Category: Matlab
Category Archives: Matlab
Can you stream and record cycling data such as power and cadence from Garmin TACX trainer using ANT+ dongle?
Hi,
I have googled it, but I am still unsure what toolbox I need and if it is feasible. I am looking for a way to stream and record cycling data measured by Garmin TACX trainer using ANT+ dongle. Also, I was wondering if the target power output can be controlled in MATLAB as well. Any advice and suggestions are appreciated.Hi,
I have googled it, but I am still unsure what toolbox I need and if it is feasible. I am looking for a way to stream and record cycling data measured by Garmin TACX trainer using ANT+ dongle. Also, I was wondering if the target power output can be controlled in MATLAB as well. Any advice and suggestions are appreciated. Hi,
I have googled it, but I am still unsure what toolbox I need and if it is feasible. I am looking for a way to stream and record cycling data measured by Garmin TACX trainer using ANT+ dongle. Also, I was wondering if the target power output can be controlled in MATLAB as well. Any advice and suggestions are appreciated. ant+ MATLAB Answers — New Questions
Invalid argument “simscapeelectricalsps”.
>> openExample(‘simscapeelectricalsps/SinglePhaseASMExample’)
while I got an error shows as following
please tell me what is causing this problem? or which toolbox is lack?
And how can I find all the toolboxes needed for examples?>> openExample(‘simscapeelectricalsps/SinglePhaseASMExample’)
while I got an error shows as following
please tell me what is causing this problem? or which toolbox is lack?
And how can I find all the toolboxes needed for examples? >> openExample(‘simscapeelectricalsps/SinglePhaseASMExample’)
while I got an error shows as following
please tell me what is causing this problem? or which toolbox is lack?
And how can I find all the toolboxes needed for examples? openexample, toolbox MATLAB Answers — New Questions
How to create Architecture model and paste the Reference example of Vehicle dynamics double lane change.
Post Content Post Content v-model, vehicle dynamics, architecture MATLAB Answers — New Questions
Population Growth Model Analysis
Below I tried to solve Population Growth problem, I provide you the mathematical solution. Also, I create MATLAB code, however i feel like that i repeat this code to each part of this code as you can see. How could I improve this code?
Assume a continuous growth model where the function that gives the rate of change is defined as . Provide the function , which represents the population size.
If the growth of a population follows the logistic law but there is also a term for population reduction (e.g., from a predator while the members of the population are the prey), the equation that expresses the population size is of the form
where the term is given here to be and it represents a term for population reduction. The constants are positive. Normalize the equation by setting and .Find the steady solutions of the problem (real and positive) and then implement a program in MATLAB to numerically solve the dimensionless form of the equation. Make the graphical representation of the numerical solution for , with initial conditions , and respectively. What do you observe?
So we have
We are going to find the function :
To find , we integrate :
where is the integration constant, which can be determined if an initial condition is given, such as .
Now we are going to normalize the equation:
Let and . Then:
We substitute into the original equation:
Then we are going to calculate the Stationary Points:
Set :
]
For :
% Define the normalized logistic equation with reduction term
% K/a = 35, ar/b = 0.4
Ka = 35;
ar_b = 0.4;
% Define the function for the normalized logistic equation
f = @(tau, N) ar_b * N * (1 – N/Ka) – N / (1 + N);
% Time span for the simulation
tspan = [0 50]; % Adjust if needed to observe long-term behavior
% Initial conditions
N0 = [40, 25, 2, 0.05];
% Prepare the figure for plotting
figure(1);
hold on;
% Colors for different trajectories
colors = [‘r’, ‘g’, ‘b’, ‘k’]; % Red, Green, Blue, Black
% Solve the equation for each initial condition and plot
for i = 1:length(N0)
[T, N] = ode45(f, tspan, N0(i));
plot(T, N, ‘Color’, colors(i), ‘LineWidth’, 2);
legendInfo{i} = [‘N_0 = ‘ num2str(N0(i))]; % Create legend entry
end
% Add graph details
title(‘Numerical solutions of the normalized logistic equation’);
xlabel(‘Normalized time τ’);
ylabel(‘Normalized population size N’);
legend(legendInfo, ‘Location’, ‘northeastoutside’);
grid on;
hold off;
% Equilibrium analysis
syms N
eqn = ar_b * N * (1 – N / Ka) – N / (1 + N) == 0;
equilibria = double(solve(eqn, N));
% Stability analysis
f = @(N) ar_b * N * (1 – N / Ka) – N / (1 + N);
J = diff(f(N), N);
eigenvalues = double(subs(J, N, equilibria));
disp(‘Equilibrium Points and their Stability:’);
for i = 1:length(equilibria)
if real(eigenvalues(i)) < 0
stability = ‘Stable’;
else
stability = ‘Unstable’;
end
fprintf(‘N = %f: Eigenvalue = %f, %sn’, equilibria(i), eigenvalues(i), stability);
end
% Bifurcation analysis setup
ar_b_values = linspace(0.1, 1, 100);
bifurcation_results = zeros(length(ar_b_values), 10); % Adjust the columns based on expected number of roots
% Compute equilibrium points for varying ar/b
for i = 1:length(ar_b_values)
eqn = ar_b_values(i) * N * (1 – N / Ka) – N / (1 + N) == 0;
solutions = double(solve(eqn, N));
bifurcation_results(i, 1:length(solutions)) = solutions’;
end
% Plot bifurcation diagram
figure;
plot(ar_b_values, bifurcation_results, ‘b*’);
title(‘Bifurcation Diagram’);
xlabel(‘ar/b’);
ylabel(‘Equilibrium Points’);
% Set up the ODE function for phase plane analysis
f = @(N) ar_b * 0.4 * N .* (1 – N / Ka) – N ./ (1 + N);
% Create a grid of N values and compute their time derivatives
N_values = linspace(0, 50, 400);
dN_dtau = f(N_values);
% Plot phase plane
figure;
plot(N_values, dN_dtau);
title(‘Phase Plane Analysis’);
xlabel(‘Population Size N’);
ylabel(‘dN/dτ’);
grid on;
hold on;
% Mark zero crossings (equilibrium points)
zero_crossings = N_values(abs(dN_dtau) < 1e-3);
for i = 1:length(zero_crossings)
plot(zero_crossings(i), 0, ‘ro’);
end
hold off;Below I tried to solve Population Growth problem, I provide you the mathematical solution. Also, I create MATLAB code, however i feel like that i repeat this code to each part of this code as you can see. How could I improve this code?
Assume a continuous growth model where the function that gives the rate of change is defined as . Provide the function , which represents the population size.
If the growth of a population follows the logistic law but there is also a term for population reduction (e.g., from a predator while the members of the population are the prey), the equation that expresses the population size is of the form
where the term is given here to be and it represents a term for population reduction. The constants are positive. Normalize the equation by setting and .Find the steady solutions of the problem (real and positive) and then implement a program in MATLAB to numerically solve the dimensionless form of the equation. Make the graphical representation of the numerical solution for , with initial conditions , and respectively. What do you observe?
So we have
We are going to find the function :
To find , we integrate :
where is the integration constant, which can be determined if an initial condition is given, such as .
Now we are going to normalize the equation:
Let and . Then:
We substitute into the original equation:
Then we are going to calculate the Stationary Points:
Set :
]
For :
% Define the normalized logistic equation with reduction term
% K/a = 35, ar/b = 0.4
Ka = 35;
ar_b = 0.4;
% Define the function for the normalized logistic equation
f = @(tau, N) ar_b * N * (1 – N/Ka) – N / (1 + N);
% Time span for the simulation
tspan = [0 50]; % Adjust if needed to observe long-term behavior
% Initial conditions
N0 = [40, 25, 2, 0.05];
% Prepare the figure for plotting
figure(1);
hold on;
% Colors for different trajectories
colors = [‘r’, ‘g’, ‘b’, ‘k’]; % Red, Green, Blue, Black
% Solve the equation for each initial condition and plot
for i = 1:length(N0)
[T, N] = ode45(f, tspan, N0(i));
plot(T, N, ‘Color’, colors(i), ‘LineWidth’, 2);
legendInfo{i} = [‘N_0 = ‘ num2str(N0(i))]; % Create legend entry
end
% Add graph details
title(‘Numerical solutions of the normalized logistic equation’);
xlabel(‘Normalized time τ’);
ylabel(‘Normalized population size N’);
legend(legendInfo, ‘Location’, ‘northeastoutside’);
grid on;
hold off;
% Equilibrium analysis
syms N
eqn = ar_b * N * (1 – N / Ka) – N / (1 + N) == 0;
equilibria = double(solve(eqn, N));
% Stability analysis
f = @(N) ar_b * N * (1 – N / Ka) – N / (1 + N);
J = diff(f(N), N);
eigenvalues = double(subs(J, N, equilibria));
disp(‘Equilibrium Points and their Stability:’);
for i = 1:length(equilibria)
if real(eigenvalues(i)) < 0
stability = ‘Stable’;
else
stability = ‘Unstable’;
end
fprintf(‘N = %f: Eigenvalue = %f, %sn’, equilibria(i), eigenvalues(i), stability);
end
% Bifurcation analysis setup
ar_b_values = linspace(0.1, 1, 100);
bifurcation_results = zeros(length(ar_b_values), 10); % Adjust the columns based on expected number of roots
% Compute equilibrium points for varying ar/b
for i = 1:length(ar_b_values)
eqn = ar_b_values(i) * N * (1 – N / Ka) – N / (1 + N) == 0;
solutions = double(solve(eqn, N));
bifurcation_results(i, 1:length(solutions)) = solutions’;
end
% Plot bifurcation diagram
figure;
plot(ar_b_values, bifurcation_results, ‘b*’);
title(‘Bifurcation Diagram’);
xlabel(‘ar/b’);
ylabel(‘Equilibrium Points’);
% Set up the ODE function for phase plane analysis
f = @(N) ar_b * 0.4 * N .* (1 – N / Ka) – N ./ (1 + N);
% Create a grid of N values and compute their time derivatives
N_values = linspace(0, 50, 400);
dN_dtau = f(N_values);
% Plot phase plane
figure;
plot(N_values, dN_dtau);
title(‘Phase Plane Analysis’);
xlabel(‘Population Size N’);
ylabel(‘dN/dτ’);
grid on;
hold on;
% Mark zero crossings (equilibrium points)
zero_crossings = N_values(abs(dN_dtau) < 1e-3);
for i = 1:length(zero_crossings)
plot(zero_crossings(i), 0, ‘ro’);
end
hold off; Below I tried to solve Population Growth problem, I provide you the mathematical solution. Also, I create MATLAB code, however i feel like that i repeat this code to each part of this code as you can see. How could I improve this code?
Assume a continuous growth model where the function that gives the rate of change is defined as . Provide the function , which represents the population size.
If the growth of a population follows the logistic law but there is also a term for population reduction (e.g., from a predator while the members of the population are the prey), the equation that expresses the population size is of the form
where the term is given here to be and it represents a term for population reduction. The constants are positive. Normalize the equation by setting and .Find the steady solutions of the problem (real and positive) and then implement a program in MATLAB to numerically solve the dimensionless form of the equation. Make the graphical representation of the numerical solution for , with initial conditions , and respectively. What do you observe?
