I have error message while trying to link Matlab with my Aspen Plus simulation:
>> CHLC
Error using COM.Apwn_Document_37_0/invoke
Invoke Error, Dispatch Exception:
Source: Aspen Plus 37.0 OLE Services
Description: Unable to open file.

Error in CHLC (line 8)
Aspen.invoke(‘InitFromFile’,[mess.Name ” Simulation_Name ‘.apwz’]);
This is the code im using:
%% Linking
Aspen = actxserver(‘Apwn.Document.37.0’);
[stat,mess]=fileattrib; % get attributes of folder (Necessary to establish the location of the simulation)
Simulation_Name = ‘CL_SCO2’;% Aspeen Plus Simulation Name
Aspen.invoke(‘InitFromArchive2’,[mess.Name ” Simulation_Name ‘.bkp’]);
Aspen.Visible = 1; % 1 —> Aspen is Visible; 0 —> Aspen is open but not visible
Aspen.SuppressDialogs = 1; % Suppress windows dialogs.
Aspen.Engine.Run2(1); % Run the simulation
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
end
Matlab version is R2022a (9.12.0.1884302) 64-bit and Aspen plus V11 (37.0.0395)
Already try saving Aspen simulation as .bkp and .apwz.
Yes, two files ( .m matlab code and .awpz / .bkp simulation) are in the same folder.I have error message while trying to link Matlab with my Aspen Plus simulation:
>> CHLC
Error using COM.Apwn_Document_37_0/invoke
Invoke Error, Dispatch Exception:
Source: Aspen Plus 37.0 OLE Services
Description: Unable to open file.

Error in CHLC (line 8)
Aspen.invoke(‘InitFromFile’,[mess.Name ” Simulation_Name ‘.apwz’]);
This is the code im using:
%% Linking
Aspen = actxserver(‘Apwn.Document.37.0’);
[stat,mess]=fileattrib; % get attributes of folder (Necessary to establish the location of the simulation)
Simulation_Name = ‘CL_SCO2’;% Aspeen Plus Simulation Name
Aspen.invoke(‘InitFromArchive2’,[mess.Name ” Simulation_Name ‘.bkp’]);
Aspen.Visible = 1; % 1 —> Aspen is Visible; 0 —> Aspen is open but not visible
Aspen.SuppressDialogs = 1; % Suppress windows dialogs.
Aspen.Engine.Run2(1); % Run the simulation
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
end
Matlab version is R2022a (9.12.0.1884302) 64-bit and Aspen plus V11 (37.0.0395)
Already try saving Aspen simulation as .bkp and .apwz.
Yes, two files ( .m matlab code and .awpz / .bkp simulation) are in the same folder. I have error message while trying to link Matlab with my Aspen Plus simulation:
>> CHLC
Error using COM.Apwn_Document_37_0/invoke
Invoke Error, Dispatch Exception:
Source: Aspen Plus 37.0 OLE Services
Description: Unable to open file.

Error in CHLC (line 8)
Aspen.invoke(‘InitFromFile’,[mess.Name ” Simulation_Name ‘.apwz’]);
This is the code im using:
%% Linking
Aspen = actxserver(‘Apwn.Document.37.0’);
[stat,mess]=fileattrib; % get attributes of folder (Necessary to establish the location of the simulation)
Simulation_Name = ‘CL_SCO2’;% Aspeen Plus Simulation Name
Aspen.invoke(‘InitFromArchive2’,[mess.Name ” Simulation_Name ‘.bkp’]);
Aspen.Visible = 1; % 1 —> Aspen is Visible; 0 —> Aspen is open but not visible
Aspen.SuppressDialogs = 1; % Suppress windows dialogs.
Aspen.Engine.Run2(1); % Run the simulation
while Aspen.Engine.IsRunning == 1 % 1 –> If Aspen is running; 0 —> If Aspen stop.
pause(0.5);
end
Matlab version is R2022a (9.12.0.1884302) 64-bit and Aspen plus V11 (37.0.0395)
Already try saving Aspen simulation as .bkp and .apwz.
Yes, two files ( .m matlab code and .awpz / .bkp simulation) are in the same folder. aspen puls, matlab, link, error MATLAB Answers — New Questions

