Category: Matlab
Category Archives: Matlab
how to draw a peak line
Hi, i’ve data and i want to draw a line that corresponds to the peaks (where there is a zero the previous value >0 must be pasted) (See black line painted)
Is there a function or graph type that does this to me?Hi, i’ve data and i want to draw a line that corresponds to the peaks (where there is a zero the previous value >0 must be pasted) (See black line painted)
Is there a function or graph type that does this to me? Hi, i’ve data and i want to draw a line that corresponds to the peaks (where there is a zero the previous value >0 must be pasted) (See black line painted)
Is there a function or graph type that does this to me? how to draw a peak line MATLAB Answers — New Questions
I continue to get this error and can’t figure out how to fix it, Unable to perform assignment because the size of the left side is 1-by-4 and the size of the right side is 1-b
% Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature);% Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature); % Given temperature-depth data
z = [0, -2.3, -4.6, -6.9, -9.2, -11.5, -13.8, -16.1]; % Depth in meters
T = [22.8, 22.7, 22.5, 20.6, 13.9, 11.7, 11.2, 11.1]; % Temperature in Celsius
% Step 1: Plot the data points
figure;
plot(T, z, ‘r*’); % Plotting temperature vs depth with red stars
xlabel(‘Temperature (°C)’);
ylabel(‘Depth (m)’);
title(‘Temperature vs Depth’);
set(gca, ‘YDir’,’reverse’); % Inverting the y-axis
grid on;
% Step 2: Sort the data since the depth must be in increasing order
[z_sorted, sortIndex] = sort(z);
T_sorted = T(sortIndex);
% Step 3: Create a cubic spline interpolation
cs = spline(z_sorted, T_sorted);
% Step 4: Evaluate the spline and its derivatives on a finer grid
z_fine = linspace(min(z_sorted), max(z_sorted), 500);
T_spline = ppval(cs, z_fine);
% Step 5: Find the first and second derivatives of the spline
[breaks,coefs,l,k,d] = unmkpp(cs); % Extracts the pieces of the cubic spline
dcoefs = coefs; % Derivative coefficients
% Each row of dcoefs will be the coefficients of the polynomial of a piece
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of derivative
csd1 = mkpp(breaks,dcoefs(:,1:k-1));
% First derivative evaluation
T_spline_deriv = ppval(csd1, z_fine);
% Find the second derivative
for j = 1:l
dcoefs(j,:) = polyder(dcoefs(j,:));
end
% Make a pp-form of second derivative
csd2 = mkpp(breaks,dcoefs(:,1:k-2));
% Second derivative evaluation
T_spline_second_deriv = ppval(csd2, z_fine);
% Step 6: Locate the thermocline by finding the depth where the second derivative
% changes sign, and the first derivative is a maximum
inflection_points = find(diff(sign(T_spline_second_deriv)) ~= 0) + 1;
[~, max_gradient_index] = max(abs(T_spline_deriv(inflection_points)));
thermocline_depth = z_fine(inflection_points(max_gradient_index));
thermocline_temperature = T_spline(inflection_points(max_gradient_index));
% Display the thermocline depth and temperature
fprintf(‘The thermocline is located at a depth of %.2f m with a temperature of %.2f°C.n’, …
thermocline_depth, thermocline_temperature); matrices MATLAB Answers — New Questions
how do i get a proper time plot for this qudratic equation?
the script seems to be not behaving properly,as it is supposed to be ?
% Define the range for x
x = linspace(-5, 5, 10); % 10 points between -5 and 5
dx1 = 2*x.^2 – 8; % Compute the derivative (velocity field)
% Create a figure
figure;
hold on;
% Plot the real line (x-axis)
plot(x, zeros(size(x)), ‘k’); % The x-axis (real line)
% Plot the vector field as arrows
for i = 1:length(x)
if dx1(i) > 0
% Arrow pointing to the right (positive dx)
quiver(x(i), 0, 0.5, 0, ‘r’, ‘MaxHeadSize’, 0.5);
elseif dx1(i) < 0
% Arrow pointing to the left (negative dx)
quiver(x(i), 0, -0.5, 0, ‘b’, ‘MaxHeadSize’, 0.5);
end
end
% Plot equilibria points
plot([-2, 2], [0, 0], ‘bo’, ‘MarkerFaceColor’, ‘b’, ‘MarkerSize’, 8); % Equilibria at x = -2 and x = 2
% Labels and title
xlabel(‘x’);
ylabel(‘x-dot’);
title(‘1D Vector Field for x-dot = 2x^2 – 8’);
ylim([-1 1]); % Set y-axis limits to keep focus on x-axis
xlim([-4 4]); % Set x-axis limits
grid on; % Add grid
% Add text annotations for equilibrium points
text(-2, 0.2, ‘Stable Equilibrium: x = -2’, ‘HorizontalAlignment’, ‘center’);
text(2, 0.2, ‘Unstable Equilibrium: x = 2’, ‘HorizontalAlignment’, ‘center’);
hold off;
t = [-5 5];
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
figure;
plot(time, x, ‘LineWidth’,0.5);
title(‘Solution of dx/dt = 2x^2 – 8 vs Time’);
xlabel(‘Time t’);
ylabel(‘x(t)’);
xlim([-t(2) t(2)]); % Adjust x-axis limits for clarity
ylim([-3 3]); % Set y-axis limits to see behavior around -2 and 2
grid on;
the first part works perfectly fine, but the 2nd part where i have to plot wrt time, is where i am having trouble? its supposed to be going from x-2 being stable , and 2 being unstable, the plot should depict that.
but I am unable to get that done properlythe script seems to be not behaving properly,as it is supposed to be ?
% Define the range for x
x = linspace(-5, 5, 10); % 10 points between -5 and 5
dx1 = 2*x.^2 – 8; % Compute the derivative (velocity field)
% Create a figure
figure;
hold on;
% Plot the real line (x-axis)
plot(x, zeros(size(x)), ‘k’); % The x-axis (real line)
% Plot the vector field as arrows
for i = 1:length(x)
if dx1(i) > 0
% Arrow pointing to the right (positive dx)
quiver(x(i), 0, 0.5, 0, ‘r’, ‘MaxHeadSize’, 0.5);
elseif dx1(i) < 0
% Arrow pointing to the left (negative dx)
quiver(x(i), 0, -0.5, 0, ‘b’, ‘MaxHeadSize’, 0.5);
end
end
% Plot equilibria points
plot([-2, 2], [0, 0], ‘bo’, ‘MarkerFaceColor’, ‘b’, ‘MarkerSize’, 8); % Equilibria at x = -2 and x = 2
% Labels and title
xlabel(‘x’);
ylabel(‘x-dot’);
title(‘1D Vector Field for x-dot = 2x^2 – 8’);
ylim([-1 1]); % Set y-axis limits to keep focus on x-axis
xlim([-4 4]); % Set x-axis limits
grid on; % Add grid
% Add text annotations for equilibrium points
text(-2, 0.2, ‘Stable Equilibrium: x = -2’, ‘HorizontalAlignment’, ‘center’);
text(2, 0.2, ‘Unstable Equilibrium: x = 2’, ‘HorizontalAlignment’, ‘center’);
hold off;
t = [-5 5];
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
figure;
plot(time, x, ‘LineWidth’,0.5);
title(‘Solution of dx/dt = 2x^2 – 8 vs Time’);
xlabel(‘Time t’);
ylabel(‘x(t)’);
xlim([-t(2) t(2)]); % Adjust x-axis limits for clarity
ylim([-3 3]); % Set y-axis limits to see behavior around -2 and 2
grid on;
the first part works perfectly fine, but the 2nd part where i have to plot wrt time, is where i am having trouble? its supposed to be going from x-2 being stable , and 2 being unstable, the plot should depict that.
but I am unable to get that done properly the script seems to be not behaving properly,as it is supposed to be ?
