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Least square fitting

Hi,

I have this graph:

x = [0.000235 0.00025 0.0003 0.0004 0.0005 0.0006 0.00075]
y = [54.9 43.1 62.7 86.3 86.3 90.2 96.1]

plot (x, y)
xlabel ('Amplitude')
ylabel ('Correct Detection [%]')
legend ('Psychometric Function')

and I want to fit it to this equation:

P(A, 'theta') = [1/sqrt(2*pi*sigma²)] * integral e^-[(A-'theta'²)/2*sigma²]

how can i do this?

thank you.
0
xety89
4/22/2010 7:45:21 AM
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"xety89 " <sow.chan@warwick.ac.uk> wrote in message <hqouqh$snf$1@fred.mathworks.com>...
> Hi,
> 
> I have this graph:
> 
> x = [0.000235 0.00025 0.0003 0.0004 0.0005 0.0006 0.00075]
> y = [54.9 43.1 62.7 86.3 86.3 90.2 96.1]
> 
> plot (x, y)
> xlabel ('Amplitude')
> ylabel ('Correct Detection [%]')
> legend ('Psychometric Function')
> 
> and I want to fit it to this equation:
> 
> P(A, 'theta') = [1/sqrt(2*pi*sigma²)] * integral e^-[(A-'theta'²)/2*sigma²]
> 
> how can i do this?

Hmm. I have a funny feeling that your expression
is wrong. Everything about that expression is a
simple, cumulative normal, except for that you
square theta, as opposed to squaring the difference
(A-theta).

Beyond that, your expression is not meaningful in
terms of the data that you have. You are talking
about doing some sort of a nonlinear least squares
fit to that expression, but that expression cannot
yield any numbers that are even reasonably close to
the data that you pose. My guess is that you wish
to scale and shift this curve, using that general
shape to model your data.

Note that a nice way to obtain this cumulative
normal shape (if you don't have the stats toolbox)
is to use erf. If you do have the stats TB, then call
normcdf. NEVER do the integration yourself using
quad.

John
0
John
4/22/2010 9:21:08 AM
You can also use the ezyfit toolbox.
http://www.mathworks.com/matlabcentral/fileexchange/10176-ezyfit-2-30

Recognising the error function, you can fit your data with
>> showfit('P(A) = 100*erf((A-theta)/sigma); sigma=1e-4; theta=1e-4');

yielding:
    sigma = 0.00040156
    theta = 4.8345e-005

Frederic.



"xety89 " <sow.chan@warwick.ac.uk> wrote in message <hqouqh$snf$1@fred.mathworks.com>...
> Hi,
> 
> I have this graph:
> 
> x = [0.000235 0.00025 0.0003 0.0004 0.0005 0.0006 0.00075]
> y = [54.9 43.1 62.7 86.3 86.3 90.2 96.1]
> 
> plot (x, y)
> xlabel ('Amplitude')
> ylabel ('Correct Detection [%]')
> legend ('Psychometric Function')
> 
> and I want to fit it to this equation:
> 
> P(A, 'theta') = [1/sqrt(2*pi*sigma²)] * integral e^-[(A-'theta'²)/2*sigma²]
> 
> how can i do this?
> 
> thank you.
0
Frederic
4/22/2010 6:45:11 PM
Reply:

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