Using the curve fitting tool to find fitting parameters of an integral.

There is a pretty advanced integral that needs to be solved numerically. There are three parameters that need to be fitted: Vt, T, and r. Below is an image of the formulas needed.

Below the image is the code I have so far. All the data needed is in the excel spreedsheet attached. Please.

clc;

%Short Channel
SVd = xlsread('W1umL60nm.xlsx', 'C4:C53');
SId0 = xlsread('W1umL60nm.xlsx', 'D56:D105');
SId2 = xlsread('W1umL60nm.xlsx', 'F56:F105');
SId4 = xlsread('W1umL60nm.xlsx', 'H56:H105');
SId6 = xlsread('W1umL60nm.xlsx', 'J56:J105');
SId8 = xlsread('W1umL60nm.xlsx', 'L56:L105');
SId10 = xlsread('W1umL60nm.xlsx', 'N56:N105');
SId12 = xlsread('W1umL60nm.xlsx', 'P56:P105');

% INITIALIZE MATLAB
close all;
clc;
clear all;

syms Vt
syms x
syms Ro
syms Vd
syms Vg
syms CoRo %Constant of p
syms p
%Vt = 0.7; %FOR NOW
%Vg = 2; %FOR NOW

CoRo = 1.652e+16;
%Vg = 0:0.2:1.2;
% Ro
p = CoRo*(Vg-Vt); %
% FUNCTIONS AND INTERVAL
a = @(SVd) 1+exp(Vd);
b = @(SVd)sqrt(a^2+4*exp(SVd)*(exp(p)-1));
u = @(SVd) log(b-(1+exp(SVd))-log(2));
f = @(u,x)sqrt(u)./(1+exp(u-x));
F = @(x)integral(@(u)f(u,x),0,inf, 'ArrayValued', true);

Answer 1

[Walter Roberson]

you cannot use integral with symbolic variables like p

Get rid of your @() and define symbolic expressions with the last one in terms of int() . That will attempt to integrate and then return back aa symbolic expression in terms of int()

Now use matlabFunction to turn it into aa numeric function handle . be sure to use the vars option so you get the right order of variables .

you can then use that in aa fittype() and lsqcurvefit

[Ausamah Hobbi]

Here is my current code. Though I am having trouble defining variables I guess.

This is the error message

"Undefined function or variable 'SVd'.

Error in Plotting_func_on_data_Short_Channel (line 37)

int(f, SVd, Vg, r, [0 inf])"

clc;

%Short Channel
SVd = xlsread('W1umL60nm.xlsx', 'C4:C53');
SId0 = xlsread('W1umL60nm.xlsx', 'D56:D105');
SId2 = xlsread('W1umL60nm.xlsx', 'F56:F105');
SId4 = xlsread('W1umL60nm.xlsx', 'H56:H105');
SId6 = xlsread('W1umL60nm.xlsx', 'J56:J105');
SId8 = xlsread('W1umL60nm.xlsx', 'L56:L105');
SId10 = xlsread('W1umL60nm.xlsx', 'N56:N105');
SId12 = xlsread('W1umL60nm.xlsx', 'P56:P105');

% INITIALIZE MATLAB
close all;
clc;
clear all;

syms Vt
syms x
syms Vd
syms Vg
syms Cop %Constant of p

%Vg = 0:0.2:1.2;

vd = @(SVd) (1/0.026).*r.*SVd; %r is the fitting parameter
% Ro (Density)
Cop = 1.652e+16; %Coefficient of Ro
p = @(Vg) Cop*(Vg-Vt);  %Vt is the fitting parameter

% Breaking down the Integral
a = @(SVd) 1+exp(vd);
b = @(SVd,Vg)sqrt(a^2+4*exp(Svd)*(exp(p)-1));
u = @(SVd,Vg) log(b-(1+exp(Svd))-log(2));
f = @(u,x)sqrt(u)./(1+exp(u-x));
%F = @(x)integral(@(u)f(u,x),0,inf, 'ArrayValued', true);
int(f, SVd, Vg, r, [0 inf])

[Walter Roberson]

Well, the only time you should ever use "clear all" is if you can logically replace it with quitting MATLAB and going back into MATLAB. You are reading in SVd and then throwing it out again.

