52 Radio Engineering for Wireless Communication and Sensor Applications
Figure 3.7 Transverse field distributions of some TE and TM wave modes of a rectangular waveguide. Solid lines represent electric field lines, dashed lines magnetic field lines.
3.5 Circular Waveguide
The analysis of a circular waveguide shown in Figure 3.8 is best carried out using the cylindrical coordinate system. The principle of analysis is similar to that of the rectangular waveguide.
The solutions of the longitudinal magnetic fields of the TE wave modes
are
Hz = AJn (k c r ) cos (nf ) |
(3.63) |
where Jn is the Bessel function of the order n . From the boundary condition
Hr (r = a ) = 0, it follows that ∂Hr /∂r (r = a ) = 0, and further that Jn′(k c a ) = 0, in which the apostrophe stands for derivative. From this we get
Figure 3.8 Circular waveguide.
Transmission Lines and Waveguides |
53 |
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p ′ |
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k cnm = |
nm |
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(3.64) |
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a |
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where p nm′ is the m th zero of Jn′. The corresponding cutoff wavelength is
l cTEnm = |
2pa |
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(3.65) |
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p nm′ |
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The solutions of the longitudinal electric fields of the TM wave modes |
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are |
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Ez = BJn (k c r ) cos (nf) |
(3.66) |
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From the boundary condition Ez (r = a ) = 0 it follows that Jn (k c a ) = 0, or
k cnm = |
p nm |
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(3.67) |
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a |
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where p nm is the m th zero of Jn . The cutoff wavelength is |
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lcTMnm = |
2pa |
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(3.68) |
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pnm |
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The subscript n denotes the number of periods in the field distribution along the f angle. The subscript m gives the number of axial field minima in the radial direction. Figure 3.9 shows the transverse field distributions of some wave modes. Table 3.3 gives the cutoff wavelengths of the wave modes having the lowest cutoff frequencies.
The relative bandwidth of the circular waveguide operating at the fundamental mode TE11 is smaller than that of the rectangular waveguide operating at the TE10 mode, as shown in Figure 3.10. Therefore, many standard waveguide sizes are needed to cover a broad frequency range.
The conductor losses of the circular waveguide are calculated from the surface currents and surface resistance the same way as in the case of the rectangular waveguide. A special feature of the TE01 wave mode is that its attenuation decreases monotonously as the frequency increases. The attenuation constant of the TE01 wave mode is
54 Radio Engineering for Wireless Communication and Sensor Applications
Figure 3.9 Transverse field distributions of some TE and TM wave modes in a circular waveguide. Solid lines represent electric field lines, dashed lines magnetic field lines.
Table 3.3
Cutoff Wavelengths of Wave Modes in a Circular Waveguide with Radius a
Wave Mode |
pnm or pn′m |
lc |
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TE11 |
p 1′1 = 1.841 |
3.41a |
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TM01 |
p 01 |
= 2.405 |
2.61a |
TE21 |
p 2′1 |
= 3.054 |
2.06a |
TE01 |
p 0′1 |
= 3.832 |
1.64a |
TM11 |
p 11 |
= 3.832 |
1.64a |
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= |
R s |
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( f c /f )2 |
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acTE 01 |
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(3.69) |
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ah |
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√1 − ( f c /f )2 |
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The attenuation of the TE01 wave mode is very low if the operating frequency is much higher than the cutoff frequency f c . However, many other modes
Transmission Lines and Waveguides |
55 |
Figure 3.10 Cutoff frequencies of the lowest wave modes of rectangular and circular waveguides.
can propagate in such an oversized waveguide. A low attenuation is achieved only if the excitation of unwanted modes is prevented.
Example 3.2
Calculate the conductor losses at 60 GHz for the TE01 wave mode in a circular waveguide made of copper. The radius of the waveguide is (a) 3.5 mm, and (b) 20 mm.
Solution
The surface resistance is
Rs = √p fm 0 /s
=√p × 60 × 109 × 4p × 10−7/58 × 106 V/m
=0.064 V/m.
The cutoff frequency of the TE01 wave mode is f c = c /lc = c /(1.64a ). (a) When a = 3.5 mm, f c = 52.3 GHz. From (3.69) we solve the attenuation constant acTE 01 = 0.0753 1/m. The attenuation of a waveguide having a length l is in decibels 20 log e al, from which we obtain an attenuation of
0.65 dB/m. (b) When a = 20 mm, f c = 9.15 GHz and a cTE 01 = 1.14 × 10−3 1/m. The attenuation is now only 0.010 dB/m or 1.0 dB/100m.