I need more precision than offered by the standard double data type in MATLAB. I am looking for a quadruple precision in MATLAB.I need more precision than offered by the standard double data type in MATLAB. I am looking for a quadruple precision in MATLAB. I need more precision than offered by the standard double data type in MATLAB. I am looking for a quadruple precision in MATLAB. 128-bit, quadruple_precision, float, double MATLAB Answers — New Questions

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I have a Simulink R2018b model which contains an S-function. The S-function is a MEX S-Function. When I try to build the model, I get the following error message:

#error Must define one of RT, NRT, MATLAB_MEX_FILE, SL_INTERNAL, or FIPXT_SHARED_MODULE
I am using Visual Studio 2017 as a compiler.I have a Simulink R2018b model which contains an S-function. The S-function is a MEX S-Function. When I try to build the model, I get the following error message:

#error Must define one of RT, NRT, MATLAB_MEX_FILE, SL_INTERNAL, or FIPXT_SHARED_MODULE
I am using Visual Studio 2017 as a compiler. I have a Simulink R2018b model which contains an S-function. The S-function is a MEX S-Function. When I try to build the model, I get the following error message:

#error Must define one of RT, NRT, MATLAB_MEX_FILE, SL_INTERNAL, or FIPXT_SHARED_MODULE
I am using Visual Studio 2017 as a compiler. external, ide, s-function, error, inlining, tlc, non-lined MATLAB Answers — New Questions

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We have an image that represents data that only makes sense when it is analyzed in numerical format.

Specifically, the data needs to be analyzed as a function of radius as shown below
Im interested specifically in hte max and min values within each of the defined areas with respect to the center point.

Ive been looking at a few examples online and it seems this should work when you pull the data from a mask.
However, the issue is that I seem to be getting values that are not realistic.
How to get pixel value inside a circle – MATLAB Answers – MATLAB Central (mathworks.com)
how to draw circle in an image? – MATLAB Answers – MATLAB Central (mathworks.com)

this is what I am doing
clear
img = double(imread(‘img121.jpg’));; %no filtration

img = -(0.0316*img) +8.3; % we did this as we cant calibeate the film, we scan the same film over and over and it changes by 80pixels
img = imrotate(img, 90);
img = imgaussfilt(img ,1.5);
figure, imagesc(img )
axis image
height2 = 3.6
caxis([0 height2])
colorbar
title(‘ ‘)
impixelinfo
%# make sure the image doesn’t disappear if we plot something else
hold on
%https://www.mathworks.com/matlabcentral/answers/1931825-how-to-get-pixel-value-inside-a-circle
%below looks like what we want
%https://www.mathworks.com/matlabcentral/answers/1931825-how-to-get-pixel-value-inside-a-circle
%# define points (in matrix coordinates)
%3"
cpx = 2050;
cpy = 2020;
inchlist = [12,10.5,9,7.5,6,4.5];
%draw lines on heel axis
for n=1:size(inchlist,2)
inch= inchlist(n)/4;
hcirc = drawcircle(‘Center’,[2050,2020],’Radius’,inch*590,’StripeColor’,’red’);
mask1 = hcirc.createMask;
maxval = (max(img(mask1)));
minval = (min(img(mask1)));
uniformity = maxval/minval
% p1 = [cpy-100,cpx+inch*590];
end

Even after getting this max and min value, I will need to remove 10 to get rid of noise. Extra credit if you can point me to a soluton for that too.

thank youWe have an image that represents data that only makes sense when it is analyzed in numerical format.

Specifically, the data needs to be analyzed as a function of radius as shown below
Im interested specifically in hte max and min values within each of the defined areas with respect to the center point.

