I got missing rate = 1 with this attached code, but my class’s auto grading system says it’s incorrect.
What part of this training and testing steps need(s) to be modified to get the correct missing rate?
[Current Code]
load("C:UsersxxooxOneDriveデスクトップMATLAB worksComputer Vision for Engineering and ScienceC2-MachineLearningForComputerVisionModule 4WoodKnotsGroundTruth.mat");

testPath = overwriteGTruthLocations(gTruthTrain);

imageLabeler(gTruthTrain)

imageLabeler(testPath)

load("C:UsersxxooxOneDriveデスクトップMATLAB worksComputer Vision for Engineering and ScienceC2-MachineLearningForComputerVisionModule 4WoodKnotsGroundTruthTest2.mat");

imdsTest = imageDatastore(testPath);

gTruth.LabelDefinitions
gTruth.DataSource
gTruth.LabelData

objectTrainingData = objectDetectorTrainingData(gTruthTrain)
acfDetector = trainACFObjectDetector(objectTrainingData)

imdsTest = imageDatastore(testPath)
bboxes = detect(acfDetector,imdsTest)

evaluateDetectionMissRate(bboxes,gTruthTest.LabelData)I got missing rate = 1 with this attached code, but my class’s auto grading system says it’s incorrect.
What part of this training and testing steps need(s) to be modified to get the correct missing rate?
[Current Code]
load("C:UsersxxooxOneDriveデスクトップMATLAB worksComputer Vision for Engineering and ScienceC2-MachineLearningForComputerVisionModule 4WoodKnotsGroundTruth.mat");

testPath = overwriteGTruthLocations(gTruthTrain);

imageLabeler(gTruthTrain)

imageLabeler(testPath)

load("C:UsersxxooxOneDriveデスクトップMATLAB worksComputer Vision for Engineering and ScienceC2-MachineLearningForComputerVisionModule 4WoodKnotsGroundTruthTest2.mat");

imdsTest = imageDatastore(testPath);

gTruth.LabelDefinitions
gTruth.DataSource
gTruth.LabelData

objectTrainingData = objectDetectorTrainingData(gTruthTrain)
acfDetector = trainACFObjectDetector(objectTrainingData)

imdsTest = imageDatastore(testPath)
bboxes = detect(acfDetector,imdsTest)

evaluateDetectionMissRate(bboxes,gTruthTest.LabelData) I got missing rate = 1 with this attached code, but my class’s auto grading system says it’s incorrect.
What part of this training and testing steps need(s) to be modified to get the correct missing rate?
[Current Code]
load("C:UsersxxooxOneDriveデスクトップMATLAB worksComputer Vision for Engineering and ScienceC2-MachineLearningForComputerVisionModule 4WoodKnotsGroundTruth.mat");

testPath = overwriteGTruthLocations(gTruthTrain);

imageLabeler(gTruthTrain)

imageLabeler(testPath)

load("C:UsersxxooxOneDriveデスクトップMATLAB worksComputer Vision for Engineering and ScienceC2-MachineLearningForComputerVisionModule 4WoodKnotsGroundTruthTest2.mat");

imdsTest = imageDatastore(testPath);

gTruth.LabelDefinitions
gTruth.DataSource
gTruth.LabelData

objectTrainingData = objectDetectorTrainingData(gTruthTrain)
acfDetector = trainACFObjectDetector(objectTrainingData)

imdsTest = imageDatastore(testPath)
bboxes = detect(acfDetector,imdsTest)

evaluateDetectionMissRate(bboxes,gTruthTest.LabelData) matlab, machine learning, testing, training, image labeling MATLAB Answers — New Questions

