Franck-Hertz Data Graphs

%matplotlib inline
import numpy as np
import math
import matplotlib.pyplot as pl


#1ST PART: Argon collected current vs accelerating voltage - peaks

#X = [1,2,3,4,5,6]
#Y = [20.7, 30.2, 40.7, 52.1, 63.6, 76.1]

#pl.plot(X,Y,'o', label = 'Accelerating Voltage at each peak number')
#pl.xlabel('Peak Number')
#pl.ylabel('Accelerating Voltage')
#pl.show()

# number of parameters for line fit (2)
npar = 2
# vector with data (Y); std of measurement error is sigma =.1
sigma = 0.1
# here is data to fit
Y = np.array([20.7, 30.2, 40.7, 52.1, 63.6, 76.1])
Yerror = 1
# here are X positions of data
X = np.array([1,2,3,4,5,6]);
# number of data points
nobs = len(X)

# create M – ones in first column, X in second column
M = np.column_stack( (np.ones(nobs),X) )

# solve
MTM = np.dot(M.transpose(),M)
MTMINV = np.linalg.inv(MTM)
MTY = np.dot(M.transpose(),Y)

P = np.dot(MTMINV,MTY)  # SOLUTION

C = MTMINV * sigma**2   # COVARIANCE MATRIX

# compute RMS and print it
Residuals = Y -	np.dot(M,P)
ChiSq =	np.dot(Residuals.transpose(),Residuals)
RMS = math.sqrt(ChiSq/nobs)
#print('RMS = %f'%(RMS))

# plot the result of the fit
pl.ion()
pl.plot(X,Y,'ro', label = 'Accelerating Voltage at each peak number')
pl.errorbar(X,Y,yerr=Yerror)
xx = np.linspace(0.5,6.5,101)
ModelY = P[0] + P[1]*xx
pl.plot(xx,ModelY,'k')
pl.xlabel('Peak Number')
pl.ylabel('Accelerating Voltage (V)')
pl.title('Accelerating Voltage at Peak Versus Peak Number')
pl.show()

# print solution
print('        Intercept= %f +/- %f'%( P[0],math.sqrt(C[0,0]) ))
print('        Slope    = %f +/- %f'%( P[1],math.sqrt(C[1,1]) ))









#2ND PART: Argon collected current vs accelerating voltage - dips

#X = [1,2,3,4,5,6]
#Y = [23.8, 35.2, 46.3, 57.6, 69, 79.6]

#pl.plot(X,Y,'o', label = 'Accelerating Voltage at each peak number')
#pl.xlabel('Peak Number')
#pl.ylabel('Accelerating Voltage')
#pl.show()

# number of parameters for line fit (2)
npar = 2
# vector with data (Y); std of measurement error is sigma =.1
sigma = 0.1
# here is data to fit
Y = np.array([23.8, 35.2, 46.3, 57.6, 69, 79.6])
Yerror = 1
# here are X positions of data
X = np.array([1,2,3,4,5,6]);
# number of data points
nobs = len(X)

# create M – ones in first column, X in second column
M = np.column_stack( (np.ones(nobs),X) )

# solve
MTM = np.dot(M.transpose(),M)
MTMINV = np.linalg.inv(MTM)
MTY = np.dot(M.transpose(),Y)

P = np.dot(MTMINV,MTY)  # SOLUTION

C = MTMINV * sigma**2   # COVARIANCE MATRIX

# compute RMS and print it
Residuals = Y -	np.dot(M,P)
ChiSq =	np.dot(Residuals.transpose(),Residuals)
RMS = math.sqrt(ChiSq/nobs)
#print('RMS = %f'%(RMS))

# plot the result of the fit
pl.ion()
pl.plot(X,Y,'ro', label = 'Accelerating Voltage at each peak number')
pl.errorbar(X,Y,yerr=Yerror)
xx = np.linspace(0.5,6.5,101)
ModelY = P[0] + P[1]*xx
pl.plot(xx,ModelY,'k')
pl.xlabel('Dip Number')
pl.ylabel('Accelerating Voltage (V)')
pl.title('Accelerating Voltage at Dip Versus Dip Number')
pl.show()

