1) For TE Mode in Circular waveguide

For TE mode EZ = 0 and HZ ≠ 0

The wave equation is

$∇^2 H_z+w^2 μEH_z=0$

Expanding $∇^2$ in cylindrical form.

$\bigg(\dfrac{∂^2}{∂ρ^2}+\dfrac1ρ \dfrac{∂}{∂ρ}+\dfrac{1}{ρ^2} \dfrac{∂^2}{∂ϕ^2} +\dfrac{∂^2}{∂z^2}\bigg) H_z+w^2 μEH_z=0 \\ \bigg(\dfrac{∂^2}{∂ρ^2}+\dfrac1ρ \dfrac{∂}{∂ρ}+\dfrac{1}{ρ^2} \dfrac{∂^2}{∂ϕ^2} +\dfrac{∂^2Hz}{∂z^2}\bigg)+w^2 μEH_z=0$

But $\dfrac{∂^2Hz}{∂z^2}=γ^2$

$\bigg(\dfrac{∂^2 H_z}{∂ρ^2}+\dfrac1ρ \dfrac{∂H_z}{∂ρ}+\dfrac{1}{ρ^2} \dfrac{∂^2 H_z}{∂ϕ^2}\bigg)+γ^2 H_z+w^2 μEH_z=0 \\ \dfrac{∂^2 H_z}{∂ρ^2}+\dfrac1ρ \dfrac{∂H_z}{∂ρ}+\dfrac{1}{ρ^2} \dfrac{∂^2 H_z}{∂ϕ^2}+(γ^2+w^2 μE)H_z=0$

But $γ^2+w^2 μE=h^2$

$\dfrac{∂^2 H_z}{∂ρ^2}+\dfrac1ρ \dfrac{∂H_z}{∂ρ}+\dfrac{1}{ρ^2} \dfrac{∂^2 H_z}{∂ϕ^2}+h^2 H_z=0$

Let solution is $H_z=R(ρ)P(ϕ)$ -------- (1)

Where, $\text{R is function of ρ} \\ \text{P is function of ϕ}$

$\dfrac{∂^2 (RP)}{∂ρ^2 }+\dfrac1ρ \dfrac{∂(RP)}{∂ρ}+\dfrac{1}{ρ^2} \dfrac{∂^2 (RP)}{∂ϕ^2}+h^2 (RP)=0 \\ \dfrac{P∂^2 R}{∂ρ^2 }+\dfrac{P}{ρ} \dfrac{∂R}{∂ρ}+\dfrac{R}{ρ^2} \dfrac{∂^2 P}{∂ϕ^2}+h^2 (RP)=0$

Divide by RP and multiply by $ρ^2$ on both sides.

$\dfrac{ρ^2}{R} \dfrac{∂^2 R}{∂ρ^2}+\dfrac{ρ}{R} \dfrac{∂R}{∂ρ}+\dfrac{1}{P} \dfrac{∂^2 P}{∂ϕ^2 }+h^2 ρ^2=0 \\ \dfrac{ρ^2}{R} \dfrac{∂^2 R}{∂ρ^2}+\dfrac{ρ}{R} \dfrac{∂R}{∂ρ}+h^2 ρ^2=-\dfrac{1}{P} \dfrac{∂^2 P}{∂ϕ^2} ---------- (2)$

As the left hand side of the equation depends on variable ρ only and the right hand side of the equation depends on variable ϕ only. Thus, each side of above equation must be constant say $K_ϕ^2$.

$-\dfrac{1}{P} \dfrac{∂^2 P}{∂ϕ^2}=K_ϕ^2 ---------- (3) \\ \dfrac{∂^2 P}{∂ϕ^2}=-K_ϕ^2 P → \dfrac{∂^2 P}{∂ϕ^2}+K_ϕ^2 P=0$

General solution of above equation is

$P(ϕ)=A \sin⁡K_ϕ ϕ+B \cos K_ϕ ϕ ---------- (4)$

Since the solution to HZ must be periodic in ϕ, $K_ϕ$ must be an integer say n. $K_ϕ=n$

Equation (4) becomes

$P(ϕ)=A \sin ⁡nϕ+B \cos nϕ ---------- (5)$

From equation (2) and (3)

$\dfrac{ρ^2}{R} \dfrac{∂^2 R}{∂ρ^2}+\dfrac{ρ}{R} \dfrac{∂R}{∂ρ}+h^2 ρ^2=K_ϕ^2 \\ \dfrac{ρ^2}{R} \dfrac{∂^2 R}{∂ρ^2}+\dfrac{ρ}{R} \dfrac{∂R}{∂ρ}+h^2 ρ^2=K_ϕ^2=0$

