Solving 3 Simultaneous Exponential Equations

x = [291.15; 293.15; 299.15];
y = [159.2543; 117.2699; 63.8384];
% % % MAPPING:  a = b(1),  b = b(2),  c = b(3)
objfcn = @(b,x) b(1).*exp(b(2)./(x - b(3)));
B0 = [0.07; 450; 230];
[B,normresid] = fminsearch(@(b) norm(y - objfcn(b,x)), B0)
xv = linspace(min(x), max(x));
figure
plot(x, y, 'p')
hold on
plot(xv, objfcn(B,xv), '-r')
hold off
grid

Values of:

a = 0.970
b = 180
c = 255

give a reasonable fit to the data.

2 个评论
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Rory Thornton 2019-12-20

Great, fits the rest of my data really well too.

Thank you so much!

Star Strider 2019-12-20

As always, my pleasure!

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Answer 2

David Goodmanson 2019-12-21

1
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HI Rory,

just for completeness

y1 = 159.2543;
y2 = 117.2699;
y3 = 63.8384;
x1 = 291.15;
x2 = 293.15;
x3 = 299.15;
A1 = log(y1/y2)/log(y2/y3);
A2 = (x2-x1)/(x3-x2);
c = (x1*A1-x3*A2)/(A1-A2);
% back substitute
b = log(y1/y2)*(x1-c)*(x2-c)/(x2-x1);
a = y1/exp(b/(x1-c));
a
b
c
a =
   10.6205
b = 
   42.5004
c = 
   275.4539
% these should be small
y1 - a*exp(b/(x1-c))
y2 - a*exp(b/(x2-c))
y3 - a*exp(b/(x3-c))
ans =
  0
ans =
 -4.2633e-14
ans =
 -7.1054e-14

The c result is fairly close to Star Strider's, but for some reason a and b differ from that result by quite a bit. The checks here show agreement at all three y points, but the best fit isn't necessarily the one that goes through all three points exactly.