Waveguides - ieeetsu

www.ece.rutgers.edu/∼orfanidi/ewa 271
272 Electromagnetic Waves & Antennas – S. J. Orfanidis – June 21, 2004
Fig. 8.6.1
Operating bandwidth in rectangular waveguides.
where
H 1 = jβ H 0 , E 0 =−η TE H 1 =−jη ω H 0 (8.7.2)
k c ω c
The Poynting vector is obtained from the general result of Eq. (8.3.7):
P z = 1
2η TE
|E T | 2 = 1
2η TE
|E y (x)| 2 = 1
2η TE
|E 0 | 2 sin 2 k c x
The transmitted power is obtained by integrating P z over the cross-sectional area
of the guide:
widest bandwidth, we also require to have the maximum power transmitted, the dimension
b must be chosen to be as large as possible, that is, b = a/2. Most practical guides
follow these side proportions.
If there is a “canonical” guide, it will have b = a/2 and be operated at a frequency
that lies in the middle of the operating band [f c , 2f c ], that is,
f = 1.5f c = 0.75 c (8.6.1)
a
Table 8.6.1 lists some standard air-filled rectangular waveguides with their naming
designations, inner side dimensions a, b in inches, cutoff frequencies in GHz, minimum
and maximum recommended operating frequencies in GHz, power ratings, and attenuations
in dB/m (the power ratings and attenuations are representative over each operating
band.) We have chosen one example from each microwave band.
name a b f c f min f max band P α
WR-510 5.10 2.55 1.16 1.45 2.20 L 9 MW 0.007
WR-284 2.84 1.34 2.08 2.60 3.95 S 2.7 MW 0.019
WR-159 1.59 0.795 3.71 4.64 7.05 C 0.9 MW 0.043
WR-90 0.90 0.40 6.56 8.20 12.50 X 250 kW 0.110
WR-62 0.622 0.311 9.49 11.90 18.00 Ku 140 kW 0.176
WR-42 0.42 0.17 14.05 17.60 26.70 K 50 kW 0.370
WR-28 0.28 0.14 21.08 26.40 40.00 Ka 27 kW 0.583
WR-15 0.148 0.074 39.87 49.80 75.80 V 7.5 kW 1.52
WR-10 0.10 0.05 59.01 73.80 112.00 W 3.5 kW 2.74
Table 8.6.1 Characteristics of some standard air-filled rectangular waveguides.
8.7 Power Transfer, Energy Density, and Group Velocity
Next, we calculate the time-averaged power transmitted in the TE 10 mode. We also calculate
the energy density of the fields and determine the velocity by which electromagnetic
energy flows down the guide and show that it is equal to the group velocity. We recall
that the non-zero field components are:
H z (x)= H 0 cos k c x, H x (x)= H 1 sin k c x, E y (x)= E 0 sin k c x (8.7.1)
Noting the definite integral,
∫ a
∫ b
P T =
0 0
1
2η TE
|E 0 | 2 sin 2 k c x dxdy
∫ a
∫ a
sin 2 k c xdx= sin
2( πx ) a dx =
0
0 a 2
√
and using η TE = ηω/βc = η/ 1 − ω 2 c/ω 2 , we obtain:
(8.7.3)
√
P T = 1 |E 0 | 2 ab = 1
4η TE 4η |E 0| 2 ab 1 − ω2 c
ω 2 (transmitted power) (8.7.4)
We may also calculate the distribution of electromagnetic energy along the guide, as
measured by the time-averaged energy density. The energy densities of the electric and
magnetic fields are:
w e = 1 2 Re( 1 2 ɛE · E∗) = 1 4 ɛ|E y| 2
w m = 1 2 Re( 1 2 µH · H∗) = 1 4 µ( |H x | 2 +|H z | 2)
Inserting the expressions for the fields, we find:
w e = 1 4 ɛ|E 0| 2 sin 2 k c x, w m = 1 4 µ( |H 1 | 2 sin 2 k c x +|H 0 | 2 cos 2 k c x )
Because these quantities represent the energy per unit volume, if we integrate them
over the cross-sectional area of the guide, we will obtain the energy distributions per
unit z-length. Using the integral (8.7.3) and an identical one for the cosine case, we find:
W ′ e = ∫ a
0
W ′ m = ∫ a
0
∫ b
0
∫ b
0
∫ a
∫ b
W e (x, y) dxdy =
0 0
1
4 ɛ|E 0| 2 sin 2 k c x dxdy = 1 8 ɛ|E 0| 2 ab
1
4 µ( |H 1 | 2 sin 2 k c x +|H 0 | 2 cos 2 k c x ) dxdy = 1 8 µ( |H 1 | 2 +|H 0 | 2) ab