信号处理项目
引入| English | 释义 | 拼音 |
|---|---|---|
| measured performance | 实测性能 | shí cè xìng néng |
| processing gain | 处理增益 | chǔ lǐ zēng yì |
Specify, build, measure
- The course ends the way real work begins: with a specification, an implementation and a measurement.
- An honest report records the gap between the three, failures included.
A verified pipeline
- The project is a tone detector: anti-alias filter, then window, then FFT, then threshold.
- Check each stage on its own before adding the next. A defect found in one stage costs an hour; found in the whole system, a week.
Put the tone detector's stages in order.
难度 3/5 · 今日首次完成可获 200 金币及少量额外奖励
Filter before sampling, window before transforming, then decide on the spectrum.
Write the specification first
- The detector must give 1 Hz bins, answer in under 1 s, and detect a tone at $-10$ dB SNR.
- Every line gets a measured performance 实测性能 figure in the report, pass or fail.
A specification line the design fails should be left out of the report.
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
Every line gets its measured value, failures included. Silent goals hide exactly what a reader needs to know.
Worked example: the specification bites back
- 1 Hz bins at $f_{s} = 1$ kHz need $N \ge 1000$, which is already 1.0 s of data, against a latency limit under 1 s.
- $N = 1024$ gives 0.977 Hz bins but a 1.024 s record, which fails. $N = 1000$ sits exactly on the line. The report records the trade.
0.5 Hz bins at f_s = 2 kHz need at least how many samples?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
$N = f_{s}/\Delta f = 2000/0.5 = 4000$, which is 2 s of data.
How long a record, in s, does that design need?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
4000 samples at 2 kHz is 2 s. Resolution always costs measurement time: $T = 1/\Delta f$.
Processing gain
- A tone's energy collects in one bin, while white noise spreads over all $N/2$ bins.
- So the FFT raises the SNR in the tone's bin by the processing gain 处理增益 $10\log_{10}(N/2)$ dB.
What processing gain does a 4096-point FFT give, in dB? Give the nearest whole number.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
$10\log_{10}(4096/2) = 10\log_{10}2048 = 33.1$ dB.
Worked example: can −10 dB be detected?
- At $N = 1024$ the processing gain is $10\log_{10}512 = 27$ dB.
- An input SNR of $-10$ dB becomes about $+17$ dB in the tone's bin. The window costs a few dB, and detection still has margin.
A 256-point FFT gives 21 dB of processing gain. Can it find a tone at −25 dB input SNR?
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
−25 + 21 = −4 dB in the bin, below the noise. A longer FFT, 4096 points or more, would be needed.
Three common slips
- Testing only the easy case. Measure the hard ones too, and report failures.
- Tuning to the test tone. A detector fitted to one test proves nothing about others.
- Skipping stage checks. Integration hides faults that a stage test would show.
Match each habit to the problem it prevents.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
Verification is the difference between a demo and a product.
Recap
- Specify first; check each stage alone; measure the whole.
- Processing gain $10\log_{10}(N/2)$ decides whether a weak tone can be found.
- Every specification line gets a measured number.