Multiplying throughout by R

$ρ^2 \dfrac{∂^2 R}{∂ρ^2}+ρ \dfrac{∂R}{∂ρ}+h^2 ρ^2 R-K_ϕ^2 R=0 \\ ρ^2 \dfrac{∂^2 R}{∂ρ^2}+ρ \dfrac{∂R}{∂ρ}+(h^2 ρ^2-K_ϕ^2 )R=0$

As $K_ϕ=n$

$ρ^2 \dfrac{∂^2 R}{∂ρ^2}+ρ \dfrac{∂R}{∂ρ}+(h^2 ρ^2-n^2 )R=0$

Which is Bessel’s differential equation.

The solution is $R(ρ)-CJ_n (hρ) -------- (6)$

From equation (1)

$H_z=R(ρ)P(ϕ) \\ H_z=CJ_n (hρ)(A \sin ⁡n ϕ+B \cos nϕ) -------- \text{(from (5) and (6))} \\ H_z=J_n (hρ)(CA \sin⁡ n ϕ+CB \cos nϕ) \\ H_z=J_n (hρ)(A \sin⁡ n ϕ+B \cos nϕ) \\ ---- \text{as c, A, B are constants.} \\ \text{CA and CB are again constants say A and b.}$

By applying the boundary condition that $E_ϕ=0$. On the wave guide wall; since $E_2=0$ we must have, $E_ϕ (ρ,ϕ)=0$ at ρ=a

For TE mode,

$E_ϕ=\dfrac{jwμ}{h^2} .\dfrac{∂H_z}{∂ρ} \\ E_ϕ=\dfrac{jwμ}{h^2} .\dfrac{∂}{∂ρ} (J_n (hρ)(A \sin⁡ n ϕ+B \cos nϕ)) ----- (from (7)) \\ E_ϕ (ρ,ϕ)=\dfrac{jwμ}{h^2} (A \sin⁡ n ϕ+B \cos nϕ) J_n' (hρ)h \\ E_ϕ (ρ,ϕ,z)=\dfrac{jwμ}{h^2} (A \sin⁡ n ϕ+B \cos nϕ) J_n' (hρ) e^{-γz} -------- (8)$

For $E_ϕ$ to vanish at ρ=a, we must have

$J_n' (ha)=0 ------ (9)$

If $P_{nm}'$ is mth root of $J_n'$ then clearly

$J_n' (P_nm' )=0$

From (8),

$ha=P_nm' \\ h=\dfrac{P'nm}{a}$

where h is cut off wave number.

As we have from equation (8)

$E_(ϕ )=\dfrac{jwμ}{h^2} (A \sin nϕ+B \cos nϕ) J_n' (hρ) e^{-γz}$

Similarly,

$E_ρ=\dfrac{-jωμ_n}{h^2 ρ} (A \cos nϕ-B \sin nϕ) J_n' (hρ) e^{-γz} \\ H_ϕ=\dfrac{-γn}{h^2 ρ} (A \cos nϕ-B \sin nϕ) J_n' (hρ) e^{-γz} \\ H_ρ=\dfrac{-γ}{h} (A \sin nϕ+B \cos nϕ) J_n' (hρ) e^{-γz}$

Similarly, for TM mode

$E_z=(Asin nϕ+Bcos nϕ) J_n (hρ) E_ϕ=\dfrac{-\sqrt n}{h^2 ρ} (A \cos nϕ-B \sin nϕ) J_n' (hρ) e^{-γz} \\ E_ρ=\dfrac{-γ}{h} (A c \sin nϕ+B \cos nϕ) J_n' (hρ) e^{-γz} \\ H_ϕ=\dfrac{-jwϵ}{h} (A \sin nϕ+B \cos nϕ) J_n' (hρ) e^{-γz} \\ H_ρ=\dfrac{-jwϵ n}{h^2 ρ} (A \cos nϕ-B \sin nϕ) J_n' (hρ) e^{-γz}$

• Cut off frequency in Circular waveguide (fc)

  We know that,

  $h^2=γ^2+ω^2 μϵ$

  But at cut off frequency, $f=f_c, γ=0 \ \ \& \ \ ω=ω_c,$

  $∴h^2=0+ω_C^2 μϵ=ω_C^2 μϵ \\ ∴h^2=(2πf_c )^2 μϵ \\ ∴(2πf_c )^2= \dfrac{h^2}{μϵ}$