I am having trouble implementing this equation into the curve fitting toolbox, both in the command and by utilizing the curve fitting GUI.
A little background that may help explain this:
I am wanting to fit experimental data of T(K) vs K_ph (W/m K) which the thermal conductivity of a material, to a model from a journal paper. I have done an okay job in python with scipy.optimize but have been trying MATLAB for a more accurate method.

I have this equation I need to fit

where

A, B, b, C_1, C_2, D, Omega_res1, Omega_res2, and L (ignore L in the exmaple code) are all coefficients I want solved for.
Omega is the independent variable in the integral (omega is solved for) while T is the range 1 to 100 in 1K steps. Or T can simply be my experimental Temperature data as is in my code.

My problems (in the GUI) are in the ways to set the values for T and use an integral in the custom equation box. As in, I am not sure how to tyep the equation in the way MATLAB accepts in the GUI for the toolbox.

My problems for the code is in the integration and lsqcurve fit it seems. I have a lot of warnings/errors such as
Warning: Derivative finite-differencing step was artificially reduced to be within bound constraints. This may
adversely affect convergence. Increasing distance between bound constraints, in dimension 6, to be at least 2e-20
may improve results.

Local minimum possible.
lsqcurvefit stopped because the size of the current step is less than
the value of the step size tolerance.

<stopping criteria details>
Fitted Parameters:
D = 1e-40
B = 1e-20
A = 1e-15
b = 100
C1 = 1e-40
C2 = 1e-40
I am not sure how to fix these, I have gone through other posts on similar matters and have changed the bounds and manually changed the optimoptions values to very low, but reached a point where it was just iterating nonstop.
Any help is appreciated!
% Data from 0T.txt
data = readmatrix(‘0T.txt’, ‘HeaderLines’, 1);
T_data = data(:, 1); % First col: Temperature (K)
k_ph_data = data(:, 2); % Second col: kappa_{xx} (W/mK)

% Constants
k_B = 1.380649e-23; % Boltzmann constant in J/K
hbar = 1.0545718e-34; % Reduced Planck constant in J·s
v_s = 4.817e3; % Sound velocity in m/s
theta_D = 235; % Debye temperature in K
L = 1e-6; % Length in m

% Define the integrand function with omega_res1 and omega_res2
integrand = @(omega, T, D, B, A, b, C1, C2, omega_res1, omega_res2) …
(omega.^4 .* exp(hbar*omega./(k_B*T)) ./ (exp(hbar*omega./(k_B*T)) – 1).^2) .* …
(v_s / L + D*omega.^4 + B*omega + A*T*omega.^3 .* exp(-theta_D / (b*T)) + …
C1 * omega.^4 ./ ((omega.^2 – omega_res1.^2).^2 + 1e-10) .* exp(-hbar*omega_res1 / (k_B * T)) ./ …
(1 + exp(-hbar*omega_res1 / (k_B * T))) + …
C2 * omega.^4 ./ ((omega.^2 – omega_res2.^2).^2 + 1e-10) .* exp(-hbar*omega_res2 / (k_B * T)) ./ …
(1 + exp(-hbar*omega_res2 / (k_B * T))));

% Define the k_ph function
k_ph_func = @(params, T) (k_B / (2 * pi^2 * v_s)) * (k_B * T / hbar)^3 .* …
integral(@(omega) integrand(omega, T, params(1), params(2), params(3), params(4), params(5), params(6), params(7), params(8)), 0, theta_D / T);

% Define the function to be minimized
fun = @(params, T) arrayfun(@(t) k_ph_func(params, t), T);

% Initial guess for parameters [D, B, A, b, C1, C2, omega_res1, omega_res2]
initial_guess = [1e-43, 1e-6, 1e-31, 1, 1e9, 1e10, 1e12, 4e12];

% Set bounds
lb = [1e-50, 1e-12, 1e-35, 1, 1e7, 1e8, 1e10, 3e12];
ub = [1e-40, 1e-3, 1e-28, 1000, 1e11, 1e12, 1e14, 5e12];

% Create opts structure
opts = optimoptions(‘lsqcurvefit’, ‘MaxFunctionEvaluations’, 1e4, ‘MaxIterations’, 1e3);

% Use lsqcurvefit for the fitting
[fitted_params, resnorm, residual, exitflag, output] = lsqcurvefit(fun, initial_guess, T_data, k_ph_data, lb, ub, opts);

% Generate fitted curve (using log spacing)
T_fit = logspace(log10(min(T_data)), log10(max(T_data)), 100);
k_ph_fit = fun(fitted_params, T_fit);I am having trouble implementing this equation into the curve fitting toolbox, both in the command and by utilizing the curve fitting GUI.
A little background that may help explain this:
I am wanting to fit experimental data of T(K) vs K_ph (W/m K) which the thermal conductivity of a material, to a model from a journal paper. I have done an okay job in python with scipy.optimize but have been trying MATLAB for a more accurate method.