Ive been looking at a few examples online and it seems this should work when you pull the data from a mask.
However, the issue is that I seem to be getting values that are not realistic.
How to get pixel value inside a circle – MATLAB Answers – MATLAB Central (mathworks.com)
how to draw circle in an image? – MATLAB Answers – MATLAB Central (mathworks.com)

this is what I am doing
clear
img = double(imread(‘img121.jpg’));; %no filtration

img = -(0.0316*img) +8.3; % we did this as we cant calibeate the film, we scan the same film over and over and it changes by 80pixels
img = imrotate(img, 90);
img = imgaussfilt(img ,1.5);
figure, imagesc(img )
axis image
height2 = 3.6
caxis([0 height2])
colorbar
title(‘ ‘)
impixelinfo
%# make sure the image doesn’t disappear if we plot something else
hold on
%https://www.mathworks.com/matlabcentral/answers/1931825-how-to-get-pixel-value-inside-a-circle
%below looks like what we want
%https://www.mathworks.com/matlabcentral/answers/1931825-how-to-get-pixel-value-inside-a-circle
%# define points (in matrix coordinates)
%3"
cpx = 2050;
cpy = 2020;
inchlist = [12,10.5,9,7.5,6,4.5];
%draw lines on heel axis
for n=1:size(inchlist,2)
inch= inchlist(n)/4;
hcirc = drawcircle(‘Center’,[2050,2020],’Radius’,inch*590,’StripeColor’,’red’);
mask1 = hcirc.createMask;
maxval = (max(img(mask1)));
minval = (min(img(mask1)));
uniformity = maxval/minval
% p1 = [cpy-100,cpx+inch*590];
end

Even after getting this max and min value, I will need to remove 10 to get rid of noise. Extra credit if you can point me to a soluton for that too.

thank you We have an image that represents data that only makes sense when it is analyzed in numerical format.

Specifically, the data needs to be analyzed as a function of radius as shown below
Im interested specifically in hte max and min values within each of the defined areas with respect to the center point.

Ive been looking at a few examples online and it seems this should work when you pull the data from a mask.
However, the issue is that I seem to be getting values that are not realistic.
How to get pixel value inside a circle – MATLAB Answers – MATLAB Central (mathworks.com)
how to draw circle in an image? – MATLAB Answers – MATLAB Central (mathworks.com)

this is what I am doing
clear
img = double(imread(‘img121.jpg’));; %no filtration

img = -(0.0316*img) +8.3; % we did this as we cant calibeate the film, we scan the same film over and over and it changes by 80pixels
img = imrotate(img, 90);
img = imgaussfilt(img ,1.5);
figure, imagesc(img )
axis image
height2 = 3.6
caxis([0 height2])
colorbar
title(‘ ‘)
impixelinfo
%# make sure the image doesn’t disappear if we plot something else
hold on
%https://www.mathworks.com/matlabcentral/answers/1931825-how-to-get-pixel-value-inside-a-circle
%below looks like what we want
%https://www.mathworks.com/matlabcentral/answers/1931825-how-to-get-pixel-value-inside-a-circle
%# define points (in matrix coordinates)
%3"
cpx = 2050;
cpy = 2020;
inchlist = [12,10.5,9,7.5,6,4.5];
%draw lines on heel axis
for n=1:size(inchlist,2)
inch= inchlist(n)/4;
hcirc = drawcircle(‘Center’,[2050,2020],’Radius’,inch*590,’StripeColor’,’red’);
mask1 = hcirc.createMask;
maxval = (max(img(mask1)));
minval = (min(img(mask1)));
uniformity = maxval/minval
% p1 = [cpy-100,cpx+inch*590];
end

Even after getting this max and min value, I will need to remove 10 to get rid of noise. Extra credit if you can point me to a soluton for that too.

thank you mask, statistics MATLAB Answers — New Questions

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I have a code, which gives a solution of a delay logistic equation with discrete delay.

tau = 1;
tspan = [0 20];
y0 = 0.5;
sol = dde23(@ddefunc, tau, y0, tspan);
% Plot the solution
figure;
plot(sol.x, sol.y, ‘LineWidth’, 2);
xlabel(‘Time (days)’);
ylabel(‘Population’);
legend(‘y’);

% Define the delay differential equation
function g = ddefunc(t, y, Z)
r = 1.5;
y_tau = Z;
g = r * y * (1 – y_tau);
end
Now I want to modify my code for distributed delay as attached below.