% Define the range for x
x = linspace(-5, 5, 10); % 10 points between -5 and 5
dx1 = 2*x.^2 – 8; % Compute the derivative (velocity field)
% Create a figure
figure;
hold on;
% Plot the real line (x-axis)
plot(x, zeros(size(x)), ‘k’); % The x-axis (real line)
% Plot the vector field as arrows
for i = 1:length(x)
if dx1(i) > 0
% Arrow pointing to the right (positive dx)
quiver(x(i), 0, 0.5, 0, ‘r’, ‘MaxHeadSize’, 0.5);
elseif dx1(i) < 0
% Arrow pointing to the left (negative dx)
quiver(x(i), 0, -0.5, 0, ‘b’, ‘MaxHeadSize’, 0.5);
end
end
% Plot equilibria points
plot([-2, 2], [0, 0], ‘bo’, ‘MarkerFaceColor’, ‘b’, ‘MarkerSize’, 8); % Equilibria at x = -2 and x = 2
% Labels and title
xlabel(‘x’);
ylabel(‘x-dot’);
title(‘1D Vector Field for x-dot = 2x^2 – 8’);
ylim([-1 1]); % Set y-axis limits to keep focus on x-axis
xlim([-4 4]); % Set x-axis limits
grid on; % Add grid
% Add text annotations for equilibrium points
text(-2, 0.2, ‘Stable Equilibrium: x = -2’, ‘HorizontalAlignment’, ‘center’);
text(2, 0.2, ‘Unstable Equilibrium: x = 2’, ‘HorizontalAlignment’, ‘center’);
hold off;
t = [-5 5];
ode_function = @(t,x) 2*x.^2-8;
x0 = [-2 2];
[time, x] = ode45(ode_function, t, x0);
figure;
plot(time, x, ‘LineWidth’,0.5);
title(‘Solution of dx/dt = 2x^2 – 8 vs Time’);
xlabel(‘Time t’);
ylabel(‘x(t)’);
xlim([-t(2) t(2)]); % Adjust x-axis limits for clarity
ylim([-3 3]); % Set y-axis limits to see behavior around -2 and 2
grid on;
the first part works perfectly fine, but the 2nd part where i have to plot wrt time, is where i am having trouble? its supposed to be going from x-2 being stable , and 2 being unstable, the plot should depict that.
but I am unable to get that done properly code generation, ode45, equation MATLAB Answers — New Questions
MATLAB Vehicle Network Toolbox cannot send CAN messages on physical channels with Vector’s hardware VN1610.
When using the vehicle network toolbox(Matlab 2024a update 6) with the vector’s hardware VN1610, if choosing to use the real physical channel instead of the virtual channel, I’m not able to turn off the canChannel.SilentMode even if the canChannel.InitializationAccess is 1, which caused the channel only able to receive messages but cannot send out any messages. I tried to turn the SilentMode off with commands below:
>> txCh_I
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 2
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 0
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: [0×0 datetime]
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 1
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 250000
SJW: 1
TSEG1: 10
TSEG2: 5
NumOfSamples: 1
Other Information
Database: []
UserData: []
>> txCh_I.SilentMode = false
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 2
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 0
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: [0×0 datetime]
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 1
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 250000
SJW: 1
TSEG1: 10
TSEG2: 5
NumOfSamples: 1
Other Information
Database: []
UserData: []
It’s wierd that the channel has opened successfully but SilentMode is set to 1 defaultly, and the initial access is true but cannot change the SilentMode without giving any error or indication. When I dig the issue further, I found that this issue only happens on the hardware VN1610 with the latest Vector driver(version 24.20.10).
After rolling-back the driver to a previous version, the issue just disappeared:
Vehicle Network Toolbox has detected a supported Vector device. To connect to this device:
1. Download and install the latest device driver from the Vector website.
2. Verify device readiness with canChannelList.
>> txCh_I
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 1
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 150
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: 28-Sep-2024 06:28:43
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 0
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 500000
SJW: 1
TSEG1: 4
TSEG2: 3
NumOfSamples: 1
Other Information
Database: []
UserData: []
Please take a look and help check this. Thanks!When using the vehicle network toolbox(Matlab 2024a update 6) with the vector’s hardware VN1610, if choosing to use the real physical channel instead of the virtual channel, I’m not able to turn off the canChannel.SilentMode even if the canChannel.InitializationAccess is 1, which caused the channel only able to receive messages but cannot send out any messages. I tried to turn the SilentMode off with commands below:
>> txCh_I
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 2
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 0
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: [0×0 datetime]
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 1
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 250000
SJW: 1
TSEG1: 10
TSEG2: 5
NumOfSamples: 1
Other Information
Database: []
UserData: []
>> txCh_I.SilentMode = false
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 2
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 0
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: [0×0 datetime]
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 1
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 250000
SJW: 1
TSEG1: 10
TSEG2: 5
NumOfSamples: 1
Other Information
Database: []
UserData: []
It’s wierd that the channel has opened successfully but SilentMode is set to 1 defaultly, and the initial access is true but cannot change the SilentMode without giving any error or indication. When I dig the issue further, I found that this issue only happens on the hardware VN1610 with the latest Vector driver(version 24.20.10).
After rolling-back the driver to a previous version, the issue just disappeared:
Vehicle Network Toolbox has detected a supported Vector device. To connect to this device:
1. Download and install the latest device driver from the Vector website.
2. Verify device readiness with canChannelList.
>> txCh_I
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 1
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 150
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: 28-Sep-2024 06:28:43
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 0
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 500000
SJW: 1
TSEG1: 4
TSEG2: 3
NumOfSamples: 1
Other Information
Database: []
UserData: []
Please take a look and help check this. Thanks! When using the vehicle network toolbox(Matlab 2024a update 6) with the vector’s hardware VN1610, if choosing to use the real physical channel instead of the virtual channel, I’m not able to turn off the canChannel.SilentMode even if the canChannel.InitializationAccess is 1, which caused the channel only able to receive messages but cannot send out any messages. I tried to turn the SilentMode off with commands below:
>> txCh_I
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 2
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 0
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: [0×0 datetime]
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 1
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 250000
SJW: 1
TSEG1: 10
TSEG2: 5
NumOfSamples: 1
Other Information
Database: []
UserData: []
>> txCh_I.SilentMode = false
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 2
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 0
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: [0×0 datetime]
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 1
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 250000
SJW: 1
TSEG1: 10
TSEG2: 5
NumOfSamples: 1
Other Information
Database: []
UserData: []
It’s wierd that the channel has opened successfully but SilentMode is set to 1 defaultly, and the initial access is true but cannot change the SilentMode without giving any error or indication. When I dig the issue further, I found that this issue only happens on the hardware VN1610 with the latest Vector driver(version 24.20.10).
After rolling-back the driver to a previous version, the issue just disappeared:
Vehicle Network Toolbox has detected a supported Vector device. To connect to this device:
1. Download and install the latest device driver from the Vector website.
2. Verify device readiness with canChannelList.
>> txCh_I
txCh_I =
Channel with properties:
Device Information
DeviceVendor: ‘Vector’
Device: ‘VN1610 1’
DeviceChannelIndex: 1
DeviceSerialNumber: 530636
ProtocolMode: ‘CAN’
Status Information
Running: 0
MessagesAvailable: 150
MessagesReceived: 0
MessagesTransmitted: 0
InitializationAccess: 1
InitialTimestamp: 28-Sep-2024 06:28:43
FilterHistory: ‘Standard ID Filter: Allow All | Extended ID Filter: Allow All’
Channel Information
BusStatus: ‘N/A’
SilentMode: 0
TransceiverName: ‘On board CAN 1051cap(Highspeed)’
TransceiverState: 0
ReceiveErrorCount: 0
TransmitErrorCount: 0
BusSpeed: 500000
SJW: 1
TSEG1: 4
TSEG2: 3
NumOfSamples: 1
Other Information
Database: []
UserData: []
Please take a look and help check this. Thanks! vehicle network toolbox, vector hardware, vn1610, can channel MATLAB Answers — New Questions
Choose role of NaN when summing two matrices
Dear all
Some time ago, I posted this question: Choosing the role of NaN elements in the sum environment of matrices. That worked great when doing the mean value, but now imagine that I just want to sum the two matrices, given by:
aa=[1 2; 3 NaN];
bb=[NaN 1; 2 NaN];
so that I obtain:
cc=[1 3; 5 NaN];
If I do something like:
cc=sum(cat(3,aa,bb),3,’omitnan’)
that gives:
cc=[1 3; 5 0];
Which way would be the best in this case?Dear all
Some time ago, I posted this question: Choosing the role of NaN elements in the sum environment of matrices. That worked great when doing the mean value, but now imagine that I just want to sum the two matrices, given by:
aa=[1 2; 3 NaN];
bb=[NaN 1; 2 NaN];
so that I obtain:
cc=[1 3; 5 NaN];
If I do something like:
cc=sum(cat(3,aa,bb),3,’omitnan’)
that gives:
cc=[1 3; 5 0];
Which way would be the best in this case? Dear all
Some time ago, I posted this question: Choosing the role of NaN elements in the sum environment of matrices. That worked great when doing the mean value, but now imagine that I just want to sum the two matrices, given by:
aa=[1 2; 3 NaN];
bb=[NaN 1; 2 NaN];
so that I obtain:
cc=[1 3; 5 NaN];
If I do something like:
cc=sum(cat(3,aa,bb),3,’omitnan’)
that gives:
cc=[1 3; 5 0];
Which way would be the best in this case? summing matrices, role of nan MATLAB Answers — New Questions
Error code line 19 with my Y=f(x), equation used is x^3+x^2-4x-4
function [x,y,err]=muller(f,x0,x1,x2,delta,epsilon,max1)
%Input – f is the object function
% – x0, x1, and x2 are the initial approximations
% – delta is the tolerance for x0, x1, and x2
% – epsilon the the tolerance for the function values y
% – max1 is the maximum number of iterations
%Output- x is the Muller approximation to the zero of f
% – y is the function value y = f(x)
% – err is the error in the approximation of x.
