dft | 易学教程

No instance of overloaded function when resizing

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问题 I got an error of "No instance of overloaded function" when I used resize(). I'm aware that what's inside of resize() might be wrong as well. I'm trying to resize x, an array of floats to all zeros. Not sure if std::vector>(0, 0) would be the right syntax for it as I am very new to C++. How do I solve the existing error? void IDFT(const std::vector<std::complex<float>> &Xf, std::vector<float> &x) { // allocate space for the vector holding the frequency bins // initialize all the values in the

Getting the frequencies associated with STFT in Librosa

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问题 When using librosa.stft() to calculate a spectrogram, how does one get back the associated frequency values? I am not interested in generating an image as in librosa.display.specshow , but rather I want to have those values in hand. y, sr = librosa.load('../recordings/high_pitch.m4a') stft = librosa.stft(y, n_fft=256, window=sig.windows.hamming) spec = np.abs(stft) spec gives me the 'amplitude' or 'power' of each frequency, but not the frequencies bins themselves. I have seen that there is a

1D Heat Equation using DFT produces incorrect results (FFTW)

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来源： https://stackoverflow.com/questions/64155448/1d-heat-equation-using-dft-produces-incorrect-results-fftw

1D Heat Equation using DFT produces incorrect results (FFTW)

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来源： https://stackoverflow.com/questions/64155448/1d-heat-equation-using-dft-produces-incorrect-results-fftw

Why does my 2D IDFT produce double the expected amplitudes? (FFTW)

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来源： https://stackoverflow.com/questions/64064195/why-does-my-2d-idft-produce-double-the-expected-amplitudes-fftw

Why does my 2D IDFT produce double the expected amplitudes? (FFTW)

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来源： https://stackoverflow.com/questions/64064195/why-does-my-2d-idft-produce-double-the-expected-amplitudes-fftw

Discrete Fourier transform

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问题 I am currently trying to write some fourier transform algorithm. I started with a simple DFT algorithm as described in the mathematical definition: public class DFT { public static Complex[] Transform(Complex[] input) { int N = input.Length; Complex[] output = new Complex[N]; double arg = -2.0 * Math.PI / (double)N; for (int n = 0; n < N; n++) { output[n] = new Complex(); for (int k = 0; k < N; k++) output[n] += input[k] * Complex.Polar(1, arg * (double)n * (double)k); } return output; } } So

一幅图弄清DFT与DTFT,DFS的关系

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很多同学学习了数字信号处理之后，被里面的几个名词搞的晕头转向，比如DFT，DTFT，DFS，FFT，FT,FS等，FT和FS属于信号与系统课程的内容，是对连续时间信号的处理，这里就不过多讨论，只解释一下前四者的关系。　　首先说明一下，我不是数字信号处理专家，因此这里只站在学生的角度以最浅显易懂的性质来解释问题，而不涉及到任何公式运算。　　学过卷积，我们都知道有时域卷积定理和频域卷积定理，在这里只需要记住两点：1.在一个域的相乘等于另一个域的卷积；2.与脉冲函数的卷积，在每个脉冲的位置上将产生一个波形的镜像。(在任何一本信号与系统课本里，此两条性质有详细公式证明) 　　下面，就用这两条性质来说明DFT，DTFT，DFS，FFT之间的联系：　　先看图片：　　首先来说图（1）和图（2），对于一个模拟信号，如图(1)所示，要分析它的频率成分，必须变换到频域，这是通过傅立叶变换即 FT(Fourier Transform) 得到的，于是有了模拟信号的频谱，如图(2)；注意1：时域和频域都是连续的！　　但是，计算机只能处理数字信号，首先需要将原模拟信号在时域离散化，即在时域对其进行采样，采样脉冲序列如图(3)所示，该采样序列的频谱如图(4)，可见它的频谱也是一系列的脉冲。所谓时域采样，就是在时域对信号进行相乘，(1)×(3)后可以得到离散时间信号x[n]，如图(5

only integers, slices (`:`), ellipsis (`…`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices

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问题 I am implementing fft as part of my homework. My problem lies in the implemention of shuffling data elements using bit reversal. I get the following warning: DeprecationWarning: using a non-integer number instead of an integer will result in an error in the future. data[x], data[y] = data[y], data[x] And the auto grading system (provided by university) returns the following: error: only integers, slices ( : ), ellipsis ( ... ), numpy.newaxis ( None ) and integer or boolean arrays are valid

only integers, slices (`:`), ellipsis (`…`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices

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