Finding a propriate function to fit "U" curve

I am trying to fit this data, but the CFT doesn't support a function that can fit my data(The curve of which is similar to the curve of Paschen's Law). Thanks a lot!

x=[0.17 0.33 0.83 1.67 2.5 3.33 5];

y=[6.62466 5.58235 4.63169 5.08213 5.4159 5.81769 6.77988];

[KSSV]

How about using polyfit ?

Answer 1

[Sam Chak]

Hi @Shilin Shi

I'm not an expert in High Voltage Engineering. Please clarify if the following rational logarithmic function describes the Paschen's law (as shown in wikipedia):

.

The local minimum point can be obtained by solving , that is

.

The singular point can be found by solving the denominator

.

There are only 7 data points. If is assumed to be the local minimum, the closest I can get using fsolve is

function F = paschen(x)

F(1) = x(1)*0.17/(log(x(2)*0.17) - log(log(1 + 1/x(3)))) - 6.62466;
F(2) = x(1)*0.83/(log(x(2)*0.83) - log(log(1 + 1/x(3)))) - 4.63169;
F(3) = x(1)*2.5/(log(x(2)*2.5) - log(log(1 + 1/x(3)))) - 5.4159;
F(4) = x(1)*5/(log(x(2)*5) - log(log(1 + 1/x(3)))) - 6.77988;
F(5) = (x(1)*(-1 + log(x(2)*0.83) - log(log(1 + 1/x(3)))))/(log(x(2)*0.83) - log(log(1 + 1/x(3))))^2;
F(6) = (1/x(2))*log((x(3) + 1)/x(3)) - 0.01;

Interpolations can be made to generate more points. However, it is preferable if you can provide more data points from your experiments. You probably need assistance from the MATLAB Data Scientists here.

[Bjoern Mulder]

Hi @Sam Chak!

Thank you for your post - I was also looking into the Paschen Law and your input helped!

I think I found an error while going through your solution

Instead of it should be .

And then to give the point where .

All was missing is an exp(1).

Have a great day!

Answer 2

[Walter Roberson]

polyfit degree 4 is not bad

[Sam Chak]

I think @Shilin Shi is trying to fit the data using the Paschen's Rational Logarithmic Function. Perhaps you can provide some insights on this matter.

[Alex Sha]

If don't mind the type of function, one like below is good enough:

y = p1*x^p2+p3*exp(x*p4)+p5*x+p6

Sum Squared Error (SSE): 0.00904817231315977
Root of Mean Square Error (RMSE): 0.0359526917592704
Correlation Coef. (R): 0.998760578765123
R-Square: 0.997522693695244

Parameter	Best Estimate     
---------	-------------     
p1       	-344.434477549634 
p2       	0.970653631107029 
p3       	-61.0490424042631 
p4       	-0.452522236917936
p5       	316.980679039058  
p6       	70.9562319738469 

if using polyfit degree 4 as suggesting by Walter Roberson, the result will be:

y = b0+b1*x+b2*x^2+b3*x^3+b4*x^4

Sum Squared Error (SSE): 0.142268105538466
Root of Mean Square Error (RMSE): 0.142562319975945
Correlation Coef. (R): 0.980330710558373
R-Square: 0.961048302063885

Parameter	Best Estimate    
---------	-------------    
b0       	7.45157976927911 
b1       	-6.56400689174901
b2       	4.88283965405082 
b3       	-1.32215012121595
b4       	0.120562079736417

if using polyfit degree 5:

y = b0+b1*x+b2*x^2+b3*x^3+b4*x^4+b5*x^5

Sum Squared Error (SSE): 0.00749681365182146
Root of Mean Square Error (RMSE): 0.0327257296149626
Correlation Coef. (R): 0.998973193543668
R-Square: 0.997947441418835

Parameter	Best Estimate      
---------	-------------      
b0       	8.10062144754325   
b1       	-10.6880037795643  
b2       	11.4448441863131   
b3       	-5.2716887492359   
b4       	1.10497426429342   
b5       	-0.0850079361618652

Answer 3

[Bruno Luong]

This is the best I come up with using BSFK FEX

x=[0.17 0.33 0.83 1.67 2.5 3.33 5];
y=[6.62466 5.58235 4.63169	5.08213	5.4159 5.81769 6.77988];

pp=BSFK(x,y,3,4);
xi=linspace(min(x),max(x));
yi=ppval(pp,xi);
plot(x,y,'or',xi,yi,'b')