Why can I no longer add a plot with a legend in R2016b same way it used to be in previews versions?

During the migration to R2016b or R2017a, the script show some problems with legend function in the plot.

An example is a simple line after a plot:
legend(‘Curve1′,’Curve2’,1);

While it used to work fine prior to R2016b, we see the following response :

Error using legend>process_inputs (line 582)
Invalid argument. Type ‘help legend’ for more information.

Error in legend>make_legend (line 340)
[autoupdate,orient,location,position,children,listen,strings,propargs] = process_inputs(ha,argin); %#ok

Error in legend (line 294)
make_legend(ha,args(arg:end),version);

Error in Script_plot (line 274)
legend(‘Curve1′,’Curve2’,1);Why can I no longer add a plot with a legend in R2016b same way it used to be in previews versions?

During the migration to R2016b or R2017a, the script show some problems with legend function in the plot.

An example is a simple line after a plot:
legend(‘Curve1′,’Curve2’,1);

While it used to work fine prior to R2016b, we see the following response :

Error using legend>process_inputs (line 582)
Invalid argument. Type ‘help legend’ for more information.

Error in legend>make_legend (line 340)
[autoupdate,orient,location,position,children,listen,strings,propargs] = process_inputs(ha,argin); %#ok

Error in legend (line 294)
make_legend(ha,args(arg:end),version);

Error in Script_plot (line 274)
legend(‘Curve1′,’Curve2’,1); Why can I no longer add a plot with a legend in R2016b same way it used to be in previews versions?

During the migration to R2016b or R2017a, the script show some problems with legend function in the plot.

An example is a simple line after a plot:
legend(‘Curve1′,’Curve2’,1);

While it used to work fine prior to R2016b, we see the following response :

Error using legend>process_inputs (line 582)
Invalid argument. Type ‘help legend’ for more information.

Error in legend>make_legend (line 340)
[autoupdate,orient,location,position,children,listen,strings,propargs] = process_inputs(ha,argin); %#ok

Error in legend (line 294)
make_legend(ha,args(arg:end),version);

Error in Script_plot (line 274)
legend(‘Curve1′,’Curve2’,1); legend, plot MATLAB Answers — New Questions

​

Hello, Im trying to downsample a 100k lines of excel and want to pick out only those rows where the column 1 values are closest to multiples of a number, say 50. So in the pic below, just the green highlighted rows.

I’ve thought about using mod but this doesn’t work:
%For example if I choose these numbers (and consider just a column vector)
n=[1;5;7;9;12;15;29;33;39]
and want to pick out those numbers closest to multiples of 10
mod(n,10)
ans =

1
5
7
9
2
5
9
3
9
If I sort these and pick out the lowest ones, the value of 29 & 39 wouldn’t be picked out.

I also considered some kind of interp, but I need to pick out the actual discrete value from the table below not an interpolated value. Normally I show my arttempt, but Im actually stuck here in where to actually start with this.Hello, Im trying to downsample a 100k lines of excel and want to pick out only those rows where the column 1 values are closest to multiples of a number, say 50. So in the pic below, just the green highlighted rows.

I’ve thought about using mod but this doesn’t work:
%For example if I choose these numbers (and consider just a column vector)
n=[1;5;7;9;12;15;29;33;39]
and want to pick out those numbers closest to multiples of 10
mod(n,10)
ans =

1
5
7
9
2
5
9
3
9
If I sort these and pick out the lowest ones, the value of 29 & 39 wouldn’t be picked out.

I also considered some kind of interp, but I need to pick out the actual discrete value from the table below not an interpolated value. Normally I show my arttempt, but Im actually stuck here in where to actually start with this. Hello, Im trying to downsample a 100k lines of excel and want to pick out only those rows where the column 1 values are closest to multiples of a number, say 50. So in the pic below, just the green highlighted rows.

I’ve thought about using mod but this doesn’t work:
%For example if I choose these numbers (and consider just a column vector)
n=[1;5;7;9;12;15;29;33;39]
and want to pick out those numbers closest to multiples of 10
mod(n,10)
ans =

1
5
7
9
2
5
9
3
9
If I sort these and pick out the lowest ones, the value of 29 & 39 wouldn’t be picked out.

I also considered some kind of interp, but I need to pick out the actual discrete value from the table below not an interpolated value. Normally I show my arttempt, but Im actually stuck here in where to actually start with this. array, mod MATLAB Answers — New Questions

​

I am solving a PDE via finite volume methods. As a result, I am left with a set of ODEs which I solve with ode15s. I have a bunch of constants which I use in the function defining the derivative, and I insert these by the inclusion of a struct to the function. When I run the ode15s I obtain an error which pertains to the inclusion of the struct as an input to the function. I don’t really want to include the constants within the function as that seems "messy", so I would prefer to do it via a struct.