​

I’m encountering a compile-time size assumption violation error when attempting to run my MATLAB code with code generation.
I am working with a column vector ffeout_lp which has the dimensions 63680×1, and an integer oversample which is set to 40. I am quite new to scripting in simulink, so what I have tried so far may not have been optimal. The ‘coder.varsize’ statements above are to initialize space for arrays, as to my knowledge dynamic initializing is not permitted for code generation. How do I go about this compile time error?
Thanks!I’m encountering a compile-time size assumption violation error when attempting to run my MATLAB code with code generation.
I am working with a column vector ffeout_lp which has the dimensions 63680×1, and an integer oversample which is set to 40. I am quite new to scripting in simulink, so what I have tried so far may not have been optimal. The ‘coder.varsize’ statements above are to initialize space for arrays, as to my knowledge dynamic initializing is not permitted for code generation. How do I go about this compile time error?
Thanks! I’m encountering a compile-time size assumption violation error when attempting to run my MATLAB code with code generation.
I am working with a column vector ffeout_lp which has the dimensions 63680×1, and an integer oversample which is set to 40. I am quite new to scripting in simulink, so what I have tried so far may not have been optimal. The ‘coder.varsize’ statements above are to initialize space for arrays, as to my knowledge dynamic initializing is not permitted for code generation. How do I go about this compile time error?
Thanks! simulink, compile time size assumption, matlab function, signal processing, code generation MATLAB Answers — New Questions

​

I’m using the train function to run a Reinforcement Learning training using a PPO agent, with a rlSimulinkEnv object defining the environment.
Regarding the rlTrainingOptions, I’m using the default option for "SimulationStorageType", which is "memory". So I expect (if I understand correctly) that all the simulation data should be saved to RAM, not to disk, whenever possible. However I noticed that Simulink is actually saving the simulation data into .dmr files stored in my TEMPDIR.
I also tried to disable the Simulink logging to TEMPDIR directly from the Simulink Data Inspector, setting the Record Mode to "View during simulation only". However, as soon as I start the Reinforcement Learning Training, the simulation data are still saved as .dmr files in my TEMPDIR.
Is there a way to actually avoid saving .dmr files to TEMPDIR when training in a Simulink environment, and instead using only RAM to store the simulation data as long as the Matlab session is active?

I’m using version R2024aI’m using the train function to run a Reinforcement Learning training using a PPO agent, with a rlSimulinkEnv object defining the environment.
Regarding the rlTrainingOptions, I’m using the default option for "SimulationStorageType", which is "memory". So I expect (if I understand correctly) that all the simulation data should be saved to RAM, not to disk, whenever possible. However I noticed that Simulink is actually saving the simulation data into .dmr files stored in my TEMPDIR.
I also tried to disable the Simulink logging to TEMPDIR directly from the Simulink Data Inspector, setting the Record Mode to "View during simulation only". However, as soon as I start the Reinforcement Learning Training, the simulation data are still saved as .dmr files in my TEMPDIR.
Is there a way to actually avoid saving .dmr files to TEMPDIR when training in a Simulink environment, and instead using only RAM to store the simulation data as long as the Matlab session is active?

I’m using version R2024a I’m using the train function to run a Reinforcement Learning training using a PPO agent, with a rlSimulinkEnv object defining the environment.
Regarding the rlTrainingOptions, I’m using the default option for "SimulationStorageType", which is "memory". So I expect (if I understand correctly) that all the simulation data should be saved to RAM, not to disk, whenever possible. However I noticed that Simulink is actually saving the simulation data into .dmr files stored in my TEMPDIR.
I also tried to disable the Simulink logging to TEMPDIR directly from the Simulink Data Inspector, setting the Record Mode to "View during simulation only". However, as soon as I start the Reinforcement Learning Training, the simulation data are still saved as .dmr files in my TEMPDIR.
Is there a way to actually avoid saving .dmr files to TEMPDIR when training in a Simulink environment, and instead using only RAM to store the simulation data as long as the Matlab session is active?

I’m using version R2024a simulink, data MATLAB Answers — New Questions

​

I have Simulink models that are different versions of a controller. The versions change static data, used throughout the model, and some blocks in the model. I am wondering if there is a way to create one Simulink model that can be changed between all these versions. A similar functionality exists in standard Simulink functions, such as with the PID Controller when you switch the form between Parallel and Ideal or the Logical Operator switching between AND and OR. Has anyone created or found similar functionality for making Simulink models with different versions?I have Simulink models that are different versions of a controller. The versions change static data, used throughout the model, and some blocks in the model. I am wondering if there is a way to create one Simulink model that can be changed between all these versions. A similar functionality exists in standard Simulink functions, such as with the PID Controller when you switch the form between Parallel and Ideal or the Logical Operator switching between AND and OR. Has anyone created or found similar functionality for making Simulink models with different versions? I have Simulink models that are different versions of a controller. The versions change static data, used throughout the model, and some blocks in the model. I am wondering if there is a way to create one Simulink model that can be changed between all these versions. A similar functionality exists in standard Simulink functions, such as with the PID Controller when you switch the form between Parallel and Ideal or the Logical Operator switching between AND and OR. Has anyone created or found similar functionality for making Simulink models with different versions? simulink, version, static data MATLAB Answers — New Questions