I have this equation I need to fit

where

A, B, b, C_1, C_2, D, Omega_res1, Omega_res2, and L (ignore L in the exmaple code) are all coefficients I want solved for.
Omega is the independent variable in the integral (omega is solved for) while T is the range 1 to 100 in 1K steps. Or T can simply be my experimental Temperature data as is in my code.

My problems (in the GUI) are in the ways to set the values for T and use an integral in the custom equation box. As in, I am not sure how to tyep the equation in the way MATLAB accepts in the GUI for the toolbox.

My problems for the code is in the integration and lsqcurve fit it seems. I have a lot of warnings/errors such as
Warning: Derivative finite-differencing step was artificially reduced to be within bound constraints. This may
adversely affect convergence. Increasing distance between bound constraints, in dimension 6, to be at least 2e-20
may improve results.

Local minimum possible.
lsqcurvefit stopped because the size of the current step is less than
the value of the step size tolerance.

<stopping criteria details>
Fitted Parameters:
D = 1e-40
B = 1e-20
A = 1e-15
b = 100
C1 = 1e-40
C2 = 1e-40
I am not sure how to fix these, I have gone through other posts on similar matters and have changed the bounds and manually changed the optimoptions values to very low, but reached a point where it was just iterating nonstop.
Any help is appreciated!
% Data from 0T.txt
data = readmatrix(‘0T.txt’, ‘HeaderLines’, 1);
T_data = data(:, 1); % First col: Temperature (K)
k_ph_data = data(:, 2); % Second col: kappa_{xx} (W/mK)

% Constants
k_B = 1.380649e-23; % Boltzmann constant in J/K
hbar = 1.0545718e-34; % Reduced Planck constant in J·s
v_s = 4.817e3; % Sound velocity in m/s
theta_D = 235; % Debye temperature in K
L = 1e-6; % Length in m

% Define the integrand function with omega_res1 and omega_res2
integrand = @(omega, T, D, B, A, b, C1, C2, omega_res1, omega_res2) …
(omega.^4 .* exp(hbar*omega./(k_B*T)) ./ (exp(hbar*omega./(k_B*T)) – 1).^2) .* …
(v_s / L + D*omega.^4 + B*omega + A*T*omega.^3 .* exp(-theta_D / (b*T)) + …
C1 * omega.^4 ./ ((omega.^2 – omega_res1.^2).^2 + 1e-10) .* exp(-hbar*omega_res1 / (k_B * T)) ./ …
(1 + exp(-hbar*omega_res1 / (k_B * T))) + …
C2 * omega.^4 ./ ((omega.^2 – omega_res2.^2).^2 + 1e-10) .* exp(-hbar*omega_res2 / (k_B * T)) ./ …
(1 + exp(-hbar*omega_res2 / (k_B * T))));

% Define the k_ph function
k_ph_func = @(params, T) (k_B / (2 * pi^2 * v_s)) * (k_B * T / hbar)^3 .* …
integral(@(omega) integrand(omega, T, params(1), params(2), params(3), params(4), params(5), params(6), params(7), params(8)), 0, theta_D / T);

% Define the function to be minimized
fun = @(params, T) arrayfun(@(t) k_ph_func(params, t), T);

% Initial guess for parameters [D, B, A, b, C1, C2, omega_res1, omega_res2]
initial_guess = [1e-43, 1e-6, 1e-31, 1, 1e9, 1e10, 1e12, 4e12];

% Set bounds
lb = [1e-50, 1e-12, 1e-35, 1, 1e7, 1e8, 1e10, 3e12];
ub = [1e-40, 1e-3, 1e-28, 1000, 1e11, 1e12, 1e14, 5e12];

% Create opts structure
opts = optimoptions(‘lsqcurvefit’, ‘MaxFunctionEvaluations’, 1e4, ‘MaxIterations’, 1e3);

% Use lsqcurvefit for the fitting
[fitted_params, resnorm, residual, exitflag, output] = lsqcurvefit(fun, initial_guess, T_data, k_ph_data, lb, ub, opts);

% Generate fitted curve (using log spacing)
T_fit = logspace(log10(min(T_data)), log10(max(T_data)), 100);
k_ph_fit = fun(fitted_params, T_fit); I am having trouble implementing this equation into the curve fitting toolbox, both in the command and by utilizing the curve fitting GUI.
A little background that may help explain this:
I am wanting to fit experimental data of T(K) vs K_ph (W/m K) which the thermal conductivity of a material, to a model from a journal paper. I have done an okay job in python with scipy.optimize but have been trying MATLAB for a more accurate method.

I have this equation I need to fit

where

A, B, b, C_1, C_2, D, Omega_res1, Omega_res2, and L (ignore L in the exmaple code) are all coefficients I want solved for.
Omega is the independent variable in the integral (omega is solved for) while T is the range 1 to 100 in 1K steps. Or T can simply be my experimental Temperature data as is in my code.

My problems (in the GUI) are in the ways to set the values for T and use an integral in the custom equation box. As in, I am not sure how to tyep the equation in the way MATLAB accepts in the GUI for the toolbox.