%This is a code to solve the isothermal sintering equations.
%%—Physical parameters—
S=struct;
S.g = 9.81; %Acceleration due to gravity.
S.K = 1; %This is the Laplace constant from the sintering potential.
S.eta_0 = 1; %The shear stress of the skeleton
S.T = 3600; %This is the time of sintering.
S.M = 0.1; %This is the total mass of the rod being sintered .
S.rho_m = 1; %This is the density of the individual metallic particles.
S.N = 251; %This is the number of cells-1
L = 1; %This is the initial length of the rod
%%—Initial conditions
%Length
X=linspace(0,L,S.N); %This is the partition of the initial length
%Density
rho_0=0.5+0.1*sin(6*pi*X);
%Amount of mass
S.h=cumtrapz(X,rho_0);
%Initial velocity
u0=zeros(S.N-1,1);
% Define the system of equations as a function
y0 = [1./rho_0′; u0];
% Finite Volume Method solution loop
tspan = [0 3]; % You can adjust this based on the time range you’re interested in
[t, y] = ode15s(@ode_system, tspan, y0);
function dydt = ode_system(t, y, S)
S.M
U=S.M/(S.rho_m*S.T);
a_1=S.T/(U*S.M);
a_2=S.T*S.rho_m/S.M^2;
a_3=S.g*S.T/U;
dh=diff(S.h);
y=NaN(1,2*S.N-2);
nu = y(1:S.N-1); % u(t)
u = y(S.N:2*S.N-2); % v(t)
%Compute the left and right specific volumes:
nu_left=15*nu(1)/8-5*nu(2)/4+3*nu(3)/8;
nu_right=15*nu(S.N-3)/8-5*nu(S.N-2)/5+3*nu(S.N-1)/8;
%Compute the fluxes on the interfaces
nu_half=NaN(1:S.N);
nu_half(2:N-1)=0.5*(nu(2:S.N-1)+nu(1:S.N-2));
nu_half(1)=nu_left;
nu_half(N)=nu_right;
%Fluxes for nu equation
nu_flux=zeros(S.N,1);
nu_flux(2:S.N-1)=0.5*(u(1:S.N-2)+u(2:S.N-1));
du_dh_left=-P_L(S,nu_left)*nu_left/zeta(nu_left);
du_dh_right=-P_L(S,nu_right)*nu_right/zeta(nu_right);
nu_flux(1)=-3*du_dh_left*dh(1)/8+9*u(1)/8-u(2)/8;
nu_flux(N)=3*du_dh_right*dh(N-1)/8-9*u(N-1)/8+u(N-2)/8;
%Fluxes for u equation
u_grad=2*(u(2:S.N-2)-u(1:S.N-2))./(dh(2:S.N-1)+dh(1:S.N-2));
u_flux=NaN(S.N,1);
u_flux(2:S.N-1)=a_1*P_L(S,nu_half(2:S.N-1))+(a_2*zeta(nu_half(2:S.N-1))./nu_half(2:S.N-1)).*u_grad;
% The system of equationsY
dnu_dt = nu_flux(2:S.N)-nu_flux(1:S.N-1);
du_dt =u_flux(2:S.N)-u_flux(1:S.N-1);

% Return the derivatives
dydt = [dnu_dt; du_dt];
end
function y=P_L(S,nu)
y=S.K./nu;
end
function y=zeta(nu)
y=(2/3)*(nu).^(-2)./(nu-1);
endI am solving a PDE via finite volume methods. As a result, I am left with a set of ODEs which I solve with ode15s. I have a bunch of constants which I use in the function defining the derivative, and I insert these by the inclusion of a struct to the function. When I run the ode15s I obtain an error which pertains to the inclusion of the struct as an input to the function. I don’t really want to include the constants within the function as that seems "messy", so I would prefer to do it via a struct.