​

% Copyright 2018, Nicholas Giacoboni
%
% Redistribution and use in source and binary forms, with or without
% modification, are permitted provided that the following conditions are met:
%
% 1. Redistributions of source code must retain the above copyright
% notice, this list of conditions and the following disclaimer.
%
% 2. Redistributions in binary form must reproduce the above copyright
% notice, this list of conditions and the following disclaimer in the
% documentation and/or other materials provided with the distribution.
%
% THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
% AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
% WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED.
% IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT,
% INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
% NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
% PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY,
% WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
% ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY
% OF SUCH DAMAGE.
%**************************************************************************
classdef basic_HX711 < matlabshared.addon.LibraryBase
properties(Access = protected)
Pins %Data and Clock pin
end
properties (Access = private, Constant = true)
READ_HX711 = hex2dec(’01’)
end
properties(Access = protected, Constant = true)
LibraryName = ‘basicHX711/basic_HX711’
DependentLibraries = {}
LibraryHeaderFiles = {}
CppHeaderFile = fullfile(arduinoio.FilePath(mfilename(‘fullpath’)), ‘src’, ‘basic_HX711.h’)
CppClassName = ‘basic_HX711’
end

methods(Hidden, Access = public)
function obj = basic_HX711(parentObj,inputPins)
obj.Parent = parentObj;
obj.Pins = getTerminalsFromPins(obj.Parent,inputPins);
configurePin(parentObj,inputPins{2},’DigitalInput’); % Data Pin
configurePin(parentObj,inputPins{3},’DigitalOutput’); % Clock Pin
end
end

methods(Access = public)
function force = read_HX711(obj)
cmdID = obj.READ_HX711;
inputs = obj.Pins;
value = sendCommand(obj, obj.LibraryName, cmdID, inputs);
value(3)=bitshift(value(3),16);
value(2)=bitshift(value(2),8);
force = bitor(value(3),bitor(value(2),value(1)));
end
end
end
%**************************************************************************

listArduinoLibraries

ans =

12×1 cell array

{‘APDS9960’ }
{‘Adafruit/MotorShieldV2’}
{‘CAN’ }
{‘I2C’ }
{‘MotorCarrier’ }
{‘RotaryEncoder’ }
{‘SPI’ }
{‘Serial’ }
{‘Servo’ }
{‘ShiftRegister’ }
{‘Ultrasonic’ }
{‘basicHX711/basic_HX711’}

a = arduino(‘COM6’, ‘Uno’, ‘Libraries’,’basicHX711/basic_HX711′,’ForceBuildOn’,’True’);
>> Loadcell = addon(a,’basicHX711/basic_HX711′,{‘D2′,’D3’})
Index exceeds the number of array elements. Index must not exceed 2.
%**************************************************************************

I’m currently working on creating a custom library for the HX711 sensor, but I’m encountering an issue with the ‘Index exceeds the number of array elements. Index must not exceed 2.’ I hope someone can help me troubleshoot.% Copyright 2018, Nicholas Giacoboni
%
% Redistribution and use in source and binary forms, with or without
% modification, are permitted provided that the following conditions are met:
%
% 1. Redistributions of source code must retain the above copyright
% notice, this list of conditions and the following disclaimer.
%
% 2. Redistributions in binary form must reproduce the above copyright
% notice, this list of conditions and the following disclaimer in the
% documentation and/or other materials provided with the distribution.
%
% THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
% AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
% WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED.
% IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT,
% INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
% NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
% PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY,
% WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
% ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY
% OF SUCH DAMAGE.
%**************************************************************************
classdef basic_HX711 < matlabshared.addon.LibraryBase
properties(Access = protected)
Pins %Data and Clock pin
end
properties (Access = private, Constant = true)
READ_HX711 = hex2dec(’01’)
end
properties(Access = protected, Constant = true)
LibraryName = ‘basicHX711/basic_HX711’
DependentLibraries = {}
LibraryHeaderFiles = {}
CppHeaderFile = fullfile(arduinoio.FilePath(mfilename(‘fullpath’)), ‘src’, ‘basic_HX711.h’)
CppClassName = ‘basic_HX711’
end