%Initalize the matrices X and Y
X=[x0 x1 x2];
Y=f(X);
%Calculate a and b in formula (15)
for k=1:max1
h0=X(1)-X(3);h1=X(2)-X(3);e0=Y(1)-Y(3);e1=Y(2)-Y(3);c=Y(3);
denom=h1*h0^2-h0*h1^2;
a=(e0*h1-e1*h0)/denom;
b=(e1*h0^2-e0*h1^2)/denom;
%Find the smallest root of (17)
if b < 0
disc=-disc;
end
z=-2*c/(b+disc);
x=X(3)+z;
%Sort the entries of X to find the two closest to x
if abs(x-X(2))<abs(x-X(1))
Q=[X(2) X(1) X(3)];
X=Q;
Y=f(X);
end
if abs(x-X(3))<abs(x-X(2))
R=[X(1) X(3) X(2)];
X=R;
Y=f(X);
end
%Replace the entry of X that was farthest from x with x
X(3)=x;
Y(3) = f(X(3));
y=Y(3);
%Determine stopping criteria
err=abs(z);
relerr=err/(abs(x)+delta);
if (err<delta)||(relerr<delta)||(abs(y)<epsilon)
break
end
endfunction [x,y,err]=muller(f,x0,x1,x2,delta,epsilon,max1)
%Input – f is the object function
% – x0, x1, and x2 are the initial approximations
% – delta is the tolerance for x0, x1, and x2
% – epsilon the the tolerance for the function values y
% – max1 is the maximum number of iterations
%Output- x is the Muller approximation to the zero of f
% – y is the function value y = f(x)
% – err is the error in the approximation of x.
%Initalize the matrices X and Y
X=[x0 x1 x2];
Y=f(X);
%Calculate a and b in formula (15)
for k=1:max1
h0=X(1)-X(3);h1=X(2)-X(3);e0=Y(1)-Y(3);e1=Y(2)-Y(3);c=Y(3);
denom=h1*h0^2-h0*h1^2;
a=(e0*h1-e1*h0)/denom;
b=(e1*h0^2-e0*h1^2)/denom;
%Find the smallest root of (17)
if b < 0
disc=-disc;
end
z=-2*c/(b+disc);
x=X(3)+z;
%Sort the entries of X to find the two closest to x
if abs(x-X(2))<abs(x-X(1))
Q=[X(2) X(1) X(3)];
X=Q;
Y=f(X);
end
if abs(x-X(3))<abs(x-X(2))
R=[X(1) X(3) X(2)];
X=R;
Y=f(X);
end
%Replace the entry of X that was farthest from x with x
X(3)=x;
Y(3) = f(X(3));
y=Y(3);
%Determine stopping criteria
err=abs(z);
relerr=err/(abs(x)+delta);
if (err<delta)||(relerr<delta)||(abs(y)<epsilon)
break
end
end function [x,y,err]=muller(f,x0,x1,x2,delta,epsilon,max1)
%Input – f is the object function
% – x0, x1, and x2 are the initial approximations
% – delta is the tolerance for x0, x1, and x2
% – epsilon the the tolerance for the function values y
% – max1 is the maximum number of iterations
%Output- x is the Muller approximation to the zero of f
% – y is the function value y = f(x)
% – err is the error in the approximation of x.
%Initalize the matrices X and Y
X=[x0 x1 x2];
Y=f(X);
%Calculate a and b in formula (15)
for k=1:max1
h0=X(1)-X(3);h1=X(2)-X(3);e0=Y(1)-Y(3);e1=Y(2)-Y(3);c=Y(3);
denom=h1*h0^2-h0*h1^2;
a=(e0*h1-e1*h0)/denom;
b=(e1*h0^2-e0*h1^2)/denom;
%Find the smallest root of (17)
if b < 0
disc=-disc;
end
z=-2*c/(b+disc);
x=X(3)+z;
%Sort the entries of X to find the two closest to x
if abs(x-X(2))<abs(x-X(1))
Q=[X(2) X(1) X(3)];
X=Q;
Y=f(X);
end
if abs(x-X(3))<abs(x-X(2))
R=[X(1) X(3) X(2)];
X=R;
Y=f(X);
end
%Replace the entry of X that was farthest from x with x
X(3)=x;
Y(3) = f(X(3));
y=Y(3);
%Determine stopping criteria
err=abs(z);
relerr=err/(abs(x)+delta);
if (err<delta)||(relerr<delta)||(abs(y)<epsilon)
break
end
end muller method MATLAB Answers — New Questions
BLDC motor driver usinf FOC simulation problem
Hi, i tried to create the simulink model of bldc motor driver using IFOC, but i just can not figure why the speed plot was not right? i tried to tune the pi controller, but seems that there might be problem with system.Hi, i tried to create the simulink model of bldc motor driver using IFOC, but i just can not figure why the speed plot was not right? i tried to tune the pi controller, but seems that there might be problem with system. Hi, i tried to create the simulink model of bldc motor driver using IFOC, but i just can not figure why the speed plot was not right? i tried to tune the pi controller, but seems that there might be problem with system. bldc foc, foc MATLAB Answers — New Questions
Hello, i need help on least square fitting for an infectious disease
An SIR model for COVID-19, Flu, and RSV
We utilize mathematical modeling based on differential equations to investigate the trans-
mission dynamics of the three respiratory infections: COVID-19, Flu, and RSV. To start, let
us consider a simple SIR model, which involves three compartments: the susceptible individ-
uals (denoted by S), the infected individuals (denoted by I), and the recovered individuals
(denoted by R).
The model is described as follows:
dS
dt = Lambda − beta SI − mu S,
dI
dt = beta SI − (gamma + w + mu )I,
dR
dt = gamma I − mu R.
The parameter Lambda is the population influx rate, beta is the transmission rates, mu is the natural
death rate for the human hosts, gamma is the rate of recovery from the infection, and w is the
disease induced death rate. For simplicity, we set Lambda = mu N, where N is the total population.
For the US, mu = 1/76.33 year−1 = 1/(76.33 ∗ 365) day−1
, and N = 345, 426, 571 people.
Other parameter values are:
COVID-19: gamma = 1/14 day−1
, w = 0.01/7 day−1
;
Flu: gamma = 1/3.5 day−1
, w = 0.006/7 day−1
;
RSV: gamma = 1/10.5 day−1
, w = 0.002/7 day−1
.
Apply this model separately to the US data on COVID-19, Flu, and RSV, and estimate the
value of the transmission rate beta in each case. The data fitting is typically done by using the
least squares method. Based on the fitted beta , calculate the basic reproduction number R0
for each.
I am supposed to consider the above instruction to write matlab codes to find the transmission rate for the flu, covid and the RSV data. Each of these diseases has six seasons starting from 2018-2019 seaons to 2023-2024 seasons.
if the matlab data for 2018-2019 flu season is this:
% Filter for the 2018-19 flu season
Flu_data_2018_2019 = Flu_data(strcmp(Flu_data.Season, ‘2018-19’), :);
% Display the filtered data for the 2018-19 season
disp(Flu_data_2018_2019);
% Display the ‘Weekly_Rate’ column (replace with the correct column name if needed)
Weekly_Rate_2018_2019 = Flu_data_2018_2019.WeeklyRate;
disp(Weekly_Rate_2018_2019);
help write a matlab code to find the transmission rate for the 2018-2019 flu season and the reproduction numberAn SIR model for COVID-19, Flu, and RSV
We utilize mathematical modeling based on differential equations to investigate the trans-
mission dynamics of the three respiratory infections: COVID-19, Flu, and RSV. To start, let
us consider a simple SIR model, which involves three compartments: the susceptible individ-
uals (denoted by S), the infected individuals (denoted by I), and the recovered individuals
(denoted by R).
