What is mode of a waveguide?
The key feature of wave-guiding structure is modal propagation. The electromagnetic energy
propagates along a waveguide in the form of some definite field patterns called modes. The
propagation of waves in the form of modes is a direct consequence of bounded nature of
the structure which a waveguide has. There are two types of modes in rectangular
waveguide.
a. Transverse Electric (TE) mode (Ez=0 and Hz≠0)
b. Transverse Magnetic (TM) mode (Hz=0 and Ez≠0)
Here, z is the direction of propagation.
1. Derive E field and H field expressions for TE/TM mode for rectangular waveguide.
Maxwell’s third and fourth equation gives
−∂ B
∇ XE=
∂t
∂D
∇ XH =J +
∂t
J is conduction current density and D is displacement current density
For a waveguide, J=0. Only displacement current exists.
∂
For time harmonic expression, replace by jω. B=μH and D=∈ E .
∂t
∇ XE=− jωμH
∇ XH = jωϵE
[ ]
i^ ^j k^
(1)
∂ ∂ ∂
=- jωμH
∂x ∂y ∂z
Ex Ey Ez
[ ]
i^ ^j k^
(2)
∂ ∂ ∂
= jω ∈ E
∂x ∂y ∂z
Hx Hy Hz
Instructions to solve the determinant
To find Hx and Ex , omit first row and first column of equation 1 and 2 repectively and do the
determinant operation on the remaining elements. To find an expression for Hy and Ey , omit
first row and second column of equation 1 and 2 repectively and do the determinant
operation on the remaining elements. To find Hy and Ey, sign on the RHS of equation 1 and 2
will be reversed. ‘+’ sign will become ‘-’ sign and vice-versa.
∂ Ez ∂ E y
− =− jωμ H x (3)
∂y ∂z
∂ Hz ∂ H y
− = jωμ E x (4)
∂y ∂z
.
∂ Ez ∂ E x
− = jωμ H y (5)
∂x ∂z
∂ Hz ∂ Hx
− =− jω∈ E y (6)
∂x ∂z
∂
Replacing - by γ , we get,
∂z
∂ Ez
+ γ E y =− jωμ H x (7)
∂y
∂ Hz
+γ H y = jωμ E x (8)
∂y
∂ Ez
+ γ E x = jωμ H y (9)
∂x
∂ Hz
+γ H x =− jω∈ E y (10)
∂x
Placing the value of Ey from equation 10 to equation 7, we get Hx=f(Ez, Hz). γ = jβ and
2 2 2
h =ω με−β .
The equations of Hx, Ex, Hy and Ey comes out to be
jωϵ ∂ E z jβ ∂ H z
H x= 2
− 2 (11)
h ∂ y h ∂x
− jωμ ∂ H z jβ ∂ Ez
E x= 2
− (12)
h ∂ y h2 ∂ x
− jωϵ ∂ E z jβ ∂ H z
H y= 2
− (13)
h ∂ x h2 ∂ y
jωμ ∂ H z jβ ∂ E z
E y= 2
− 2 (14)
h ∂x h ∂y
After these 4 equations 11,12,13 and 14, proceed to derive for E-field and H-field equations
of TE or TM mode which is earlier mentioned in the notes.