Your a ignores its input parameter. Is that deliberate? It also tries to take the exponential of the function handle vd .

Your b tries to square the function handle a and take the exponential of the function handle p.

Your vd refers to the undefined r.

Your u is defined with two parameters. Your f has u as a dummy variable. Is there a connection?

You try to int() a function handle that expects two parameters, and then the int() call lists three variable names after the handle, and then has what appears to be a bounds of integration. What is that intended to mean??

[Walter Roberson]

It does not make sense to me that the image defines u as a formula but then integrates using u as the variable of integration.

[Ausamah Hobbi]

Here is an updated code.

% INITIALIZE MATLAB
close all;
clc;

syms Vt
syms x
syms Vd
syms Vg
syms Cop %Constant of p
syms r

%Short Channel
SVd = xlsread('W1umL60nm.xlsx', 'C4:C53');
SId0 = xlsread('W1umL60nm.xlsx', 'D56:D105');
SId2 = xlsread('W1umL60nm.xlsx', 'F56:F105');
SId4 = xlsread('W1umL60nm.xlsx', 'H56:H105');
SId6 = xlsread('W1umL60nm.xlsx', 'J56:J105');
SId8 = xlsread('W1umL60nm.xlsx', 'L56:L105');
SId10 = xlsread('W1umL60nm.xlsx', 'N56:N105');
SId12 = xlsread('W1umL60nm.xlsx', 'P56:P105');


%Vg = 0:0.2:1.2;

vd = @(SVd, r) (1/0.026).*r.*SVd; %r is the fitting parameter
% Ro (Density)
Cop = 1.652e+16; %Coefficient of Ro
p = @(Vg, Vt) Cop*(Vg-Vt);  %Vt is the fitting parameter

% Breaking down the Integral
a = @(SVd,r) 1+exp(vd(SVd, r));
b = @(SVd,Vg,Vt,r)sqrt(a(SVd,r)^2+4*exp(vd(SVd,r))*(exp(p(Vg, Vt))-1));
u = @(SVd,Vg,Vt,r) log(b(SVd,Vg,Vt)-(1+exp(vd(SVd,r)))-log(2));
f = @(SVd,Vg,Vt,r,x)sqrt(u(SVd,Vg,Vt,r))./(1+exp(u(SVd,Vg,Vt,r)-x));
%F = @(x)integral(@(u)f(u,x),0,inf, 'ArrayValued', true);
int(f, SVd, Vg, r, [0 inf])

[Walter Roberson]

What is the intention of

int(f, SVd, Vg, r, [0 inf])

f is a function handle that takes four parameters. What are you integrating with respect to, and what is that list of values doing there? Your first parameter, SVd, is a vector: what does it mean to int() passing in a vector of values in a place that is not the array bounds?

syms Vt
syms x
syms Vd
syms Vg
syms Cop %Constant of p
syms r
%Short Channel

syms SVd

%Vg = 0:0.2:1.2;
vd = @(SVd, r) (1/0.026).*r.*SVd; %r is the fitting parameter
% Ro (Density)
Cop = 1.652e+16; %Coefficient of Ro
p = @(Vg, Vt) Cop*(Vg-Vt);  %Vt is the fitting parameter
% Breaking down the Integral
a = @(SVd,r) 1+exp(vd(SVd, r));
b = @(SVd,Vg,Vt,r)sqrt(a(SVd,r)^2+4*exp(vd(SVd,r))*(exp(p(Vg, Vt))-1));
u = @(SVd,Vg,Vt,r) log(b(SVd,Vg,Vt,r)-(1+exp(vd(SVd,r)))-log(2));
f = @(SVd,Vg,Vt,r,x)sqrt(u(SVd,Vg,Vt,r))./(1+exp(u(SVd,Vg,Vt,r)-x));
%F = @(x)integral(@(u)f(u,x),0,inf, 'ArrayValued', true);
fint = int(f(SVd, Vg, Vt, r, x), SVd, [0 inf]);

