Waveguides and antennas
Introduit| English | Sens |
|---|---|
| cutoff frequency/ˈkʌtɒf ˈfriːkwənsi/ | cutoff frequency |
| guide wavelength | guide wavelength |
| dipole/ˈdaɪpəʊl/ | dipole |
| Friis equation | équation de Friis |
Pipes and the last metre
- Above about 10 GHz, coax loses too much signal. Radar and microwave links move into hollow metal pipes instead.
- At each end of any radio link sits an antenna, and one formula prices the whole link. This lesson covers both.
The waveguide's cutoff
- A hollow rectangular pipe carries waves only above its cutoff frequency 截止频率. For the main mode, half a wavelength must fit across the wide side $a$: $f_{c} = \dfrac{c}{2a}$.
- Below $f_{c}$ the wave does not travel; it dies away within a short distance.
WR-90 waveguide is 22.86 mm wide. Its cutoff frequency is about
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$f_{c} = \dfrac{c}{2a} = \dfrac{3 \times 10^{8}}{0.04572} = 6.56$ GHz.
A 5 GHz signal travels along WR-90 waveguide.
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5 GHz is below the 6.56 GHz cutoff, so the wave dies away within a short distance instead of travelling.
The guide wavelength stretches
- Above cutoff the wave zigzags between the walls, so its pattern along the pipe is longer than in free space. The guide wavelength 波导波长 is $\lambda_{g} = \dfrac{\lambda}{\sqrt{1 - (f_{c}/f)^{2}}}$.
- Near cutoff $\lambda_{g}$ becomes very long; far above it, $\lambda_{g}$ approaches $\lambda$.
Worked example: WR-90
- WR-90 has $a = 22.86\,\text{mm}$: $f_{c} = \dfrac{3 \times 10^{8}}{2 \times 0.02286} = 6.56\,\text{GHz}$. Its working band, 8.2 to 12.4 GHz, lies above that.
- At 10 GHz, $\lambda = 30\,\text{mm}$ and $\lambda_{g} = \dfrac{30}{\sqrt{1 - 0.656^{2}}} = 39.8\,\text{mm}$: 33% longer than in free space.
In WR-90 at 10 GHz, what is the guide wavelength, in mm?
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$\lambda = 30$ mm and $\dfrac{f_{c}}{f} = 0.656$: $\lambda_{g} = \dfrac{30}{\sqrt{1 - 0.430}} = 39.8$ mm.
Antennas and gain
- An antenna turns guided waves into radiated ones. The half-wave dipole 偶极子天线 has an input resistance of about 73 Ω and a gain of 2.15 dBi.
- Gain in dBi compares the strongest direction with an isotropic antenna that radiates equally everywhere.
How long is a half-wave dipole for 100 MHz, in metres?
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$\lambda = \dfrac{3 \times 10^{8}}{10^{8}} = 3$ m, so half a wavelength is 1.5 m.
The Friis equation
- The Friis equation 弗里斯方程 prices a link: $\dfrac{P_{r}}{P_{t}} = G_{t}G_{r}\left(\dfrac{\lambda}{4\pi R}\right)^{2}$.
- The last factor is pure geometry: the power spreads over a sphere, and a receiving antenna catches only its share.
Two 10 dBi antennas ($G = 10$) at 3 GHz ($\lambda = 0.1\,\text{m}$). At 1 m: $100 \times \left(\dfrac{0.1}{4\pi}\right)^{2} = 6.3 \times 10^{-3}$, a 22 dB loss. At 1 km the distance is a thousand times greater, so the loss grows by $10^{6}$, or 60 dB, to 82 dB.
A radio link is moved from 1 km to 10 km apart. The received power falls by
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Received power goes as $\dfrac{1}{R^{2}}$: ten times the distance is one hundred times less power, 20 dB.
Link budget in dB: transmit 20 dBm, both antennas 10 dBi, path loss 82 dB. What is the received power, in dBm?
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$20 + 10 + 10 - 82 = -42$ dBm. In decibels the Friis chain is simple addition.
Three common slips
- The narrow side for cutoff. $f_{c} = \dfrac{c}{2a}$ uses the wide dimension.
- $\lambda_{g}$ in the Friis equation. Friis uses the free-space wavelength.
- dB without the reference. A dipole is 2.15 dBi, relative to isotropic; relative to a dipole it is 0 dB.
Match each quantity to what sets it.
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Each quantity has one controlling input; using the wrong one is the usual slip.
Recap
- Waveguide: $f_{c} = \dfrac{c}{2a}$; above it, $\lambda_{g} = \dfrac{\lambda}{\sqrt{1 - (f_{c}/f)^{2}}}$.
- Half-wave dipole: about 73 Ω, 2.15 dBi.
- Friis: $\dfrac{P_{r}}{P_{t}} = G_{t}G_{r}\left(\dfrac{\lambda}{4\pi R}\right)^{2}$; ten times the distance costs 20 dB.