​

I’m going to make a plot like this picture.
I interpolated the data using the interp1 function. However, contourf can only contain values of 2*2 or more, so the function cannot be executed. How do I solve this?I’m going to make a plot like this picture.
I interpolated the data using the interp1 function. However, contourf can only contain values of 2*2 or more, so the function cannot be executed. How do I solve this? I’m going to make a plot like this picture.
I interpolated the data using the interp1 function. However, contourf can only contain values of 2*2 or more, so the function cannot be executed. How do I solve this? #argo, #interp1, #ocean, #pressure, #temperature, #salinty, #contourf MATLAB Answers — New Questions

​

Hello,
I am working with the IEEE 34-bus distribution network in MATLAB-Simulink. When running the load flow analysis after setting up the buses, I encountered the following error message:
"Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.218374e-17."
Given that this is an unbalanced system, could you please advise on the potential causes of this issue?How to solve it? Additionally, are there alternative methods or tools within MATLAB-Simulink to handle load flow analysis for such systems?
Thank you for your assistance.
Best regards,Hello,
I am working with the IEEE 34-bus distribution network in MATLAB-Simulink. When running the load flow analysis after setting up the buses, I encountered the following error message:
"Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.218374e-17."
Given that this is an unbalanced system, could you please advise on the potential causes of this issue?How to solve it? Additionally, are there alternative methods or tools within MATLAB-Simulink to handle load flow analysis for such systems?
Thank you for your assistance.
Best regards, Hello,
I am working with the IEEE 34-bus distribution network in MATLAB-Simulink. When running the load flow analysis after setting up the buses, I encountered the following error message:
"Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.218374e-17."
Given that this is an unbalanced system, could you please advise on the potential causes of this issue?How to solve it? Additionally, are there alternative methods or tools within MATLAB-Simulink to handle load flow analysis for such systems?
Thank you for your assistance.
Best regards, ieee 34, load flow, matlab, simulink, unbalanced s MATLAB Answers — New Questions