The model is described as follows:
dS
dt = Lambda − beta SI − mu S,
dI
dt = beta SI − (gamma + w + mu )I,
dR
dt = gamma I − mu R.
The parameter Lambda is the population influx rate, beta is the transmission rates, mu is the natural
death rate for the human hosts, gamma is the rate of recovery from the infection, and w is the
disease induced death rate. For simplicity, we set Lambda = mu N, where N is the total population.
For the US, mu = 1/76.33 year−1 = 1/(76.33 ∗ 365) day−1
, and N = 345, 426, 571 people.
Other parameter values are:
COVID-19: gamma = 1/14 day−1
, w = 0.01/7 day−1
;
Flu: gamma = 1/3.5 day−1
, w = 0.006/7 day−1
;
RSV: gamma = 1/10.5 day−1
, w = 0.002/7 day−1
.
Apply this model separately to the US data on COVID-19, Flu, and RSV, and estimate the
value of the transmission rate beta in each case. The data fitting is typically done by using the
least squares method. Based on the fitted beta , calculate the basic reproduction number R0
for each.
I am supposed to consider the above instruction to write matlab codes to find the transmission rate for the flu, covid and the RSV data. Each of these diseases has six seasons starting from 2018-2019 seaons to 2023-2024 seasons.
if the matlab data for 2018-2019 flu season is this:
% Filter for the 2018-19 flu season
Flu_data_2018_2019 = Flu_data(strcmp(Flu_data.Season, ‘2018-19’), :);
% Display the filtered data for the 2018-19 season
disp(Flu_data_2018_2019);
% Display the ‘Weekly_Rate’ column (replace with the correct column name if needed)
Weekly_Rate_2018_2019 = Flu_data_2018_2019.WeeklyRate;
disp(Weekly_Rate_2018_2019);
help write a matlab code to find the transmission rate for the 2018-2019 flu season and the reproduction number An SIR model for COVID-19, Flu, and RSV
We utilize mathematical modeling based on differential equations to investigate the trans-
mission dynamics of the three respiratory infections: COVID-19, Flu, and RSV. To start, let
us consider a simple SIR model, which involves three compartments: the susceptible individ-
uals (denoted by S), the infected individuals (denoted by I), and the recovered individuals
(denoted by R).
The model is described as follows:
dS
dt = Lambda − beta SI − mu S,
dI
dt = beta SI − (gamma + w + mu )I,
dR
dt = gamma I − mu R.
The parameter Lambda is the population influx rate, beta is the transmission rates, mu is the natural
death rate for the human hosts, gamma is the rate of recovery from the infection, and w is the
disease induced death rate. For simplicity, we set Lambda = mu N, where N is the total population.
For the US, mu = 1/76.33 year−1 = 1/(76.33 ∗ 365) day−1
, and N = 345, 426, 571 people.
Other parameter values are:
COVID-19: gamma = 1/14 day−1
, w = 0.01/7 day−1
;
Flu: gamma = 1/3.5 day−1
, w = 0.006/7 day−1
;
RSV: gamma = 1/10.5 day−1
, w = 0.002/7 day−1
.
Apply this model separately to the US data on COVID-19, Flu, and RSV, and estimate the
value of the transmission rate beta in each case. The data fitting is typically done by using the
least squares method. Based on the fitted beta , calculate the basic reproduction number R0
for each.
I am supposed to consider the above instruction to write matlab codes to find the transmission rate for the flu, covid and the RSV data. Each of these diseases has six seasons starting from 2018-2019 seaons to 2023-2024 seasons.
if the matlab data for 2018-2019 flu season is this:
% Filter for the 2018-19 flu season
Flu_data_2018_2019 = Flu_data(strcmp(Flu_data.Season, ‘2018-19’), :);
% Display the filtered data for the 2018-19 season
disp(Flu_data_2018_2019);
% Display the ‘Weekly_Rate’ column (replace with the correct column name if needed)
Weekly_Rate_2018_2019 = Flu_data_2018_2019.WeeklyRate;
disp(Weekly_Rate_2018_2019);
help write a matlab code to find the transmission rate for the 2018-2019 flu season and the reproduction number matlab, least square fitting MATLAB Answers — New Questions
Real-Time error after the last update windows 10
Hi,
My Matlab / Simulink after the latest Windows 10 update (19041.329) can’t simulate in real time, the error message is:
An error occurred while running the simulation and the simulation was terminated
Caused by:
Error reported by S-function ‘sldrtsync’ in ‘ForcesCompensation_vRo_Ce_Be_Ga_v3/Real-Time Synchronization’: Hardware timer cannot be allocated. Real-time kernel cannot run.
Yesterday before the update everything works fine.
I tried to reinstall the following packages:
– Simulink in real time.
– Real-time desktop simulation.
– MATLAB support for the MinGW-w64 C / C ++ compiler.
And reinstall the real-time kernel with "sldrtkernel -install" command.
But the problem was not resolved.
I have the Matlab 2018b and Matlab 2019b and both have the same problem.
Can anyone help me?
Thanks.Hi,
My Matlab / Simulink after the latest Windows 10 update (19041.329) can’t simulate in real time, the error message is:
An error occurred while running the simulation and the simulation was terminated
Caused by:
Error reported by S-function ‘sldrtsync’ in ‘ForcesCompensation_vRo_Ce_Be_Ga_v3/Real-Time Synchronization’: Hardware timer cannot be allocated. Real-time kernel cannot run.
Yesterday before the update everything works fine.
I tried to reinstall the following packages:
– Simulink in real time.
– Real-time desktop simulation.
– MATLAB support for the MinGW-w64 C / C ++ compiler.
And reinstall the real-time kernel with "sldrtkernel -install" command.
But the problem was not resolved.
I have the Matlab 2018b and Matlab 2019b and both have the same problem.
Can anyone help me?
Thanks. Hi,
My Matlab / Simulink after the latest Windows 10 update (19041.329) can’t simulate in real time, the error message is:
An error occurred while running the simulation and the simulation was terminated
Caused by:
Error reported by S-function ‘sldrtsync’ in ‘ForcesCompensation_vRo_Ce_Be_Ga_v3/Real-Time Synchronization’: Hardware timer cannot be allocated. Real-time kernel cannot run.
Yesterday before the update everything works fine.
I tried to reinstall the following packages:
– Simulink in real time.
– Real-time desktop simulation.
– MATLAB support for the MinGW-w64 C / C ++ compiler.
And reinstall the real-time kernel with "sldrtkernel -install" command.
But the problem was not resolved.
I have the Matlab 2018b and Matlab 2019b and both have the same problem.
Can anyone help me?
Thanks. real-time, windows 10 update, simulink, matlab2018b, matlab2019b MATLAB Answers — New Questions
the neuro fuzzy sistem modify the membership function in input or in output?
Hello,
I wanted to understand what happens to the membership function (in input/output) or rules when I insert in the anfis editor: Sugeno file and a training set.Hello,
I wanted to understand what happens to the membership function (in input/output) or rules when I insert in the anfis editor: Sugeno file and a training set. Hello,
I wanted to understand what happens to the membership function (in input/output) or rules when I insert in the anfis editor: Sugeno file and a training set. sugeno, anfis editor, neuro fuzzy MATLAB Answers — New Questions
Thingspeak – Expired License message & upload stopped working
After login I got sweet message:
Expired License
Your trial license has expired as of September 4, 2019.
In account settings:
License Type: Free
Expiration Date: 04 Dec 2019
Available Remaining
Messages 3 000 000 2 851 627
Channels 4 2
Upload to channels stopped working on september 4th…After login I got sweet message:
Expired License
Your trial license has expired as of September 4, 2019.
In account settings:
License Type: Free
Expiration Date: 04 Dec 2019
Available Remaining
Messages 3 000 000 2 851 627
Channels 4 2
Upload to channels stopped working on september 4th… After login I got sweet message:
Expired License
Your trial license has expired as of September 4, 2019.