So we have
We are going to find the function :
To find , we integrate :
where is the integration constant, which can be determined if an initial condition is given, such as .
Now we are going to normalize the equation:
Let and . Then:
We substitute into the original equation:
Then we are going to calculate the Stationary Points:
Set :
]
For :
% Define the normalized logistic equation with reduction term
% K/a = 35, ar/b = 0.4
Ka = 35;
ar_b = 0.4;
% Define the function for the normalized logistic equation
f = @(tau, N) ar_b * N * (1 – N/Ka) – N / (1 + N);
% Time span for the simulation
tspan = [0 50]; % Adjust if needed to observe long-term behavior
% Initial conditions
N0 = [40, 25, 2, 0.05];
% Prepare the figure for plotting
figure(1);
hold on;
% Colors for different trajectories
colors = [‘r’, ‘g’, ‘b’, ‘k’]; % Red, Green, Blue, Black
% Solve the equation for each initial condition and plot
for i = 1:length(N0)
[T, N] = ode45(f, tspan, N0(i));
plot(T, N, ‘Color’, colors(i), ‘LineWidth’, 2);
legendInfo{i} = [‘N_0 = ‘ num2str(N0(i))]; % Create legend entry
end
% Add graph details
title(‘Numerical solutions of the normalized logistic equation’);
xlabel(‘Normalized time τ’);
ylabel(‘Normalized population size N’);
legend(legendInfo, ‘Location’, ‘northeastoutside’);
grid on;
hold off;
% Equilibrium analysis
syms N
eqn = ar_b * N * (1 – N / Ka) – N / (1 + N) == 0;
equilibria = double(solve(eqn, N));
% Stability analysis
f = @(N) ar_b * N * (1 – N / Ka) – N / (1 + N);
J = diff(f(N), N);
eigenvalues = double(subs(J, N, equilibria));
disp(‘Equilibrium Points and their Stability:’);
for i = 1:length(equilibria)
if real(eigenvalues(i)) < 0
stability = ‘Stable’;
else
stability = ‘Unstable’;
end
fprintf(‘N = %f: Eigenvalue = %f, %sn’, equilibria(i), eigenvalues(i), stability);
end
% Bifurcation analysis setup
ar_b_values = linspace(0.1, 1, 100);
bifurcation_results = zeros(length(ar_b_values), 10); % Adjust the columns based on expected number of roots
% Compute equilibrium points for varying ar/b
for i = 1:length(ar_b_values)
eqn = ar_b_values(i) * N * (1 – N / Ka) – N / (1 + N) == 0;
solutions = double(solve(eqn, N));
bifurcation_results(i, 1:length(solutions)) = solutions’;
end
% Plot bifurcation diagram
figure;
plot(ar_b_values, bifurcation_results, ‘b*’);
title(‘Bifurcation Diagram’);
xlabel(‘ar/b’);
ylabel(‘Equilibrium Points’);
% Set up the ODE function for phase plane analysis
f = @(N) ar_b * 0.4 * N .* (1 – N / Ka) – N ./ (1 + N);
% Create a grid of N values and compute their time derivatives
N_values = linspace(0, 50, 400);
dN_dtau = f(N_values);
% Plot phase plane
figure;
plot(N_values, dN_dtau);
title(‘Phase Plane Analysis’);
xlabel(‘Population Size N’);
ylabel(‘dN/dτ’);
grid on;
hold on;
% Mark zero crossings (equilibrium points)
zero_crossings = N_values(abs(dN_dtau) < 1e-3);
for i = 1:length(zero_crossings)
plot(zero_crossings(i), 0, ‘ro’);
end
hold off; differential equations, equation, plotting, plot, ode, homework MATLAB Answers — New Questions
How to solve SIR model with using DTM (Differential Transform Method)
I am trying to use dtm for solving SIR model, Although my code is run but I think the DTM part is wrong. I need help for DTM part
here is my code
function sir_model_dtm()
% Parameters
alpha = 0.3; % Infection rate
beta = 0.1; % Recovery rate
N = 1000; % Total population
I0 = 1; % Initial number of infected individuals
R0 = 0; % Initial number of recovered individuals
S0 = N – I0 – R0; % Initial number of susceptible individuals
% Time parameters
T = 100; % Total simulation time
dt = 1; % Time step
% Initialize arrays to store results
S = zeros(1, T);
I = zeros(1, T);
R = zeros(1, T);
% Initial conditions
S(1) = S0;
I(1) = I0;
R(1) = R0;
% Simulate the SIR model using DTM
for t = 2:T
% Compute new values
S(t) = S(t-1) – alpha*S(t-1)*I(t-1)/N * dt;
I(t) = I(t-1) + (alpha*S(t-1)*I(t-1)/N – beta*I(t-1)) * dt;
R(t) = R(t-1) + beta*I(t-1) * dt;
end
% Plot the results
t = 0:dt:T-dt;
plot(t, S, ‘b’, t, I, ‘r’, t, R, ‘g’, ‘LineWidth’, 2);
legend(‘Susceptible’, ‘Infectious’, ‘Recovered’);
xlabel(‘Time’);
ylabel(‘Number of individuals’);
title(‘SIR Model’);
endI am trying to use dtm for solving SIR model, Although my code is run but I think the DTM part is wrong. I need help for DTM part
here is my code
function sir_model_dtm()
% Parameters
alpha = 0.3; % Infection rate
beta = 0.1; % Recovery rate
N = 1000; % Total population
I0 = 1; % Initial number of infected individuals
R0 = 0; % Initial number of recovered individuals
S0 = N – I0 – R0; % Initial number of susceptible individuals
% Time parameters
T = 100; % Total simulation time
dt = 1; % Time step
% Initialize arrays to store results
S = zeros(1, T);
I = zeros(1, T);
R = zeros(1, T);
% Initial conditions
S(1) = S0;
I(1) = I0;
R(1) = R0;
% Simulate the SIR model using DTM
for t = 2:T
% Compute new values
S(t) = S(t-1) – alpha*S(t-1)*I(t-1)/N * dt;
I(t) = I(t-1) + (alpha*S(t-1)*I(t-1)/N – beta*I(t-1)) * dt;
R(t) = R(t-1) + beta*I(t-1) * dt;
end
% Plot the results
t = 0:dt:T-dt;
plot(t, S, ‘b’, t, I, ‘r’, t, R, ‘g’, ‘LineWidth’, 2);
legend(‘Susceptible’, ‘Infectious’, ‘Recovered’);
xlabel(‘Time’);
ylabel(‘Number of individuals’);
title(‘SIR Model’);
end I am trying to use dtm for solving SIR model, Although my code is run but I think the DTM part is wrong. I need help for DTM part
here is my code
function sir_model_dtm()
% Parameters
alpha = 0.3; % Infection rate
beta = 0.1; % Recovery rate
N = 1000; % Total population
I0 = 1; % Initial number of infected individuals
R0 = 0; % Initial number of recovered individuals
S0 = N – I0 – R0; % Initial number of susceptible individuals
% Time parameters
T = 100; % Total simulation time
dt = 1; % Time step
% Initialize arrays to store results
S = zeros(1, T);
I = zeros(1, T);
R = zeros(1, T);
% Initial conditions
S(1) = S0;
I(1) = I0;
R(1) = R0;
% Simulate the SIR model using DTM
for t = 2:T
% Compute new values
S(t) = S(t-1) – alpha*S(t-1)*I(t-1)/N * dt;
I(t) = I(t-1) + (alpha*S(t-1)*I(t-1)/N – beta*I(t-1)) * dt;
R(t) = R(t-1) + beta*I(t-1) * dt;
end
% Plot the results
t = 0:dt:T-dt;
plot(t, S, ‘b’, t, I, ‘r’, t, R, ‘g’, ‘LineWidth’, 2);
legend(‘Susceptible’, ‘Infectious’, ‘Recovered’);
xlabel(‘Time’);
ylabel(‘Number of individuals’);
title(‘SIR Model’);
end sir, matlab, dtm, differential equations, transform method, nonlinear, covid19 MATLAB Answers — New Questions
Why does my ode45 return a solution with the first values as zero despite I supplying non-zero initial conditions?
In the following code to simulate Pond volume and water temperature changes the code returns the solution (Y) with the first values of pondvolume and water temperature as 0,0. What could be the problem and ho can I solve it?