Can someone guide me how to deal with distributed delayI have a code, which gives a solution of a delay logistic equation with discrete delay.

tau = 1;
tspan = [0 20];
y0 = 0.5;
sol = dde23(@ddefunc, tau, y0, tspan);
% Plot the solution
figure;
plot(sol.x, sol.y, ‘LineWidth’, 2);
xlabel(‘Time (days)’);
ylabel(‘Population’);
legend(‘y’);

% Define the delay differential equation
function g = ddefunc(t, y, Z)
r = 1.5;
y_tau = Z;
g = r * y * (1 – y_tau);
end
Now I want to modify my code for distributed delay as attached below.

Can someone guide me how to deal with distributed delay I have a code, which gives a solution of a delay logistic equation with discrete delay.

tau = 1;
tspan = [0 20];
y0 = 0.5;
sol = dde23(@ddefunc, tau, y0, tspan);
% Plot the solution
figure;
plot(sol.x, sol.y, ‘LineWidth’, 2);
xlabel(‘Time (days)’);
ylabel(‘Population’);
legend(‘y’);

% Define the delay differential equation
function g = ddefunc(t, y, Z)
r = 1.5;
y_tau = Z;
g = r * y * (1 – y_tau);
end
Now I want to modify my code for distributed delay as attached below.

Can someone guide me how to deal with distributed delay distributed delay, delay differentia equations, solve MATLAB Answers — New Questions

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Hi all, im trying to do simple ERP study using EEGLAB. To RESHAPE number of elements must not change, such an error messahe was thrown. let me know how to fix it. ThanksHi all, im trying to do simple ERP study using EEGLAB. To RESHAPE number of elements must not change, such an error messahe was thrown. let me know how to fix it. Thanks Hi all, im trying to do simple ERP study using EEGLAB. To RESHAPE number of elements must not change, such an error messahe was thrown. let me know how to fix it. Thanks please help MATLAB Answers — New Questions

​

Hi guys,
I’m trying to train a lstm using sequential data to predict classes, and I’m a little confused by the format of input data and labels.
For the sake of simplicity, I’ll use an example to mimic my situation.
let’s say I’m trying to use temperature data to predict 3 cities: A, B, and C.
Within each city, i have temperature readings from 10 therometers over 2 seconds at a sample frequency of 100 hz.
So far, at each observation, I have a 200 by 10 matrix (time point by therometer).
temperature_matrix = randi(40, 200, 10) % pseudodata
We collected the temperature data 40 times throughout the day at each city, and this will give us 120 observations (3 cities * 40). Within each observation, I have a 200 by 10 matrix.
As for my input format, I now have a 120 by 1 cell array, and again within each cell array is a 200 by 10 matrix.
temperature_input = cell(120,1)
for ii = 1:length(temperature_input)
temperature_input{ii} = randi(40, 200, 10)
end
labels = [repmat("city A", 40,1); repmat("city B", 40,1); repmat("city C", 40,1)]
Per my undstanding, if I were to have a time step of 10, i should make a sliding window with a size of 5, and move it down the time dimenssion at a moving step of 1. That is to say, for each 200 by 10 temperature_matrix, I now slice it into 196 2D arrays, where each array is 5 by 10 (window size by therometer).
My question is how this sliding window plays a part in the input format? the sliding window create the fourth dimension in my example. The other three dimension is observation, time, and therometer. I think my overall structure is still a 120 by 1 cell array, but the dimenssions within each entry, I dont know how to organize them.
Also, out of curiosity, will it mess up the structure i transpose the time point by therometer matrice? I’m only asking between I’ve seen examples on the sequencce either in row or column.