In account settings:
License Type: Free
Expiration Date: 04 Dec 2019
Available Remaining
Messages 3 000 000 2 851 627
Channels 4 2
Upload to channels stopped working on september 4th… thingspeak licence expired, thingspeak MATLAB Answers — New Questions
extracting table data from a plot picture question
Hello,I have many plots like this in the data sheet.
there is a java tools which manually allow to extract the data of each plot.
Is the a way in matlab to do it?
Thanks.Hello,I have many plots like this in the data sheet.
there is a java tools which manually allow to extract the data of each plot.
Is the a way in matlab to do it?
Thanks. Hello,I have many plots like this in the data sheet.
there is a java tools which manually allow to extract the data of each plot.
Is the a way in matlab to do it?
Thanks. plot, photo MATLAB Answers — New Questions
Matlab plots step response vs timestep number instead of time in seconds
s = tf(‘s’); %Laplace variable
t = 0:0.01:10; %time vector start, dt , end
K = 1;
KI = 0;
sysT = (K * s + KI) / ( ( s^3 + 3*s^2 + (2+K)*s + KI) );
[y,t] = step(sysT,t);
plot(y);
Above snippet shows issue. The issue is that plot(y); plots the step response vs the number of timesteps instead of vs time in secs. For instance if dt = 0.01, the abcissa is in the 100’s. If dt = 0.001, the abcissa is in the 1000’s. If dt = 0.0001, the abcissa is in the 10000’s. How can I just plot vs time in seconds??s = tf(‘s’); %Laplace variable
t = 0:0.01:10; %time vector start, dt , end
K = 1;
KI = 0;
sysT = (K * s + KI) / ( ( s^3 + 3*s^2 + (2+K)*s + KI) );
[y,t] = step(sysT,t);
plot(y);
Above snippet shows issue. The issue is that plot(y); plots the step response vs the number of timesteps instead of vs time in secs. For instance if dt = 0.01, the abcissa is in the 100’s. If dt = 0.001, the abcissa is in the 1000’s. If dt = 0.0001, the abcissa is in the 10000’s. How can I just plot vs time in seconds?? s = tf(‘s’); %Laplace variable
t = 0:0.01:10; %time vector start, dt , end
K = 1;
KI = 0;
sysT = (K * s + KI) / ( ( s^3 + 3*s^2 + (2+K)*s + KI) );
[y,t] = step(sysT,t);
plot(y);
Above snippet shows issue. The issue is that plot(y); plots the step response vs the number of timesteps instead of vs time in secs. For instance if dt = 0.01, the abcissa is in the 100’s. If dt = 0.001, the abcissa is in the 1000’s. If dt = 0.0001, the abcissa is in the 10000’s. How can I just plot vs time in seconds?? plot vs timestep number MATLAB Answers — New Questions
embedding inset plots in subplots matlab
Hi, I am trying to plots an insets plots in aubplots but only the last subplot and inset plots are shown. Please can someone help with this?
% Define xi_select and slope_select for main and inset plots
xi_select = [1, 10, 11, 12];
slope_select = [1, 3, 4, 5, 7, 9];
% Create a tiled layout with 2 rows and 3 columns of subplots
t = tiledlayout(2, 3, ‘TileSpacing’, ‘compact’, ‘Padding’, ‘compact’);
% Define custom colors (green to red gradient)
num_colors = length(xi_select);
custom_colors = [linspace(0, 1, num_colors)’, linspace(1, 0, num_colors)’, zeros(num_colors, 1)];
custom_linestyles = repmat({‘–‘}, 1, num_colors); % All dashed lines
% Initialize arrays for legend
legend_labels = {}; % Cell array to store legend labels
legend_index = 0; % Initialize legend index
plot_handles = []; % Array for storing plot handles
for j = 1:length(slope_select) % Loop through selected ramp_slope
% Create the main subplot
ax = nexttile; % Move to the next tile for each subplot
hold(ax, ‘on’); % Hold on to plot multiple lines in the same subplot
% Loop through selected xi values for the main plots
for i = 1:length(xi_select)
% Find the switching times for the current xi and ramp_slope
transitions = find(switching(:, xi_select(i), slope_select(j)));
if ~isempty(transitions)
% Plot the uxlt data at the switching times for this xi and ramp_slope
for t_idx = transitions
% Extract the uxlt data for the current time point
uxlt_data = squeeze(uxlt(ym, :, xi_select(i), slope_select(j))); % Dimension: y, xi, ramp_slope
% Proper LaTeX syntax for the legend entry, omitting time
display_name = sprintf(‘$\xi=%.2f$’, xi(xi_select(i)));
% Plot the data with specified color and linestyle
h = plot(ax, tarray, uxlt_data, ‘LineWidth’, 1.5, …
‘Color’, custom_colors(i, :), ‘LineStyle’, custom_linestyles{i});
% Only add the label and handle once to the legend
if ~ismember(display_name, legend_labels)
legend_index = legend_index + 1;
plot_handles(legend_index) = h(1); % Store only the first handle
legend_labels{legend_index} = display_name; % Append label
end
end
end
end
% Add labels and title to each subplot
xlabel(ax, ‘Time (s)’, ‘Interpreter’, ‘latex’);
ylabel(ax, ‘$int u_L , mathrm{dz}$’, ‘Interpreter’, ‘latex’);
xlim(ax, [min(tarray), max(tarray)]);
ylim(ax, [0, 2.5e-04]);
% Add subplot numbering and S_{rf} value
title(ax, sprintf(‘(%c) $S_{rf} = %.2f$’, ‘a’ + j – 1, ramp_slope(slope_select(j))), ‘Interpreter’, ‘latex’);
box(ax, ‘on’); % Place a box around the current subplot
% Add inset plots
inset_pos = ax.Position; % Get the position of the current subplot
if j == 1
% The first inset plot on the top left of the first subplot
inset_pos = [inset_pos(1) + 0.1 * inset_pos(3), inset_pos(2) + 0.65 * inset_pos(4), 0.15, 0.15];
else
% Remaining inset plots on the top right of the respective subplots
inset_pos = [inset_pos(1) + 0.6 * inset_pos(3), inset_pos(2) + 0.65 * inset_pos(4), 0.15, 0.15];
end
% Create new axes for inset plots, without attaching them to the main subplot axes
inset_ax = axes(‘Position’, inset_pos); % Create inset axes with adjusted position
hold(inset_ax, ‘on’); % Hold on inset axes to plot multiple lines
% Loop through selected xi values for the inset plots
for i = 1:length(xi_select)
% Find the switching times for the current xi and ramp_slope
transitions = find(switching(:, xi_select(i), slope_select(j)));
if ~isempty(transitions)
% Plot the uxl data in the inset plot
for t_idx = transitions
% Extract the uxl data for the current time point
uxl_data = squeeze(uxl(:, xi_select(i), slope_select(j))); % Dimension: y, xi, ramp_slope
% Plot the data in the inset with color and linestyle
plot(inset_ax, y, uxl_data, ‘LineWidth’, 1.5, …
‘Color’, custom_colors(i, :), ‘LineStyle’, custom_linestyles{i});
end
end
end
% Add labels to inset
xlabel(inset_ax, ‘$z$’, ‘Interpreter’, ‘latex’);
ylabel(inset_ax, ‘$int u_L , mathrm{dz}$’, ‘Interpreter’, ‘latex’);
xlim(inset_ax, [min(y), max(y)]);
hold(inset_ax, ‘off’); % Release hold on the inset axes
end
% Create a global legend (optional, if needed)
lgd = legend(plot_handles, legend_labels, ‘Interpreter’, ‘latex’, ‘NumColumns’, 2);
% Ensure that the entire figure has a box around it
box on;Hi, I am trying to plots an insets plots in aubplots but only the last subplot and inset plots are shown. Please can someone help with this?
