1-(ω_{m n}/ω)^2)}$