My problems for the code is in the integration and lsqcurve fit it seems. I have a lot of warnings/errors such as
Warning: Derivative finite-differencing step was artificially reduced to be within bound constraints. This may
adversely affect convergence. Increasing distance between bound constraints, in dimension 6, to be at least 2e-20
may improve results.

Local minimum possible.
lsqcurvefit stopped because the size of the current step is less than
the value of the step size tolerance.

<stopping criteria details>
Fitted Parameters:
D = 1e-40
B = 1e-20
A = 1e-15
b = 100
C1 = 1e-40
C2 = 1e-40
I am not sure how to fix these, I have gone through other posts on similar matters and have changed the bounds and manually changed the optimoptions values to very low, but reached a point where it was just iterating nonstop.
Any help is appreciated!
% Data from 0T.txt
data = readmatrix(‘0T.txt’, ‘HeaderLines’, 1);
T_data = data(:, 1); % First col: Temperature (K)
k_ph_data = data(:, 2); % Second col: kappa_{xx} (W/mK)

% Constants
k_B = 1.380649e-23; % Boltzmann constant in J/K
hbar = 1.0545718e-34; % Reduced Planck constant in J·s
v_s = 4.817e3; % Sound velocity in m/s
theta_D = 235; % Debye temperature in K
L = 1e-6; % Length in m

% Define the integrand function with omega_res1 and omega_res2
integrand = @(omega, T, D, B, A, b, C1, C2, omega_res1, omega_res2) …
(omega.^4 .* exp(hbar*omega./(k_B*T)) ./ (exp(hbar*omega./(k_B*T)) – 1).^2) .* …
(v_s / L + D*omega.^4 + B*omega + A*T*omega.^3 .* exp(-theta_D / (b*T)) + …
C1 * omega.^4 ./ ((omega.^2 – omega_res1.^2).^2 + 1e-10) .* exp(-hbar*omega_res1 / (k_B * T)) ./ …
(1 + exp(-hbar*omega_res1 / (k_B * T))) + …
C2 * omega.^4 ./ ((omega.^2 – omega_res2.^2).^2 + 1e-10) .* exp(-hbar*omega_res2 / (k_B * T)) ./ …
(1 + exp(-hbar*omega_res2 / (k_B * T))));

% Define the k_ph function
k_ph_func = @(params, T) (k_B / (2 * pi^2 * v_s)) * (k_B * T / hbar)^3 .* …
integral(@(omega) integrand(omega, T, params(1), params(2), params(3), params(4), params(5), params(6), params(7), params(8)), 0, theta_D / T);

% Define the function to be minimized
fun = @(params, T) arrayfun(@(t) k_ph_func(params, t), T);

% Initial guess for parameters [D, B, A, b, C1, C2, omega_res1, omega_res2]
initial_guess = [1e-43, 1e-6, 1e-31, 1, 1e9, 1e10, 1e12, 4e12];

% Set bounds
lb = [1e-50, 1e-12, 1e-35, 1, 1e7, 1e8, 1e10, 3e12];
ub = [1e-40, 1e-3, 1e-28, 1000, 1e11, 1e12, 1e14, 5e12];

% Create opts structure
opts = optimoptions(‘lsqcurvefit’, ‘MaxFunctionEvaluations’, 1e4, ‘MaxIterations’, 1e3);

% Use lsqcurvefit for the fitting
[fitted_params, resnorm, residual, exitflag, output] = lsqcurvefit(fun, initial_guess, T_data, k_ph_data, lb, ub, opts);

% Generate fitted curve (using log spacing)
T_fit = logspace(log10(min(T_data)), log10(max(T_data)), 100);
k_ph_fit = fun(fitted_params, T_fit); curve fitting MATLAB Answers — New Questions

​

Problem:
When I adjust the latent heat, there’s no change in the temperature curve of the XPS material. I anticipate a stable temperature, ranging between 305K and 320K. So the problem is that the material does not release the absorbed heat.
equations
assert(Cps > 0 && Cpl > 0)
assert(T > 0, ‘Temperature must be greater than absolute zero’)
T == M.T;
if T >= Tmelt_start && T <= Tmelt_end
phaseChangeRate == {0.001, ‘1/s’};
else
phaseChangeRate == {0, ‘1/s’};
end
if phase < 1 && phase > 0
Q == m * Cps * T.der + m * L * phaseChangeRate;
phase.der == phaseChangeRate;
else
Q == m * Cpl * T.der;
phase.der == {0, ‘1/s’};
end
end

explanation of the full code:

This code describes a thermal component that models internal energy storage in a thermal network. Here’s a breakdown of the main parts of this code:

1. **Nodes**: These are points where other components can connect. Two thermal nodes are defined: `M` (top) and `N` (bottom).

2. **Inputs**: These values can be supplied from outside. `Mdot` represents the mass flow rate (in kg/s) and `T_in` is the input temperature (in Kelvin).