% Define xi_select and slope_select for main and inset plots
xi_select = [1, 10, 11, 12];
slope_select = [1, 3, 4, 5, 7, 9];
% Create a tiled layout with 2 rows and 3 columns of subplots
t = tiledlayout(2, 3, ‘TileSpacing’, ‘compact’, ‘Padding’, ‘compact’);
% Define custom colors (green to red gradient)
num_colors = length(xi_select);
custom_colors = [linspace(0, 1, num_colors)’, linspace(1, 0, num_colors)’, zeros(num_colors, 1)];
custom_linestyles = repmat({‘–‘}, 1, num_colors); % All dashed lines
% Initialize arrays for legend
legend_labels = {}; % Cell array to store legend labels
legend_index = 0; % Initialize legend index
plot_handles = []; % Array for storing plot handles
for j = 1:length(slope_select) % Loop through selected ramp_slope
% Create the main subplot
ax = nexttile; % Move to the next tile for each subplot
hold(ax, ‘on’); % Hold on to plot multiple lines in the same subplot
% Loop through selected xi values for the main plots
for i = 1:length(xi_select)
% Find the switching times for the current xi and ramp_slope
transitions = find(switching(:, xi_select(i), slope_select(j)));
if ~isempty(transitions)
% Plot the uxlt data at the switching times for this xi and ramp_slope
for t_idx = transitions
% Extract the uxlt data for the current time point
uxlt_data = squeeze(uxlt(ym, :, xi_select(i), slope_select(j))); % Dimension: y, xi, ramp_slope
% Proper LaTeX syntax for the legend entry, omitting time
display_name = sprintf(‘$\xi=%.2f$’, xi(xi_select(i)));
% Plot the data with specified color and linestyle
h = plot(ax, tarray, uxlt_data, ‘LineWidth’, 1.5, …
‘Color’, custom_colors(i, :), ‘LineStyle’, custom_linestyles{i});
% Only add the label and handle once to the legend
if ~ismember(display_name, legend_labels)
legend_index = legend_index + 1;
plot_handles(legend_index) = h(1); % Store only the first handle
legend_labels{legend_index} = display_name; % Append label
end
end
end
end
% Add labels and title to each subplot
xlabel(ax, ‘Time (s)’, ‘Interpreter’, ‘latex’);
ylabel(ax, ‘$int u_L , mathrm{dz}$’, ‘Interpreter’, ‘latex’);
xlim(ax, [min(tarray), max(tarray)]);
ylim(ax, [0, 2.5e-04]);
% Add subplot numbering and S_{rf} value
title(ax, sprintf(‘(%c) $S_{rf} = %.2f$’, ‘a’ + j – 1, ramp_slope(slope_select(j))), ‘Interpreter’, ‘latex’);
box(ax, ‘on’); % Place a box around the current subplot
% Add inset plots
inset_pos = ax.Position; % Get the position of the current subplot
if j == 1
% The first inset plot on the top left of the first subplot
inset_pos = [inset_pos(1) + 0.1 * inset_pos(3), inset_pos(2) + 0.65 * inset_pos(4), 0.15, 0.15];
else
% Remaining inset plots on the top right of the respective subplots
inset_pos = [inset_pos(1) + 0.6 * inset_pos(3), inset_pos(2) + 0.65 * inset_pos(4), 0.15, 0.15];
end
% Create new axes for inset plots, without attaching them to the main subplot axes
inset_ax = axes(‘Position’, inset_pos); % Create inset axes with adjusted position
hold(inset_ax, ‘on’); % Hold on inset axes to plot multiple lines
% Loop through selected xi values for the inset plots
for i = 1:length(xi_select)
% Find the switching times for the current xi and ramp_slope
transitions = find(switching(:, xi_select(i), slope_select(j)));
if ~isempty(transitions)
% Plot the uxl data in the inset plot
for t_idx = transitions
% Extract the uxl data for the current time point
uxl_data = squeeze(uxl(:, xi_select(i), slope_select(j))); % Dimension: y, xi, ramp_slope
% Plot the data in the inset with color and linestyle
plot(inset_ax, y, uxl_data, ‘LineWidth’, 1.5, …
‘Color’, custom_colors(i, :), ‘LineStyle’, custom_linestyles{i});
end
end
end
% Add labels to inset
xlabel(inset_ax, ‘$z$’, ‘Interpreter’, ‘latex’);
ylabel(inset_ax, ‘$int u_L , mathrm{dz}$’, ‘Interpreter’, ‘latex’);
xlim(inset_ax, [min(y), max(y)]);
hold(inset_ax, ‘off’); % Release hold on the inset axes
end
% Create a global legend (optional, if needed)
lgd = legend(plot_handles, legend_labels, ‘Interpreter’, ‘latex’, ‘NumColumns’, 2);
% Ensure that the entire figure has a box around it
box on; Hi, I am trying to plots an insets plots in aubplots but only the last subplot and inset plots are shown. Please can someone help with this?
% Define xi_select and slope_select for main and inset plots
xi_select = [1, 10, 11, 12];
slope_select = [1, 3, 4, 5, 7, 9];
% Create a tiled layout with 2 rows and 3 columns of subplots
t = tiledlayout(2, 3, ‘TileSpacing’, ‘compact’, ‘Padding’, ‘compact’);
% Define custom colors (green to red gradient)
num_colors = length(xi_select);
custom_colors = [linspace(0, 1, num_colors)’, linspace(1, 0, num_colors)’, zeros(num_colors, 1)];
custom_linestyles = repmat({‘–‘}, 1, num_colors); % All dashed lines
% Initialize arrays for legend
legend_labels = {}; % Cell array to store legend labels
legend_index = 0; % Initialize legend index
plot_handles = []; % Array for storing plot handles
for j = 1:length(slope_select) % Loop through selected ramp_slope
% Create the main subplot
ax = nexttile; % Move to the next tile for each subplot
hold(ax, ‘on’); % Hold on to plot multiple lines in the same subplot
% Loop through selected xi values for the main plots
for i = 1:length(xi_select)
% Find the switching times for the current xi and ramp_slope
transitions = find(switching(:, xi_select(i), slope_select(j)));
if ~isempty(transitions)
% Plot the uxlt data at the switching times for this xi and ramp_slope
for t_idx = transitions
% Extract the uxlt data for the current time point
uxlt_data = squeeze(uxlt(ym, :, xi_select(i), slope_select(j))); % Dimension: y, xi, ramp_slope
% Proper LaTeX syntax for the legend entry, omitting time
display_name = sprintf(‘$\xi=%.2f$’, xi(xi_select(i)));
% Plot the data with specified color and linestyle
h = plot(ax, tarray, uxlt_data, ‘LineWidth’, 1.5, …
‘Color’, custom_colors(i, :), ‘LineStyle’, custom_linestyles{i});
% Only add the label and handle once to the legend
if ~ismember(display_name, legend_labels)
legend_index = legend_index + 1;
plot_handles(legend_index) = h(1); % Store only the first handle
legend_labels{legend_index} = display_name; % Append label
end
end
end
end
% Add labels and title to each subplot
xlabel(ax, ‘Time (s)’, ‘Interpreter’, ‘latex’);
ylabel(ax, ‘$int u_L , mathrm{dz}$’, ‘Interpreter’, ‘latex’);
xlim(ax, [min(tarray), max(tarray)]);
ylim(ax, [0, 2.5e-04]);
% Add subplot numbering and S_{rf} value
title(ax, sprintf(‘(%c) $S_{rf} = %.2f$’, ‘a’ + j – 1, ramp_slope(slope_select(j))), ‘Interpreter’, ‘latex’);
box(ax, ‘on’); % Place a box around the current subplot
% Add inset plots
inset_pos = ax.Position; % Get the position of the current subplot
if j == 1
% The first inset plot on the top left of the first subplot
inset_pos = [inset_pos(1) + 0.1 * inset_pos(3), inset_pos(2) + 0.65 * inset_pos(4), 0.15, 0.15];
else
% Remaining inset plots on the top right of the respective subplots
inset_pos = [inset_pos(1) + 0.6 * inset_pos(3), inset_pos(2) + 0.65 * inset_pos(4), 0.15, 0.15];
end
% Create new axes for inset plots, without attaching them to the main subplot axes
inset_ax = axes(‘Position’, inset_pos); % Create inset axes with adjusted position
hold(inset_ax, ‘on’); % Hold on inset axes to plot multiple lines
% Loop through selected xi values for the inset plots
for i = 1:length(xi_select)
% Find the switching times for the current xi and ramp_slope
transitions = find(switching(:, xi_select(i), slope_select(j)));
if ~isempty(transitions)
% Plot the uxl data in the inset plot
for t_idx = transitions
% Extract the uxl data for the current time point
uxl_data = squeeze(uxl(:, xi_select(i), slope_select(j))); % Dimension: y, xi, ramp_slope
% Plot the data in the inset with color and linestyle
plot(inset_ax, y, uxl_data, ‘LineWidth’, 1.5, …
‘Color’, custom_colors(i, :), ‘LineStyle’, custom_linestyles{i});
end
end
end
% Add labels to inset
xlabel(inset_ax, ‘$z$’, ‘Interpreter’, ‘latex’);
ylabel(inset_ax, ‘$int u_L , mathrm{dz}$’, ‘Interpreter’, ‘latex’);
xlim(inset_ax, [min(y), max(y)]);
hold(inset_ax, ‘off’); % Release hold on the inset axes
end
% Create a global legend (optional, if needed)
lgd = legend(plot_handles, legend_labels, ‘Interpreter’, ‘latex’, ‘NumColumns’, 2);
% Ensure that the entire figure has a box around it
box on; insets, subplots, figures MATLAB Answers — New Questions
How to write code for repetitive process?
