WAVEGUIDES x
z
y
µ, ε
Maxwell's Equation
∇2E + ω2µεE = 0 (A)
∇2H + ω2µεH = 0 (B)
For a waveguide with arbitrary cross section as shown in the above figure, we assume a plane wave solution and as a first trial, we set Ez = 0. This defines the TE modes. ∂H ∇ × = −µ , we have From E ∂t ∂ ∂ ∂ Ez Ey µ Hx ⇒ β ωµ ∂y - ∂z = - ∂t +j zEy = -j Hx (1) ∂ ∂ ∂ Ex Ez µ Hy ⇒ β ωµ ∂z - ∂x = - ∂t -j zEx = -j Hy (2) ∂ ∂ ∂ ∂ ∂ Ey Ex µ Hz ⇒ Ey Ex ωµ ∂x - ∂y = - ∂t ∂x - ∂y = -j Hz (3)
From ∇ × E = − jωµH, we get
xyzˆˆˆ ∂ ∂ ∂ jωεE = ∂xxz∂ ∂ Hx Hy Hz ∂ ∂ ∂ Hz Hy ωε ⇒ Hz β ωε ∂y - ∂z = j Ex ∂y + j zHy = j Ex (4) ∂ ∂ ∂ Hx Hz ωε ⇒ β Hz ωε ∂z - ∂x = j Ey -j zHx - ∂x = j Ey (5) - 2 -
∂Hy ∂Hx ∂x - ∂y = 0 (6)
We want to express all quantities in terms of Hz.
From (2), we have
β E H = z x (7) y ωµ in (4) ∂H E z + jβ 2 x = jωεE (8) ∂y z ωµ x
Solving for Ex
jωµ ∂Hz Ex = ∂y (9) βz2-ω2µε
From (1)
-β E H = z y (10) x ωµ
in (5)
β 2E ∂Hz +j z y - = jωεE (11) ωµ ∂x y
so that
-jωµ ∂Hz Ey = ∂x (12) βz2-ω2µε
jβz ∂Hz Hx = ∂x (13) βz2-ω2µε
jβz ∂Hz Hy = ∂y (14) βz2-ω2µε
Ez = 0 (15) - 3 -
Combining solutions for Ex and Ey into (3) gives
∂2Hz ∂2Hz + = [βz2-ω2µε]Hz (16) ∂x2 ∂y2
RECTANGULAR WAVEGUIDES
y
x z µ, ε b
a
If the cross section of the waveguide is a rectangle, we have a rectangular waveguide and the boundary conditions are such that the tangential electric field is zero on all the PEC walls.
TE Modes
The general solution for TE modes with Ez=0 is obtained from (16)
−−ββ + β − β + β =+jzzx jx jx x jy y+ jy y Hz e[] Ae Be[] Ce De (17)
βωµ −−+−+jzβββββ jx jx jy jy E= x ezxxyy[]−+ Ae Be[] Ce+ De (18) y βωµε22− z βωµ j y −−jzββ jx + jx β − jy β + jy β E = eAeBeCeDezx[]+ x[]−+ y y (19) x βωµε22− z
At y=0, Ex=0 which leads to C=D
At x=0 Ey=0 which leads to A=B
−β = jzz ββ HHezo cosxy x cos y (20)
βωµ − jzβ EHexy= x z sinββ cos (21) y βωµε22− o xy z βωµ j y − jzβ E = Hez cosββ x sin y (22) x βωµε22− o xy z - 4 -
π β m At x=a, Ey=0; this leads to x = a
π β n At y=b, Ex=0; this leads to y = b
The dispersion relation is obtained by placing (20) in (16)
βββωµε222++= 2 xyz (23)
mππ 22 n + +=βωµε22 (2 4) a b z
and
mππ 22 n βωµε=−2 − (25) z a b
The guidance condition is
mππ 22 n ωµε2 > + (26) a b or f > fc where fc is the cutoff frequency of the TEmn mode given by the relation
1 m 22 n f = + c µε 2 a b (27)
The TEmn mode will not propagate unless f is greater than fc. Obviously, different modes will have different cutoff frequencies.
TM Modes
The transverse magnetic modes for a general waveguide are obtained by assuming Hz =0. By duality with the TE modes, we have
∂2Ez ∂2Ez + =[βz2-ω2µε]Ez (28) ∂x2 ∂y2 - 5 - with general solution
−−jzββ jx + jx β − jy β + jy β E=+ ezx[] Ae Be x[] Ce y+ De y z (29)
The boundary conditions are
At x=0, Ez=0 which leads to A=-B
At y=0, Ez=0 which leads to C=-D
mπ At x=a, E β z=0 which leads to x = a
π β n At y=b, Ez=0 which leads to y = b
so that the generating equation for the TMmn modes is
−β mxππny = jzz EEezo sin sin a b (30)
NOTE: THE DISPERSION RELATION, GUIDANCE CONDITION AND CUTOFF EQUA- TIONS FOR A RECTANGULAR WAVEGUIDE ARE THE SAME FOR TE AND TM MODES.
For additional information on the field equations see Rao, page 539 Table 9.1.
There is no TE00 mode
There are no TMm0 or TM0n modes
The first TE mode is the TE10 mode
The first TM mode is the TM11 mode
Impedance of a Waveguide
For a TE mode, we define the transverse impedance as
ωµ η -Ey Ex gTE = H = H = x y βz - 6 -
From the relationship for βz and using
1 m 22 n f 2 = + c 4πµε2 a b (31)
we get η η = gTE (32) f 2 1 − c f 2
µ where η is the intrinsic impedance η = . Analogously, for TM modes, it can be shown that ε
f 2 ηη=−1 c (33) gTM f 2
Power Flow in a Rectangular Waveguide (TE10)
The time-average Poynting vector for the TE10 mode in a rectangular waveguide is given by
2 1 E β πx PEH*z=×Re[]= ˆ o z sin2 (34) 22ωµ a
2 a b E β πx Power = ∫ ∫ o z sin2 dxdy (35) 0 0 2 ωµ a
22 E β ab Eab Power ==o z o (36) 44ωµ η gTE10
Therefore the time-average power flow in a waveguide is proportional to its cross-section area.