% Solar constant [MJ m-2 min-1]
Gsc = 0.08;
lat = 0; % Latitude
% Define simulation period
startdate = datetime(2023, 01, 01, ‘Format’, ‘uuuu-MM-dd’);
finishdate = datetime(2023, 06, 30, ‘Format’, ‘uuuu-MM-dd’);
tspans = (1:1:days(finishdate – startdate) + 1);
% Initialize arrays to store MM, doy, and dd values
MM_array_s = zeros(1, days(finishdate – startdate) + 1);
doy_array_s = zeros(1, days(finishdate – startdate) + 1);
dd_array_s = zeros(1, days(finishdate – startdate) + 1);
% Convert date to datetime object
for idx = 1:days(finishdate – startdate) + 1
current_date = startdate + days(idx – 1);
MM = month(current_date);
dd = day(current_date);
doy = days(current_date – datetime(year(current_date), 1, 1)) + 1;
% Store values in arrays
MM_array_s(idx) = MM;
doy_array_s(idx) = doy;
dd_array_s(idx) = dd;
end
CWR = 20; % To be read from Aquacrop
A_Irr = 200;
Irr_data = A_Irr * CWR / 1000*ones(length(tspans),1);
T_a_data = 25 *ones(length(tspans),1); % Later read from file/ it should be a data series
U2_data = 1.3*ones(length(tspans),1); % Read from file
RH_data = 0.75*ones(length(tspans),1); % Read from file
Y= zeros(length(tspans), 2);
Rho_net_values =zeros(length(tspans), 1);
for t=1:length(tspans)-1
T_a =T_a_data(t);
U2 = U2_data(t);
RH = RH_data(t);
Irr = Irr_data(t);
doy_array = doy_array_s(t);
% Convert latitude to radians
phi = deg2rad(lat);
% Calculate inverse relative distance Earth-Sun (dr)
dr = 1 + 0.033 * cos(2 .* pi .* doy_array ./ 365);
% Calculate solar declination (delta)
delta = 0.409 * sin(((2 * pi ./ 365)* doy_array) – 1.39);
% Calculate sunset hour angle (ws)
ws = acos(-tan(phi) * tan(delta));
% Calculate extraterrestrial radiation (Ra)
Ra = 1000*extraterrestrialRadiation(Gsc, dr, pi, phi, ws, delta);
% Display the result
% Constants and parameters
As = 0.06;
as = 0.25;
bs = 0.5;
%N = (2 / 15) .* acos(-tand(lat) .* tand(rad2deg(delta)));
N = (24/pi)*ws;
n = 11; %to be provided as a data series
Rs = (as + (bs .* (n ./ N))) .* Ra;
% Calculate net shortwave solar radiation (and photoperiod/24(phi_Light))
pi_light = N / 24;
Rho_SWR = Rs * (1 – As);
% Calculate rho_an (the difference between the incident and reflected components of long-wave radiation)
sigma = 4.896 * 10^(-6);
r = 0.36; % Change the value as necessary
C_c = 0.5; % Read from file/enter
epslon_a = 0.937 * 10^-5 * (273+T_a).^2 * (1 + 0.17 * C_c^2);
Rho_an = (1 – r) * epslon_a * sigma .* (273+T_a).^4;
% Calculate Water Surface Radiation (HO2_heat_Emm)
Initial_T_w = 27;
T_w = Initial_T_w; % To be simulated first
epslon_w = 0.97;
Rho_HO2_heat_Emm = epslon_w * sigma * (273+T_w).^4;
% Calculate Evaporative Heat Loss (Evap_HLoss)
bo = 368.61;
lambda = 311.02;
es = 25.37 * exp(17.62 – 5271 ./ (273+T_w));
ea = RH .* 25.37 .* exp(17.62 – 5271 ./ (273+T_a));
z = 1000; % Entered in the app
P = 760 / (10^(z / 19748.2));
T_wv = ((273+T_w) / (1.0 – (0.378 * es / P)));
T_av = ((273+T_a) / (1.0 – (0.378 * ea / P)));
Rho_Evap_HLoss = (es – ea) * (lambda * ((T_wv – T_av))^(1 / 3) + bo * U2);
% Calculate Conductive Heat Loss or Gain (Heat_cond)
Rho_Heat_cond = Rho_Evap_HLoss * 0.61 * 10^-3 * P * (((273+T_w) – (273+T_a)) / (es – ea));
Rho_net= Rho_SWR + Rho_an – Rho_HO2_heat_Emm – Rho_Evap_HLoss + Rho_Heat_cond;
% Define Rho_net as an anonymous function of time
%Computation of water balance components
A = 100;
pond_depth = 1;
V = A * pond_depth;
dmin = 0.8;
dcurr = 0.6;
dmax = pond_depth;
Ir = 12.923;
if Ir == 0
Qi = A * (dmin – dcurr) / tspans(2);
else
Qi = Ir / 100 * V;
end
Er = 10;
if Er == 0
Qe = A * (dcurr – dmax) / tspans(2);
else
Qe = Er / 100 * V;
end
pd = 10; % To be read from a file
PCP = A * pd / 1000;
Density_w = 1000;
Latent_vapw = 2260;
Evap = A * Rho_Evap_HLoss / (Density_w * Latent_vapw);
Sr = 5;
Seep = A * Sr / 1000;
% Initial conditions for the ODEs
T_win = 27; % Initial water temperature
T_wout = 27;
Cpw = 4.18;
% Define your ODEs
dVdt = @(t, Y) Qi + PCP – Qe + Evap + Seep – Irr;
dTdt = @(t, Y) Qi * T_win / Y(1) – Qe * T_wout / Y(1) + Rho_net / (Density_w * Cpw * pond_depth) – Y(2) / Y(1) * dVdt(t,Y);
% Combine the ODEs into a single function
dYdt = @(t, Y) [dVdt(t, Y); dTdt(t, Y)];
% Set initial conditions
Initial_conditions10 = [V; Initial_T_w]; % Ensure that this is a column vector
% Solve the system of ODEs using ode45
options = odeset(‘AbsTol’, 1e-6, ‘RelTol’, 1e-6);
[t_integrated10,Y_integrated]=ode45(@(t,Y) dYdt(t,Y),tspans(t:t+1),Initial_conditions10);
Y(t+1,:) = Y_integrated(end,:);
Initial_conditions10=Y(t+1,:);
Rho_net_values(t+1) = Rho_net;
end
disp([‘Extraterrestrial Radiation (Ra): ‘ num2str(Ra(1)) ‘ kJ/m^2/day’]); % Displaying only the first value
figure
plot(tspans(1):tspans(end), Rho_net_values(tspans(1):tspans(end)), ‘.’, ‘markersize’, 3), grid on,
title(‘rho_{net}’)
% Extract results
V_solution = Y(:, 1);
T_w_solution = Y(:, 2);
% Display or use the solutions as needed
disp(‘Volume (V) solution:’);
disp(V_solution);
disp(‘Water Temperature (T_w) solution:’);
disp(T_w_solution);
% Plot the results
figure;
subplot(2, 1, 1);
plot(tspans, V_solution); grid on
title(‘Volume (V) vs Time’);
xlabel(‘Time’);
ylabel(‘Volume (V)’);
subplot(2, 1, 2);
plot(tspans, T_w_solution); grid on
title(‘Water Temperature (T_w) vs Time’);
xlabel(‘Time’);
ylabel(‘Water Temperature (T_w)’);
function Ra = extraterrestrialRadiation(Gsc, dr, pi, phi, ws, delta)
% Calculate extraterrestrial radiation (Ra)
Ra = (24 * 60 / pi) * Gsc * dr .* (ws .* sin(phi) .* sin(delta) + cos(phi) .* cos(delta) .* sin(ws));
endIn the following code to simulate Pond volume and water temperature changes the code returns the solution (Y) with the first values of pondvolume and water temperature as 0,0. What could be the problem and ho can I solve it?
% Solar constant [MJ m-2 min-1]
Gsc = 0.08;
lat = 0; % Latitude
% Define simulation period
startdate = datetime(2023, 01, 01, ‘Format’, ‘uuuu-MM-dd’);
finishdate = datetime(2023, 06, 30, ‘Format’, ‘uuuu-MM-dd’);
tspans = (1:1:days(finishdate – startdate) + 1);
% Initialize arrays to store MM, doy, and dd values
MM_array_s = zeros(1, days(finishdate – startdate) + 1);
doy_array_s = zeros(1, days(finishdate – startdate) + 1);
dd_array_s = zeros(1, days(finishdate – startdate) + 1);
% Convert date to datetime object
for idx = 1:days(finishdate – startdate) + 1
current_date = startdate + days(idx – 1);
MM = month(current_date);
dd = day(current_date);
doy = days(current_date – datetime(year(current_date), 1, 1)) + 1;
% Store values in arrays
MM_array_s(idx) = MM;
doy_array_s(idx) = doy;
dd_array_s(idx) = dd;
end
CWR = 20; % To be read from Aquacrop
A_Irr = 200;
Irr_data = A_Irr * CWR / 1000*ones(length(tspans),1);
T_a_data = 25 *ones(length(tspans),1); % Later read from file/ it should be a data series
U2_data = 1.3*ones(length(tspans),1); % Read from file
RH_data = 0.75*ones(length(tspans),1); % Read from file
Y= zeros(length(tspans), 2);
Rho_net_values =zeros(length(tspans), 1);
for t=1:length(tspans)-1
T_a =T_a_data(t);
U2 = U2_data(t);
RH = RH_data(t);
Irr = Irr_data(t);
doy_array = doy_array_s(t);
% Convert latitude to radians
phi = deg2rad(lat);
% Calculate inverse relative distance Earth-Sun (dr)
dr = 1 + 0.033 * cos(2 .* pi .* doy_array ./ 365);
% Calculate solar declination (delta)
delta = 0.409 * sin(((2 * pi ./ 365)* doy_array) – 1.39);
% Calculate sunset hour angle (ws)
ws = acos(-tan(phi) * tan(delta));
% Calculate extraterrestrial radiation (Ra)
Ra = 1000*extraterrestrialRadiation(Gsc, dr, pi, phi, ws, delta);
% Display the result
% Constants and parameters
As = 0.06;
as = 0.25;
bs = 0.5;
%N = (2 / 15) .* acos(-tand(lat) .* tand(rad2deg(delta)));
N = (24/pi)*ws;
n = 11; %to be provided as a data series
Rs = (as + (bs .* (n ./ N))) .* Ra;
% Calculate net shortwave solar radiation (and photoperiod/24(phi_Light))
pi_light = N / 24;
Rho_SWR = Rs * (1 – As);
% Calculate rho_an (the difference between the incident and reflected components of long-wave radiation)
sigma = 4.896 * 10^(-6);
r = 0.36; % Change the value as necessary
C_c = 0.5; % Read from file/enter
epslon_a = 0.937 * 10^-5 * (273+T_a).^2 * (1 + 0.17 * C_c^2);
Rho_an = (1 – r) * epslon_a * sigma .* (273+T_a).^4;
% Calculate Water Surface Radiation (HO2_heat_Emm)
Initial_T_w = 27;
T_w = Initial_T_w; % To be simulated first
epslon_w = 0.97;
Rho_HO2_heat_Emm = epslon_w * sigma * (273+T_w).^4;
% Calculate Evaporative Heat Loss (Evap_HLoss)
bo = 368.61;
lambda = 311.02;
es = 25.37 * exp(17.62 – 5271 ./ (273+T_w));
ea = RH .* 25.37 .* exp(17.62 – 5271 ./ (273+T_a));
z = 1000; % Entered in the app
P = 760 / (10^(z / 19748.2));
T_wv = ((273+T_w) / (1.0 – (0.378 * es / P)));
T_av = ((273+T_a) / (1.0 – (0.378 * ea / P)));
Rho_Evap_HLoss = (es – ea) * (lambda * ((T_wv – T_av))^(1 / 3) + bo * U2);
% Calculate Conductive Heat Loss or Gain (Heat_cond)
Rho_Heat_cond = Rho_Evap_HLoss * 0.61 * 10^-3 * P * (((273+T_w) – (273+T_a)) / (es – ea));
Rho_net= Rho_SWR + Rho_an – Rho_HO2_heat_Emm – Rho_Evap_HLoss + Rho_Heat_cond;
% Define Rho_net as an anonymous function of time
%Computation of water balance components
A = 100;
pond_depth = 1;
V = A * pond_depth;
dmin = 0.8;
dcurr = 0.6;
dmax = pond_depth;
Ir = 12.923;
if Ir == 0
Qi = A * (dmin – dcurr) / tspans(2);
else
Qi = Ir / 100 * V;
end
Er = 10;
if Er == 0
Qe = A * (dcurr – dmax) / tspans(2);
else
Qe = Er / 100 * V;
end
pd = 10; % To be read from a file
PCP = A * pd / 1000;
Density_w = 1000;
Latent_vapw = 2260;
Evap = A * Rho_Evap_HLoss / (Density_w * Latent_vapw);
Sr = 5;
Seep = A * Sr / 1000;
% Initial conditions for the ODEs
T_win = 27; % Initial water temperature
T_wout = 27;
Cpw = 4.18;
% Define your ODEs
dVdt = @(t, Y) Qi + PCP – Qe + Evap + Seep – Irr;
dTdt = @(t, Y) Qi * T_win / Y(1) – Qe * T_wout / Y(1) + Rho_net / (Density_w * Cpw * pond_depth) – Y(2) / Y(1) * dVdt(t,Y);
% Combine the ODEs into a single function
dYdt = @(t, Y) [dVdt(t, Y); dTdt(t, Y)];
% Set initial conditions
Initial_conditions10 = [V; Initial_T_w]; % Ensure that this is a column vector
% Solve the system of ODEs using ode45
options = odeset(‘AbsTol’, 1e-6, ‘RelTol’, 1e-6);
[t_integrated10,Y_integrated]=ode45(@(t,Y) dYdt(t,Y),tspans(t:t+1),Initial_conditions10);
Y(t+1,:) = Y_integrated(end,:);
Initial_conditions10=Y(t+1,:);
Rho_net_values(t+1) = Rho_net;
end
disp([‘Extraterrestrial Radiation (Ra): ‘ num2str(Ra(1)) ‘ kJ/m^2/day’]); % Displaying only the first value
figure
plot(tspans(1):tspans(end), Rho_net_values(tspans(1):tspans(end)), ‘.’, ‘markersize’, 3), grid on,
title(‘rho_{net}’)
% Extract results
V_solution = Y(:, 1);
T_w_solution = Y(:, 2);
% Display or use the solutions as needed
disp(‘Volume (V) solution:’);
disp(V_solution);
disp(‘Water Temperature (T_w) solution:’);
disp(T_w_solution);
% Plot the results
figure;
subplot(2, 1, 1);
plot(tspans, V_solution); grid on
title(‘Volume (V) vs Time’);
xlabel(‘Time’);
ylabel(‘Volume (V)’);
subplot(2, 1, 2);
plot(tspans, T_w_solution); grid on
title(‘Water Temperature (T_w) vs Time’);
xlabel(‘Time’);
ylabel(‘Water Temperature (T_w)’);
function Ra = extraterrestrialRadiation(Gsc, dr, pi, phi, ws, delta)
% Calculate extraterrestrial radiation (Ra)
Ra = (24 * 60 / pi) * Gsc * dr .* (ws .* sin(phi) .* sin(delta) + cos(phi) .* cos(delta) .* sin(ws));
end In the following code to simulate Pond volume and water temperature changes the code returns the solution (Y) with the first values of pondvolume and water temperature as 0,0. What could be the problem and ho can I solve it?