SVd_n = xlsread('W1umL60nm.xlsx', 'C4:C53');     %name change
SId0 = xlsread('W1umL60nm.xlsx', 'D56:D105');
SId2 = xlsread('W1umL60nm.xlsx', 'F56:F105');
SId4 = xlsread('W1umL60nm.xlsx', 'H56:H105');
SId6 = xlsread('W1umL60nm.xlsx', 'J56:J105');
SId8 = xlsread('W1umL60nm.xlsx', 'L56:L105');
SId10 = xlsread('W1umL60nm.xlsx', 'N56:N105');
SId12 = xlsread('W1umL60nm.xlsx', 'P56:P105');

F = subs(fint, SVd, SVd_n);

perhaps... which will generate an array of 50 unresolved int() calls.

To go further you would probably use matlabFunction(fint) and use the 'vars' option to specify the order of variables. After that... probably a bunch of arrayfun() with various SVd and r and Vt values.

[Ausamah Hobbi]

Hey Walter.

I've been playing around and this is what I have so far. I keep, however keep getting errors.

% INITIALIZE MATLAB
close all;
clc;
clear all;

syms Vt
syms x
syms Vd
syms Vg
syms Cop %Constant of p
syms r

Z = 1e-6;
q = 1.6e-19;
kT = 4.11e-21;
m0 = 9.1e-31;
hbar = 1.054e-34;
mt = .2 * m0;
mv = 2.5;
e0 = 8.85e-12;
er = 3.7;
Ci = .0163725;

%Defining Vg
Vg0 = 0;
Vg2 = 0.2;
Vg4 = 0.4;
Vg6 = 0.6;
Vg8 = 0.8;
Vg10 = 1.0;
Vg12 = 1.2;

%Short Channel
SVd = xlsread('W1umL60nm.xlsx', 'C4:C53');
SId0 = xlsread('W1umL60nm.xlsx', 'D56:D105');
SId2 = xlsread('W1umL60nm.xlsx', 'F56:F105');
SId4 = xlsread('W1umL60nm.xlsx', 'H56:H105');
SId6 = xlsread('W1umL60nm.xlsx', 'J56:J105');
SId8 = xlsread('W1umL60nm.xlsx', 'L56:L105');
SId10 = xlsread('W1umL60nm.xlsx', 'N56:N105');
SId12 = xlsread('W1umL60nm.xlsx', 'P56:P105');

Io = (sqrt(2).*q.*((kT)^1.5).*mv.*sqrt(mt))/((pi.^2).*(hbar).^2);

vd = @(Vd) (q.*r.*SVd)/(kT);

Ro = @(Vg) (2.*pi.*(hbar.^2).*Ci.*(Vg-Vt))./(q.*kT.*mt.*mv);

fermi_half = @(x) (x.^0.5)/(1+exp.^(x-a));

fermi_half_vd = @(x) (x.^0.5)/(1+exp.^(x-a+vd(Vd)));

%To get u(Vd, Vg)
a = @(Vd, Vg) sqrt((1 + exp.^(vd(Vd))).^2 + 4.*exp(vd(Vd)).*(math.exp(p(Vg)) - 1));
b = @(Vd)(1 + exp.^(vd(Vd)));
c = log(2);

u = @(Vd, Vg) log(a-b)-c

fermi_half_int = quadgk(fermi_half(x), 0, inf)

fermi_half_vd_int = quadgk(fermi_half_vd(x), 0, inf)

[Walter Roberson]

fermi_half_int = quadgk(fermi_half(x), 0, inf)

At that point, x refers to syms x, so fermi_half would be executed passing in symbolic variable f, returning a symbolic expression. Then you try to quadgk with that symbolic expression as the first parameter. quadgk is strictly for function handles, not symbolic. Perhaps

fermi_half_int = quadgk(fermi_half, 0, inf)