3. **Parameters**: These values determine the behavior of the component. They include:
– `mass_type`: Type of mass (constant or variable).
– `mass`: The mass of the thermal component.
– `Cps` and `Cpl`: Specific heat capacity for solid and liquid, respectively.
– `Tmelt_start` and `Tmelt_end`: The temperature range where the material starts melting and stops melting.
– `L`: Specific latent heat.
– `mass_min`: Minimum mass.
– `num_ports`: The number of graphical ports.

4. **Annotations**: These are metadata that provide additional information about the component or dictate how it is displayed in a GUI. Here, it is primarily used to indicate which parameters can be modified externally and to define the component’s icon.

5. **Variables**: These represent the internal states of the component. They include:
– `m`: The mass.
– `T`: The temperature.
– `phase`: The phase (0 for solid, 1 for liquid, and values in between for a mixed phase).
– `phaseChangeRate`: The rate of phase change.
– `Q`: Heat flow rate.

6. **Branches**: These are connections between different parts of the network. Here, the heat flow `Q` from node `M` to another part of the network is defined.

7. **Equations**: These describe the mathematical relationships between the different variables and parameters. They include:
– Relationships between `T`, `Q`, `m`, and other variables, especially around phase change.
– Constraints on certain values (like that `Cps`, `Cpl`, and `T` must be positive).

8. **Connections**: These are the actual connections between the different nodes within the component. Here, node `M` is connected to node `N`.

The essence of this component is that it models a thermal mass that can store and release energy. It also accounts for phase changes, meaning it can melt or solidify within a certain temperature range.
component derdecomponent
% Thermal Mass
% This block models internal energy storage in a thermal network.

nodes
M = foundation.thermal.thermal; % :top
end

inputs(ExternalAccess = none)
Mdot = {0, ‘kg/s’}; % Mdot:bottom
T_in = {300, ‘K’}; % Tin:bottom
end

nodes(ExternalAccess = none)
N = foundation.thermal.thermal; % :bottom
end

parameters
mass_type = foundation.enum.constant_variable.constant; % Mass type
end

parameters (ExternalAccess = none)
mass = {1, ‘kg’}; % Mass
end

parameters
Cps = {200, ‘J/(kg*K)’}; % Specific heat solid
Cpl = {1000, ‘J/(kg*K)’}; % Specific heat liquid
Tmelt_start = {320, ‘K’}; % Start Melting Temperature
Tmelt_end = {305, ‘K’}; % End Melting Temperature
L = {100, ‘J/(kg)’}; % Specific Latent Heat
end

parameters (ExternalAccess = none)
mass_min = {1e-6,’kg’}; % Minimum mass
end

parameters
num_ports = foundation.enum.numPorts2.one; % Number of graphical ports
end

if num_ports == 2
annotations
N : ExternalAccess=modify
end
if mass_type == foundation.enum.constant_variable.constant
annotations
% Icon = ‘mass2.svg’
end
else
annotations
% Icon = ‘mass4.svg’
end
end
else
if mass_type == foundation.enum.constant_variable.variable
annotations
% Icon = ‘mass3.svg’
end
end
end

if mass_type == foundation.enum.constant_variable.constant
annotations
mass : ExternalAccess=modify
end
equations
m == {1,’kg’};
end
else
annotations
[Mdot, T_in, m, mass_min] : ExternalAccess=modify
end
equations
assert(mass_min > 0)
end
end

variables (ExternalAccess=none)
m = {value = {1,’kg’}, priority=priority.high}; % Mass
end

variables
T = {value = {300, ‘K’}, priority = priority.high}; % Temperature
phase = {value = {0, ‘1’}, priority = priority.high}; % Phase
phaseChangeRate = {0, ‘1/s’}; % Rate of phase change
end

variables (Access=private)
Q = {0, ‘W’}; % Heat flow rate
end

branches
Q : M.Q -> *;
end

equations
assert(Cps > 0 && Cpl > 0)
assert(T > 0, ‘Temperature must be greater than absolute zero’)

T == M.T;

if T >= Tmelt_start && T <= Tmelt_end
phaseChangeRate == {0.001, ‘1/s’};
else
phaseChangeRate == {0, ‘1/s’};
end

if phase < 1 && phase > 0
Q == m * Cps * T.der + m * L * phaseChangeRate;
phase.der == phaseChangeRate;
else
Q == m * Cpl * T.der;
phase.der == {0, ‘1/s’};
end
end

connections
connect(M,N)
end

endProblem:
When I adjust the latent heat, there’s no change in the temperature curve of the XPS material. I anticipate a stable temperature, ranging between 305K and 320K. So the problem is that the material does not release the absorbed heat.
equations
assert(Cps > 0 && Cpl > 0)
assert(T > 0, ‘Temperature must be greater than absolute zero’)
T == M.T;
if T >= Tmelt_start && T <= Tmelt_end
phaseChangeRate == {0.001, ‘1/s’};
else
phaseChangeRate == {0, ‘1/s’};
end
if phase < 1 && phase > 0
Q == m * Cps * T.der + m * L * phaseChangeRate;
phase.der == phaseChangeRate;
else
Q == m * Cpl * T.der;
phase.der == {0, ‘1/s’};
end
end

explanation of the full code:

This code describes a thermal component that models internal energy storage in a thermal network. Here’s a breakdown of the main parts of this code:

1. **Nodes**: These are points where other components can connect. Two thermal nodes are defined: `M` (top) and `N` (bottom).