% Solar constant [MJ m-2 min-1]
Gsc = 0.08;
lat = 0; % Latitude
% Define simulation period
startdate = datetime(2023, 01, 01, ‘Format’, ‘uuuu-MM-dd’);
finishdate = datetime(2023, 06, 30, ‘Format’, ‘uuuu-MM-dd’);
tspans = (1:1:days(finishdate – startdate) + 1);
% Initialize arrays to store MM, doy, and dd values
MM_array_s = zeros(1, days(finishdate – startdate) + 1);
doy_array_s = zeros(1, days(finishdate – startdate) + 1);
dd_array_s = zeros(1, days(finishdate – startdate) + 1);
% Convert date to datetime object
for idx = 1:days(finishdate – startdate) + 1
current_date = startdate + days(idx – 1);
MM = month(current_date);
dd = day(current_date);
doy = days(current_date – datetime(year(current_date), 1, 1)) + 1;
% Store values in arrays
MM_array_s(idx) = MM;
doy_array_s(idx) = doy;
dd_array_s(idx) = dd;
end
CWR = 20; % To be read from Aquacrop
A_Irr = 200;
Irr_data = A_Irr * CWR / 1000*ones(length(tspans),1);
T_a_data = 25 *ones(length(tspans),1); % Later read from file/ it should be a data series
U2_data = 1.3*ones(length(tspans),1); % Read from file
RH_data = 0.75*ones(length(tspans),1); % Read from file
Y= zeros(length(tspans), 2);
Rho_net_values =zeros(length(tspans), 1);
for t=1:length(tspans)-1
T_a =T_a_data(t);
U2 = U2_data(t);
RH = RH_data(t);
Irr = Irr_data(t);
doy_array = doy_array_s(t);
% Convert latitude to radians
phi = deg2rad(lat);
% Calculate inverse relative distance Earth-Sun (dr)
dr = 1 + 0.033 * cos(2 .* pi .* doy_array ./ 365);
% Calculate solar declination (delta)
delta = 0.409 * sin(((2 * pi ./ 365)* doy_array) – 1.39);
% Calculate sunset hour angle (ws)
ws = acos(-tan(phi) * tan(delta));
% Calculate extraterrestrial radiation (Ra)
Ra = 1000*extraterrestrialRadiation(Gsc, dr, pi, phi, ws, delta);
% Display the result
% Constants and parameters
As = 0.06;
as = 0.25;
bs = 0.5;
%N = (2 / 15) .* acos(-tand(lat) .* tand(rad2deg(delta)));
N = (24/pi)*ws;
n = 11; %to be provided as a data series
Rs = (as + (bs .* (n ./ N))) .* Ra;
% Calculate net shortwave solar radiation (and photoperiod/24(phi_Light))
pi_light = N / 24;
Rho_SWR = Rs * (1 – As);
% Calculate rho_an (the difference between the incident and reflected components of long-wave radiation)
sigma = 4.896 * 10^(-6);
r = 0.36; % Change the value as necessary
C_c = 0.5; % Read from file/enter
epslon_a = 0.937 * 10^-5 * (273+T_a).^2 * (1 + 0.17 * C_c^2);
Rho_an = (1 – r) * epslon_a * sigma .* (273+T_a).^4;
% Calculate Water Surface Radiation (HO2_heat_Emm)
Initial_T_w = 27;
T_w = Initial_T_w; % To be simulated first
epslon_w = 0.97;
Rho_HO2_heat_Emm = epslon_w * sigma * (273+T_w).^4;
% Calculate Evaporative Heat Loss (Evap_HLoss)
bo = 368.61;
lambda = 311.02;
es = 25.37 * exp(17.62 – 5271 ./ (273+T_w));
ea = RH .* 25.37 .* exp(17.62 – 5271 ./ (273+T_a));
z = 1000; % Entered in the app
P = 760 / (10^(z / 19748.2));
T_wv = ((273+T_w) / (1.0 – (0.378 * es / P)));
T_av = ((273+T_a) / (1.0 – (0.378 * ea / P)));
Rho_Evap_HLoss = (es – ea) * (lambda * ((T_wv – T_av))^(1 / 3) + bo * U2);
% Calculate Conductive Heat Loss or Gain (Heat_cond)
Rho_Heat_cond = Rho_Evap_HLoss * 0.61 * 10^-3 * P * (((273+T_w) – (273+T_a)) / (es – ea));
Rho_net= Rho_SWR + Rho_an – Rho_HO2_heat_Emm – Rho_Evap_HLoss + Rho_Heat_cond;
% Define Rho_net as an anonymous function of time
%Computation of water balance components
A = 100;
pond_depth = 1;
V = A * pond_depth;
dmin = 0.8;
dcurr = 0.6;
dmax = pond_depth;
Ir = 12.923;
if Ir == 0
Qi = A * (dmin – dcurr) / tspans(2);
else
Qi = Ir / 100 * V;
end
Er = 10;
if Er == 0
Qe = A * (dcurr – dmax) / tspans(2);
else
Qe = Er / 100 * V;
end
pd = 10; % To be read from a file
PCP = A * pd / 1000;
Density_w = 1000;
Latent_vapw = 2260;
Evap = A * Rho_Evap_HLoss / (Density_w * Latent_vapw);
Sr = 5;
Seep = A * Sr / 1000;
% Initial conditions for the ODEs
T_win = 27; % Initial water temperature
T_wout = 27;
Cpw = 4.18;
% Define your ODEs
dVdt = @(t, Y) Qi + PCP – Qe + Evap + Seep – Irr;
dTdt = @(t, Y) Qi * T_win / Y(1) – Qe * T_wout / Y(1) + Rho_net / (Density_w * Cpw * pond_depth) – Y(2) / Y(1) * dVdt(t,Y);
% Combine the ODEs into a single function
dYdt = @(t, Y) [dVdt(t, Y); dTdt(t, Y)];
% Set initial conditions
Initial_conditions10 = [V; Initial_T_w]; % Ensure that this is a column vector
% Solve the system of ODEs using ode45
options = odeset(‘AbsTol’, 1e-6, ‘RelTol’, 1e-6);
[t_integrated10,Y_integrated]=ode45(@(t,Y) dYdt(t,Y),tspans(t:t+1),Initial_conditions10);
Y(t+1,:) = Y_integrated(end,:);
Initial_conditions10=Y(t+1,:);
Rho_net_values(t+1) = Rho_net;
end
disp([‘Extraterrestrial Radiation (Ra): ‘ num2str(Ra(1)) ‘ kJ/m^2/day’]); % Displaying only the first value
figure
plot(tspans(1):tspans(end), Rho_net_values(tspans(1):tspans(end)), ‘.’, ‘markersize’, 3), grid on,
title(‘rho_{net}’)
% Extract results
V_solution = Y(:, 1);
T_w_solution = Y(:, 2);
% Display or use the solutions as needed
disp(‘Volume (V) solution:’);
disp(V_solution);
disp(‘Water Temperature (T_w) solution:’);
disp(T_w_solution);
% Plot the results
figure;
subplot(2, 1, 1);
plot(tspans, V_solution); grid on
title(‘Volume (V) vs Time’);
xlabel(‘Time’);
ylabel(‘Volume (V)’);
subplot(2, 1, 2);
plot(tspans, T_w_solution); grid on
title(‘Water Temperature (T_w) vs Time’);
xlabel(‘Time’);
ylabel(‘Water Temperature (T_w)’);
function Ra = extraterrestrialRadiation(Gsc, dr, pi, phi, ws, delta)
% Calculate extraterrestrial radiation (Ra)
Ra = (24 * 60 / pi) * Gsc * dr .* (ws .* sin(phi) .* sin(delta) + cos(phi) .* cos(delta) .* sin(ws));
end ode45 MATLAB Answers — New Questions
3D U-Net Semantic Segmentation on custom CT dataset
Hello:
I am trying to apply the concept and sample code of the tutorial "3D brain tumor segmentation using deep learning" found on the below link –
https://in.mathworks.com/help/deeplearning/examples/segment-3d-brain-tumor-using-deep-learning.html
to a custom CT dataset in order to segment the lymph nodes. In my use case also there are 2 classes i.e background and lymph nodes. The current dataset has less data points (70 in total) and the split after preprocessing is 56 for train, 10 for validation and 4 for test.
I have modified the preprocessing code to accept the custom dataset and am cropping the ROI (i.e. the cubiod region around the abdomen which has the ground truth for the lumph nodes from the full CT Torso Volume)
I have visualized the patches that are extracted by the randomPatchExtractorDatastore and found that there is a class imbalance i.e. 4 out of 5 times the extracted patch has a Lymph node portion and 1 out of 5 times – it has the background portion. So, this may be causing an issue is what I am suspecting. (The dice accuracy suggest the same – as mentioed at the later part of this text)
In my case the volume image are 1 channel where as in the tutorial the volume image are 4 channel, so I have changed the input_layer to [64 64 64 1]
The code is working and there are no errors. The training process is also working but the accuracy starts at 100% and the loss starts at approx 1. So, this is unlike the graph shown in the tutorial where the accuracy gradually increase and reaches about 72 %. To train 1 EPOCH it takes like 75 mins on RTX 2080Ti so I tried training for 10 epochs as I was suspicious about the class imbalance of the randomly extracted patches.
