FFT, Frequency Bins, and Nyquist
FFT：把一段波形变成各频率的振幅与相位
这是全库共用的知识点。FFT（Fast Fourier Transform，快速傅里叶变换）是计算 DFT（离散傅里叶变换）的快速方法；它们描述同一项变换，不是两种不同的物理分析。 Nyquist（奈奎斯特）频率是采样率的一半，它给出通常无混叠频带的边界，下面用采样图解释。
来自探针诊断页？读完可 返回第 4 节继续读功率谱。
1. 先看图：一个复杂波形可以由几个简单波相加
取一个单位为任意信号单位的例子：
两个成分的频率分别为 1 和 3 kHz，振幅分别为 1 和 0.5。 为了在图中看清相位，这里特意给 3 kHz 成分设置 \(\pi/3\) 的初相位； 探针诊断页的双频示例仍使用它原来的初相位。
左上：两个成分；右上：相加后的信号与采样点，绘出前 2 ms。 左下：用完整 8 ms 记录计算的单边振幅谱，两个峰分别为 1 和 0.5。 右下：相应的复数系数乘以 2/N 后画成箭头，长度给出振幅，角度给出相位。 图中的 Re、Im 分别表示实部和虚部。
时域图问“每个时刻的数值是什么”，频域图问“每个频率的成分有多强、相位是多少”。 FFT 并不只保留峰的位置，还保留一组复数。 用 \(z=R(\cos\phi+i\sin\phi)=Re^{i\phi}\) 表示一个复数， 其中 \(i^2=-1\)，长度 \(R=|z|\) 与角度 \(\phi=\arg z\) 就能同时装下这两个信息。
2. 频点是什么？为什么间隔为 fs/N？
采样率 \(f_s\) 的单位是 Hz，即每秒取得的样本数。一段有 \(N\) 个样本， 时刻为 \(t_n=n/f_s\)，\(n=0,\ldots,N-1\)。 DFT 把这 N 个数看作一段长度为 \(T_{\rm seg}=N/f_s\) 的周期记录； 最后一个实际样本在 \((N-1)/f_s\)，不再重复采集段终点。 在这段时间内正好转 m 圈的频率是
这些用于比较的离散频率叫作 频点，\(m\) 是频点序号。 它不是说真实波只能有这些频率，而是说一次 DFT 在这些频率上给出系数。 本例 \(f_s=32000\ \mathrm{Hz}\)、\(N=256\)，因此 \(T_{\rm seg}=8\ \mathrm{ms}\)、\(\Delta f=125\ \mathrm{Hz}\)。 1 kHz 对应 \(m=8\)，3 kHz 对应 \(m=24\)。
3. 求和为什么能取出振幅和相位？
NumPy 默认的正变换是
先用一个以候选频率反向旋转的复数乘每个样本，再相加。 如果信号正好含有这个频率，对应项在相乘后停止旋转，因而反复累加； 其他整数频点的项在一圈圈旋转中相互抵消。这里比较的是整段数据，不是某一个峰。
具体地，若 \(x[n]=A\cos(2\pi mn/N+\phi)\)，利用 \(\cos\alpha=(e^{i\alpha}+e^{-i\alpha})/2\) 得到
第一项加了 N 次；第二项在普通正频率点上恰好抵消。因此 \(A=2|X[m]|/N\)、\(\phi=\arg X[m]\)。 这个简单结果要求频率恰好在频点上，并且不是直流（0 Hz）或 Nyquist 端点。 未加窗是这里采用的条件。若不满足整数周期条件，请读 谱泄漏。 谱值接近零时，计算出的相角通常也没有可靠的物理意义。
探针页写作 \(X_j[m]\) 时，额外的 \(j\) 只是区分第 1、2 路信号； 每一路都独立执行同一项变换。定义和正负号采用 NumPy 官方 FFT 约定。
4. Nyquist 频率限制了什么？
Nyquist 频率 定义为 \(f_{\rm Nyq}=f_s/2\)，对应每周期两个样本。 若想由均匀采样唯一恢复一个一般的带限信号，需要其频率成分严格低于这个边界， 并排除采样前的带外成分。边界本身是特殊点，不能简单理解为“两点就足够恢复任意相位”。
为看清采样点，这张图另用 fs=8 Hz，因此 Nyquist 频率是 4 Hz。 蓝色 3 Hz 波和橙色 5 Hz 波在连续时间中不同，却经过完全相同的黑色采样点。 只保留这些样本，就分不清它们；5 Hz 折到了 3 Hz，这叫时间混叠。
本页的 32 kHz 采样对应 16 kHz 的 Nyquist 频率，所以 1 和 3 kHz 都在时间采样允许的范围内。 时间 Nyquist 的单位是 Hz；探针间距决定的空间 Nyquist 波数单位是 rad/m，二者不是同一个量。
5. 为什么 rfft 可以不返回负频率？
实数记录的系数满足共轭对称关系 \(X[N-m]=X[m]^*\)，这里的
\(N-m\) 位置代表相应的负频率。共轭就是把复数的虚部变号；