methods(Hidden, Access = public)
function obj = basic_HX711(parentObj,inputPins)
obj.Parent = parentObj;
obj.Pins = getTerminalsFromPins(obj.Parent,inputPins);
configurePin(parentObj,inputPins{2},’DigitalInput’); % Data Pin
configurePin(parentObj,inputPins{3},’DigitalOutput’); % Clock Pin
end
end

methods(Access = public)
function force = read_HX711(obj)
cmdID = obj.READ_HX711;
inputs = obj.Pins;
value = sendCommand(obj, obj.LibraryName, cmdID, inputs);
value(3)=bitshift(value(3),16);
value(2)=bitshift(value(2),8);
force = bitor(value(3),bitor(value(2),value(1)));
end
end
end
%**************************************************************************

listArduinoLibraries

ans =

12×1 cell array

{‘APDS9960’ }
{‘Adafruit/MotorShieldV2’}
{‘CAN’ }
{‘I2C’ }
{‘MotorCarrier’ }
{‘RotaryEncoder’ }
{‘SPI’ }
{‘Serial’ }
{‘Servo’ }
{‘ShiftRegister’ }
{‘Ultrasonic’ }
{‘basicHX711/basic_HX711’}

a = arduino(‘COM6’, ‘Uno’, ‘Libraries’,’basicHX711/basic_HX711′,’ForceBuildOn’,’True’);
>> Loadcell = addon(a,’basicHX711/basic_HX711′,{‘D2′,’D3’})
Index exceeds the number of array elements. Index must not exceed 2.
%**************************************************************************

I’m currently working on creating a custom library for the HX711 sensor, but I’m encountering an issue with the ‘Index exceeds the number of array elements. Index must not exceed 2.’ I hope someone can help me troubleshoot. % Copyright 2018, Nicholas Giacoboni
%
% Redistribution and use in source and binary forms, with or without
% modification, are permitted provided that the following conditions are met:
%
% 1. Redistributions of source code must retain the above copyright
% notice, this list of conditions and the following disclaimer.
%
% 2. Redistributions in binary form must reproduce the above copyright
% notice, this list of conditions and the following disclaimer in the
% documentation and/or other materials provided with the distribution.
%
% THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
% AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
% WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED.
% IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT,
% INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
% NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
% PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY,
% WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
% ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY
% OF SUCH DAMAGE.
%**************************************************************************
classdef basic_HX711 < matlabshared.addon.LibraryBase
properties(Access = protected)
Pins %Data and Clock pin
end
properties (Access = private, Constant = true)
READ_HX711 = hex2dec(’01’)
end
properties(Access = protected, Constant = true)
LibraryName = ‘basicHX711/basic_HX711’
DependentLibraries = {}
LibraryHeaderFiles = {}
CppHeaderFile = fullfile(arduinoio.FilePath(mfilename(‘fullpath’)), ‘src’, ‘basic_HX711.h’)
CppClassName = ‘basic_HX711’
end

methods(Hidden, Access = public)
function obj = basic_HX711(parentObj,inputPins)
obj.Parent = parentObj;
obj.Pins = getTerminalsFromPins(obj.Parent,inputPins);
configurePin(parentObj,inputPins{2},’DigitalInput’); % Data Pin
configurePin(parentObj,inputPins{3},’DigitalOutput’); % Clock Pin
end
end

methods(Access = public)
function force = read_HX711(obj)
cmdID = obj.READ_HX711;
inputs = obj.Pins;
value = sendCommand(obj, obj.LibraryName, cmdID, inputs);
value(3)=bitshift(value(3),16);
value(2)=bitshift(value(2),8);
force = bitor(value(3),bitor(value(2),value(1)));
end
end
end
%**************************************************************************

