【发布时间】：2016-11-09 03:37:07
【问题描述】：
我正在一个使用 Raspberry Pi 3 B 的项目中工作，我通过 ADC MPC3008 从 IR 传感器（Sharp GP2Y0A21YK0F）获取数据并使用 PyQtgraph 库实时显示。
ADC 的数据表显示在 5.0V 时，采样率为 200khz。但是我每秒只能获得大约 30 个样本。
使用树莓派可以达到 200khz 吗？
如果是，我应该学习或实施什么才能获得它？
如果不是，我应该怎么做才能获得可能的最高采样率？如何找出最高采样率？
这是我的代码：
# -*- coding: utf-8 -*-
import time
import Adafruit_GPIO.SPI as SPI
import Adafruit_MCP3008
from collections import deque
import serial
import pyqtgraph as pg
from pyqtgraph.Qt import QtCore, QtGui
import numpy as np
SPI_PORT = 0
SPI_DEVICE = 0
mcp = Adafruit_MCP3008.MCP3008(spi=SPI.SpiDev(SPI_PORT, SPI_DEVICE))
win = pg.GraphicsWindow()
win.setWindowTitle('pyqtgraph example: Scrolling Plots')
nsamples=600 #tamanho das matrizes para os dados
tx_aq = 0 #velocidade da aquisição
intervalo_sp = 0.5 #intervalo para secao de poincare
# 1) Simplest approach -- update data in the array such that plot appears to scroll
# In these examples, the array size is fixed.
p1 = win.addPlot()
p1.setRange(yRange=[0,35])
p2 = win.addPlot()
p2.setRange(yRange=[-100,100])
p3 = win.addPlot()
p3.setRange(yRange=[-100,100])
p3.setRange(xRange=[-0,35])
#p3.plot(np.random.normal(size=100), pen=(200,200,200), symbolBrush=(255,0,0), symbolPen='w')
'''
p3.setDownsampling(mode='peak')
p3.setClipToView(True)
p3.setRange(xRange=[-100, 0])
p3.setLimits(xMax=0)
'''
data1= np.zeros((nsamples,2),float) #ARMAZENAR POSICAO
vec_0=deque()
vec_1=deque()
vec_2=deque()
ptr1 = 0
data2= np.zeros((nsamples,2),float) #ARMAZENAR VELOCIDADE
diff=np.zeros((2,2),float)
diff_v=deque()
data3= np.zeros((nsamples,2),float)
data3_sp=np.zeros((1,2),float)
ptr3=0
curve1 = p1.plot(data1)
curve2 = p2.plot(data2)
curve3 = p3.plot(data3)
#Coeficientes da calibração do IR
c1=-7.246
c2=44.17
c3=-95.88
c4=85.28
tlast=time.clock()
tlast_sp=time.clock()
#print tlast
def getdata():
global vec_0, vec_1, vec_2, tlast
timenow=time.clock()
if timenow-tlast>=tx_aq:
#name=input("HUGO")
tlast=timenow
t0=float(time.clock())
str_0 =mcp.read_adc(0)
t1=float(time.clock())
str_1 =mcp.read_adc(0)
t2=float(time.clock())
str_2 =mcp.read_adc(0)
d0x=(float(str_0))*(3.3/1023)
d0= c1*d0x**3+c2*d0x**2+c3*d0x+c4
vec_0=(t0, d0)
d1x=(float(str_1))*(3.3/1023)
d1= c1*d1x**3+c2*d1x**2+c3*d1x+c4
vec_1=(t1, d1)
d2x=(float(str_2))*(3.3/1023)
d2= c1*d2x**3+c2*d2x**2+c3*d2x+c4
vec_2=(t2, d2)
functions()
def diferenciar():
global data2
diff=(data1[-1,1]-data1[-3,1])/(data1[-1,0]-data1[-3,0])
data2[:-1] = data2[1:]
data2[-1,1] = diff
data2[-1,0] = data1[-2,0]
def organizar():
global data1, data3
data1[:-1] = data1[1:]
vec_x1=np.array(vec_1)
data1[-1]=vec_x1
def EF(): #ESPACO DE FASE
global data3, ptr3
data3[:-1] = data3[1:]
data3[-1,0]=data1[-1,1]
data3[-1,1]=data2[-1,1]
def SP():
global timenow_sp, tlast_sp
timenow_sp=time.clock()
if timenow_sp-tlast_sp>=intervalo_sp:
tlast_sp=timenow_sp
data3_sp[0,0]=data3[-2,0]
data3_sp[0,1]=data3[-2,1]
p3.plot(data3_sp, pen=None, symbol='o', symbolPen=None, symbolSize=4, symbolBrush=('r'))
#print data3_sp
def plotar():
global ptr1
curve1.setData(data1)
ptr1 += 1
curve2.setData(data2)
#curve2.setPos(ptr1, 0)
#p3.plot(data3)
def functions():
diferenciar()
organizar()
EF()
SP()
plotar()
def update1():
global data1, curve1, ptr1
getdata()
# update all plots
def update():
update1()
timer = pg.QtCore.QTimer()
timer.timeout.connect(update)
timer.start(50)
## Start Qt event loop unless running in interactive mode or using pyside.
if __name__ == '__main__':
import sys
if (sys.flags.interactive != 1) or not hasattr(QtCore, 'PYQT_VERSION'):
QtGui.QApplication.instance().exec_()
我正在尝试找到解决它的方法，但到目前为止我失败了。
你们能帮我解决这个问题，或者至少指出我在哪里可以找到有关这方面的信息吗？
【问题讨论】：
标签： python raspberry-pi pyqtgraph adc