2. **Inputs**: These values can be supplied from outside. `Mdot` represents the mass flow rate (in kg/s) and `T_in` is the input temperature (in Kelvin).

3. **Parameters**: These values determine the behavior of the component. They include:
– `mass_type`: Type of mass (constant or variable).
– `mass`: The mass of the thermal component.
– `Cps` and `Cpl`: Specific heat capacity for solid and liquid, respectively.
– `Tmelt_start` and `Tmelt_end`: The temperature range where the material starts melting and stops melting.
– `L`: Specific latent heat.
– `mass_min`: Minimum mass.
– `num_ports`: The number of graphical ports.

4. **Annotations**: These are metadata that provide additional information about the component or dictate how it is displayed in a GUI. Here, it is primarily used to indicate which parameters can be modified externally and to define the component’s icon.

5. **Variables**: These represent the internal states of the component. They include:
– `m`: The mass.
– `T`: The temperature.
– `phase`: The phase (0 for solid, 1 for liquid, and values in between for a mixed phase).
– `phaseChangeRate`: The rate of phase change.
– `Q`: Heat flow rate.

6. **Branches**: These are connections between different parts of the network. Here, the heat flow `Q` from node `M` to another part of the network is defined.

7. **Equations**: These describe the mathematical relationships between the different variables and parameters. They include:
– Relationships between `T`, `Q`, `m`, and other variables, especially around phase change.
– Constraints on certain values (like that `Cps`, `Cpl`, and `T` must be positive).

8. **Connections**: These are the actual connections between the different nodes within the component. Here, node `M` is connected to node `N`.

The essence of this component is that it models a thermal mass that can store and release energy. It also accounts for phase changes, meaning it can melt or solidify within a certain temperature range.
component derdecomponent
% Thermal Mass
% This block models internal energy storage in a thermal network.

nodes
M = foundation.thermal.thermal; % :top
end

inputs(ExternalAccess = none)
Mdot = {0, ‘kg/s’}; % Mdot:bottom
T_in = {300, ‘K’}; % Tin:bottom
end

nodes(ExternalAccess = none)
N = foundation.thermal.thermal; % :bottom
end

parameters
mass_type = foundation.enum.constant_variable.constant; % Mass type
end

parameters (ExternalAccess = none)
mass = {1, ‘kg’}; % Mass
end

parameters
Cps = {200, ‘J/(kg*K)’}; % Specific heat solid
Cpl = {1000, ‘J/(kg*K)’}; % Specific heat liquid
Tmelt_start = {320, ‘K’}; % Start Melting Temperature
Tmelt_end = {305, ‘K’}; % End Melting Temperature
L = {100, ‘J/(kg)’}; % Specific Latent Heat
end

parameters (ExternalAccess = none)
mass_min = {1e-6,’kg’}; % Minimum mass
end

parameters
num_ports = foundation.enum.numPorts2.one; % Number of graphical ports
end

if num_ports == 2
annotations
N : ExternalAccess=modify
end
if mass_type == foundation.enum.constant_variable.constant
annotations
% Icon = ‘mass2.svg’
end
else
annotations
% Icon = ‘mass4.svg’
end
end
else
if mass_type == foundation.enum.constant_variable.variable
annotations
% Icon = ‘mass3.svg’
end
end
end

if mass_type == foundation.enum.constant_variable.constant
annotations
mass : ExternalAccess=modify
end
equations
m == {1,’kg’};
end
else
annotations
[Mdot, T_in, m, mass_min] : ExternalAccess=modify
end
equations
assert(mass_min > 0)
end
end

variables (ExternalAccess=none)
m = {value = {1,’kg’}, priority=priority.high}; % Mass
end

variables
T = {value = {300, ‘K’}, priority = priority.high}; % Temperature
phase = {value = {0, ‘1’}, priority = priority.high}; % Phase
phaseChangeRate = {0, ‘1/s’}; % Rate of phase change
end

variables (Access=private)
Q = {0, ‘W’}; % Heat flow rate
end

branches
Q : M.Q -> *;
end

equations
assert(Cps > 0 && Cpl > 0)
assert(T > 0, ‘Temperature must be greater than absolute zero’)