listArduinoLibraries

ans =

12×1 cell array

{‘APDS9960’ }
{‘Adafruit/MotorShieldV2’}
{‘CAN’ }
{‘I2C’ }
{‘MotorCarrier’ }
{‘RotaryEncoder’ }
{‘SPI’ }
{‘Serial’ }
{‘Servo’ }
{‘ShiftRegister’ }
{‘Ultrasonic’ }
{‘basicHX711/basic_HX711’}

a = arduino(‘COM6’, ‘Uno’, ‘Libraries’,’basicHX711/basic_HX711′,’ForceBuildOn’,’True’);
>> Loadcell = addon(a,’basicHX711/basic_HX711′,{‘D2′,’D3’})
Index exceeds the number of array elements. Index must not exceed 2.
%**************************************************************************

I’m currently working on creating a custom library for the HX711 sensor, but I’m encountering an issue with the ‘Index exceeds the number of array elements. Index must not exceed 2.’ I hope someone can help me troubleshoot. matlab addon, hx711 MATLAB Answers — New Questions

​

I am trying to plot a pcolor plot for Sea surface temperature data set and to represent the coastaline in the plot i have downloadwed global coastline shapefile and when i am using geoshow to read and overlay on the pcolor plot it is not matching perfectlyI am trying to plot a pcolor plot for Sea surface temperature data set and to represent the coastaline in the plot i have downloadwed global coastline shapefile and when i am using geoshow to read and overlay on the pcolor plot it is not matching perfectly I am trying to plot a pcolor plot for Sea surface temperature data set and to represent the coastaline in the plot i have downloadwed global coastline shapefile and when i am using geoshow to read and overlay on the pcolor plot it is not matching perfectly geoshow, shapefile, pcolor plot MATLAB Answers — New Questions

​

sltest.harness.create(‘Harness_Test’,’Source’,’Signal editor’,’Sink’,’Outport’);

error:
Expected input to match one of these values:

‘Inport’, ‘Signal Editor’, ‘From Workspace’, ‘From File’, ‘Test Sequence’, ‘Chart’, ‘Constant’, ‘Ground’, ‘None’, ‘Custom’

The input, ‘Signal Builder’, did not match any of the valid values.

This means I can’t assign ‘Signal Builder’ to ‘Source’, matlab version is 2022b, but there is no error in 2022a(Tested many times).
Is this issue related to matlab version and how to solve it? Still need to use ‘Signal Builder’.sltest.harness.create(‘Harness_Test’,’Source’,’Signal editor’,’Sink’,’Outport’);

error:
Expected input to match one of these values:

‘Inport’, ‘Signal Editor’, ‘From Workspace’, ‘From File’, ‘Test Sequence’, ‘Chart’, ‘Constant’, ‘Ground’, ‘None’, ‘Custom’

The input, ‘Signal Builder’, did not match any of the valid values.

This means I can’t assign ‘Signal Builder’ to ‘Source’, matlab version is 2022b, but there is no error in 2022a(Tested many times).
Is this issue related to matlab version and how to solve it? Still need to use ‘Signal Builder’. sltest.harness.create(‘Harness_Test’,’Source’,’Signal editor’,’Sink’,’Outport’);

error:
Expected input to match one of these values:

‘Inport’, ‘Signal Editor’, ‘From Workspace’, ‘From File’, ‘Test Sequence’, ‘Chart’, ‘Constant’, ‘Ground’, ‘None’, ‘Custom’

The input, ‘Signal Builder’, did not match any of the valid values.

