离散时间傅里叶变换与离散傅里叶变换
引入| English | 释义 | 拼音 |
|---|---|---|
| DTFT | 离散时间傅里叶变换 | lí sàn shí jiān fù lǐ yè biàn huàn |
| DFT | 离散傅里叶变换 | lí sàn fù lǐ yè biàn huàn |
| resolution/ˌrezəˈluːʃn/ | 分辨率 | fēn biàn lǜ |
| leakage | 频谱泄漏 | pín pǔ xiè lòu |
| Hamming window | 海明窗 | hǎi míng chuāng |
| interpolates | 插值 | chā zhí |
A spectrum from a finite list
- A computer holds $N$ samples, not an integral. The spectrum must become a finite list of numbers too.
- The ways that list can mislead you are the most practical topic in this course.
From DTFT to DFT
- The DTFT 离散时间傅里叶变换 $X(e^{j\omega}) = \sum x[n]e^{-j\omega n}$ is continuous in $\omega$ and repeats every $2\pi$.
- The DFT 离散傅里叶变换 samples it at $N$ points, $\omega_{k} = 2\pi k/N$. In return, it treats the $N$ samples as repeating for ever.
The DFT treats its N samples as one period of a signal that repeats for ever.
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Sampling the spectrum makes the time signal periodic. That is where leakage comes from.
Resolution
- The bins are $\Delta f = f_{s}/N$ apart, and bin $k$ is the frequency $k f_{s}/N$.
- Resolution 分辨率 is set by measurement time: 1 Hz bins at 1 kHz need $N = 1000$, one whole second of data.
A 256-point DFT runs at f_s = 8 kHz. What frequency is bin 32, in Hz?
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$32 \times 8000/256 = 1000$ Hz.
Two tones 4 Hz apart must land in different bins. At least how long a record is needed, in s?
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The bins must be 4 Hz or narrower, and $\Delta f = 1/T$, so $T \ge 0.25$ s, whatever the sample rate.
Leakage and windows
- A tone between bins does not fit a whole number of periods, so the repeated copies do not join. Its energy spreads into every bin: leakage 频谱泄漏.
- A window tapers the ends. The rectangle's sidelobes are only $-13$ dB down; the Hamming window 海明窗 pushes them to $-43$ dB, at the cost of a wider main lobe.
A weak tone sits 30 dB below a strong one nearby. Which window lets you see it?
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The rectangle's −13 dB sidelobes bury anything 30 dB down. Hamming's −43 dB sidelobes leave it visible.
Worked example: on and off the bin
- An 8-point DFT of a cosine exactly on bin 1: $|X| = 4$ at bins 1 and 7, zero elsewhere.
- Move it half a bin: the magnitudes become $1.0, 2.40, 2.80, 1.19, 1.0, \ldots$ in every bin. A small tone nearby would be lost in that spread.
A 16-point DFT takes a unit cosine exactly on bin 2. What is |X| at bin 2?
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An on-bin cosine of amplitude 1 gives N/2 at its bin and its mirror (bins 2 and 14), and zero elsewhere.
Worked example: the resolution budget
- At $f_{s} = 1$ kHz, $N = 100$ gives 10 Hz bins: two tones 6 Hz apart show as one bump.
- Zero-padding to 4000 points draws that bump more smoothly. It interpolates 插值 the spectrum, but cannot split the two tones.
You zero-pad a 100-point record (f_s = 1 kHz) to 4000 points. Two tones are 6 Hz apart. What happens?
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The record is still 0.1 s, so the true resolution is still 10 Hz. Padding only interpolates.
Three common slips
- Bins as hertz. Bin $k$ is $k f_{s}/N$ Hz.
- Zero-padding for resolution. Only a longer record resolves closer tones.
- Forgetting the repeat. The jump from the last sample back to the first is itself a broadband signal.
Match each choice to what it decides.
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Only measurement time buys resolution.
Recap
- The DFT samples the DTFT; the time record is treated as periodic.
- $\Delta f = f_{s}/N$ is bought with measurement time.
- Off-bin tones leak; windows trade sidelobes for main-lobe width.