Best,
FYHi guys,
I’m trying to train a lstm using sequential data to predict classes, and I’m a little confused by the format of input data and labels.
For the sake of simplicity, I’ll use an example to mimic my situation.
let’s say I’m trying to use temperature data to predict 3 cities: A, B, and C.
Within each city, i have temperature readings from 10 therometers over 2 seconds at a sample frequency of 100 hz.
So far, at each observation, I have a 200 by 10 matrix (time point by therometer).
temperature_matrix = randi(40, 200, 10) % pseudodata
We collected the temperature data 40 times throughout the day at each city, and this will give us 120 observations (3 cities * 40). Within each observation, I have a 200 by 10 matrix.
As for my input format, I now have a 120 by 1 cell array, and again within each cell array is a 200 by 10 matrix.
temperature_input = cell(120,1)
for ii = 1:length(temperature_input)
temperature_input{ii} = randi(40, 200, 10)
end
labels = [repmat("city A", 40,1); repmat("city B", 40,1); repmat("city C", 40,1)]
Per my undstanding, if I were to have a time step of 10, i should make a sliding window with a size of 5, and move it down the time dimenssion at a moving step of 1. That is to say, for each 200 by 10 temperature_matrix, I now slice it into 196 2D arrays, where each array is 5 by 10 (window size by therometer).
My question is how this sliding window plays a part in the input format? the sliding window create the fourth dimension in my example. The other three dimension is observation, time, and therometer. I think my overall structure is still a 120 by 1 cell array, but the dimenssions within each entry, I dont know how to organize them.
Also, out of curiosity, will it mess up the structure i transpose the time point by therometer matrice? I’m only asking between I’ve seen examples on the sequencce either in row or column.

Best,
FY Hi guys,
I’m trying to train a lstm using sequential data to predict classes, and I’m a little confused by the format of input data and labels.
For the sake of simplicity, I’ll use an example to mimic my situation.
let’s say I’m trying to use temperature data to predict 3 cities: A, B, and C.
Within each city, i have temperature readings from 10 therometers over 2 seconds at a sample frequency of 100 hz.
So far, at each observation, I have a 200 by 10 matrix (time point by therometer).
temperature_matrix = randi(40, 200, 10) % pseudodata
We collected the temperature data 40 times throughout the day at each city, and this will give us 120 observations (3 cities * 40). Within each observation, I have a 200 by 10 matrix.
As for my input format, I now have a 120 by 1 cell array, and again within each cell array is a 200 by 10 matrix.
temperature_input = cell(120,1)
for ii = 1:length(temperature_input)
temperature_input{ii} = randi(40, 200, 10)
end
labels = [repmat("city A", 40,1); repmat("city B", 40,1); repmat("city C", 40,1)]
Per my undstanding, if I were to have a time step of 10, i should make a sliding window with a size of 5, and move it down the time dimenssion at a moving step of 1. That is to say, for each 200 by 10 temperature_matrix, I now slice it into 196 2D arrays, where each array is 5 by 10 (window size by therometer).
My question is how this sliding window plays a part in the input format? the sliding window create the fourth dimension in my example. The other three dimension is observation, time, and therometer. I think my overall structure is still a 120 by 1 cell array, but the dimenssions within each entry, I dont know how to organize them.
Also, out of curiosity, will it mess up the structure i transpose the time point by therometer matrice? I’m only asking between I’ve seen examples on the sequencce either in row or column.

Best,
FY lstm, input, dimension MATLAB Answers — New Questions

​

There is an issue using CAN Unpack to decode CAN message with length greater than 8. Is there any other recommanded way to decode CAN data? For starter, I have used bit extract and bitwise AND Shift Arithmetic to extract bits in simulink. I’m not sure if matlab function in simulink can also handle this if it’s possible.There is an issue using CAN Unpack to decode CAN message with length greater than 8. Is there any other recommanded way to decode CAN data? For starter, I have used bit extract and bitwise AND Shift Arithmetic to extract bits in simulink. I’m not sure if matlab function in simulink can also handle this if it’s possible. There is an issue using CAN Unpack to decode CAN message with length greater than 8. Is there any other recommanded way to decode CAN data? For starter, I have used bit extract and bitwise AND Shift Arithmetic to extract bits in simulink. I’m not sure if matlab function in simulink can also handle this if it’s possible. data, control MATLAB Answers — New Questions

​

Hello, I want to extrapolate a point (longitude,latitude) from the coordinates of a surface file (attached here as "slab_strike") as it is empty (Strike=NaN) when using interp2 as the points are outside the boundary. Despsite I used the option "nearest" is empty anyway.