# print solution
print('        Intercept= %f +/- %f'%( P[0],math.sqrt(C[0,0]) ))
print('        Slope    = %f +/- %f'%( P[1],math.sqrt(C[1,1]) ))







#3RD PART: Neon collected current vs accelerating voltage - peaks

#X = [1,2,3,4]
#Y = [19.2,38.3,61.6,80.6]

#pl.plot(X,Y,'o', label = 'Accelerating Voltage at each peak number')
#pl.xlabel('Peak Number')
#pl.ylabel('Accelerating Voltage')
#pl.show()

# number of parameters for line fit (2)
npar = 2
# vector with data (Y); std of measurement error is sigma =.1
sigma = 0.1
# here is data to fit
Y = np.array([19.2,38.3,61.6,80.6])
Yerror = 1
# here are X positions of data
X = np.array([1,2,3,4]);
# number of data points
nobs = len(X)

# create M – ones in first column, X in second column
M = np.column_stack( (np.ones(nobs),X) )

# solve
MTM = np.dot(M.transpose(),M)
MTMINV = np.linalg.inv(MTM)
MTY = np.dot(M.transpose(),Y)

P = np.dot(MTMINV,MTY)  # SOLUTION

C = MTMINV * sigma**2   # COVARIANCE MATRIX

# compute RMS and print it
Residuals = Y -	np.dot(M,P)
ChiSq =	np.dot(Residuals.transpose(),Residuals)
RMS = math.sqrt(ChiSq/nobs)
#print('RMS = %f'%(RMS))

# plot the result of the fit
pl.ion()
pl.plot(X,Y,'ro', label = 'Accelerating Voltage at each peak number')
pl.errorbar(X,Y,yerr=Yerror)
xx = np.linspace(0.99,4.01,101)
ModelY = P[0] + P[1]*xx
pl.plot(xx,ModelY,'k')
pl.xlabel('Peak Number')
pl.ylabel('Accelerating Voltage (V)')
pl.title('Accelerating Voltage at Peak Versus Peak Number (Neon)')
pl.show()

# print solution
print('        Intercept= %f +/- %f'%( P[0],math.sqrt(C[0,0]) ))
print('        Slope    = %f +/- %f'%( P[1],math.sqrt(C[1,1]) ))







#4TH PART: Neon collected current vs accelerating voltage - dips

#X = [1,2,3]
#Y = [25.5,46.7,69.4]

#pl.plot(X,Y,'o', label = 'Accelerating Voltage at each peak number')
#pl.xlabel('Peak Number')
#pl.ylabel('Accelerating Voltage')
#pl.show()

# number of parameters for line fit (2)
npar = 2
# vector with data (Y); std of measurement error is sigma =.1
sigma = 0.1
# here is data to fit
Y = np.array([25.5,46.7,69.4])
Yerror = 1
# here are X positions of data
X = np.array([1,2,3]);
# number of data points
nobs = len(X)

# create M – ones in first column, X in second column
M = np.column_stack( (np.ones(nobs),X) )

# solve
MTM = np.dot(M.transpose(),M)
MTMINV = np.linalg.inv(MTM)
MTY = np.dot(M.transpose(),Y)

P = np.dot(MTMINV,MTY)  # SOLUTION

C = MTMINV * sigma**2   # COVARIANCE MATRIX

# compute RMS and print it
Residuals = Y -	np.dot(M,P)
ChiSq =	np.dot(Residuals.transpose(),Residuals)
RMS = math.sqrt(ChiSq/nobs)
#print('RMS = %f'%(RMS))

# plot the result of the fit
pl.ion()
pl.plot(X,Y,'ro', label = 'Accelerating Voltage at each peak number')
pl.errorbar(X,Y,yerr=Yerror)
xx = np.linspace(0.99,3.01,101)
ModelY = P[0] + P[1]*xx
pl.plot(xx,ModelY,'k')
pl.xlabel('Dip Number')
pl.ylabel('Accelerating Voltage (V)')
pl.title('Accelerating Voltage at Dip Versus Dip Number (Neon)')
pl.show()

# print solution
print('        Intercept= %f +/- %f'%( P[0],math.sqrt(C[0,0]) ))
print('        Slope    = %f +/- %f'%( P[1],math.sqrt(C[1,1]) ))