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用matplotlib绘制正弦波形的傅里叶变换

continuous-fourier fft scipy matplotlib python

Sam · 技术社区 · 4 年前

我正试图绘制一个基于傅里叶变换的符号波 scipy documentation

import numpy as np
import matplotlib.pyplot as plt
import scipy.fft

def sinWav(amp, freq, time, phase=0):
    return amp * np.sin(2 * np.pi * (freq * time - phase))

def plotFFT(f, speriod, time):
    """Plots a fast fourier transform

    Args:
        f (np.arr): A signal wave
        speriod (int): Number of samples per second
        time ([type]): total seconds in wave
    """

    N = speriod * time
    # sample spacing
    T = 1.0 / 800.0
    x = np.linspace(0.0, N*T, N, endpoint=False)

    yf = scipy.fft.fft(f)
    xf = scipy.fft.fftfreq(N, T)[:N//2]
    plt.plot(xf, 2.0/N * np.abs(yf[0:N//2]))
    plt.grid()
    plt.xlim([1,3])
    plt.show()


speriod = 1000
time  = {
    0: np.arange(0, 4, 1/speriod),
    1: np.arange(4, 8, 1/speriod),
    2: np.arange(8, 12, 1/speriod)
}

signal = np.concatenate([
    sinWav(amp=0.25, freq=2, time=time[0]),
    sinWav(amp=1, freq=2, time=time[1]),
    sinWav(amp=0.5, freq=2, time=time[2])
])   # generate signal

plotFFT(signal, speriod, 12)

我想得到一个像这样的傅里叶变换图

电流输出

额外的

这就是我正在处理的罪恶波

1 回复 | 直到 4 年前

Synthaze 4 年前

import numpy as np
import matplotlib.pyplot as plt
import scipy.fft

def sinWav(amp, freq, time, phase=0):
    return amp * np.sin(2 * np.pi * (freq * time - phase))

def plotFFT(f, speriod, time):
    """Plots a fast fourier transform

    Args:
        f (np.arr): A signal wave
        speriod (int): Number of samples per second
        time ([type]): total seconds in wave
    """

    N = speriod * time
    # sample spacing
    T = 1.0 / 800.0
    x = np.linspace(0.0, N*T, N, endpoint=False)

    yf = scipy.fft.fft(f)
    xf = scipy.fft.fftfreq(N, T)[:N//2]

    amplitudes = 1/speriod* np.abs(yf[:N//2])
  
    plt.plot(xf, amplitudes)
    plt.grid()
    plt.xlim([1,3])
    plt.show()


speriod = 800
time  = {
    0: np.arange(0, 4, 1/speriod),
    1: np.arange(4, 8, 1/speriod),
    2: np.arange(8, 12, 1/speriod)
}

signal = np.concatenate([
    sinWav(amp=0.25, freq=2, time=time[0]),
    sinWav(amp=1, freq=2, time=time[1]),
    sinWav(amp=0.5, freq=2, time=time[2])
])   # generate signal

plotFFT(signal, speriod, 12)

你应该得到你想要的。你的振幅计算不正确,因为你的分辨率和速度不一致。

更长的数据采集时间:

import numpy as np
import matplotlib.pyplot as plt
import scipy.fft

def sinWav(amp, freq, time, phase=0):
    return amp * np.sin(2 * np.pi * (freq * time - phase))

def plotFFT(f, speriod, time):
    """Plots a fast fourier transform

    Args:
        f (np.arr): A signal wave
        speriod (int): Number of samples per second
        time ([type]): total seconds in wave
    """

    N = speriod * time
    # sample spacing
    T = 1.0 / 800.0
    x = np.linspace(0.0, N*T, N, endpoint=False)

    yf = scipy.fft.fft(f)
    xf = scipy.fft.fftfreq(N, T)[:N//2]

    amplitudes = 1/(speriod*4)* np.abs(yf[:N//2])
  
    plt.plot(xf, amplitudes)
    plt.grid()
    plt.xlim([1,3])
    plt.show()


speriod = 800
time  = {
    0: np.arange(0, 4*4, 1/speriod),
    1: np.arange(4*4, 8*4, 1/speriod),
    2: np.arange(8*4, 12*4, 1/speriod)
}

signal = np.concatenate([
    sinWav(amp=0.25, freq=2, time=time[0]),
    sinWav(amp=1, freq=2, time=time[1]),
    sinWav(amp=0.5, freq=2, time=time[2])
])   # generate signal

plotFFT(signal, speriod, 48)