This means I can’t assign ‘Signal Builder’ to ‘Source’, matlab version is 2022b, but there is no error in 2022a(Tested many times).
Is this issue related to matlab version and how to solve it? Still need to use ‘Signal Builder’. sltest.harness.create MATLAB Answers — New Questions

​

Hi,everyone.I got a problem of doing numeric integral on the matrix H (which is at the bottom part of my code).
The code works fine but I would like to speed it up? Am I doing something stupid to make it slow?Thanks for your help.
clear

L=0.1;
section=50;
a=L./2;
b=L./section;
v=3e8;
f1=1.2e9;
f2=4.8e9;
f3=7.5e9;
f4=10e9;
w1=(2.*pi.*f1);
w2=(2.*pi.*f2);
w3=(2.*pi.*f3);
w4=(2.*pi.*f4);
Z01=50;
Z02=75;
a0=(log(Z02./Z01))./(2..*L);
T=a;
Ub=a;
Lb=-a;
ub=a;
lb=-a;
k=5;

syms x y w m

%%% Revised Algorithm

l=0:(2.*L)./(2.*k):L % Approximate sections dvided

% Approximate terms
sec=numel(l)-1;

Z1=[50 55 60 65 70 80 75]; % Set the impedance of the end of each sections

ac=10; % Choose the accuracy of dividing a section

for s=1:1:sec

c=(l(s+1)-l(s))./ac;

app_cos=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*cos(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

app_sin=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*sin(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

amp=0.75;

Ga(s)=(app_cos-(1i.*amp.*app_sin));

end

Ta=(1-(abs(Ga(1:1:end)).^2));

%%% Turn closed loop terms into matrix presentation

for t=1:1:sec

for r=1:1:sec

E(t,r)=Ga(t).*Ga(r);

end

end

E=triu(E,1);

for t=1:1:sec

for r=1:1:sec

D(t,r)=abs(E(r,t));

end

end

C1=sum(D);

% Construct approximate loop terms

for t=1:1:sec

D1(t)=prod(Ta(1:t))./(1+(sum(C1(1:t))));

end

D=[1 D1(2:end)];

Dm=transpose(D).*D;

% Cmn

m=transpose(1:5);

n=1:5;

g=(v.^2.*w.^2.*cos((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x).*cos(20.*sym(pi).*n.*y) + 10.*sym(pi).*m.*v.^3.*w.*sin(20.*sym(pi).*m.*x).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*n.*y) – 10.*sym(pi).*n.*v.^3.*w.*sin(20.*sym(pi).*n.*y).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x) + 100.*m.*n.*v.^4.*sym(pi).^2.*sin(20.*sym(pi).*m.*x).*sin(20.*sym(pi).*n.*y).*cos((2.*w.*(- y + x))./v))./(4.*(w.^2 – 100.*sym(pi).^2.*m.^2.*v.^2).*(w.^2 – 100.*sym(pi).^2.*n.^2.*v.^2));

for p=1:numel(l)-1

x1=l(p);

x2=l(p+1);

for q=1:numel(l)-1

y1=l(q);

y2=l(p+1);

Ht{p,q}=matlabFunction(((subs(g,[x y],[x2 y2])-subs(g,[x y],[x1 y1]))*Dm(p,q)));

end

end

Hi=Ht{1};

for m=2:1:numel(Ht)

Hi=Hi+vpa(Ht{m},3);

end

H=integral(Hi,w1,w2,"ArrayValued",true)Hi,everyone.I got a problem of doing numeric integral on the matrix H (which is at the bottom part of my code).
The code works fine but I would like to speed it up? Am I doing something stupid to make it slow?Thanks for your help.
clear

L=0.1;
section=50;
a=L./2;
b=L./section;
v=3e8;
f1=1.2e9;
f2=4.8e9;
f3=7.5e9;
f4=10e9;
w1=(2.*pi.*f1);
w2=(2.*pi.*f2);
w3=(2.*pi.*f3);
w4=(2.*pi.*f4);
Z01=50;
Z02=75;
a0=(log(Z02./Z01))./(2..*L);
T=a;
Ub=a;
Lb=-a;
ub=a;
lb=-a;
k=5;

syms x y w m

%%% Revised Algorithm

l=0:(2.*L)./(2.*k):L % Approximate sections dvided

% Approximate terms
sec=numel(l)-1;

Z1=[50 55 60 65 70 80 75]; % Set the impedance of the end of each sections

ac=10; % Choose the accuracy of dividing a section

for s=1:1:sec

c=(l(s+1)-l(s))./ac;

app_cos=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*cos(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

app_sin=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*sin(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

amp=0.75;

Ga(s)=(app_cos-(1i.*amp.*app_sin));

end

Ta=(1-(abs(Ga(1:1:end)).^2));

