Sensitivity estimate example

Sensitivity estimate example#

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

import sys
sys.path.append('../scripts')
from linear_regression import linreg

from IPython.core.display import Image 
Image(filename='../img/sensitivity_error_example.png',width=400) 
%pylab is deprecated, use %matplotlib inline and import the required libraries.
Populating the interactive namespace from numpy and matplotlib
../_images/2539a693cea3ca7b3efae72fe0b808757ac88799ecadcf02ba89b2800798f88e.png
x = np.array([0.5, 1.0, 2.0, 5.0, 10.0, 20.0, 50.0, 100.0])
y = np.array([0.4, 1.0, 2.3, 6.9, 15.8, 36.4, 110.1, 253.2])
pl.plot(x,y,'--o')
pl.xlabel('$x$ [cm]')
pl.ylabel('$y$ [V]')
pl.title('Calibration curve')
Text(0.5, 1.0, 'Calibration curve')
../_images/5e643608a86d31bb93057fe3d79bbe37e0ebd7dfd9d71a4b2f6d33302367e09c.png

Sensitivity, \(K\) is:

\( K_i = \left( \frac{\partial y}{\partial x} \right)_{x_i} \)

K = np.diff(y)/np.diff(x)
print (K)
[1.2        1.3        1.53333333 1.78       2.06       2.45666667
 2.862     ]
pl.plot(x[1:],K,'--o')
pl.xlabel('$x$ [cm]')
pl.ylabel('$K$ [V/cm]')
pl.title('Sensitivity')
Text(0.5, 1.0, 'Sensitivity')
../_images/9c86682cbeec92b3449d78a714994709bc60bf846118339c2b6336207cc7bd23.png

Instead of working with non-linear curve of sensitivity we can use the usual trick: the logarithmic scale

pl.loglog(x,y,'--o')
pl.xlabel('$x$ [cm]')
pl.ylabel('$y$ [V]')
pl.title('Logarithmic scale')
Text(0.5, 1.0, 'Logarithmic scale')
../_images/3a5107316eb08e93c91055a3de61a757d2ae7b1fc8d80a2a16b74bf3f6972992.png
logK = np.diff(np.log10(y))/np.diff(np.log10(x))
print( logK)
pl.plot(x[1:],logK,'--o')
pl.xlabel('$x$ [cm]')
pl.ylabel('$K$ [V/cm]')
pl.title('Logarithmic sensitivity')
pl.plot([x[1],x[-1]],[1.2,1.2],'r--')
[1.32192809 1.20163386 1.19897785 1.19525629 1.20401389 1.20793568
 1.20146294]
[<matplotlib.lines.Line2D at 0x7e2b14505d90>]
../_images/27af7231e552c064161a187b6a82fe8d2949e1cf4becc787fa504c314c8a506f.png
pl.loglog(x,y,'o',x,x**(1.2))
pl.xlabel('$x$ [cm]')
pl.ylabel('$y$ [V]')
pl.title('Logarithmic scale')
pl.legend(('$y$','$x^{1.2}$'),loc='best')
<matplotlib.legend.Legend at 0x7e2b145ca150>
../_images/6793c9adcf284f98595aedfd7713de66da94430753804aa21ffa150e69274cff.png
pl.plot(x,y-x**(1.2),'o')
pl.xlabel('$x$ [cm]')
pl.ylabel('$y - y_c$ [V]')
pl.title('Deviation plot')
# pl.legend(('$y$','$x^{1.2}$'),loc='best')
Text(0.5, 1.0, 'Deviation plot')
../_images/5de35756692121ac1f6e63cf57b4d13bdc2700b29328b594757524326fd1672f.png

Regression analysis#

Following the recipe of http://www.answermysearches.com/how-to-do-a-simple-linear-regression-in-python/124/

print (linreg(np.log10(x),np.log10(y)))
Estimate: y = ax + b
N = 8
Degrees of freedom $\nu$ = 6 
a = 1.21 $\pm$ 0.005
b = -0.01 $\pm$ 0.005
R^2 = 1.000
Syx = 0.011
y = 1.21 x + -0.01 $\pm$ 0.010 V
(np.float64(1.2103157469888082), np.float64(-0.012527809481276199), np.float64(0.9998888247342179), np.float64(0.011222369359282008))
pl.loglog(x,y,'o',x,x**(1.21)-0.01252)
pl.xlabel('$x$ [cm]')
pl.ylabel('$y$ [V]')
pl.title('Logarithmic scale')
pl.legend(('$y$','$x^{1.2}$'),loc='best')
<matplotlib.legend.Legend at 0x7e2b14337990>
../_images/3c4ba417a616481d9bac00b5f5e8f51bc516b361a4600fd369f45d826a557a2e.png
pl.plot(x,y-(x**(1.21)-0.01252),'o')
pl.xlabel('$x$ [cm]')
pl.ylabel('$y - y_c$ [V]')
pl.title('Deviation plot');
# pl.legend(('$y$','$x^{1.2}$'),loc='best')
../_images/b3186240df3b6829b7a417748d1b9c3eb3ca33ee40e117aa8610c204f5c50ebc.png