T == M.T;

if T >= Tmelt_start && T <= Tmelt_end
phaseChangeRate == {0.001, ‘1/s’};
else
phaseChangeRate == {0, ‘1/s’};
end

if phase < 1 && phase > 0
Q == m * Cps * T.der + m * L * phaseChangeRate;
phase.der == phaseChangeRate;
else
Q == m * Cpl * T.der;
phase.der == {0, ‘1/s’};
end
end

connections
connect(M,N)
end

end Problem:
When I adjust the latent heat, there’s no change in the temperature curve of the XPS material. I anticipate a stable temperature, ranging between 305K and 320K. So the problem is that the material does not release the absorbed heat.
equations
assert(Cps > 0 && Cpl > 0)
assert(T > 0, ‘Temperature must be greater than absolute zero’)
T == M.T;
if T >= Tmelt_start && T <= Tmelt_end
phaseChangeRate == {0.001, ‘1/s’};
else
phaseChangeRate == {0, ‘1/s’};
end
if phase < 1 && phase > 0
Q == m * Cps * T.der + m * L * phaseChangeRate;
phase.der == phaseChangeRate;
else
Q == m * Cpl * T.der;
phase.der == {0, ‘1/s’};
end
end

explanation of the full code:

This code describes a thermal component that models internal energy storage in a thermal network. Here’s a breakdown of the main parts of this code:

1. **Nodes**: These are points where other components can connect. Two thermal nodes are defined: `M` (top) and `N` (bottom).

2. **Inputs**: These values can be supplied from outside. `Mdot` represents the mass flow rate (in kg/s) and `T_in` is the input temperature (in Kelvin).

3. **Parameters**: These values determine the behavior of the component. They include:
– `mass_type`: Type of mass (constant or variable).
– `mass`: The mass of the thermal component.
– `Cps` and `Cpl`: Specific heat capacity for solid and liquid, respectively.
– `Tmelt_start` and `Tmelt_end`: The temperature range where the material starts melting and stops melting.
– `L`: Specific latent heat.
– `mass_min`: Minimum mass.
– `num_ports`: The number of graphical ports.

4. **Annotations**: These are metadata that provide additional information about the component or dictate how it is displayed in a GUI. Here, it is primarily used to indicate which parameters can be modified externally and to define the component’s icon.

5. **Variables**: These represent the internal states of the component. They include:
– `m`: The mass.
– `T`: The temperature.
– `phase`: The phase (0 for solid, 1 for liquid, and values in between for a mixed phase).
– `phaseChangeRate`: The rate of phase change.
– `Q`: Heat flow rate.

6. **Branches**: These are connections between different parts of the network. Here, the heat flow `Q` from node `M` to another part of the network is defined.

7. **Equations**: These describe the mathematical relationships between the different variables and parameters. They include:
– Relationships between `T`, `Q`, `m`, and other variables, especially around phase change.
– Constraints on certain values (like that `Cps`, `Cpl`, and `T` must be positive).

8. **Connections**: These are the actual connections between the different nodes within the component. Here, node `M` is connected to node `N`.

The essence of this component is that it models a thermal mass that can store and release energy. It also accounts for phase changes, meaning it can melt or solidify within a certain temperature range.
component derdecomponent
% Thermal Mass
% This block models internal energy storage in a thermal network.

nodes
M = foundation.thermal.thermal; % :top
end

inputs(ExternalAccess = none)
Mdot = {0, ‘kg/s’}; % Mdot:bottom
T_in = {300, ‘K’}; % Tin:bottom
end

nodes(ExternalAccess = none)
N = foundation.thermal.thermal; % :bottom
end

parameters
mass_type = foundation.enum.constant_variable.constant; % Mass type
end

parameters (ExternalAccess = none)
mass = {1, ‘kg’}; % Mass
end

parameters
Cps = {200, ‘J/(kg*K)’}; % Specific heat solid
Cpl = {1000, ‘J/(kg*K)’}; % Specific heat liquid
Tmelt_start = {320, ‘K’}; % Start Melting Temperature
Tmelt_end = {305, ‘K’}; % End Melting Temperature
L = {100, ‘J/(kg)’}; % Specific Latent Heat
end

parameters (ExternalAccess = none)
mass_min = {1e-6,’kg’}; % Minimum mass
end

parameters
num_ports = foundation.enum.numPorts2.one; % Number of graphical ports
end

if num_ports == 2
annotations
N : ExternalAccess=modify
end
if mass_type == foundation.enum.constant_variable.constant
annotations
% Icon = ‘mass2.svg’
end
else
annotations
% Icon = ‘mass4.svg’
end
end
else
if mass_type == foundation.enum.constant_variable.variable
annotations
% Icon = ‘mass3.svg’
end
end
end

if mass_type == foundation.enum.constant_variable.constant
annotations
mass : ExternalAccess=modify
end
equations
m == {1,’kg’};
end
else
annotations
[Mdot, T_in, m, mass_min] : ExternalAccess=modify
end
equations
assert(mass_min > 0)
end
end

variables (ExternalAccess=none)
m = {value = {1,’kg’}, priority=priority.high}; % Mass
end

variables
T = {value = {300, ‘K’}, priority = priority.high}; % Temperature
phase = {value = {0, ‘1’}, priority = priority.high}; % Phase
phaseChangeRate = {0, ‘1/s’}; % Rate of phase change
end

variables (Access=private)
Q = {0, ‘W’}; % Heat flow rate
end

branches
Q : M.Q -> *;
end

equations
assert(Cps > 0 && Cpl > 0)
assert(T > 0, ‘Temperature must be greater than absolute zero’)

