我想使用python
从MPU6050传感器中提取的数据阵列进行FFT请在下面找到数据示例
0.13,0.04,1.03
0.14,0.01,1.02
0.15,-0.04,1.05
0.16,0.02,1.05
0.14,0.01,1.02
0.16,-0.03,1.04
0.15,-0.00,1.04
0.14,0.03,1.02
0.14,0.01,1.03
0.17,0.02,1.05
0.15,0.03,1.03
0.14,0.00,1.02
0.17,-0.02,1.05
0.16,0.01,1.04
0.14,0.02,1.01
0.15,0.00,1.03
0.16,0.03,1.05
0.11,0.03,1.01
0.15,-0.01,1.03
0.16,0.01,1.05
0.14,0.02,1.03
0.13,0.01,1.02
0.15,0.02,1.05
0.13,0.00,1.03
0.08,0.01,1.03
0.09,-0.01,1.03
0.09,-0.02,1.03
0.07,0.01,1.03
0.06,0.00,1.05
0.04,0.00,1.04
0.01,0.01,1.02
0.03,-0.05,1.02
-0.03,-0.05,1.03
-0.05,-0.02,1.02
我采用了第一列(x轴(并保存在数组中
参考:https://hackaday.io/project/12109-Open-source-fft-spectrum-analyzer/details由此,我参与了FFT的一部分,代码如下
from scipy.signal import filtfilt, iirfilter, butter, lfilter
from scipy import fftpack, arange
import numpy as np
import string
import matplotlib.pyplot as plt
sample_rate = 0.2
accx_list_MPU=[]
outputfile1='C:/Users/Meena/Desktop/SensorData.txt'
def fftfunction(array):
n=len(array)
print('The length is....',n)
k=arange(n)
fs=sample_rate/1.0
T=n/fs
freq=k/T
freq=freq[range(n//2)]
Y = fftpack.fft(array)/n
Y = Y[range(n//2)]
pyl.plot(freq, abs(Y))
pyl.grid()
ply.show()
with open(outputfile1) as f:
string1=f.readlines()
N1=len(string1)
for i in range (10,N1):
if (i%2==0):
new_list=string1[i].split(',')
l=len(new_list)
if (l==3):
accx_list_MPU.append(float(new_list[0]))
fftfunction(accx_list_MPU)
我已经获得了FFT的输出,如fftOutput
我不明白该图是否正确。这是我第一次使用FFT,我们如何将其与数据联系起来
这是我在建议的更改之后得到的:fftNew
这是您的fftfunction
的一些返工:
def fftfunction(array):
N = len(array)
amp_spec = abs(fftpack.fft(array)) / N
freq = np.linspace(0, 1, num=N, endpoint=False)
plt.plot(freq, amp_spec, "o-", markerfacecolor="none")
plt.xlim(0, 0.6) # easy way to hide datapoints
plt.margins(0.05, 0.05)
plt.xlabel("Frequency $f/f_{sample}$")
plt.ylabel("Amplitude spectrum")
plt.minorticks_on()
plt.grid(True, which="both")
fftfunction(X)
具体而言,它删除了fs=sample_rate/1.0
部分 - 是否应该是逆的?
该图基本上告诉您哪种频率(相对于样本频率(的强度有多强。查看您的图像,在f=0
上,您的信号偏移量或平均值约为0.12。对于其余部分,没有太多的事情发生,没有任何峰表明测量数据中存在某种频率。