​

I would like to speed up the following code which solves a discrete dynamic programming problem using the method of successive approximations, as described e.g. in Bertsekas.
The algorithm is made of two steps. In step 1, I precompute the payoff array R(a’,a,z) where a’ is the action and (a,z) are the states. In step 2 I compute the value function using the method of successive approximations: I guess V0, then I compute an updated V1 and finally I check if ||V1-V0|| is less than a tolerance level. If it is, I stop, otherwise I set V0=V1 and go on.
I profiled the code (see below a MWE) and the two most time-consuming lines are the following ones:
(1) RHS = Ret+beta*permute(EV,[1,3,2]);
(2) [max_val,max_ind] = max(RHS,[],1);
Line (1) takes up 63% of the total running time, line (2) takes 32%. As you can see I have already vectorized all loops.
I would be very grateful for any suggestion. I post below a MWE.
%% Solve income fluctuation problem CPU
clear;clc;close all
%% Economic parameters
sigma = 2;
r = 0.03;
beta = 0.96;
PZ = [0.60 0.40;
0.05 0.95];
z_grid = [0.5 1.0]’;
n_z = length(z_grid);
b = 0; %lower bound for asset holdings a
grid_max = 4;
n_a = 500; % IN PRACTICE THIS IS EQUAL TO 5000-10000
R = 1+r;
a_grid = linspace(-b,grid_max,n_a)’;
if sigma==1
fun_u = @(c) log(c);
else
fun_u = @(c) c.^(1-sigma)/(1-sigma);
end
%% Computational parameters
verbose = 0;
tiny = 1e-8; %very small positive number
tol = 1e-6; %tolerance for VFI and TI
max_iter = 500; %maximum num. of iterations for both VFI and TI
%% Start timing
tic
%% STEP 1- Precompute current payoff array R(a’,a,z)
a_tomorrow = a_grid; %(a’,1,1)
a_today = a_grid’; %(1,a,1)
z_today = shiftdim(z_grid,-2); %(1,1,z)
cons = (1+r)*a_today+z_today-a_tomorrow;
Ret = fun_u(cons); %size: [n_a,n_a,n_z]
Ret(cons<=0) = -inf;
%% STEP 2 – Value function iteration
iter = 1;
err = tol+1;
V0 = zeros(n_a,n_z);
while err>tol && iter<=max_iter
EV = V0*PZ’; %(a’,z)
RHS = Ret+beta*permute(EV,[1,3,2]);
[max_val,max_ind] = max(RHS,[],1);
V1 = squeeze(max_val);
pol_ind_ap = squeeze(max_ind);
err = max(abs(V0(:)-V1(:)));
if verbose==1
fprintf(‘iter = %d, err = %f n’,iter,err)
end
iter = iter+1;
V0 = V1;
end
if err>tol
error(‘VFI did not converge!’)
else
fprintf(‘VFI converged after = %d iterations n’,iter)
end
pol_ap = a_grid(pol_ind_ap);
pol_c = (1+r)*a_grid+z_grid’-pol_ap;
%% End timing
toc
%% Figures
figure
plot(a_grid,pol_c(:,1),’linewidth’,2)
hold on
plot(a_grid,pol_c(:,2),’linewidth’,2)
legend(‘Low shock’,’High shock’,’Location’,’NorthWest’)
xlabel(‘asset level’)
ylabel(‘consumption’)
title(‘Consumption Policy Function’)
figure
plot(a_grid,a_grid,’–‘,’linewidth’,2)
hold on
plot(a_grid,pol_ap(:,1),’linewidth’,2)
hold on
plot(a_grid,pol_ap(:,2),’linewidth’,2)
legend(’45 line’,’Low shock’,’High shock’,’Location’,’NorthWest’)
xlabel(‘Current period assets’)
ylabel(‘Next-period assets’)
title(‘Assets Policy Function’)I would like to speed up the following code which solves a discrete dynamic programming problem using the method of successive approximations, as described e.g. in Bertsekas.
The algorithm is made of two steps. In step 1, I precompute the payoff array R(a’,a,z) where a’ is the action and (a,z) are the states. In step 2 I compute the value function using the method of successive approximations: I guess V0, then I compute an updated V1 and finally I check if ||V1-V0|| is less than a tolerance level. If it is, I stop, otherwise I set V0=V1 and go on.
I profiled the code (see below a MWE) and the two most time-consuming lines are the following ones:
(1) RHS = Ret+beta*permute(EV,[1,3,2]);
(2) [max_val,max_ind] = max(RHS,[],1);
Line (1) takes up 63% of the total running time, line (2) takes 32%. As you can see I have already vectorized all loops.
I would be very grateful for any suggestion. I post below a MWE.
%% Solve income fluctuation problem CPU
clear;clc;close all
%% Economic parameters
sigma = 2;
r = 0.03;
beta = 0.96;
PZ = [0.60 0.40;
0.05 0.95];
z_grid = [0.5 1.0]’;
n_z = length(z_grid);
b = 0; %lower bound for asset holdings a
grid_max = 4;
n_a = 500; % IN PRACTICE THIS IS EQUAL TO 5000-10000
R = 1+r;
a_grid = linspace(-b,grid_max,n_a)’;
if sigma==1
fun_u = @(c) log(c);
else
fun_u = @(c) c.^(1-sigma)/(1-sigma);
end
%% Computational parameters
verbose = 0;
tiny = 1e-8; %very small positive number
tol = 1e-6; %tolerance for VFI and TI