I used PatchPerImage = 32 and the initial learning rate at 0.001 for the 1st 5 EPOCHs. The validation freq is 10.
Training Graph is as below –
The segmentation test shows that the model is just identifing the background and not learning to segment the lymph node. The Quantify Segmentation Accuracy section of the code suggest the same thing that the lymph nodes are not identified by the newtork.
Average Dice Score of the background across 4 test volumes = 0.99685 and
Average Dice Score of the lymph nodes across 4 test volumes = 0
Box plot of the Dice Score is as below –
I need some help and guidance on the following –
So, what can be the reason for this. Is class imbalance the reason or lack of training or few datapoint ?
How can i ensure that the patches in the randomPatchExtractorDatastore are equally balance accross both the classes i.e. Background and Lymph Nodes ?
Also, should i train the network for more epochs (like 50 or 100) before evaluating its efficacy ?
Is the limited data point the problem. I only have 175 total datapoints in the comprehensive dataset. So, should i use a higher patchPerImage value like 64 (instead of 16 as used in the tutorial)
Thanks for your guidance.Hello:
I am trying to apply the concept and sample code of the tutorial "3D brain tumor segmentation using deep learning" found on the below link –
https://in.mathworks.com/help/deeplearning/examples/segment-3d-brain-tumor-using-deep-learning.html
to a custom CT dataset in order to segment the lymph nodes. In my use case also there are 2 classes i.e background and lymph nodes. The current dataset has less data points (70 in total) and the split after preprocessing is 56 for train, 10 for validation and 4 for test.
I have modified the preprocessing code to accept the custom dataset and am cropping the ROI (i.e. the cubiod region around the abdomen which has the ground truth for the lumph nodes from the full CT Torso Volume)
I have visualized the patches that are extracted by the randomPatchExtractorDatastore and found that there is a class imbalance i.e. 4 out of 5 times the extracted patch has a Lymph node portion and 1 out of 5 times – it has the background portion. So, this may be causing an issue is what I am suspecting. (The dice accuracy suggest the same – as mentioed at the later part of this text)
In my case the volume image are 1 channel where as in the tutorial the volume image are 4 channel, so I have changed the input_layer to [64 64 64 1]
The code is working and there are no errors. The training process is also working but the accuracy starts at 100% and the loss starts at approx 1. So, this is unlike the graph shown in the tutorial where the accuracy gradually increase and reaches about 72 %. To train 1 EPOCH it takes like 75 mins on RTX 2080Ti so I tried training for 10 epochs as I was suspicious about the class imbalance of the randomly extracted patches.
I used PatchPerImage = 32 and the initial learning rate at 0.001 for the 1st 5 EPOCHs. The validation freq is 10.
Training Graph is as below –
The segmentation test shows that the model is just identifing the background and not learning to segment the lymph node. The Quantify Segmentation Accuracy section of the code suggest the same thing that the lymph nodes are not identified by the newtork.
Average Dice Score of the background across 4 test volumes = 0.99685 and
Average Dice Score of the lymph nodes across 4 test volumes = 0
Box plot of the Dice Score is as below –
I need some help and guidance on the following –
So, what can be the reason for this. Is class imbalance the reason or lack of training or few datapoint ?
How can i ensure that the patches in the randomPatchExtractorDatastore are equally balance accross both the classes i.e. Background and Lymph Nodes ?
Also, should i train the network for more epochs (like 50 or 100) before evaluating its efficacy ?
Is the limited data point the problem. I only have 175 total datapoints in the comprehensive dataset. So, should i use a higher patchPerImage value like 64 (instead of 16 as used in the tutorial)
Thanks for your guidance. Hello:
I am trying to apply the concept and sample code of the tutorial "3D brain tumor segmentation using deep learning" found on the below link –
https://in.mathworks.com/help/deeplearning/examples/segment-3d-brain-tumor-using-deep-learning.html
to a custom CT dataset in order to segment the lymph nodes. In my use case also there are 2 classes i.e background and lymph nodes. The current dataset has less data points (70 in total) and the split after preprocessing is 56 for train, 10 for validation and 4 for test.
I have modified the preprocessing code to accept the custom dataset and am cropping the ROI (i.e. the cubiod region around the abdomen which has the ground truth for the lumph nodes from the full CT Torso Volume)
I have visualized the patches that are extracted by the randomPatchExtractorDatastore and found that there is a class imbalance i.e. 4 out of 5 times the extracted patch has a Lymph node portion and 1 out of 5 times – it has the background portion. So, this may be causing an issue is what I am suspecting. (The dice accuracy suggest the same – as mentioed at the later part of this text)
In my case the volume image are 1 channel where as in the tutorial the volume image are 4 channel, so I have changed the input_layer to [64 64 64 1]
The code is working and there are no errors. The training process is also working but the accuracy starts at 100% and the loss starts at approx 1. So, this is unlike the graph shown in the tutorial where the accuracy gradually increase and reaches about 72 %. To train 1 EPOCH it takes like 75 mins on RTX 2080Ti so I tried training for 10 epochs as I was suspicious about the class imbalance of the randomly extracted patches.
I used PatchPerImage = 32 and the initial learning rate at 0.001 for the 1st 5 EPOCHs. The validation freq is 10.
Training Graph is as below –
The segmentation test shows that the model is just identifing the background and not learning to segment the lymph node. The Quantify Segmentation Accuracy section of the code suggest the same thing that the lymph nodes are not identified by the newtork.
Average Dice Score of the background across 4 test volumes = 0.99685 and
Average Dice Score of the lymph nodes across 4 test volumes = 0
Box plot of the Dice Score is as below –
I need some help and guidance on the following –
So, what can be the reason for this. Is class imbalance the reason or lack of training or few datapoint ?
How can i ensure that the patches in the randomPatchExtractorDatastore are equally balance accross both the classes i.e. Background and Lymph Nodes ?
Also, should i train the network for more epochs (like 50 or 100) before evaluating its efficacy ?
Is the limited data point the problem. I only have 175 total datapoints in the comprehensive dataset. So, should i use a higher patchPerImage value like 64 (instead of 16 as used in the tutorial)
Thanks for your guidance. 3d, unet, semantic segmentation, deep learning, custom dataset, own dataset, class imbalance, randompatchextractordatastore, random patch extractor datastore MATLAB Answers — New Questions
Loaded quaternion but still cannot find slerp
After installing the Sensor Fusion toolbox, Matlab reopens. I used the following code:
q1 = quaternion([75,-20,-10],"eulerd","ZYZ","frame");
q2 = quaternion([-45,20,30],"eulerd","ZYZ","frame");
T = 0.5;
b = slerp(q1,q2,T);
But MATLAB returns error:
Unrecognized function or variable ‘slerp’.
When I look it up, the only posts say quaternion must be loaded, and it is indeed loaded since I can see Quaternion objects for q1 and q2. Then, I also installed navigation toolbox, but the problem remains.After installing the Sensor Fusion toolbox, Matlab reopens. I used the following code:
q1 = quaternion([75,-20,-10],"eulerd","ZYZ","frame");
q2 = quaternion([-45,20,30],"eulerd","ZYZ","frame");
T = 0.5;
b = slerp(q1,q2,T);
But MATLAB returns error:
Unrecognized function or variable ‘slerp’.
When I look it up, the only posts say quaternion must be loaded, and it is indeed loaded since I can see Quaternion objects for q1 and q2. Then, I also installed navigation toolbox, but the problem remains. After installing the Sensor Fusion toolbox, Matlab reopens. I used the following code:
q1 = quaternion([75,-20,-10],"eulerd","ZYZ","frame");
q2 = quaternion([-45,20,30],"eulerd","ZYZ","frame");
T = 0.5;
b = slerp(q1,q2,T);
But MATLAB returns error:
Unrecognized function or variable ‘slerp’.
When I look it up, the only posts say quaternion must be loaded, and it is indeed loaded since I can see Quaternion objects for q1 and q2. Then, I also installed navigation toolbox, but the problem remains. toolbox, installation, slerp MATLAB Answers — New Questions
my netcdf.open can not open the url I want ;help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
error:
…
READDAP_New: did not work at 98 try: lets try again.
READDAP_New: did not work at 99 try: lets try again.
READDAP_New: did not work at 100 try: lets try again.
READDAP_New: did not work at 101 try: lets try again.
错误使用 readdap (line 45)
READDAP_New: repeated failures after 100 queries
出错 get_ECCO_subgrid (line 45)
lon=readdap(path,’LONGITUDE_T’,[]);
出错 download_ECCO (line 65)
get_ECCO_subgrid(fname,lonmin,lonmax,latmin,latmax);
出错 make_OGCM (line 122)
eval([‘download_’,OGCM,'(Ymin,Ymax,Mmin,Mmax,lonmin,lonmax,latmin,latmax,’,…
readdap (line 45):
if nargin <2
disp([‘not engough input argments’]);
elseif nargin <3 || isempty(query)
disp([‘READDAP_New: Extract : ‘, varname])
while isempty(data)
if ntry>nmax
error([‘READDAP_New: repeated failures after ‘,num2str(nmax),’ queries’])
end
ntry=ntry+1;
try
ncid = netcdf.open ( url,’NOWRITE’ );
varid = netcdf.inqVarID(ncid,varname);
data = netcdf.getVar(ncid,varid,’double’);
netcdf.close (ncid);
catch
data=[];
disp([‘READDAP_New: did not work at ‘,num2str(ntry),’ try: lets try again.’])
end
end
else
…….
the problem:error:
…
READDAP_New: did not work at 98 try: lets try again.
READDAP_New: did not work at 99 try: lets try again.
READDAP_New: did not work at 100 try: lets try again.
READDAP_New: did not work at 101 try: lets try again.
错误使用 readdap (line 45)
READDAP_New: repeated failures after 100 queries
出错 get_ECCO_subgrid (line 45)
lon=readdap(path,’LONGITUDE_T’,[]);
出错 download_ECCO (line 65)
get_ECCO_subgrid(fname,lonmin,lonmax,latmin,latmax);
出错 make_OGCM (line 122)
eval([‘download_’,OGCM,'(Ymin,Ymax,Mmin,Mmax,lonmin,lonmax,latmin,latmax,’,…
readdap (line 45):
if nargin <2
disp([‘not engough input argments’]);
elseif nargin <3 || isempty(query)
disp([‘READDAP_New: Extract : ‘, varname])
while isempty(data)
if ntry>nmax
error([‘READDAP_New: repeated failures after ‘,num2str(nmax),’ queries’])
end
ntry=ntry+1;
try
ncid = netcdf.open ( url,’NOWRITE’ );
varid = netcdf.inqVarID(ncid,varname);
data = netcdf.getVar(ncid,varid,’double’);
netcdf.close (ncid);
catch
data=[];
disp([‘READDAP_New: did not work at ‘,num2str(ntry),’ try: lets try again.’])
end
end
else
…….
the problem: error:
…
READDAP_New: did not work at 98 try: lets try again.
READDAP_New: did not work at 99 try: lets try again.
READDAP_New: did not work at 100 try: lets try again.