Hello.
Given:
Variable components: species_1 and species_2;
Initial values: species_1=1000 g and species_2=200 g;
Process parameters: timecut=2 hour, timecon=1hour, kf=0.1 1/hour
The components are related to each other by a reaction according to the law of mass action:
species_1 -> species_2: kf*species_1
In the time interval 0<=time<=timecut, the mass of the component species_1 decreases while the mass of the component species_2 increases. At the point time=timecut, the masses of the components return to their initial values. Two sawtooth patterns are formed on the charts: one sawtooth on chart for species_1 and another inverted sawtooth on chart for species_2.
Then, during the timecon, the masses of components retain their original values: "shelves" are formed. Codes in the Simbiology Builder:
Trigger: time>=timecut
Event FCNS:
kf=0
species_1=1000
species_2=200
If time>=timecut+timecon and kf=0.1, the mass of the species_1 component decreases and the mass of the species_2 component increases. No saw teeth and shelves are formed.
How should the code be written in Simbiology so that the "tooth-shelf" process are repeated n times?Hello.
Given:
Variable components: species_1 and species_2;
Initial values: species_1=1000 g and species_2=200 g;
Process parameters: timecut=2 hour, timecon=1hour, kf=0.1 1/hour
The components are related to each other by a reaction according to the law of mass action:
species_1 -> species_2: kf*species_1
In the time interval 0<=time<=timecut, the mass of the component species_1 decreases while the mass of the component species_2 increases. At the point time=timecut, the masses of the components return to their initial values. Two sawtooth patterns are formed on the charts: one sawtooth on chart for species_1 and another inverted sawtooth on chart for species_2.
Then, during the timecon, the masses of components retain their original values: "shelves" are formed. Codes in the Simbiology Builder:
Trigger: time>=timecut
Event FCNS:
kf=0
species_1=1000
species_2=200
If time>=timecut+timecon and kf=0.1, the mass of the species_1 component decreases and the mass of the species_2 component increases. No saw teeth and shelves are formed.
How should the code be written in Simbiology so that the "tooth-shelf" process are repeated n times? Hello.
Given:
Variable components: species_1 and species_2;
Initial values: species_1=1000 g and species_2=200 g;
Process parameters: timecut=2 hour, timecon=1hour, kf=0.1 1/hour
The components are related to each other by a reaction according to the law of mass action:
species_1 -> species_2: kf*species_1
In the time interval 0<=time<=timecut, the mass of the component species_1 decreases while the mass of the component species_2 increases. At the point time=timecut, the masses of the components return to their initial values. Two sawtooth patterns are formed on the charts: one sawtooth on chart for species_1 and another inverted sawtooth on chart for species_2.
Then, during the timecon, the masses of components retain their original values: "shelves" are formed. Codes in the Simbiology Builder:
Trigger: time>=timecut
Event FCNS:
kf=0
species_1=1000
species_2=200
If time>=timecut+timecon and kf=0.1, the mass of the species_1 component decreases and the mass of the species_2 component increases. No saw teeth and shelves are formed.
How should the code be written in Simbiology so that the "tooth-shelf" process are repeated n times? programming, cycle MATLAB Answers — New Questions
Check tcpclient connection status
Hi, I’m currently using tcpclient command to establish tcpip communication with an external device (not tcpip() from instrument toolbox).
Sometimes, the connection is not stable and when I write data it returns error: ‘An existing connection was forcibly closed by the remote host’.
Therefore, I want to implement a method to check the connection before I write / read from the connection. However, I just couldn’t find much information regarding tcpclient.
I have attempted the following:
% connect
t=tcpclient(‘172.1.1.102′,50000,’Timeout’,1,’ConnectTimeout’,3);
Here, before running the I deliberately disconnected Ethernet cable just want to test trigger the error:
% My intention: try a write / read to check if an error returns
try
write(t,0);
read(t);
disp(‘send succesfull’);
catch ME
disp(‘connection lost’);
disp(ME.identifier);
end
For the first run ‘An existing connection was forcibly closed by the remote host’ still appears in the command window and the ‘send successful’ message is printed, catch statement is skipped. However, a second run will jump into catch statement though.
If I run the try-catch step by step, when it reaches ‘write(t,0)’ statement it returns the ”An existing … remote host”, and continue to ‘read(t)’ statement, and jumps into catch statement.
I couldn’t quite understand why this happened.
Thanks for your help very much!Hi, I’m currently using tcpclient command to establish tcpip communication with an external device (not tcpip() from instrument toolbox).
Sometimes, the connection is not stable and when I write data it returns error: ‘An existing connection was forcibly closed by the remote host’.
Therefore, I want to implement a method to check the connection before I write / read from the connection. However, I just couldn’t find much information regarding tcpclient.
I have attempted the following:
% connect
t=tcpclient(‘172.1.1.102′,50000,’Timeout’,1,’ConnectTimeout’,3);
Here, before running the I deliberately disconnected Ethernet cable just want to test trigger the error:
% My intention: try a write / read to check if an error returns
try
write(t,0);
read(t);
disp(‘send succesfull’);
catch ME
disp(‘connection lost’);
disp(ME.identifier);
end
For the first run ‘An existing connection was forcibly closed by the remote host’ still appears in the command window and the ‘send successful’ message is printed, catch statement is skipped. However, a second run will jump into catch statement though.
If I run the try-catch step by step, when it reaches ‘write(t,0)’ statement it returns the ”An existing … remote host”, and continue to ‘read(t)’ statement, and jumps into catch statement.
I couldn’t quite understand why this happened.
Thanks for your help very much! Hi, I’m currently using tcpclient command to establish tcpip communication with an external device (not tcpip() from instrument toolbox).
Sometimes, the connection is not stable and when I write data it returns error: ‘An existing connection was forcibly closed by the remote host’.
Therefore, I want to implement a method to check the connection before I write / read from the connection. However, I just couldn’t find much information regarding tcpclient.
I have attempted the following:
% connect
t=tcpclient(‘172.1.1.102′,50000,’Timeout’,1,’ConnectTimeout’,3);
Here, before running the I deliberately disconnected Ethernet cable just want to test trigger the error:
% My intention: try a write / read to check if an error returns
try
write(t,0);
read(t);
disp(‘send succesfull’);
catch ME
disp(‘connection lost’);
disp(ME.identifier);
end
For the first run ‘An existing connection was forcibly closed by the remote host’ still appears in the command window and the ‘send successful’ message is printed, catch statement is skipped. However, a second run will jump into catch statement though.
If I run the try-catch step by step, when it reaches ‘write(t,0)’ statement it returns the ”An existing … remote host”, and continue to ‘read(t)’ statement, and jumps into catch statement.
I couldn’t quite understand why this happened.
Thanks for your help very much! matlab, tcpip, tcpclient, connection MATLAB Answers — New Questions
How to export 500 images in one file
I have around 500 images which I want to export into one .xls .word or .pdf in 3 collums. I tried putting them all in one figure but then they become extremely small. If I create many separete figures by 3, I don´t know how to export them all into one file. I tried putting all of the images into table but then the table doesn´t generate because I run out of memorry. If I were to guess, it most likely tries to put every single value of the picture into a separated cell.
