帮我了解FFT函数(Matlab) [英] Help me understand FFT function (Matlab)

问题描述

1)除负频率外，FFT函数提供的最小频率是多少?是零吗?
2)如果为零，我们如何在对数刻度上绘制零?
3)结果总是对称的?还是看起来是对称的?
4)如果我使用abs(fft(y))比较2个信号，我会失去一些准确性吗?

1) Besides the negative frequencies, which is the minimum frequency provided by the FFT function? Is it zero?
2) If it is zero how do we plot zero on a logarithmic scale?
3) The result is always symmetrical? Or it just appears to be symmetrical?
4) If I use abs(fft(y)) to compare 2 signals, may I lose some accuracy?

推荐答案

1)除负频率外，FFT函数提供的最小频率是多少?是零吗?

1) Besides the negative frequencies, which is the minimum frequency provided by the FFT function? Is it zero?

fft(y)返回一个向量，该向量具有y的DFT的第0至第(N-1)个样本，其中y(t)应该认为是在0 ... N-1上定义的(因此，可以将y(t)的周期性重复"视为在Z上定义的周期性信号).

fft(y) returns a vector with the 0-th to (N-1)-th samples of the DFT of y, where y(t) should be thought of as defined on 0 ... N-1 (hence, the 'periodic repetition' of y(t) can be thought of as a periodic signal defined over Z).

fft(y)的第一个样本对应于频率0. 实时，离散时间，周期信号的傅立叶变换也具有离散域，并且是周期和Hermitian的(请参见下文).因此，负频率的变换是正频率对应样本的共轭.

The first sample of fft(y) corresponds to the frequency 0. The Fourier transform of real, discrete-time, periodic signals has also discrete domain, and it is periodic and Hermitian (see below). Hence, the transform for negative frequencies is the conjugate of the corresponding samples for positive frequencies.

例如，如果您将y的(周期性重复)解释为在Z上定义的周期性实信号(采样周期== 1)，则fft(y)的域应解释为N个等距点0， 2π/N ...2π(N-1)/N.负频率-&pi处的变换样本； …-π/N是在π频率上的样本的共轭. ...π/N，等于频率下的样本 π ... 2π(N-1)/N.

For example, if you interpret (the periodic repetition of) y as a periodic real signal defined over Z (sampling period == 1), then the domain of fft(y) should be interpreted as N equispaced points 0, 2π/N ... 2π(N-1)/N. The samples of the transform at the negative frequencies -π ... -π/N are the conjugates of the samples at frequencies π ... π/N, and are equal to the samples at frequencies π ... 2π(N-1)/N.

2)如果为零，我们如何在对数刻度上绘制零?

2) If it is zero how do we plot zero on a logarithmic scale?

如果您想绘制某种 Bode绘图，则可以只为正图绘制变换频率，而忽略了与最低频率(尤其是0)相对应的样本.

If you want to draw some sort of Bode plot you may plot the transform only for positive frequencies, ignoring the samples corresponding to the lowest frequencies (in particular 0).

3)结果始终是对称的?还是看起来是对称的?

3) The result is always symmetrical? Or it just appears to be symmetrical?

如果y是真实的，则具有 Hermitian对称:其实部是对称的，虚部是反对称的.换句话说，它的幅度是对称的，相位是反对称的.

It has Hermitian symmetry if y is real: Its real part is symmetric, its imaginary part is anti-symmetric. Stated another way, its amplitude is symmetric and its phase anti-symmetric.

4)如果我使用abs(fft(y))比较2个信号，我会失去一些准确性吗?

4) If I use abs(fft(y)) to compare 2 signals, may I lose some accuracy?

如果您的意思是abs(fft(x - y))，则可以，您可以使用它来了解差值的频率分布(如果x是y的估计，则为误差).如果您的意思是abs(fft(x)) - abs(fft(y))(???)，则至少会丢失相位信息.

If you mean abs(fft(x - y)), this is OK and you can use it to get an idea of the frequency distribution of the difference (or error, if x is an estimate of y). If you mean abs(fft(x)) - abs(fft(y)) (???) you lose at least phase information.

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