%%% Turn closed loop terms into matrix presentation

for t=1:1:sec

for r=1:1:sec

E(t,r)=Ga(t).*Ga(r);

end

end

E=triu(E,1);

for t=1:1:sec

for r=1:1:sec

D(t,r)=abs(E(r,t));

end

end

C1=sum(D);

% Construct approximate loop terms

for t=1:1:sec

D1(t)=prod(Ta(1:t))./(1+(sum(C1(1:t))));

end

D=[1 D1(2:end)];

Dm=transpose(D).*D;

% Cmn

m=transpose(1:5);

n=1:5;

g=(v.^2.*w.^2.*cos((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x).*cos(20.*sym(pi).*n.*y) + 10.*sym(pi).*m.*v.^3.*w.*sin(20.*sym(pi).*m.*x).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*n.*y) – 10.*sym(pi).*n.*v.^3.*w.*sin(20.*sym(pi).*n.*y).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x) + 100.*m.*n.*v.^4.*sym(pi).^2.*sin(20.*sym(pi).*m.*x).*sin(20.*sym(pi).*n.*y).*cos((2.*w.*(- y + x))./v))./(4.*(w.^2 – 100.*sym(pi).^2.*m.^2.*v.^2).*(w.^2 – 100.*sym(pi).^2.*n.^2.*v.^2));

for p=1:numel(l)-1

x1=l(p);

x2=l(p+1);

for q=1:numel(l)-1

y1=l(q);

y2=l(p+1);

Ht{p,q}=matlabFunction(((subs(g,[x y],[x2 y2])-subs(g,[x y],[x1 y1]))*Dm(p,q)));

end

end

Hi=Ht{1};

for m=2:1:numel(Ht)

Hi=Hi+vpa(Ht{m},3);

end

H=integral(Hi,w1,w2,"ArrayValued",true) Hi,everyone.I got a problem of doing numeric integral on the matrix H (which is at the bottom part of my code).
The code works fine but I would like to speed it up? Am I doing something stupid to make it slow?Thanks for your help.
clear

L=0.1;
section=50;
a=L./2;
b=L./section;
v=3e8;
f1=1.2e9;
f2=4.8e9;
f3=7.5e9;
f4=10e9;
w1=(2.*pi.*f1);
w2=(2.*pi.*f2);
w3=(2.*pi.*f3);
w4=(2.*pi.*f4);
Z01=50;
Z02=75;
a0=(log(Z02./Z01))./(2..*L);
T=a;
Ub=a;
Lb=-a;
ub=a;
lb=-a;
k=5;

syms x y w m

%%% Revised Algorithm

l=0:(2.*L)./(2.*k):L % Approximate sections dvided

% Approximate terms
sec=numel(l)-1;

Z1=[50 55 60 65 70 80 75]; % Set the impedance of the end of each sections

ac=10; % Choose the accuracy of dividing a section

for s=1:1:sec

c=(l(s+1)-l(s))./ac;

app_cos=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*cos(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

app_sin=0.5./ac.*log(Z1(s+1)./Z1(s)).*sin(ac./2.*(w./v).*c).*sin(((ac+1)./2).*(w./v).*c)./sin(((c.*w)./(2.*v)));

amp=0.75;

Ga(s)=(app_cos-(1i.*amp.*app_sin));

end

Ta=(1-(abs(Ga(1:1:end)).^2));

%%% Turn closed loop terms into matrix presentation

for t=1:1:sec

for r=1:1:sec

E(t,r)=Ga(t).*Ga(r);

end

end

E=triu(E,1);

for t=1:1:sec

for r=1:1:sec

D(t,r)=abs(E(r,t));

end

end

C1=sum(D);

% Construct approximate loop terms

for t=1:1:sec

D1(t)=prod(Ta(1:t))./(1+(sum(C1(1:t))));

end

D=[1 D1(2:end)];

Dm=transpose(D).*D;