T == M.T;

if T >= Tmelt_start && T <= Tmelt_end
phaseChangeRate == {0.001, ‘1/s’};
else
phaseChangeRate == {0, ‘1/s’};
end

if phase < 1 && phase > 0
Q == m * Cps * T.der + m * L * phaseChangeRate;
phase.der == phaseChangeRate;
else
Q == m * Cpl * T.der;
phase.der == {0, ‘1/s’};
end
end

connections
connect(M,N)
end

end phasechangingmaterial, heattransfer MATLAB Answers — New Questions

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I am compiling a file that contains a caller block, and that c caller block is calling c file that have many header files included. But while compiling its giving error Unknown "file extension ‘ ‘ ". Can anyone tell me how to solve this one?I am compiling a file that contains a caller block, and that c caller block is calling c file that have many header files included. But while compiling its giving error Unknown "file extension ‘ ‘ ". Can anyone tell me how to solve this one? I am compiling a file that contains a caller block, and that c caller block is calling c file that have many header files included. But while compiling its giving error Unknown "file extension ‘ ‘ ". Can anyone tell me how to solve this one? c caller, mex, error, unknown file MATLAB Answers — New Questions

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Hello!
I would like to calculate the mean value of a defined number of values of the columns in a matrix.
e.g.
Mean value every 2 values column by column of A.
A = [1,2,3;4,6,8;7,12,7;14,4,23];
result = [2.5,4,5.5;10.5,8,15]
Thanks for the answers in advance.Hello!
I would like to calculate the mean value of a defined number of values of the columns in a matrix.
e.g.
Mean value every 2 values column by column of A.
A = [1,2,3;4,6,8;7,12,7;14,4,23];
result = [2.5,4,5.5;10.5,8,15]
Thanks for the answers in advance. Hello!
I would like to calculate the mean value of a defined number of values of the columns in a matrix.
e.g.
Mean value every 2 values column by column of A.
A = [1,2,3;4,6,8;7,12,7;14,4,23];
result = [2.5,4,5.5;10.5,8,15]
Thanks for the answers in advance. mean, matlab, reshape MATLAB Answers — New Questions

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Hello, I have been checking and trying the GPS data decode example found in this link:
https://www.mathworks.com/help/satcom/ug/gps-data-decode.html
At the end of the example, it includes a further exploration which says that:
"Also, this example does not update the bit synchronization status back to the tracking loop. You can also use the bit synchronization value to increase the phase locked loop (PLL) integration time to enhance the PLL performance at low SNR values"

I am interested to do and apply this in the GPS Software receiver I am using. Is the sync index the location of start of navigation data bit? how do I use this to increase the PLL integration time?

Thank you for your help.Hello, I have been checking and trying the GPS data decode example found in this link:
https://www.mathworks.com/help/satcom/ug/gps-data-decode.html
At the end of the example, it includes a further exploration which says that:
"Also, this example does not update the bit synchronization status back to the tracking loop. You can also use the bit synchronization value to increase the phase locked loop (PLL) integration time to enhance the PLL performance at low SNR values"

I am interested to do and apply this in the GPS Software receiver I am using. Is the sync index the location of start of navigation data bit? how do I use this to increase the PLL integration time?

Thank you for your help. Hello, I have been checking and trying the GPS data decode example found in this link:
https://www.mathworks.com/help/satcom/ug/gps-data-decode.html
At the end of the example, it includes a further exploration which says that:
"Also, this example does not update the bit synchronization status back to the tracking loop. You can also use the bit synchronization value to increase the phase locked loop (PLL) integration time to enhance the PLL performance at low SNR values"

I am interested to do and apply this in the GPS Software receiver I am using. Is the sync index the location of start of navigation data bit? how do I use this to increase the PLL integration time?

Thank you for your help. gps MATLAB Answers — New Questions

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I tried using set(gca…………), but that is for plot. i am stuck at this.I tried using set(gca…………), but that is for plot. i am stuck at this. I tried using set(gca…………), but that is for plot. i am stuck at this. heatmap, colormap MATLAB Answers — New Questions

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I am saving data in a .txt file from c++ code which is continuously updating in real time.can i get this real time data loaded in to matrix in matlab in real time?? Please helpI am saving data in a .txt file from c++ code which is continuously updating in real time.can i get this real time data loaded in to matrix in matlab in real time?? Please help I am saving data in a .txt file from c++ code which is continuously updating in real time.can i get this real time data loaded in to matrix in matlab in real time?? Please help export MATLAB Answers — New Questions