% Coordinates of the points:
lat_GMM= -17.8990;
lon_GMM=-73.5295;
% The surface plot
load slab_strike % Loading the slab strike
Slab_strike.x=x
Slab_strike.y=y
Slab_strike.z=z
Strike = interp2(Slab_strike.x,Slab_strike.y,Slab_strike.z,lon_GMM,lat_GMM)
As Strike=NaN, there is a way I can choose the closest point value from the surface instead.
I would appreciate the helpHello, I want to extrapolate a point (longitude,latitude) from the coordinates of a surface file (attached here as "slab_strike") as it is empty (Strike=NaN) when using interp2 as the points are outside the boundary. Despsite I used the option "nearest" is empty anyway.

% Coordinates of the points:
lat_GMM= -17.8990;
lon_GMM=-73.5295;
% The surface plot
load slab_strike % Loading the slab strike
Slab_strike.x=x
Slab_strike.y=y
Slab_strike.z=z
Strike = interp2(Slab_strike.x,Slab_strike.y,Slab_strike.z,lon_GMM,lat_GMM)
As Strike=NaN, there is a way I can choose the closest point value from the surface instead.
I would appreciate the help Hello, I want to extrapolate a point (longitude,latitude) from the coordinates of a surface file (attached here as "slab_strike") as it is empty (Strike=NaN) when using interp2 as the points are outside the boundary. Despsite I used the option "nearest" is empty anyway.

% Coordinates of the points:
lat_GMM= -17.8990;
lon_GMM=-73.5295;
% The surface plot
load slab_strike % Loading the slab strike
Slab_strike.x=x
Slab_strike.y=y
Slab_strike.z=z
Strike = interp2(Slab_strike.x,Slab_strike.y,Slab_strike.z,lon_GMM,lat_GMM)
As Strike=NaN, there is a way I can choose the closest point value from the surface instead.
I would appreciate the help selecting the closest point reference MATLAB Answers — New Questions

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How can I create an offset of a boundary?
rng(‘default’)
x = rand(30,1);
y = rand(30,1);
plot(x,y,’.’)
xlim([-0.2 1.2])
ylim([-0.2 1.2])
k = boundary(x,y);
hold on;
plot(x(k),y(k));

My desired output:How can I create an offset of a boundary?
rng(‘default’)
x = rand(30,1);
y = rand(30,1);
plot(x,y,’.’)
xlim([-0.2 1.2])
ylim([-0.2 1.2])
k = boundary(x,y);
hold on;
plot(x(k),y(k));

My desired output: How can I create an offset of a boundary?
rng(‘default’)
x = rand(30,1);
y = rand(30,1);
plot(x,y,’.’)
xlim([-0.2 1.2])
ylim([-0.2 1.2])
k = boundary(x,y);
hold on;
plot(x(k),y(k));

My desired output: offset, boundary MATLAB Answers — New Questions

​

function HTP()
clc
clear all format long % hybrid Carreau
% Define constants
J1 = 0.1;
J2 = 0.1;
J3 = 0.1;
J4 = 0.1;
JS = 0.1;
z = 0.1;
S = 0.1;
GC = 0.1;
Gr = 0.1;
H = 0.1;
a = 0.1;
m = 1;
G = 0.5;
phi = 0.1;

% Define time vector
t = linspace(0, 5, 100); % 100 points between 0 and 5
%t= 1;
% Calculate u1 and u2
u1 = exp(-t) – 1;

% Compute u2 with corrected parentheses and mathematical operations
exp_t = exp(t); % Compute exp(t) once for efficiency
term1 = -33 / J1 * (1 – exp_t – z / 3 * S);
term2 = 2 * GC * J3 + 2 * Gr * J2;
term3 = (2 * J4 * H * a^2) / (1 + m^2);
term4 = 3 * G * exp_t + 6 * exp_t / ((1 – phi)^2.5);
term5 = 2 * GC * JS * exp_t + 2 * Gr * J2 * exp_t;
u2 = (term1 + term2 – term3 – term4 – term5 + term3 * (exp_t – m + m * exp_t)) / (6 * a);