max_iter = 500; %maximum num. of iterations for both VFI and TI
%% Start timing
tic
%% STEP 1- Precompute current payoff array R(a’,a,z)
a_tomorrow = a_grid; %(a’,1,1)
a_today = a_grid’; %(1,a,1)
z_today = shiftdim(z_grid,-2); %(1,1,z)
cons = (1+r)*a_today+z_today-a_tomorrow;
Ret = fun_u(cons); %size: [n_a,n_a,n_z]
Ret(cons<=0) = -inf;
%% STEP 2 – Value function iteration
iter = 1;
err = tol+1;
V0 = zeros(n_a,n_z);
while err>tol && iter<=max_iter
EV = V0*PZ’; %(a’,z)
RHS = Ret+beta*permute(EV,[1,3,2]);
[max_val,max_ind] = max(RHS,[],1);
V1 = squeeze(max_val);
pol_ind_ap = squeeze(max_ind);
err = max(abs(V0(:)-V1(:)));
if verbose==1
fprintf(‘iter = %d, err = %f n’,iter,err)
end
iter = iter+1;
V0 = V1;
end
if err>tol
error(‘VFI did not converge!’)
else
fprintf(‘VFI converged after = %d iterations n’,iter)
end
pol_ap = a_grid(pol_ind_ap);
pol_c = (1+r)*a_grid+z_grid’-pol_ap;
%% End timing
toc
%% Figures
figure
plot(a_grid,pol_c(:,1),’linewidth’,2)
hold on
plot(a_grid,pol_c(:,2),’linewidth’,2)
legend(‘Low shock’,’High shock’,’Location’,’NorthWest’)
xlabel(‘asset level’)
ylabel(‘consumption’)
title(‘Consumption Policy Function’)
figure
plot(a_grid,a_grid,’–‘,’linewidth’,2)
hold on
plot(a_grid,pol_ap(:,1),’linewidth’,2)
hold on
plot(a_grid,pol_ap(:,2),’linewidth’,2)
legend(’45 line’,’Low shock’,’High shock’,’Location’,’NorthWest’)
xlabel(‘Current period assets’)
ylabel(‘Next-period assets’)
title(‘Assets Policy Function’) I would like to speed up the following code which solves a discrete dynamic programming problem using the method of successive approximations, as described e.g. in Bertsekas.
The algorithm is made of two steps. In step 1, I precompute the payoff array R(a’,a,z) where a’ is the action and (a,z) are the states. In step 2 I compute the value function using the method of successive approximations: I guess V0, then I compute an updated V1 and finally I check if ||V1-V0|| is less than a tolerance level. If it is, I stop, otherwise I set V0=V1 and go on.
I profiled the code (see below a MWE) and the two most time-consuming lines are the following ones:
(1) RHS = Ret+beta*permute(EV,[1,3,2]);
(2) [max_val,max_ind] = max(RHS,[],1);
Line (1) takes up 63% of the total running time, line (2) takes 32%. As you can see I have already vectorized all loops.
I would be very grateful for any suggestion. I post below a MWE.
%% Solve income fluctuation problem CPU
clear;clc;close all
%% Economic parameters
sigma = 2;
r = 0.03;
beta = 0.96;
PZ = [0.60 0.40;
0.05 0.95];
z_grid = [0.5 1.0]’;
n_z = length(z_grid);
b = 0; %lower bound for asset holdings a
grid_max = 4;
n_a = 500; % IN PRACTICE THIS IS EQUAL TO 5000-10000
R = 1+r;
a_grid = linspace(-b,grid_max,n_a)’;
if sigma==1
fun_u = @(c) log(c);
else
fun_u = @(c) c.^(1-sigma)/(1-sigma);
end
%% Computational parameters
verbose = 0;
tiny = 1e-8; %very small positive number
tol = 1e-6; %tolerance for VFI and TI
max_iter = 500; %maximum num. of iterations for both VFI and TI
%% Start timing
tic
%% STEP 1- Precompute current payoff array R(a’,a,z)
a_tomorrow = a_grid; %(a’,1,1)
a_today = a_grid’; %(1,a,1)
z_today = shiftdim(z_grid,-2); %(1,1,z)
cons = (1+r)*a_today+z_today-a_tomorrow;
Ret = fun_u(cons); %size: [n_a,n_a,n_z]
Ret(cons<=0) = -inf;
%% STEP 2 – Value function iteration
iter = 1;
err = tol+1;
V0 = zeros(n_a,n_z);
while err>tol && iter<=max_iter
EV = V0*PZ’; %(a’,z)
RHS = Ret+beta*permute(EV,[1,3,2]);
[max_val,max_ind] = max(RHS,[],1);
V1 = squeeze(max_val);
pol_ind_ap = squeeze(max_ind);
err = max(abs(V0(:)-V1(:)));
if verbose==1
fprintf(‘iter = %d, err = %f n’,iter,err)
end
iter = iter+1;
V0 = V1;
end
if err>tol
error(‘VFI did not converge!’)
else
fprintf(‘VFI converged after = %d iterations n’,iter)
end
pol_ap = a_grid(pol_ind_ap);
pol_c = (1+r)*a_grid+z_grid’-pol_ap;
%% End timing
toc
%% Figures
figure
plot(a_grid,pol_c(:,1),’linewidth’,2)
hold on
plot(a_grid,pol_c(:,2),’linewidth’,2)
legend(‘Low shock’,’High shock’,’Location’,’NorthWest’)
xlabel(‘asset level’)
ylabel(‘consumption’)
title(‘Consumption Policy Function’)
figure
plot(a_grid,a_grid,’–‘,’linewidth’,2)
hold on
plot(a_grid,pol_ap(:,1),’linewidth’,2)
hold on
plot(a_grid,pol_ap(:,2),’linewidth’,2)
legend(’45 line’,’Low shock’,’High shock’,’Location’,’NorthWest’)
xlabel(‘Current period assets’)
ylabel(‘Next-period assets’)
title(‘Assets Policy Function’) vectorized, performance MATLAB Answers — New Questions

