Finding the vertical offset of a gaussian fit

I apologize for posting this question again, as I believe I miss presented my question earlier.

I have the following data points as shown in the picture, and I am in the quest to obtain the vertical offset (v0) for the best gaussian fit to my data point. Equation for gaussian fit: gaus = amp*exp(-(((x-mu).^2)/(2*sig.^2)))+ v0.

I currently have the datasets x and y and I am stuck at that, any help is appreciated and sorry for posting this again

[Yazan]

You did not upload the picture nor the data. Do so to increase your chances of getting support.

[Matt J]

sorry for posting this again

Indeed, you Accepted an answer to the same question here:

https://www.mathworks.com/matlabcentral/answers/887569-obtaining-the-gaussian-equation-for-a-distribution?s_tid=srchtitle

Why do you need a different answer now?

[Adam Danz]

Also, the previous answer you accepted and then unaccepted also includes a vertical offset term and even lists the fit parameter values in the axes title.

You also copied the guassian equation from the previous answer into your current question above. I don't see how this question differs from the previous question.

Answer 1

[Matt J]

This File Exchange routine does gaussian+constant fitting:

https://www.mathworks.com/matlabcentral/fileexchange/69116-gaussfitn

mu=5;
sig=2.3;
amp=4;
v0=1.7;

x=linspace(0,10,100);
y=amp*exp(-(((x-mu).^2)/(2*sig.^2)))+ v0;
 y=y+randn(size(y))*0.1;
 
p=gaussfitn(x(:),y(:));

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

[v0,amp,mu,var]=p{:}, sig=sqrt(var)

v0 = 1.6697

amp = 4.0421

mu = 4.9868

var = 5.4130

sig = 2.3266

yfit=v0+amp*exp(-(((x-mu).^2)/(2*var)));
 
 plot(x,y,'o',x,yfit)

[Ahmed Abdulla]

Thank you very much this is perfect :)