% Cmn

m=transpose(1:5);

n=1:5;

g=(v.^2.*w.^2.*cos((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x).*cos(20.*sym(pi).*n.*y) + 10.*sym(pi).*m.*v.^3.*w.*sin(20.*sym(pi).*m.*x).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*n.*y) – 10.*sym(pi).*n.*v.^3.*w.*sin(20.*sym(pi).*n.*y).*sin((2.*w.*(- y + x))./v).*cos(20.*sym(pi).*m.*x) + 100.*m.*n.*v.^4.*sym(pi).^2.*sin(20.*sym(pi).*m.*x).*sin(20.*sym(pi).*n.*y).*cos((2.*w.*(- y + x))./v))./(4.*(w.^2 – 100.*sym(pi).^2.*m.^2.*v.^2).*(w.^2 – 100.*sym(pi).^2.*n.^2.*v.^2));

for p=1:numel(l)-1

x1=l(p);

x2=l(p+1);

for q=1:numel(l)-1

y1=l(q);

y2=l(p+1);

Ht{p,q}=matlabFunction(((subs(g,[x y],[x2 y2])-subs(g,[x y],[x1 y1]))*Dm(p,q)));

end

end

Hi=Ht{1};

for m=2:1:numel(Ht)

Hi=Hi+vpa(Ht{m},3);

end

H=integral(Hi,w1,w2,"ArrayValued",true) integral MATLAB Answers — New Questions

​

Hello,
I am wondering if there is a function to identify and fill outliers in a 2d matrix. I know that the filloutliers function does work on a matrix, but it only works on each column separately. I am hoping to look at outliers in both directions.
There is a function fillmissing2 that does exactly this, but for missing values rather than outliers. Does anyone know if there is such a function? Or perhaps a workaround?
Thanks in advance,
SeanHello,
I am wondering if there is a function to identify and fill outliers in a 2d matrix. I know that the filloutliers function does work on a matrix, but it only works on each column separately. I am hoping to look at outliers in both directions.
There is a function fillmissing2 that does exactly this, but for missing values rather than outliers. Does anyone know if there is such a function? Or perhaps a workaround?
Thanks in advance,
Sean Hello,
I am wondering if there is a function to identify and fill outliers in a 2d matrix. I know that the filloutliers function does work on a matrix, but it only works on each column separately. I am hoping to look at outliers in both directions.
There is a function fillmissing2 that does exactly this, but for missing values rather than outliers. Does anyone know if there is such a function? Or perhaps a workaround?
Thanks in advance,
Sean filloutliers, matrix, interpolation MATLAB Answers — New Questions

​

load post.mat
x1 = decomposition(CA,’qr’)b_f;

[qq2,rr2,pp2] = qr(CA);
x2= pp2 * (rr2(qq2’*b_f));

[qq3,rr3,pp3] = qr(CA,"econ","vector");
x3(pp3,:) = rr3(qq3’*b_f);

any(x1-x2~=0)
any(x3-x2~=0)

I know that the CA matrix is ill-conditioned. But that doesn’t explain the difference in solution, right?
Also, solving the system using decomposition(CA,’lu’) and lu(CA) produce the same results. So why not the ‘qr’ pair?load post.mat
x1 = decomposition(CA,’qr’)b_f;

[qq2,rr2,pp2] = qr(CA);
x2= pp2 * (rr2(qq2’*b_f));

[qq3,rr3,pp3] = qr(CA,"econ","vector");
x3(pp3,:) = rr3(qq3’*b_f);

any(x1-x2~=0)
any(x3-x2~=0)

I know that the CA matrix is ill-conditioned. But that doesn’t explain the difference in solution, right?
Also, solving the system using decomposition(CA,’lu’) and lu(CA) produce the same results. So why not the ‘qr’ pair? load post.mat
x1 = decomposition(CA,’qr’)b_f;

[qq2,rr2,pp2] = qr(CA);
x2= pp2 * (rr2(qq2’*b_f));

[qq3,rr3,pp3] = qr(CA,"econ","vector");
x3(pp3,:) = rr3(qq3’*b_f);

any(x1-x2~=0)
any(x3-x2~=0)

I know that the CA matrix is ill-conditioned. But that doesn’t explain the difference in solution, right?
Also, solving the system using decomposition(CA,’lu’) and lu(CA) produce the same results. So why not the ‘qr’ pair? qr decomposition, linear algebra, linear system MATLAB Answers — New Questions

​