​

Hi All,

I tried to run PREP pipeline on my EEG data. I managed to install the plug-in, and when click OK to run the pipeline, nothing happens (I check this from windows task manager CPU utilization percentage). I am using MATLAB R2023a. I had first installed the latest version (0.56.0) of PREP, but after it did not work, I went to install an earlier version (0.55.4) of it. But, it still is not working. I would appereciate it if you could help me figure out the problem here.Hi All,

I tried to run PREP pipeline on my EEG data. I managed to install the plug-in, and when click OK to run the pipeline, nothing happens (I check this from windows task manager CPU utilization percentage). I am using MATLAB R2023a. I had first installed the latest version (0.56.0) of PREP, but after it did not work, I went to install an earlier version (0.55.4) of it. But, it still is not working. I would appereciate it if you could help me figure out the problem here. Hi All,

I tried to run PREP pipeline on my EEG data. I managed to install the plug-in, and when click OK to run the pipeline, nothing happens (I check this from windows task manager CPU utilization percentage). I am using MATLAB R2023a. I had first installed the latest version (0.56.0) of PREP, but after it did not work, I went to install an earlier version (0.55.4) of it. But, it still is not working. I would appereciate it if you could help me figure out the problem here. eeglab, eeg, prep MATLAB Answers — New Questions

​

Hi,

I have an uiaxes on a GUI (appdesigner) on which I draw a 2D surface with axis equal. When I zoom in by selecting an area, I’d like the area displayed to correspond exactly to the one selected (no more, no less). I had this behavior with MATLAB 2019/2020. However, since upgrading to version 2024, I’ve been able to zoom in, but the initial width/length aspect ratio of the displayed surface remains unchanged. As the background isn’t clickable, elongated surfaces are very difficult to manage.

Here’s an illustration and an attached .fig. I’ve tried changing the daspect and DataAspectRatioMode, but couldn’t get it right.

If anyone has a solution, I’d love to hear from you!

Thanks in advance.Hi,

I have an uiaxes on a GUI (appdesigner) on which I draw a 2D surface with axis equal. When I zoom in by selecting an area, I’d like the area displayed to correspond exactly to the one selected (no more, no less). I had this behavior with MATLAB 2019/2020. However, since upgrading to version 2024, I’ve been able to zoom in, but the initial width/length aspect ratio of the displayed surface remains unchanged. As the background isn’t clickable, elongated surfaces are very difficult to manage.

Here’s an illustration and an attached .fig. I’ve tried changing the daspect and DataAspectRatioMode, but couldn’t get it right.

If anyone has a solution, I’d love to hear from you!

Thanks in advance. Hi,

I have an uiaxes on a GUI (appdesigner) on which I draw a 2D surface with axis equal. When I zoom in by selecting an area, I’d like the area displayed to correspond exactly to the one selected (no more, no less). I had this behavior with MATLAB 2019/2020. However, since upgrading to version 2024, I’ve been able to zoom in, but the initial width/length aspect ratio of the displayed surface remains unchanged. As the background isn’t clickable, elongated surfaces are very difficult to manage.

Here’s an illustration and an attached .fig. I’ve tried changing the daspect and DataAspectRatioMode, but couldn’t get it right.

If anyone has a solution, I’d love to hear from you!