READDAP_New: did not work at 101 try: lets try again.
错误使用 readdap (line 45)
READDAP_New: repeated failures after 100 queries
出错 get_ECCO_subgrid (line 45)
lon=readdap(path,’LONGITUDE_T’,[]);
出错 download_ECCO (line 65)
get_ECCO_subgrid(fname,lonmin,lonmax,latmin,latmax);
出错 make_OGCM (line 122)
eval([‘download_’,OGCM,'(Ymin,Ymax,Mmin,Mmax,lonmin,lonmax,latmin,latmax,’,…
readdap (line 45):
if nargin <2
disp([‘not engough input argments’]);
elseif nargin <3 || isempty(query)
disp([‘READDAP_New: Extract : ‘, varname])
while isempty(data)
if ntry>nmax
error([‘READDAP_New: repeated failures after ‘,num2str(nmax),’ queries’])
end
ntry=ntry+1;
try
ncid = netcdf.open ( url,’NOWRITE’ );
varid = netcdf.inqVarID(ncid,varname);
data = netcdf.getVar(ncid,varid,’double’);
netcdf.close (ncid);
catch
data=[];
disp([‘READDAP_New: did not work at ‘,num2str(ntry),’ try: lets try again.’])
end
end
else
…….
the problem: netcdf.open MATLAB Answers — New Questions
Not able to use ‘urROSNode’ or ‘universalrobot’ API due to ROS action server unavailable.
Facing following error message while setting up ROS VM and using ‘universalrobot’ API from ‘Robotics System Toolbox Support Package for Universal Robots UR Series Manipulators’.
Error:
Error using codertarget.manipROSInterface.manipulatorROSInterface/initializeROSActionClient (line 359)
The ROS action server "/pos_joint_traj_controller/follow_joint_trajectory" is currently unavailable.Facing following error message while setting up ROS VM and using ‘universalrobot’ API from ‘Robotics System Toolbox Support Package for Universal Robots UR Series Manipulators’.
Error:
Error using codertarget.manipROSInterface.manipulatorROSInterface/initializeROSActionClient (line 359)
The ROS action server "/pos_joint_traj_controller/follow_joint_trajectory" is currently unavailable. Facing following error message while setting up ROS VM and using ‘universalrobot’ API from ‘Robotics System Toolbox Support Package for Universal Robots UR Series Manipulators’.
Error:
Error using codertarget.manipROSInterface.manipulatorROSInterface/initializeROSActionClient (line 359)
The ROS action server "/pos_joint_traj_controller/follow_joint_trajectory" is currently unavailable. #universalrobot, #urrosnode MATLAB Answers — New Questions
MATLAB third package Simulink for Arduino take too long time ?
MATLAB third package Simulink for Arduino take too long time ? what can i do ?MATLAB third package Simulink for Arduino take too long time ? what can i do ? MATLAB third package Simulink for Arduino take too long time ? what can i do ? matlab support, simulink arduino MATLAB Answers — New Questions
I want to classify linear, parabolic and exponential functions using machine learning.
I need to use machine learning for this project. 10 x and y values of the functions will be given and using these values, it will be understood whether the other functions to be written are linear parabolic or exponential functions. I need help with where to start and where to go after. Not sure what to do.I need to use machine learning for this project. 10 x and y values of the functions will be given and using these values, it will be understood whether the other functions to be written are linear parabolic or exponential functions. I need help with where to start and where to go after. Not sure what to do. I need to use machine learning for this project. 10 x and y values of the functions will be given and using these values, it will be understood whether the other functions to be written are linear parabolic or exponential functions. I need help with where to start and where to go after. Not sure what to do. machine learning MATLAB Answers — New Questions
How to calculate Bootstrap confidence interval
Hi all. I wonder how to calculate bootstrapped confidence interval for following data sent. The conventiona confidence interval is quite large and I wonder if it may make more sense with bootstrapped CIs.
x = [6 10 14 20 26 34 38];
y = [122 107 119 120 105 95 92];
I am using a*exp(-x/b) for fitting which gives a = 127, b = 130. I want to calculated the CIs for bootstrapped sample.Hi all. I wonder how to calculate bootstrapped confidence interval for following data sent. The conventiona confidence interval is quite large and I wonder if it may make more sense with bootstrapped CIs.
x = [6 10 14 20 26 34 38];
y = [122 107 119 120 105 95 92];
I am using a*exp(-x/b) for fitting which gives a = 127, b = 130. I want to calculated the CIs for bootstrapped sample. Hi all. I wonder how to calculate bootstrapped confidence interval for following data sent. The conventiona confidence interval is quite large and I wonder if it may make more sense with bootstrapped CIs.
x = [6 10 14 20 26 34 38];
y = [122 107 119 120 105 95 92];
I am using a*exp(-x/b) for fitting which gives a = 127, b = 130. I want to calculated the CIs for bootstrapped sample. confidence interval, boostrapped MATLAB Answers — New Questions
Temperature profile in a pipe flow with Matlab pdepe solver
I’m currently trying to solve the heat equation for the Ohmic heating of a pipe flow using the MATLAB pdepe solver. The problem is described in the following paper: https://doi.org/10.4236/ojmsi.2021.91002. This paper also contains the solution for the axial bulk temperature profile. Unfortunately, I don’t find the same solution, and I wonder if anyone could help me find the error.
The simplified heat equation for this case becomes:
𝜌𝐶𝑝𝑣𝑧∂𝑇∂𝑧=𝑘𝑟∂∂𝑟(𝑟∂𝑇∂𝑟)+𝑄̂.𝑄̂=𝜎𝐸2 is the heat source due to the Ohmic heating (this is also defined in the paper). 𝑣𝑧=2𝑉𝑎𝑣𝑔(1−(𝑥𝐷/2)2) is the classical solution for the axial velocity of the fluid in the pipe.
Please find the values of the constants in the code below.
The boundary conditions are:
𝑇(𝑧=0,𝑟)=40°𝐶,∂𝑇∂𝑟|𝑧,𝑟=0=0,−𝑘∂𝑇∂𝑟|𝑧,𝑟=𝐷/2=ℎ(𝑇(𝑧,𝑟=𝐷/2)−𝑇∞).
Please find my Matlab code below. Note that in order to be able to use the Matlab pdepe solver, the time argument in the solver corresponds with the axial coordinate 𝑧. Another important remark is that I want to solve it with the option 𝑚=0.
Any help/comments/remarks will be greatly appreciated. Thanks in advance!
%% Main code
m=0;
r=linspace(0,0.025,1000); % m r corresponds with x
z=linspace(0,2.5,1000); % m z corresponds with t
sol=pdepe(m,@pdex1pde,@ic1pde,@bc1pde,r,z);
u=sol(:,:,1); % u=T
% Figure I’m trying to recreate
figure
plot(z,u(:,1))
title(‘Axial bulk temperature distribution’)
xlabel(‘z’)
ylabel(‘T(z,r=0)’)
%% Functions
% Definition of the PDE
function [c,f,s]=pdex1pde(x,t,u,DuDx)
% Parameters
T_0=40; %°C
k=0.6178; % W/m*°C
rho=992.2; % kg/m^3
C_p=4182; % J/kg*°C
m=82.5/3600; % kg/s
R=25*10^(-3); % m
D=50*10^(-3); % m
A=pi*R^2; % m^2
V_gem=m/(rho*A); % kg m
k_0=0.015; % 1/°C^3 /s
elec_cond=0.998*(1+k_0*(u-T_0)); % S/m
v_z=2*V_gem*(1-(x^2/(D/2)^2)); % kg m^3 /s
Volt=2304; % V
L=2.5; % m
Q=elec_cond*(Volt/L)^2*x/k;
c=(rho*C_p*v_z*x/k);
f=x*DuDx;
s=Q;
end
% "Initial" condition
function u0=ic1pde(x)
u0=40;
end
% Boundary conditions
function [pl,ql,pr,qr]= bc1pde(xl,ul,xr,ur,t)
% Parameters
T_amb=20; % °C
T_0=40; % °C
h=10; % W/m^2*°C
k=0.6178; % W/m*°C
pl=0; % left is ul=0
ql=1; % p+qf=0
pr=h*(ur-T_amb);
qr=k/xr;
endI’m currently trying to solve the heat equation for the Ohmic heating of a pipe flow using the MATLAB pdepe solver. The problem is described in the following paper: https://doi.org/10.4236/ojmsi.2021.91002. This paper also contains the solution for the axial bulk temperature profile. Unfortunately, I don’t find the same solution, and I wonder if anyone could help me find the error.
The simplified heat equation for this case becomes:
𝜌𝐶𝑝𝑣𝑧∂𝑇∂𝑧=𝑘𝑟∂∂𝑟(𝑟∂𝑇∂𝑟)+𝑄̂.𝑄̂=𝜎𝐸2 is the heat source due to the Ohmic heating (this is also defined in the paper). 𝑣𝑧=2𝑉𝑎𝑣𝑔(1−(𝑥𝐷/2)2) is the classical solution for the axial velocity of the fluid in the pipe.
Please find the values of the constants in the code below.
The boundary conditions are:
𝑇(𝑧=0,𝑟)=40°𝐶,∂𝑇∂𝑟|𝑧,𝑟=0=0,−𝑘∂𝑇∂𝑟|𝑧,𝑟=𝐷/2=ℎ(𝑇(𝑧,𝑟=𝐷/2)−𝑇∞).
Please find my Matlab code below. Note that in order to be able to use the Matlab pdepe solver, the time argument in the solver corresponds with the axial coordinate 𝑧. Another important remark is that I want to solve it with the option 𝑚=0.
Any help/comments/remarks will be greatly appreciated. Thanks in advance!
%% Main code
m=0;
r=linspace(0,0.025,1000); % m r corresponds with x
z=linspace(0,2.5,1000); % m z corresponds with t
sol=pdepe(m,@pdex1pde,@ic1pde,@bc1pde,r,z);
u=sol(:,:,1); % u=T
% Figure I’m trying to recreate
figure
plot(z,u(:,1))
title(‘Axial bulk temperature distribution’)
xlabel(‘z’)
ylabel(‘T(z,r=0)’)
%% Functions
% Definition of the PDE
function [c,f,s]=pdex1pde(x,t,u,DuDx)
% Parameters
T_0=40; %°C
k=0.6178; % W/m*°C
rho=992.2; % kg/m^3
C_p=4182; % J/kg*°C
m=82.5/3600; % kg/s
R=25*10^(-3); % m
D=50*10^(-3); % m
A=pi*R^2; % m^2
V_gem=m/(rho*A); % kg m
k_0=0.015; % 1/°C^3 /s
elec_cond=0.998*(1+k_0*(u-T_0)); % S/m
v_z=2*V_gem*(1-(x^2/(D/2)^2)); % kg m^3 /s
Volt=2304; % V
L=2.5; % m
Q=elec_cond*(Volt/L)^2*x/k;
c=(rho*C_p*v_z*x/k);
f=x*DuDx;
s=Q;
end
% "Initial" condition
function u0=ic1pde(x)
u0=40;
end
% Boundary conditions
function [pl,ql,pr,qr]= bc1pde(xl,ul,xr,ur,t)
% Parameters
T_amb=20; % °C
T_0=40; % °C
h=10; % W/m^2*°C
k=0.6178; % W/m*°C
pl=0; % left is ul=0
ql=1; % p+qf=0
pr=h*(ur-T_amb);
qr=k/xr;
end I’m currently trying to solve the heat equation for the Ohmic heating of a pipe flow using the MATLAB pdepe solver. The problem is described in the following paper: https://doi.org/10.4236/ojmsi.2021.91002. This paper also contains the solution for the axial bulk temperature profile. Unfortunately, I don’t find the same solution, and I wonder if anyone could help me find the error.