负频率部分可以从正频率部分算回来，所以 rfft 只存非负频率。
偶数 N 时结果有 \(N/2+1\) 项，含 0 Hz 和 \(f_s/2\)；
奇数 N 时最高频点略低于 \(f_s/2\)。
直流和偶数 N 的 Nyquist 点没有另一项独立的负频率伙伴，不能像普通正频率那样乘 2。 在 Nyquist 点，\(\cos(\pi n+\phi)=(-1)^n\cos\phi\)， 因而采样不能独立保留这个波的全部振幅与相位信息。
下面就是生成第一张图的计算代码。amplitude 为单边振幅显示，phase 单位为 rad；
它不是功率谱密度，不能混用归一化。
fs, N = 32000.0, 256
t = np.arange(N) / fs
x = np.cos(2*np.pi*1000*t) + 0.5*np.cos(2*np.pi*3000*t + np.pi/3)
X = np.fft.rfft(x)
f = np.fft.rfftfreq(N, d=1/fs)
amplitude = 2*np.abs(X)/N
amplitude[0] /= 2
amplitude[-1] /= 2 # N is even: do not double the Nyquist bin
phase = np.angle(X)
本例两个有信号的频点返回振幅 1、0.5，以及相位 0、π/3。 返回探针诊断第 4 节：继续读功率谱 · 继续了解谱泄漏 · 知识点目录
FFT: turn a finite record into amplitudes and phases
This note is shared across the library. The FFT (Fast Fourier Transform) efficiently computes the DFT (Discrete Fourier Transform); these are not different physical analyses. The Nyquist frequency is half the sampling rate, the boundary of the usual unaliased band. A sampling diagram below explains it.
Arriving from probe diagnostics? Afterwards, return to section 4 and power spectra.
1. Start with the pictures: add simple components to make a complicated trace
In arbitrary signal units, choose
The components are at 1 and 3 kHz, with amplitudes 1 and 0.5. To make phase visible, this note gives the 3 kHz tone initial phase \(\pi/3\); the probe tutorial’s example retains its original initial phase.
Top left: components. Top right: their sum and samples, showing the first 2 ms. Bottom left: one-sided amplitudes from the full 8 ms record, with peaks 1 and 0.5. Bottom right: arrows for coefficients multiplied by 2/N; length is amplitude and angle is phase. Re and Im mean real and imaginary parts.
A time trace gives the value at each time; a spectrum gives the strength and phase at each frequency. The FFT retains complex coefficients, not just peak locations. A complex number \(z=R(\cos\phi+i\sin\phi)=Re^{i\phi}\), where \(i^2=-1\), stores both its length \(R=|z|\) and its angle \(\phi=\arg z\).