Thanks in advance. aspect ratio, zoom, uiaxes, axes, surface, daspect, aspect, image MATLAB Answers — New Questions

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I am using the UAV Package Delivery Example to fly a drone on top of the buildings in the US City Block. Are the heights of the buildings have a finite value or they are assumed to be inifinity? How do i change the altitude of the UAV such that I can fly above the buildings?I am using the UAV Package Delivery Example to fly a drone on top of the buildings in the US City Block. Are the heights of the buildings have a finite value or they are assumed to be inifinity? How do i change the altitude of the UAV such that I can fly above the buildings? I am using the UAV Package Delivery Example to fly a drone on top of the buildings in the US City Block. Are the heights of the buildings have a finite value or they are assumed to be inifinity? How do i change the altitude of the UAV such that I can fly above the buildings? uav, package, delivery, simulink MATLAB Answers — New Questions

​

Hi,
This is a very basic question but I coudln’t find information regarind my inquery.
I have multiple slider switches on the app designer and each of them has label/name. What I wanted to do is to count the number of slider switches that are on the app, then create the for loop to obtain the slider’s name then store them. Once it is stored then will display on the UITable.
I believe Matlab should have the ability to count them but I couldn’t find a document for it.Hi,
This is a very basic question but I coudln’t find information regarind my inquery.
I have multiple slider switches on the app designer and each of them has label/name. What I wanted to do is to count the number of slider switches that are on the app, then create the for loop to obtain the slider’s name then store them. Once it is stored then will display on the UITable.
I believe Matlab should have the ability to count them but I couldn’t find a document for it. Hi,
This is a very basic question but I coudln’t find information regarind my inquery.
I have multiple slider switches on the app designer and each of them has label/name. What I wanted to do is to count the number of slider switches that are on the app, then create the for loop to obtain the slider’s name then store them. Once it is stored then will display on the UITable.
I believe Matlab should have the ability to count them but I couldn’t find a document for it. appdesigner, matlab MATLAB Answers — New Questions

​

I am trying to change a Gain parameter during my Simulink Real-Time (SLRT) simulation, but I get the following error when I use "setparam":

&gt;&gt; tg = slrealtime;
&gt;&gt; tg.setparam(‘MyModel/Gain’,’Gain’,-2)
Error using slrealtime.Target/throwErrorWithCause
Unable to set ‘Gain’ parameter value on target computer ‘TargetPC1’: Incorrect data type.

Error in slrealtime.Target/setparam (line 99)
        this.throwErrorWithCause(‘slrealtime:target:setparamError’, …

Caused by:
    Error using slrealtime.internal.ParameterTuning/setParamWork
    Incorrect data type.
When I try changing the parameter in the block mask while using the external mode ("Run on Target"), I see warnings such as the following in the MATLAB prompt:

Warning: Parameter precision loss occurred for ‘Gain’ of ‘MyModel/Gain’. The
original value of the parameter, 1, cannot be represented exactly using the run-time data type
sfix32_En31. The value is quantized to 0.9999999995343387. Quantization error occurred with an absolute
difference of 4.656612873077393e-10 and a relative difference of 4.65661287307739e-08%.
Suggested Actions:
• To disable this warning or error for all parameters in the model, set the ‘Detect precision loss’
option to ‘none’. – Apply
• Inspect details in the Parameter Quantization Advisor. – Open
• – Suppress

&gt; In slrealtime/Instrument/dataAvailableFromObserverViaTarget (line 1413)
In slrealtime.Target/dataAvailableFromObserver (line 66)
In slrealtime.Instrument&gt;@(varargin)tg.dataAvailableFromObserver(varargin{:}) (line 1332)
In slrealtime.internal.instrument.StreamingCallBack.NewData (line 25)I am trying to change a Gain parameter during my Simulink Real-Time (SLRT) simulation, but I get the following error when I use "setparam":

&gt;&gt; tg = slrealtime;
&gt;&gt; tg.setparam(‘MyModel/Gain’,’Gain’,-2)
Error using slrealtime.Target/throwErrorWithCause
Unable to set ‘Gain’ parameter value on target computer ‘TargetPC1’: Incorrect data type.