The simplified heat equation for this case becomes:
𝜌𝐶𝑝𝑣𝑧∂𝑇∂𝑧=𝑘𝑟∂∂𝑟(𝑟∂𝑇∂𝑟)+𝑄̂.𝑄̂=𝜎𝐸2 is the heat source due to the Ohmic heating (this is also defined in the paper). 𝑣𝑧=2𝑉𝑎𝑣𝑔(1−(𝑥𝐷/2)2) is the classical solution for the axial velocity of the fluid in the pipe.
Please find the values of the constants in the code below.
The boundary conditions are:
𝑇(𝑧=0,𝑟)=40°𝐶,∂𝑇∂𝑟|𝑧,𝑟=0=0,−𝑘∂𝑇∂𝑟|𝑧,𝑟=𝐷/2=ℎ(𝑇(𝑧,𝑟=𝐷/2)−𝑇∞).
Please find my Matlab code below. Note that in order to be able to use the Matlab pdepe solver, the time argument in the solver corresponds with the axial coordinate 𝑧. Another important remark is that I want to solve it with the option 𝑚=0.
Any help/comments/remarks will be greatly appreciated. Thanks in advance!
%% Main code
m=0;
r=linspace(0,0.025,1000); % m r corresponds with x
z=linspace(0,2.5,1000); % m z corresponds with t
sol=pdepe(m,@pdex1pde,@ic1pde,@bc1pde,r,z);
u=sol(:,:,1); % u=T
% Figure I’m trying to recreate
figure
plot(z,u(:,1))
title(‘Axial bulk temperature distribution’)
xlabel(‘z’)
ylabel(‘T(z,r=0)’)
%% Functions
% Definition of the PDE
function [c,f,s]=pdex1pde(x,t,u,DuDx)
% Parameters
T_0=40; %°C
k=0.6178; % W/m*°C
rho=992.2; % kg/m^3
C_p=4182; % J/kg*°C
m=82.5/3600; % kg/s
R=25*10^(-3); % m
D=50*10^(-3); % m
A=pi*R^2; % m^2
V_gem=m/(rho*A); % kg m
k_0=0.015; % 1/°C^3 /s
elec_cond=0.998*(1+k_0*(u-T_0)); % S/m
v_z=2*V_gem*(1-(x^2/(D/2)^2)); % kg m^3 /s
Volt=2304; % V
L=2.5; % m
Q=elec_cond*(Volt/L)^2*x/k;
c=(rho*C_p*v_z*x/k);
f=x*DuDx;
s=Q;
end
% "Initial" condition
function u0=ic1pde(x)
u0=40;
end
% Boundary conditions
function [pl,ql,pr,qr]= bc1pde(xl,ul,xr,ur,t)
% Parameters
T_amb=20; % °C
T_0=40; % °C
h=10; % W/m^2*°C
k=0.6178; % W/m*°C
pl=0; % left is ul=0
ql=1; % p+qf=0
pr=h*(ur-T_amb);
qr=k/xr;
end fluid meganics, ohmic heating proces, heat transfer, pdepe solver MATLAB Answers — New Questions
How to create a surface plot having 3 vectors?
I have 3 vectors X, Y, Z and each vector is 1×200.
I need to show the relationship between them on a plot. I used "plot3" which works, but it doesn’t show the relationship that well.
So, I was wondering if there is a way to create a surface plot. I tried to do:
surf(X,Y,Z) but it does not work and Matlab wants Z to be a matrix where Z depends on X and Y which is not the case in my calculations.
So, is there a way to get a surface plot with X, Y, and Z?
P.S.: I looked at other questions like this, but I am still confused how their answers can be applied to my case.I have 3 vectors X, Y, Z and each vector is 1×200.
I need to show the relationship between them on a plot. I used "plot3" which works, but it doesn’t show the relationship that well.
So, I was wondering if there is a way to create a surface plot. I tried to do:
surf(X,Y,Z) but it does not work and Matlab wants Z to be a matrix where Z depends on X and Y which is not the case in my calculations.
So, is there a way to get a surface plot with X, Y, and Z?
P.S.: I looked at other questions like this, but I am still confused how their answers can be applied to my case. I have 3 vectors X, Y, Z and each vector is 1×200.
I need to show the relationship between them on a plot. I used "plot3" which works, but it doesn’t show the relationship that well.
So, I was wondering if there is a way to create a surface plot. I tried to do:
surf(X,Y,Z) but it does not work and Matlab wants Z to be a matrix where Z depends on X and Y which is not the case in my calculations.
So, is there a way to get a surface plot with X, Y, and Z?
P.S.: I looked at other questions like this, but I am still confused how their answers can be applied to my case. surface, plotting, 3d plots MATLAB Answers — New Questions
Converting MATLAB code to AutoCAD
I am new to MATLAB. I have been trying to convert the code to excel, so I can convert that into autocad, but nothing seems to be working. It is possible to do because I need it for school and past students have done it. This is my code:
t = linspace(0, 2*pi, 201); % this is used to create 201 evenly spaced
% points between 0 and 2pi.
r = sqrt(abs(7*sin(5*t))); % 7 controls the size of the shape.
% 5 controls the amount of petals on each
% side.
% abs is used to ignore all imaginary
% values.
% sin and sqrt are used to give a lower
% value.
[x, y] = pol2cart(t, r); % this command transforms corresponding elements of
% the polar coordinate arrays of theta and ro to 2-D
% cartesian coordinates.
z = 3 * (x.^2 + y.^2); % The (x.^2 + y.^2) is used to control how curved
% up the flower is. The (3) controls the size of
% the flower.
% Plotting the 3D surface
figure(1); % this is used to create a new figure window
fill3(x, y, z, ‘c’); % this is used to plot on a 3-D axis.
% the c is used to fill the shape with a cyan color.
grid on; % this is used to show gridlines on the plot.
% Creating a triangulation object
tri = delaunay(x(:), y(:)); % Triangulate the surface
% Save as an STL file
vertices = [x(:), y(:), z(:)]; % Create vertices matrix
stlwrite(tri,’flower.stl’); % Save as an STL file using triangulation objectt = linspace(0, 2*pi, 201); % this is used to create 201 evenly spaced
% points between 0 and 2pi.
r = sqrt(abs(7*sin(5*t))); % 7 controls the size of the shape.
% 5 controls the amount of petals on each
% side.
% abs is used to ignore all imaginary
% values.
% sin and sqrt are used to give a lower
% value.
[x, y] = pol2cart(t, r); % this command transforms corresponding elements of
% the polar coordinate arrays of theta and ro to 2-D
% cartesian coordinates.
z = 3 * (x.^2 + y.^2); % The (x.^2 + y.^2) is used to control how curved
% up the flower is. The (3) controls the size of
% the flower.
% Plotting the 3D surface
figure(1); % this is used to create a new figure window
fill3(x, y, z, ‘c’); % this is used to plot on a 3-D axis.
% the c is used to fill the shape with a cyan color.
grid on; % this is used to show gridlines on the plot.
i have since then produced the following code, but when I try to open it with excel, an error comes up.
t = linspace(0, 2*pi, 201); % 201 evenly spaced points between 0 and 2pi
r = sqrt(abs(7*sin(5*t))); % Polar coordinates
[x, y] = pol2cart(t, r); % Convert polar to Cartesian coordinates
z = 3 * (x.^2 + y.^2); % Calculate z coordinates
data = [x(:), y(:), z(:)]; % Create a matrix where each row is [x, y, z]
filename = ‘flower_data.xlsx’; % Define the filename
writematrix(data, filename); % Write data to Excel
the error is:
excel cannot open the file ‘flower_data.xlsx’ because the file format or file extension is not valid. verify that the file has not been corrupted and that the file extension matches the format of the file
can anyone please help?I am new to MATLAB. I have been trying to convert the code to excel, so I can convert that into autocad, but nothing seems to be working. It is possible to do because I need it for school and past students have done it. This is my code:
t = linspace(0, 2*pi, 201); % this is used to create 201 evenly spaced
% points between 0 and 2pi.
r = sqrt(abs(7*sin(5*t))); % 7 controls the size of the shape.
% 5 controls the amount of petals on each
% side.
% abs is used to ignore all imaginary
% values.
% sin and sqrt are used to give a lower
% value.
[x, y] = pol2cart(t, r); % this command transforms corresponding elements of
% the polar coordinate arrays of theta and ro to 2-D
% cartesian coordinates.
z = 3 * (x.^2 + y.^2); % The (x.^2 + y.^2) is used to control how curved
% up the flower is. The (3) controls the size of
% the flower.
% Plotting the 3D surface
figure(1); % this is used to create a new figure window
fill3(x, y, z, ‘c’); % this is used to plot on a 3-D axis.
% the c is used to fill the shape with a cyan color.
grid on; % this is used to show gridlines on the plot.
% Creating a triangulation object
tri = delaunay(x(:), y(:)); % Triangulate the surface
% Save as an STL file
vertices = [x(:), y(:), z(:)]; % Create vertices matrix
stlwrite(tri,’flower.stl’); % Save as an STL file using triangulation objectt = linspace(0, 2*pi, 201); % this is used to create 201 evenly spaced
% points between 0 and 2pi.
r = sqrt(abs(7*sin(5*t))); % 7 controls the size of the shape.
% 5 controls the amount of petals on each
% side.
% abs is used to ignore all imaginary
% values.
% sin and sqrt are used to give a lower
% value.
[x, y] = pol2cart(t, r); % this command transforms corresponding elements of
% the polar coordinate arrays of theta and ro to 2-D
% cartesian coordinates.
z = 3 * (x.^2 + y.^2); % The (x.^2 + y.^2) is used to control how curved
% up the flower is. The (3) controls the size of
% the flower.
% Plotting the 3D surface
figure(1); % this is used to create a new figure window
fill3(x, y, z, ‘c’); % this is used to plot on a 3-D axis.