How do I make pictures in the table to be saved in their pictural form?
clear all; clc; close all;
dir_img_AP=dir(‘**/NUSCH*AP/**/*.jpg’);
dir_img_JS=dir(‘**/NUSCH*JS/**/*.jpg’);
dir_img_LZ=dir(‘**/NUSCH*LZ/**/*.jpg’);
n_img_AP=numel(dir_img_AP);
n_img_JS=numel(dir_img_JS);
n_img_LZ=numel(dir_img_LZ);
n_img=max([n_img_AP,n_img_JS,n_img_LZ]);
T=cell(n_img,1);
for i=1:n_img_AP
file_path_AP=[dir_img_AP(i).folder, ‘/’,dir_img_AP(i).name];
filename_AP = dir_img_AP(i).name;
pic_AP = imread(filename_AP);
T{i,1} = pic_AP;
end
for i=1:n_img_JS
file_path_JS=[dir_img_JS(i).folder, ‘/’,dir_img_JS(i).name];
filename_JS = dir_img_JS(i).name;
pic_JS = imread(filename_JS);
T{i,2} = pic_JS;
end
for i=1:n_img_LZ
file_path_LZ=[dir_img_LZ(i).folder, ‘/’,dir_img_LZ(i).name];
filename_LZ = dir_img_LZ(i).name;
pic_LZ = imread(filename_LZ);
T{i,3} = pic_LZ;
end
table=cell2table(T);
arr = table2array(table);
% imshow(arr);
f_name=’pic.xls’;
% writetable(table,f_name)I have around 500 images which I want to export into one .xls .word or .pdf in 3 collums. I tried putting them all in one figure but then they become extremely small. If I create many separete figures by 3, I don´t know how to export them all into one file. I tried putting all of the images into table but then the table doesn´t generate because I run out of memorry. If I were to guess, it most likely tries to put every single value of the picture into a separated cell.
How do I make pictures in the table to be saved in their pictural form?
clear all; clc; close all;
dir_img_AP=dir(‘**/NUSCH*AP/**/*.jpg’);
dir_img_JS=dir(‘**/NUSCH*JS/**/*.jpg’);
dir_img_LZ=dir(‘**/NUSCH*LZ/**/*.jpg’);
n_img_AP=numel(dir_img_AP);
n_img_JS=numel(dir_img_JS);
n_img_LZ=numel(dir_img_LZ);
n_img=max([n_img_AP,n_img_JS,n_img_LZ]);
T=cell(n_img,1);
for i=1:n_img_AP
file_path_AP=[dir_img_AP(i).folder, ‘/’,dir_img_AP(i).name];
filename_AP = dir_img_AP(i).name;
pic_AP = imread(filename_AP);
T{i,1} = pic_AP;
end
for i=1:n_img_JS
file_path_JS=[dir_img_JS(i).folder, ‘/’,dir_img_JS(i).name];
filename_JS = dir_img_JS(i).name;
pic_JS = imread(filename_JS);
T{i,2} = pic_JS;
end
for i=1:n_img_LZ
file_path_LZ=[dir_img_LZ(i).folder, ‘/’,dir_img_LZ(i).name];
filename_LZ = dir_img_LZ(i).name;
pic_LZ = imread(filename_LZ);
T{i,3} = pic_LZ;
end
table=cell2table(T);
arr = table2array(table);
% imshow(arr);
f_name=’pic.xls’;
% writetable(table,f_name) I have around 500 images which I want to export into one .xls .word or .pdf in 3 collums. I tried putting them all in one figure but then they become extremely small. If I create many separete figures by 3, I don´t know how to export them all into one file. I tried putting all of the images into table but then the table doesn´t generate because I run out of memorry. If I were to guess, it most likely tries to put every single value of the picture into a separated cell.
How do I make pictures in the table to be saved in their pictural form?
clear all; clc; close all;
dir_img_AP=dir(‘**/NUSCH*AP/**/*.jpg’);
dir_img_JS=dir(‘**/NUSCH*JS/**/*.jpg’);
dir_img_LZ=dir(‘**/NUSCH*LZ/**/*.jpg’);
n_img_AP=numel(dir_img_AP);
n_img_JS=numel(dir_img_JS);
n_img_LZ=numel(dir_img_LZ);
n_img=max([n_img_AP,n_img_JS,n_img_LZ]);
T=cell(n_img,1);
for i=1:n_img_AP
file_path_AP=[dir_img_AP(i).folder, ‘/’,dir_img_AP(i).name];
filename_AP = dir_img_AP(i).name;
pic_AP = imread(filename_AP);
T{i,1} = pic_AP;
end
for i=1:n_img_JS
file_path_JS=[dir_img_JS(i).folder, ‘/’,dir_img_JS(i).name];
filename_JS = dir_img_JS(i).name;
pic_JS = imread(filename_JS);
T{i,2} = pic_JS;
end
for i=1:n_img_LZ
file_path_LZ=[dir_img_LZ(i).folder, ‘/’,dir_img_LZ(i).name];
filename_LZ = dir_img_LZ(i).name;
pic_LZ = imread(filename_LZ);
T{i,3} = pic_LZ;
end
table=cell2table(T);
arr = table2array(table);
% imshow(arr);
f_name=’pic.xls’;
% writetable(table,f_name) database, export, image MATLAB Answers — New Questions
store images in a table
Hello,
I have three images and want to store them in a table. I wrote this in command line:
dinfo = dir(‘*.jpg’);
T = table();
for K = 1 : length(dinfo)
filename = dinfo(K).name;
filecontent = imread(filename);
T{filename,1} = filecontent;
end
It gives the error:
To assign to or create a variable in a table, the number of rows must match the height of the table.
Secondly, I want to make my three images equal in size, as we have to make the size of variables in rows equal for tables.
Help needed. Thanks in advance.Hello,
I have three images and want to store them in a table. I wrote this in command line:
dinfo = dir(‘*.jpg’);
T = table();
for K = 1 : length(dinfo)
filename = dinfo(K).name;
filecontent = imread(filename);
T{filename,1} = filecontent;
end
It gives the error:
To assign to or create a variable in a table, the number of rows must match the height of the table.
Secondly, I want to make my three images equal in size, as we have to make the size of variables in rows equal for tables.
Help needed. Thanks in advance. Hello,
I have three images and want to store them in a table. I wrote this in command line:
dinfo = dir(‘*.jpg’);
T = table();
for K = 1 : length(dinfo)
filename = dinfo(K).name;
filecontent = imread(filename);
T{filename,1} = filecontent;
end
It gives the error:
To assign to or create a variable in a table, the number of rows must match the height of the table.
Secondly, I want to make my three images equal in size, as we have to make the size of variables in rows equal for tables.
Help needed. Thanks in advance. matlab tables, table, database MATLAB Answers — New Questions
Cannot run kernel – Simulink Desktop Real-time
When I try to use the Desktop-Real-Time/Stream input in the simulink, and start a simulation, it apears a error message like "cannot acess kernel". Why do this happen? How it can be solved? Thanks!
The error message is: "Hardware time cannot be allocated. Real time kernel cannot run"When I try to use the Desktop-Real-Time/Stream input in the simulink, and start a simulation, it apears a error message like "cannot acess kernel". Why do this happen? How it can be solved? Thanks!
The error message is: "Hardware time cannot be allocated. Real time kernel cannot run" When I try to use the Desktop-Real-Time/Stream input in the simulink, and start a simulation, it apears a error message like "cannot acess kernel". Why do this happen? How it can be solved? Thanks!
The error message is: "Hardware time cannot be allocated. Real time kernel cannot run" simulink, kernel MATLAB Answers — New Questions
problem in simulink (FOPID controller)
Derivative of state ‘1’ in block ‘FOPIDC/Fractional PID controller1/Fractional derivative1/Transfer
Fcn1′ at time 3.2188803386168852 is not finite. The simulation will be stopped. There may be a
singularity in the solution. If not, try reducing the step size (either by reducing the fixed step
size or by tightening the error tolerances)
how can I solve this errorDerivative of state ‘1’ in block ‘FOPIDC/Fractional PID controller1/Fractional derivative1/Transfer
Fcn1′ at time 3.2188803386168852 is not finite. The simulation will be stopped. There may be a
singularity in the solution. If not, try reducing the step size (either by reducing the fixed step
size or by tightening the error tolerances)
how can I solve this error Derivative of state ‘1’ in block ‘FOPIDC/Fractional PID controller1/Fractional derivative1/Transfer
Fcn1′ at time 3.2188803386168852 is not finite. The simulation will be stopped. There may be a
singularity in the solution. If not, try reducing the step size (either by reducing the fixed step
size or by tightening the error tolerances)
how can I solve this error fopid MATLAB Answers — New Questions