% Compute u
y = 1; % Define y as 1 or another constant value; adjust as needed
u = u1 * y + u2 * y.^2;

% Plot the result
phi = 0.1
figure ;
plot(t, u);
phi= 0.2
figure ;
plot(t, u);
phi = 0.3
figure ;
plot(t, u);
phi = 0.4
figure ;
plot(t, u);
xlabel(‘Time (t)’);
ylabel(‘u(t)’);
title(‘Plot of u(t)’);
axis([0 5 min(u) max(u)]);
grid on;
endfunction HTP()
clc
clear all format long % hybrid Carreau
% Define constants
J1 = 0.1;
J2 = 0.1;
J3 = 0.1;
J4 = 0.1;
JS = 0.1;
z = 0.1;
S = 0.1;
GC = 0.1;
Gr = 0.1;
H = 0.1;
a = 0.1;
m = 1;
G = 0.5;
phi = 0.1;

% Define time vector
t = linspace(0, 5, 100); % 100 points between 0 and 5
%t= 1;
% Calculate u1 and u2
u1 = exp(-t) – 1;

% Compute u2 with corrected parentheses and mathematical operations
exp_t = exp(t); % Compute exp(t) once for efficiency
term1 = -33 / J1 * (1 – exp_t – z / 3 * S);
term2 = 2 * GC * J3 + 2 * Gr * J2;
term3 = (2 * J4 * H * a^2) / (1 + m^2);
term4 = 3 * G * exp_t + 6 * exp_t / ((1 – phi)^2.5);
term5 = 2 * GC * JS * exp_t + 2 * Gr * J2 * exp_t;
u2 = (term1 + term2 – term3 – term4 – term5 + term3 * (exp_t – m + m * exp_t)) / (6 * a);

% Compute u
y = 1; % Define y as 1 or another constant value; adjust as needed
u = u1 * y + u2 * y.^2;

% Plot the result
phi = 0.1
figure ;
plot(t, u);
phi= 0.2
figure ;
plot(t, u);
phi = 0.3
figure ;
plot(t, u);
phi = 0.4
figure ;
plot(t, u);
xlabel(‘Time (t)’);
ylabel(‘u(t)’);
title(‘Plot of u(t)’);
axis([0 5 min(u) max(u)]);
grid on;
end function HTP()
clc
clear all format long % hybrid Carreau
% Define constants
J1 = 0.1;
J2 = 0.1;
J3 = 0.1;
J4 = 0.1;
JS = 0.1;
z = 0.1;
S = 0.1;
GC = 0.1;
Gr = 0.1;
H = 0.1;
a = 0.1;
m = 1;
G = 0.5;
phi = 0.1;

% Define time vector
t = linspace(0, 5, 100); % 100 points between 0 and 5
%t= 1;
% Calculate u1 and u2
u1 = exp(-t) – 1;

% Compute u2 with corrected parentheses and mathematical operations
exp_t = exp(t); % Compute exp(t) once for efficiency
term1 = -33 / J1 * (1 – exp_t – z / 3 * S);
term2 = 2 * GC * J3 + 2 * Gr * J2;
term3 = (2 * J4 * H * a^2) / (1 + m^2);
term4 = 3 * G * exp_t + 6 * exp_t / ((1 – phi)^2.5);
term5 = 2 * GC * JS * exp_t + 2 * Gr * J2 * exp_t;
u2 = (term1 + term2 – term3 – term4 – term5 + term3 * (exp_t – m + m * exp_t)) / (6 * a);

% Compute u
y = 1; % Define y as 1 or another constant value; adjust as needed
u = u1 * y + u2 * y.^2;

% Plot the result
phi = 0.1
figure ;
plot(t, u);
phi= 0.2
figure ;
plot(t, u);
phi = 0.3
figure ;
plot(t, u);
phi = 0.4
figure ;
plot(t, u);
xlabel(‘Time (t)’);
ylabel(‘u(t)’);
title(‘Plot of u(t)’);
axis([0 5 min(u) max(u)]);
grid on;
end multiple lines in one graph MATLAB Answers — New Questions

​