噪声与信噪比
引入| English | 释义 | 拼音 |
|---|---|---|
| noise figure | 噪声系数 | zào shēng xì shù |
A floor you cannot dig below
- Every radio link ends at a noise floor. Even a perfect amplifier cannot go below the thermal noise of its own resistance.
- Each receiver stage decides how far above that floor the signal arrives.
The thermal floor
- At room temperature the noise power is $-174$ dBm in each hertz of bandwidth.
- In a bandwidth $B$ the floor is $-174 + 10\log_{10}B$ dBm: 10 dB more for every ten times the bandwidth.
What is the thermal noise floor in a 1 MHz bandwidth, in dBm?
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$-174 + 10\log_{10}(10^{6}) = -174 + 60 = -114$ dBm.
The bandwidth falls from 1 MHz to 10 kHz. By how many dB does the floor drop?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
A factor of 100 in bandwidth is 20 dB, so the floor moves to −134 dBm.
Noise figure and Friis
- The noise figure 噪声系数 $F$ says how much noise a stage adds. It is a ratio; in dB it is $10\log_{10}F$.
- For a chain, $F = F_{1} + \dfrac{F_{2} - 1}{G_{1}} + \dfrac{F_{3} - 1}{G_{1}G_{2}} + \cdots$. The first stage's gain shrinks every later stage's noise.
A stage with F = 2 and gain 20 feeds a stage with F = 10. What is the cascade noise figure, as a ratio? Give 2 decimal places.
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$F = 2 + (10 - 1)/20 = 2.45$, which is 3.89 dB. The second stage adds only 0.45.
Work in ratios
- Convert each dB value to a ratio first: 1.76 dB is 1.5, 4.77 dB is 3, and 10 dB is 10.
- Apply Friis to the ratios, then convert the answer back to dB once.
The noise figures of two stages in cascade can be added in dB.
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Friis works with ratios. Convert from dB, combine, and convert back once.
Worked example: the LNA goes first
- LNA ($F = 1.5$, $G = 10$) then mixer ($F = 3$): $F = 1.5 + \dfrac{3 - 1}{10} = 1.70$, or 2.30 dB.
- Swap them: $3 + \dfrac{1.5 - 1}{10} = 3.05$, or 4.84 dB. The same parts in the wrong order are 2.5 dB worse.
Which stage should come first in a receiver?
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The first stage's gain divides every later stage's added noise, so it must be quiet and have plenty of gain.
Worked example: the link budget
- In a 180 kHz FM channel the floor is $-174 + 10\log_{10}(180\,000) = -121.4$ dBm. The 2.3 dB receiver lifts it to $-119.1$ dBm.
- A $-20$ dBm signal then arrives $-20 - (-119.1) = 99.1$ dB above the floor.
A −90 dBm signal arrives in a 1 MHz channel at a receiver with a 3 dB noise figure. What is the SNR, in dB?
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The floor is $-114 + 3 = -111$ dBm, and $-90 - (-111) = 21$ dB.
Three common slips
- Adding noise figures in dB. Convert to ratios, use Friis, convert back.
- Forgetting bandwidth. The floor rises 10 dB for every ten times the bandwidth.
- The noisy stage first. A low-noise, high-gain stage must lead the chain.
Match each value in dB to its ratio.
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A ratio r is $10\log_{10} r$ dB.
Recap
- Floor $= -174 + 10\log_{10}B$ dBm.
- $F = F_{1} + (F_{2} - 1)/G_{1} + \cdots$: the first stage rules.
- Do the arithmetic in ratios, report in dB.