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Hi All,
I found the following two ways of predicting performs differently:
(1).
for i=1:size(XTest,2)
[trainedNet,YTest(:,i)]=PredictAndUpdateState(trainedNet,XTest(:,i));
end
(2).
for i=1:size(XTest,2)
[YTest(:,i),state]=predict(trainedNet,XTest(:,i));
trainedNet.State=state;
end
I wonder that what is the reseason for this phenomenon? Or, what is the difference between the ways of predict() and PredictAndUpdateState() to update networks?
Any help will be appreciated!Hi All,
I found the following two ways of predicting performs differently:
(1).
for i=1:size(XTest,2)
[trainedNet,YTest(:,i)]=PredictAndUpdateState(trainedNet,XTest(:,i));
end
(2).
for i=1:size(XTest,2)
[YTest(:,i),state]=predict(trainedNet,XTest(:,i));
trainedNet.State=state;
end
I wonder that what is the reseason for this phenomenon? Or, what is the difference between the ways of predict() and PredictAndUpdateState() to update networks?
Any help will be appreciated! Hi All,
I found the following two ways of predicting performs differently:
(1).
for i=1:size(XTest,2)
[trainedNet,YTest(:,i)]=PredictAndUpdateState(trainedNet,XTest(:,i));
end
(2).
for i=1:size(XTest,2)
[YTest(:,i),state]=predict(trainedNet,XTest(:,i));
trainedNet.State=state;
end
I wonder that what is the reseason for this phenomenon? Or, what is the difference between the ways of predict() and PredictAndUpdateState() to update networks?
Any help will be appreciated! lstm, predictandupdatestate, predict, prediction, deep learning MATLAB Answers — New Questions

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I want to convert RT structure set from DICOM format to uncompressed NIfTI (.nii) format.I want to convert RT structure set from DICOM format to uncompressed NIfTI (.nii) format. I want to convert RT structure set from DICOM format to uncompressed NIfTI (.nii) format. rt structure set, dicom, nifti MATLAB Answers — New Questions

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I have an area bounded by a data set in the (-) plane as shown below. Is it mathematically possible to represent this area (blue color) in (-) plane?
clear; clc;

data = [1.0000 0.9000 0.8000 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0 -0.1000 -0.2000 -0.3000 -0.4000 -0.5000 -0.6000 -0.6000 -0.6000 -0.6000 -0.6000 -0.5000 -0.4000 -0.3000 -0.2000 -0.1000 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000
-2.1351 -2.1892 -2.2432 -2.2973 -2.3243 -2.3243 -2.3243 -2.3514 -2.2703 -2.2432 -2.2432 -2.2432 -2.2703 -2.2973 -2.2162 -2.1081 -2.0000 -1.0541 0.0270 1.1081 1.3243 2.6486 4.0270 5.2703 6.5676 7.5405 8.2973 9.1081 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000];

x = data(1, :);
y = data(2, :);

figure
fill(x, y, ‘b’,’FaceAlpha’,0.5)
xlabel(‘Q_2/Q_4+1’)
ylabel(‘Q/Q_4+5’)
set(gca,’FontName’,’times’,’fontsize’,14)I have an area bounded by a data set in the (-) plane as shown below. Is it mathematically possible to represent this area (blue color) in (-) plane?
clear; clc;

data = [1.0000 0.9000 0.8000 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0 -0.1000 -0.2000 -0.3000 -0.4000 -0.5000 -0.6000 -0.6000 -0.6000 -0.6000 -0.6000 -0.5000 -0.4000 -0.3000 -0.2000 -0.1000 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000
-2.1351 -2.1892 -2.2432 -2.2973 -2.3243 -2.3243 -2.3243 -2.3514 -2.2703 -2.2432 -2.2432 -2.2432 -2.2703 -2.2973 -2.2162 -2.1081 -2.0000 -1.0541 0.0270 1.1081 1.3243 2.6486 4.0270 5.2703 6.5676 7.5405 8.2973 9.1081 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000];

x = data(1, :);
y = data(2, :);

figure
fill(x, y, ‘b’,’FaceAlpha’,0.5)
xlabel(‘Q_2/Q_4+1’)
ylabel(‘Q/Q_4+5’)
set(gca,’FontName’,’times’,’fontsize’,14) I have an area bounded by a data set in the (-) plane as shown below. Is it mathematically possible to represent this area (blue color) in (-) plane?
clear; clc;

data = [1.0000 0.9000 0.8000 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0 -0.1000 -0.2000 -0.3000 -0.4000 -0.5000 -0.6000 -0.6000 -0.6000 -0.6000 -0.6000 -0.5000 -0.4000 -0.3000 -0.2000 -0.1000 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000
-2.1351 -2.1892 -2.2432 -2.2973 -2.3243 -2.3243 -2.3243 -2.3514 -2.2703 -2.2432 -2.2432 -2.2432 -2.2703 -2.2973 -2.2162 -2.1081 -2.0000 -1.0541 0.0270 1.1081 1.3243 2.6486 4.0270 5.2703 6.5676 7.5405 8.2973 9.1081 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000 10.0000];

x = data(1, :);
y = data(2, :);

figure
fill(x, y, ‘b’,’FaceAlpha’,0.5)
xlabel(‘Q_2/Q_4+1’)
ylabel(‘Q/Q_4+5’)
set(gca,’FontName’,’times’,’fontsize’,14) logic, math, mathematics MATLAB Answers — New Questions

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