% the c is used to fill the shape with a cyan color.
grid on; % this is used to show gridlines on the plot.
i have since then produced the following code, but when I try to open it with excel, an error comes up.
t = linspace(0, 2*pi, 201); % 201 evenly spaced points between 0 and 2pi
r = sqrt(abs(7*sin(5*t))); % Polar coordinates
[x, y] = pol2cart(t, r); % Convert polar to Cartesian coordinates
z = 3 * (x.^2 + y.^2); % Calculate z coordinates
data = [x(:), y(:), z(:)]; % Create a matrix where each row is [x, y, z]
filename = ‘flower_data.xlsx’; % Define the filename
writematrix(data, filename); % Write data to Excel
the error is:
excel cannot open the file ‘flower_data.xlsx’ because the file format or file extension is not valid. verify that the file has not been corrupted and that the file extension matches the format of the file
can anyone please help? I am new to MATLAB. I have been trying to convert the code to excel, so I can convert that into autocad, but nothing seems to be working. It is possible to do because I need it for school and past students have done it. This is my code:
t = linspace(0, 2*pi, 201); % this is used to create 201 evenly spaced
% points between 0 and 2pi.
r = sqrt(abs(7*sin(5*t))); % 7 controls the size of the shape.
% 5 controls the amount of petals on each
% side.
% abs is used to ignore all imaginary
% values.
% sin and sqrt are used to give a lower
% value.
[x, y] = pol2cart(t, r); % this command transforms corresponding elements of
% the polar coordinate arrays of theta and ro to 2-D
% cartesian coordinates.
z = 3 * (x.^2 + y.^2); % The (x.^2 + y.^2) is used to control how curved
% up the flower is. The (3) controls the size of
% the flower.
% Plotting the 3D surface
figure(1); % this is used to create a new figure window
fill3(x, y, z, ‘c’); % this is used to plot on a 3-D axis.
% the c is used to fill the shape with a cyan color.
grid on; % this is used to show gridlines on the plot.
% Creating a triangulation object
tri = delaunay(x(:), y(:)); % Triangulate the surface
% Save as an STL file
vertices = [x(:), y(:), z(:)]; % Create vertices matrix
stlwrite(tri,’flower.stl’); % Save as an STL file using triangulation objectt = linspace(0, 2*pi, 201); % this is used to create 201 evenly spaced
% points between 0 and 2pi.
r = sqrt(abs(7*sin(5*t))); % 7 controls the size of the shape.
% 5 controls the amount of petals on each
% side.
% abs is used to ignore all imaginary
% values.
% sin and sqrt are used to give a lower
% value.
[x, y] = pol2cart(t, r); % this command transforms corresponding elements of
% the polar coordinate arrays of theta and ro to 2-D
% cartesian coordinates.
z = 3 * (x.^2 + y.^2); % The (x.^2 + y.^2) is used to control how curved
% up the flower is. The (3) controls the size of
% the flower.
% Plotting the 3D surface
figure(1); % this is used to create a new figure window
fill3(x, y, z, ‘c’); % this is used to plot on a 3-D axis.
% the c is used to fill the shape with a cyan color.
grid on; % this is used to show gridlines on the plot.
i have since then produced the following code, but when I try to open it with excel, an error comes up.
t = linspace(0, 2*pi, 201); % 201 evenly spaced points between 0 and 2pi
r = sqrt(abs(7*sin(5*t))); % Polar coordinates
[x, y] = pol2cart(t, r); % Convert polar to Cartesian coordinates
z = 3 * (x.^2 + y.^2); % Calculate z coordinates
data = [x(:), y(:), z(:)]; % Create a matrix where each row is [x, y, z]
filename = ‘flower_data.xlsx’; % Define the filename
writematrix(data, filename); % Write data to Excel
the error is:
excel cannot open the file ‘flower_data.xlsx’ because the file format or file extension is not valid. verify that the file has not been corrupted and that the file extension matches the format of the file
can anyone please help? matlab, autocad, excel, convert, .xlsx MATLAB Answers — New Questions
Understanding implicit surface plot behavior
Here is some code that plots an implicit infinite, tilted cylinder.
Q=diag([0,1,1]./3^2);
axis([-5 5 -5 5 -5 5])
h1=fimplicit3(@(x,y,z) quadform(x,y,z,Q) ,’EdgeColor’,’none’,’FaceAlpha’,0.3);
view([-2,11])
xlabel X, ylabel Y, zlabel Z
T=makehgtform(‘axisrotate’,[0 1 0],pi/4);
h1.Parent=hgtransform(‘Matrix’,T);
My first question is, why doesn’t the cylinder, which is infinite, extend to the axis limits?
My second question is, below is the same code again, except at the very end, we double the extent of the axes. Why does the cylinder become invisible when this is done?
figure;
axis([-5 5 -5 5 -5 5])
h2=fimplicit3(@(x,y,z) quadform(x,y,z,Q) ,’EdgeColor’,’none’,’FaceAlpha’,0.3);
view([-2,11])
xlabel X, ylabel Y, zlabel Z
T=makehgtform(‘axisrotate’,[0 1 0],pi/4);
h2.Parent=hgtransform(‘Matrix’,T);
axis([-5 5 -5 5 -5 5]*2) %<—-makes the cylinder disappear!
function d=quadform(x,y,z, Q)
xyz=[x(:).’; y(:).’; z(:).’];
d=reshape( (sum((Q*xyz).*xyz)-1), size(x));
endHere is some code that plots an implicit infinite, tilted cylinder.
Q=diag([0,1,1]./3^2);
axis([-5 5 -5 5 -5 5])
h1=fimplicit3(@(x,y,z) quadform(x,y,z,Q) ,’EdgeColor’,’none’,’FaceAlpha’,0.3);
view([-2,11])
xlabel X, ylabel Y, zlabel Z
T=makehgtform(‘axisrotate’,[0 1 0],pi/4);
h1.Parent=hgtransform(‘Matrix’,T);
My first question is, why doesn’t the cylinder, which is infinite, extend to the axis limits?
My second question is, below is the same code again, except at the very end, we double the extent of the axes. Why does the cylinder become invisible when this is done?
figure;
axis([-5 5 -5 5 -5 5])
h2=fimplicit3(@(x,y,z) quadform(x,y,z,Q) ,’EdgeColor’,’none’,’FaceAlpha’,0.3);
view([-2,11])
xlabel X, ylabel Y, zlabel Z
T=makehgtform(‘axisrotate’,[0 1 0],pi/4);
h2.Parent=hgtransform(‘Matrix’,T);
axis([-5 5 -5 5 -5 5]*2) %<—-makes the cylinder disappear!
function d=quadform(x,y,z, Q)
xyz=[x(:).’; y(:).’; z(:).’];
d=reshape( (sum((Q*xyz).*xyz)-1), size(x));
end Here is some code that plots an implicit infinite, tilted cylinder.
Q=diag([0,1,1]./3^2);
axis([-5 5 -5 5 -5 5])
h1=fimplicit3(@(x,y,z) quadform(x,y,z,Q) ,’EdgeColor’,’none’,’FaceAlpha’,0.3);
view([-2,11])
xlabel X, ylabel Y, zlabel Z
T=makehgtform(‘axisrotate’,[0 1 0],pi/4);
h1.Parent=hgtransform(‘Matrix’,T);
My first question is, why doesn’t the cylinder, which is infinite, extend to the axis limits?
My second question is, below is the same code again, except at the very end, we double the extent of the axes. Why does the cylinder become invisible when this is done?
figure;
axis([-5 5 -5 5 -5 5])
h2=fimplicit3(@(x,y,z) quadform(x,y,z,Q) ,’EdgeColor’,’none’,’FaceAlpha’,0.3);
view([-2,11])
xlabel X, ylabel Y, zlabel Z
T=makehgtform(‘axisrotate’,[0 1 0],pi/4);
h2.Parent=hgtransform(‘Matrix’,T);
axis([-5 5 -5 5 -5 5]*2) %<—-makes the cylinder disappear!
function d=quadform(x,y,z, Q)
xyz=[x(:).’; y(:).’; z(:).’];
d=reshape( (sum((Q*xyz).*xyz)-1), size(x));
end graphics, implicit surface, implicitfunctionsurface, bug? MATLAB Answers — New Questions
Display in the command window
my code looks as follows but i don’t understand why the values of x and z won’t display on the command window. The values are supposed to plot a semicircle.
startUpHeight = 299.07;
startUpValue = 100;
radius = 15;
minConstraintX = 150;
maxConstraintX = 50;
minConstraintZ = 10;
maxConstraintZ = 300;
for x = (startUpValue – radius):0.25:(startUpValue + radius)
if ((maxConstraintX >= x) && (x >= minConstraintX))
z = -1*((sqrt(radius^2)-((x-(3*radius))^2)) – startUpHeight);
if ((maxConstraintZ >= z) && (z >= minConstraintZ))
disp([‘x: ‘, num2str(x), ‘, z: ‘, num2str(z)]);
end
end
endmy code looks as follows but i don’t understand why the values of x and z won’t display on the command window. The values are supposed to plot a semicircle.
startUpHeight = 299.07;
startUpValue = 100;
radius = 15;
minConstraintX = 150;
maxConstraintX = 50;
minConstraintZ = 10;
maxConstraintZ = 300;
for x = (startUpValue – radius):0.25:(startUpValue + radius)
if ((maxConstraintX >= x) && (x >= minConstraintX))
z = -1*((sqrt(radius^2)-((x-(3*radius))^2)) – startUpHeight);
if ((maxConstraintZ >= z) && (z >= minConstraintZ))
disp([‘x: ‘, num2str(x), ‘, z: ‘, num2str(z)]);
end
end
end my code looks as follows but i don’t understand why the values of x and z won’t display on the command window. The values are supposed to plot a semicircle.
startUpHeight = 299.07;
startUpValue = 100;
radius = 15;
minConstraintX = 150;
maxConstraintX = 50;
minConstraintZ = 10;
maxConstraintZ = 300;
for x = (startUpValue – radius):0.25:(startUpValue + radius)
if ((maxConstraintX >= x) && (x >= minConstraintX))
z = -1*((sqrt(radius^2)-((x-(3*radius))^2)) – startUpHeight);
if ((maxConstraintZ >= z) && (z >= minConstraintZ))
disp([‘x: ‘, num2str(x), ‘, z: ‘, num2str(z)]);
end
end
end display MATLAB Answers — New Questions
Solving Differential Equation by Laplace Transform
Post Content Post Content laplace transform MATLAB Answers — New Questions
Roadrunner opendrive 1.4 import
When I try to load a .xodr file in Roadrunner nothing happens. I dragged my opendrive 1.4 file (generated with sumo netconvert) into the Library Browser Assets, set the projection mode to Full Projection, and set a Custom Projection (Use Tranverse Mercator), just as was shown in the tutorial video. However, when I then try to drag the file to load it nothing happens, no network spawns. How do I resolve this?When I try to load a .xodr file in Roadrunner nothing happens. I dragged my opendrive 1.4 file (generated with sumo netconvert) into the Library Browser Assets, set the projection mode to Full Projection, and set a Custom Projection (Use Tranverse Mercator), just as was shown in the tutorial video. However, when I then try to drag the file to load it nothing happens, no network spawns. How do I resolve this? When I try to load a .xodr file in Roadrunner nothing happens. I dragged my opendrive 1.4 file (generated with sumo netconvert) into the Library Browser Assets, set the projection mode to Full Projection, and set a Custom Projection (Use Tranverse Mercator), just as was shown in the tutorial video. However, when I then try to drag the file to load it nothing happens, no network spawns. How do I resolve this? roadrunner, opendrive MATLAB Answers — New Questions