Error in slrealtime.Target/setparam (line 99)
        this.throwErrorWithCause(‘slrealtime:target:setparamError’, …

Caused by:
    Error using slrealtime.internal.ParameterTuning/setParamWork
    Incorrect data type.
When I try changing the parameter in the block mask while using the external mode ("Run on Target"), I see warnings such as the following in the MATLAB prompt:

Warning: Parameter precision loss occurred for ‘Gain’ of ‘MyModel/Gain’. The
original value of the parameter, 1, cannot be represented exactly using the run-time data type
sfix32_En31. The value is quantized to 0.9999999995343387. Quantization error occurred with an absolute
difference of 4.656612873077393e-10 and a relative difference of 4.65661287307739e-08%.
Suggested Actions:
• To disable this warning or error for all parameters in the model, set the ‘Detect precision loss’
option to ‘none’. – Apply
• Inspect details in the Parameter Quantization Advisor. – Open
• – Suppress

&gt; In slrealtime/Instrument/dataAvailableFromObserverViaTarget (line 1413)
In slrealtime.Target/dataAvailableFromObserver (line 66)
In slrealtime.Instrument&gt;@(varargin)tg.dataAvailableFromObserver(varargin{:}) (line 1332)
In slrealtime.internal.instrument.StreamingCallBack.NewData (line 25) I am trying to change a Gain parameter during my Simulink Real-Time (SLRT) simulation, but I get the following error when I use "setparam":

&gt;&gt; tg = slrealtime;
&gt;&gt; tg.setparam(‘MyModel/Gain’,’Gain’,-2)
Error using slrealtime.Target/throwErrorWithCause
Unable to set ‘Gain’ parameter value on target computer ‘TargetPC1’: Incorrect data type.

Error in slrealtime.Target/setparam (line 99)
        this.throwErrorWithCause(‘slrealtime:target:setparamError’, …

Caused by:
    Error using slrealtime.internal.ParameterTuning/setParamWork
    Incorrect data type.
When I try changing the parameter in the block mask while using the external mode ("Run on Target"), I see warnings such as the following in the MATLAB prompt:

Warning: Parameter precision loss occurred for ‘Gain’ of ‘MyModel/Gain’. The
original value of the parameter, 1, cannot be represented exactly using the run-time data type
sfix32_En31. The value is quantized to 0.9999999995343387. Quantization error occurred with an absolute
difference of 4.656612873077393e-10 and a relative difference of 4.65661287307739e-08%.
Suggested Actions:
• To disable this warning or error for all parameters in the model, set the ‘Detect precision loss’
option to ‘none’. – Apply
• Inspect details in the Parameter Quantization Advisor. – Open
• – Suppress

&gt; In slrealtime/Instrument/dataAvailableFromObserverViaTarget (line 1413)
In slrealtime.Target/dataAvailableFromObserver (line 66)
In slrealtime.Instrument&gt;@(varargin)tg.dataAvailableFromObserver(varargin{:}) (line 1332)
In slrealtime.internal.instrument.StreamingCallBack.NewData (line 25) slrt, setparam, getparam MATLAB Answers — New Questions

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Question 3 Create a mask for each frame of this video that isolates the nearly-white flame exiting the rocket. Do this by converting each frame to grayscale. Then label all pixels with intensities less than or equal to 245 as false, and all pixels with intensities greater than 245 as true.
What is the ratio of area of the flame between the last frame and the first? (Use the number of true pixels for area.)
% Load the video file ‘shuttle.avi’
videoFile = ‘shuttle.avi’;
v = VideoReader(videoFile);

% Read the first frame
firstFrame = readFrame(v);

% Move to the last frame
while hasFrame(v)
lastFrame = readFrame(v);
end

% Convert both frames to grayscale
firstGray = rgb2gray(firstFrame);
lastGray = rgb2gray(lastFrame);

% Create masks where pixel intensities > 245 are true, the rest are false
firstMask = firstGray > 245;
lastMask = lastGray > 245;

% Calculate the flame area (number of ‘true’ pixels) for both frames
areaFirst = sum(firstMask(:));
areaLast = sum(lastMask(:));

% Calculate the ratio of the flame area between the last and the first frame
if areaFirst == 0
error(‘No flame detected in the first frame.’);
else
ratio = areaLast / areaFirst;
end

% Display the results
fprintf(‘Area of the flame in the first frame: %d pixelsn’, areaFirst);
fprintf(‘Area of the flame in the last frame: %d pixelsn’, areaLast);
fprintf(‘Ratio of flame area (last frame / first frame): %.2fn’, ratio);