2. What are frequency bins, and why are they spaced by fs/N?
Sampling rate \(f_s\) is samples per second, in Hz. A segment contains \(N\) values at \(t_n=n/f_s\), \(n=0,\ldots,N-1\). The DFT treats them as one period of a record lasting \(T_{\rm seg}=N/f_s\). The last actual sample is at \((N-1)/f_s\); the endpoint is not sampled twice. A frequency completing exactly m cycles in this interval is
These discrete comparison frequencies are bins, indexed by \(m\). Real waves need not have these frequencies; a DFT reports coefficients at these locations. Here \(f_s=32000\ \mathrm{Hz}\) and \(N=256\) give \(T_{\rm seg}=8\ \mathrm{ms}\) and \(\Delta f=125\ \mathrm{Hz}\). The 1 and 3 kHz tones occupy bins 8 and 24.
3. How does the sum recover amplitude and phase?
NumPy’s default forward transform is
Multiply samples by a complex reference rotating backwards at the candidate frequency, then add them. A matching component stops rotating and accumulates, while components at other integer bins cancel over complete turns. This compares the entire segment.
For \(x[n]=A\cos(2\pi mn/N+\phi)\), expand \(\cos\alpha=(e^{i\alpha}+e^{-i\alpha})/2\):
The first term accumulates N times; the second cancels at ordinary positive bins. Consequently \(A=2|X[m]|/N\) and \(\phi=\arg X[m]\). This simple result assumes an unwindowed, bin-aligned tone away from DC (0 Hz) and the Nyquist endpoint. For non-integer cycles see spectral leakage. The angle of a near-zero coefficient usually has no reliable physical interpretation.
In the probe tutorial’s \(X_j[m]\), the extra \(j\) only identifies the signal; both probes independently use the same transform. Definitions and signs follow the official NumPy FFT convention.
4. What does the Nyquist frequency limit?
The Nyquist frequency is \(f_{\rm Nyq}=f_s/2\), or two samples per cycle. Unique recovery of a general band-limited signal from uniform samples requires its components to lie strictly below this boundary and excludes out-of-band content before sampling. The boundary itself is special: two points do not recover arbitrary phase.
To expose individual samples this separate illustration uses fs=8 Hz and Nyquist frequency 4 Hz. The blue 3 Hz and orange 5 Hz curves differ continuously but pass through the same black samples. Samples alone cannot distinguish them: 5 Hz aliases to 3 Hz. This is temporal aliasing.
The main example’s 32 kHz sampling has Nyquist frequency 16 kHz, so both 1 and 3 kHz lie within its temporal band. Temporal Nyquist frequency is in Hz; the spatial Nyquist wavenumber set by probe spacing is in rad/m. They are different quantities.
5. Why can rfft omit negative frequencies?
A real record obeys \(X[N-m]=X[m]^*\), where index \(N-m\) represents the matching
negative frequency. Conjugation reverses the imaginary part. Negative coefficients can
therefore be reconstructed, so rfft stores only nonnegative frequencies. For even N
there are \(N/2+1\) entries including DC and \(f_s/2\); for odd N the highest bin
is slightly below \(f_s/2\).
DC and the even-N Nyquist bin have no separate negative-frequency partner and must not be doubled like ordinary positive bins. At Nyquist, \(\cos(\pi n+\phi)=(-1)^n\cos\phi\), so samples cannot independently retain all amplitude and phase information of that component.
The following calculation generates the first figure. amplitude is a one-sided
amplitude display and phase is in radians; this is not a power spectral density.
fs, N = 32000.0, 256
t = np.arange(N) / fs
x = np.cos(2*np.pi*1000*t) + 0.5*np.cos(2*np.pi*3000*t + np.pi/3)
X = np.fft.rfft(x)
f = np.fft.rfftfreq(N, d=1/fs)
amplitude = 2*np.abs(X)/N
amplitude[0] /= 2
amplitude[-1] /= 2 # N is even: do not double the Nyquist bin
phase = np.angle(X)
The populated bins return amplitudes 1 and 0.5 and phases 0 and π/3. Return to probe diagnostics section 4: power spectra · Continue to spectral leakage · Knowledge index