I have 4 values of B corresponding to 4 values of A. I need 100 values of B for 100 values of A. There is no well defined interval between two values of A but values of A are within the range 0.049317 to 0.160588. Can anyone help me to find out

A =        
[0.049317,      
0.0892293,      
0.120183,      
0.160588]      

B =               
[3.3976e+05, 
3.4549e+04,
8.1311e+03,
4.4150e+03]

Answer 1

[Image Analyst]

Since you don't have uniform spacing in the A/x values then you can't use interp1() to get the new A/x values or B/y values. Try spline or scatteredInterpolant. My spline demo is attached. Spline will give you evenly spaced values for the new A's but your original A's will be at the same locations (with uneven spacing). I don't know how you can guess at the B values for those new A values unless you have some formula relating the two.

Answer 2

[Star Strider]

A power function actually fits it reasonably well, but without knowing the process that created your data, any function approximation is a leap of faith:

yf = @(b,x) b(1).*x.^b(2);
p = nlinfit(A, B, yf, [B(1); 1])
x = linspace(min(A), max(A));
ye = yf(p,x);

figure(1)
plot(A, B, '*b')
hold on
plot(x, ye, '-r')
hold off

The function with fitted parameters becomes:

B = 2.7*A^(-3.9)

Answer 3

[Matt J]

How about

   Bmore = interp1(A,B,linspace(A(1), A(end),100))

[Image Analyst]

That will give what he called "well defined intervals" which I think should be fine. Maybe he meant that only for the initial A and not for the bigger A. If he really needs to have "no well defined intervals" inthe new, bigger A I guess he could use rand instead of linspace, or just include the original 4 A values

newXValues = sort([A, linspace(A(1), A(end), 96)]);
Bmore = interp1(A,B, newXValues)

Answer 4

[Roger Stafford]

With just four pairs you just as well use a Lagrange interpolating polynomial. In your case it would be a cubic. It has no requirements of sorting or range.

   http://en.wikipedia.org/wiki/Lagrange_polynomial