% Optional: Display the original frames and the masks for visual verification
figure;
subplot(2,2,1), imshow(firstFrame), title(‘First Frame’);
subplot(2,2,2), imshow(lastFrame), title(‘Last Frame’);
subplot(2,2,3), imshow(firstMask), title(‘First Frame Mask’);
subplot(2,2,4), imshow(lastMask), title(‘Last Frame Mask’);Question 3 Create a mask for each frame of this video that isolates the nearly-white flame exiting the rocket. Do this by converting each frame to grayscale. Then label all pixels with intensities less than or equal to 245 as false, and all pixels with intensities greater than 245 as true.
What is the ratio of area of the flame between the last frame and the first? (Use the number of true pixels for area.)
% Load the video file ‘shuttle.avi’
videoFile = ‘shuttle.avi’;
v = VideoReader(videoFile);

% Read the first frame
firstFrame = readFrame(v);

% Move to the last frame
while hasFrame(v)
lastFrame = readFrame(v);
end

% Convert both frames to grayscale
firstGray = rgb2gray(firstFrame);
lastGray = rgb2gray(lastFrame);

% Create masks where pixel intensities > 245 are true, the rest are false
firstMask = firstGray > 245;
lastMask = lastGray > 245;

% Calculate the flame area (number of ‘true’ pixels) for both frames
areaFirst = sum(firstMask(:));
areaLast = sum(lastMask(:));

% Calculate the ratio of the flame area between the last and the first frame
if areaFirst == 0
error(‘No flame detected in the first frame.’);
else
ratio = areaLast / areaFirst;
end

% Display the results
fprintf(‘Area of the flame in the first frame: %d pixelsn’, areaFirst);
fprintf(‘Area of the flame in the last frame: %d pixelsn’, areaLast);
fprintf(‘Ratio of flame area (last frame / first frame): %.2fn’, ratio);

% Optional: Display the original frames and the masks for visual verification
figure;
subplot(2,2,1), imshow(firstFrame), title(‘First Frame’);
subplot(2,2,2), imshow(lastFrame), title(‘Last Frame’);
subplot(2,2,3), imshow(firstMask), title(‘First Frame Mask’);
subplot(2,2,4), imshow(lastMask), title(‘Last Frame Mask’); Question 3 Create a mask for each frame of this video that isolates the nearly-white flame exiting the rocket. Do this by converting each frame to grayscale. Then label all pixels with intensities less than or equal to 245 as false, and all pixels with intensities greater than 245 as true.
What is the ratio of area of the flame between the last frame and the first? (Use the number of true pixels for area.)
% Load the video file ‘shuttle.avi’
videoFile = ‘shuttle.avi’;
v = VideoReader(videoFile);

% Read the first frame
firstFrame = readFrame(v);

% Move to the last frame
while hasFrame(v)
lastFrame = readFrame(v);
end

% Convert both frames to grayscale
firstGray = rgb2gray(firstFrame);
lastGray = rgb2gray(lastFrame);

% Create masks where pixel intensities > 245 are true, the rest are false
firstMask = firstGray > 245;
lastMask = lastGray > 245;

% Calculate the flame area (number of ‘true’ pixels) for both frames
areaFirst = sum(firstMask(:));
areaLast = sum(lastMask(:));

% Calculate the ratio of the flame area between the last and the first frame
if areaFirst == 0
error(‘No flame detected in the first frame.’);
else
ratio = areaLast / areaFirst;
end

% Display the results
fprintf(‘Area of the flame in the first frame: %d pixelsn’, areaFirst);
fprintf(‘Area of the flame in the last frame: %d pixelsn’, areaLast);
fprintf(‘Ratio of flame area (last frame / first frame): %.2fn’, ratio);

% Optional: Display the original frames and the masks for visual verification
figure;
subplot(2,2,1), imshow(firstFrame), title(‘First Frame’);
subplot(2,2,2), imshow(lastFrame), title(‘Last Frame’);
subplot(2,2,3), imshow(firstMask), title(‘First Frame Mask’);
subplot(2,2,4), imshow(lastMask), title(‘Last Frame Mask’); image processing MATLAB Answers — New Questions

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