How to do exponential curve fitting in the form of y=a*exp(-b*x)+ a constant

I have the following date

x=1:119

y=

11.45339775

12.78518772

11.98611355

12.25247192

12.78518772

10.92068195

13.31790447

9.322532654

11.71975613

11.98611355

11.98611355

10.65432358

10.92068195

11.71975613

10.65432358

12.5188303

11.98611355

11.45339775

12.5188303

12.5188303

10.92068195

12.25247192

14.91605282

10.12160683

9.855249405

15.18241119

11.71975613

12.5188303

14.91605282

15.71512699

9.855249405

14.38333702

11.45339775

12.5188303

11.18703938

12.5188303

13.31790447

12.5188303

13.31790447

13.85062027

13.31790447

10.65432358

12.5188303

14.91605282

12.25247192

13.31790447

14.11697865

11.45339775

13.0515461

12.5188303

13.58426189

14.38333702

9.855249405

11.71975613

13.0515461

11.71975613

11.45339775

13.31790447

11.98611355

14.38333702

13.0515461

14.91605282

11.71975613

14.38333702

14.64969444

13.58426189

12.5188303

13.0515461

14.91605282

13.58426189

15.98148537

13.0515461

14.38333702

16.51420212

11.45339775

12.25247192

10.3879652

15.18241119

10.92068195

14.91605282

14.38333702

12.25247192

10.65432358

13.58426189

10.65432358

13.31790447

15.44876862

12.25247192

12.78518772

13.58426189

13.0515461

13.31790447

14.64969444

14.11697865

13.85062027

14.11697865

13.0515461

11.98611355

14.11697865

14.11697865

14.38333702

13.31790447

14.64969444

11.71975613

14.38333702

14.64969444

14.38333702

13.0515461

15.44876862

12.5188303

15.18241119

15.44876862

13.0515461

14.64969444

12.25247192

13.85062027

13.31790447

13.58426189

13.31790447

10.65432358

7.990742683

9.588891029

10.3879652

12.78518772

11.18703938

9.855249405

9.056175232

9.855249405

10.65432358

9.855249405

11.18703938

10.12160683

I'd like to to have a curve fitting like y=a*exp(b*x) + a constant. I am wondering if someone could help me with this please! :)

Thanks

[KALYAN ACHARJYA]

Both are having different sizes??

>> whos x
  Name      Size             Bytes  Class     Attributes
  x         1x119              952  double              
>> whos y
  Name        Size            Bytes  Class     Attributes
  y         132x1              1056  double          

Answer 1

[Star Strider]

Note that ‘x’ and ‘y’ are not the same lengths, so this adjusts ‘y’ to correct for that:

yt = y(1:numel(x));                                             % Trim ‘y’ To Match ‘x’

objfcn = @(b,x) b(1).*exp(b(2).*x)+b(3);
B = fminsearch(@(b) norm(yt - objfcn(b,x(:))), rand(3,1));

figure
plot(x, yt, 'p')
hold on
plot(x, objfcn(B,x), '-r')
hold off
grid

producing:

B =
   6.196520631377196
   0.002516305503884
   5.724451530242829

and the plot.

[Star Strider]

Note that in my code:

B(1) = a
B(2) = b
B(3) = ‘a constant’

Answer 2

[John D'Errico]

Sigh. You give us x as:

x = 1:119;

Then a vector y, of length 132.

Since x was so simply created, I'll assume it is just 1:32 to match that y vector. PLOT YOUR DATA. THINK!!!!!!!

plot(1:132,y,'o')

If what you see does not make sense, then think harder. Does that model even remotely allow you to fit that data? NO!

Here is the model you suggest.

y=a*exp(-b*x)+ c

Are you kidding? Seriously? If you think it does, then think again. And, then, think harder. Just for kicks, think about what a exponential looks like:

fplot(@(x) exp(-x))

Now, does the curve I drew look even remotely like the data in your plot? (NO!) Those constants in the model do not change the fundamental shape of the curve. They just stretch it out, scale it, raise it up or down, etc.

So, while you COULD use a curve fitting tool to estimate parmeters for that model to that data, the result will be utterly meaningless crapola. Complete computer generated drivel.