%This is a code to solve the isothermal sintering equations.
%%—Physical parameters—
S=struct;
S.g = 9.81; %Acceleration due to gravity.
S.K = 1; %This is the Laplace constant from the sintering potential.
S.eta_0 = 1; %The shear stress of the skeleton
S.T = 3600; %This is the time of sintering.
S.M = 0.1; %This is the total mass of the rod being sintered .
S.rho_m = 1; %This is the density of the individual metallic particles.
S.N = 251; %This is the number of cells-1
L = 1; %This is the initial length of the rod
%%—Initial conditions
%Length
X=linspace(0,L,S.N); %This is the partition of the initial length
%Density
rho_0=0.5+0.1*sin(6*pi*X);
%Amount of mass
S.h=cumtrapz(X,rho_0);
%Initial velocity
u0=zeros(S.N-1,1);
% Define the system of equations as a function
y0 = [1./rho_0′; u0];
% Finite Volume Method solution loop
tspan = [0 3]; % You can adjust this based on the time range you’re interested in
[t, y] = ode15s(@ode_system, tspan, y0);
function dydt = ode_system(t, y, S)
S.M
U=S.M/(S.rho_m*S.T);
a_1=S.T/(U*S.M);
a_2=S.T*S.rho_m/S.M^2;
a_3=S.g*S.T/U;
dh=diff(S.h);
y=NaN(1,2*S.N-2);
nu = y(1:S.N-1); % u(t)
u = y(S.N:2*S.N-2); % v(t)
%Compute the left and right specific volumes:
nu_left=15*nu(1)/8-5*nu(2)/4+3*nu(3)/8;
nu_right=15*nu(S.N-3)/8-5*nu(S.N-2)/5+3*nu(S.N-1)/8;
%Compute the fluxes on the interfaces
nu_half=NaN(1:S.N);
nu_half(2:N-1)=0.5*(nu(2:S.N-1)+nu(1:S.N-2));
nu_half(1)=nu_left;
nu_half(N)=nu_right;
%Fluxes for nu equation
nu_flux=zeros(S.N,1);
nu_flux(2:S.N-1)=0.5*(u(1:S.N-2)+u(2:S.N-1));
du_dh_left=-P_L(S,nu_left)*nu_left/zeta(nu_left);
du_dh_right=-P_L(S,nu_right)*nu_right/zeta(nu_right);
nu_flux(1)=-3*du_dh_left*dh(1)/8+9*u(1)/8-u(2)/8;
nu_flux(N)=3*du_dh_right*dh(N-1)/8-9*u(N-1)/8+u(N-2)/8;
%Fluxes for u equation
u_grad=2*(u(2:S.N-2)-u(1:S.N-2))./(dh(2:S.N-1)+dh(1:S.N-2));
u_flux=NaN(S.N,1);
u_flux(2:S.N-1)=a_1*P_L(S,nu_half(2:S.N-1))+(a_2*zeta(nu_half(2:S.N-1))./nu_half(2:S.N-1)).*u_grad;
% The system of equationsY
dnu_dt = nu_flux(2:S.N)-nu_flux(1:S.N-1);
du_dt =u_flux(2:S.N)-u_flux(1:S.N-1);

% Return the derivatives
dydt = [dnu_dt; du_dt];
end
function y=P_L(S,nu)
y=S.K./nu;
end
function y=zeta(nu)
y=(2/3)*(nu).^(-2)./(nu-1);
end I am solving a PDE via finite volume methods. As a result, I am left with a set of ODEs which I solve with ode15s. I have a bunch of constants which I use in the function defining the derivative, and I insert these by the inclusion of a struct to the function. When I run the ode15s I obtain an error which pertains to the inclusion of the struct as an input to the function. I don’t really want to include the constants within the function as that seems "messy", so I would prefer to do it via a struct.

%This is a code to solve the isothermal sintering equations.
%%—Physical parameters—
S=struct;
S.g = 9.81; %Acceleration due to gravity.
S.K = 1; %This is the Laplace constant from the sintering potential.
S.eta_0 = 1; %The shear stress of the skeleton
S.T = 3600; %This is the time of sintering.
S.M = 0.1; %This is the total mass of the rod being sintered .
S.rho_m = 1; %This is the density of the individual metallic particles.
S.N = 251; %This is the number of cells-1
L = 1; %This is the initial length of the rod
%%—Initial conditions
%Length
X=linspace(0,L,S.N); %This is the partition of the initial length
%Density
rho_0=0.5+0.1*sin(6*pi*X);
%Amount of mass
S.h=cumtrapz(X,rho_0);
%Initial velocity
u0=zeros(S.N-1,1);
% Define the system of equations as a function
y0 = [1./rho_0′; u0];
% Finite Volume Method solution loop
tspan = [0 3]; % You can adjust this based on the time range you’re interested in
[t, y] = ode15s(@ode_system, tspan, y0);
function dydt = ode_system(t, y, S)
S.M
U=S.M/(S.rho_m*S.T);
a_1=S.T/(U*S.M);
a_2=S.T*S.rho_m/S.M^2;
a_3=S.g*S.T/U;
dh=diff(S.h);
y=NaN(1,2*S.N-2);
nu = y(1:S.N-1); % u(t)
u = y(S.N:2*S.N-2); % v(t)
%Compute the left and right specific volumes:
nu_left=15*nu(1)/8-5*nu(2)/4+3*nu(3)/8;
nu_right=15*nu(S.N-3)/8-5*nu(S.N-2)/5+3*nu(S.N-1)/8;
%Compute the fluxes on the interfaces
nu_half=NaN(1:S.N);
nu_half(2:N-1)=0.5*(nu(2:S.N-1)+nu(1:S.N-2));
nu_half(1)=nu_left;
nu_half(N)=nu_right;
%Fluxes for nu equation
nu_flux=zeros(S.N,1);
nu_flux(2:S.N-1)=0.5*(u(1:S.N-2)+u(2:S.N-1));
du_dh_left=-P_L(S,nu_left)*nu_left/zeta(nu_left);
du_dh_right=-P_L(S,nu_right)*nu_right/zeta(nu_right);
nu_flux(1)=-3*du_dh_left*dh(1)/8+9*u(1)/8-u(2)/8;
nu_flux(N)=3*du_dh_right*dh(N-1)/8-9*u(N-1)/8+u(N-2)/8;
%Fluxes for u equation
u_grad=2*(u(2:S.N-2)-u(1:S.N-2))./(dh(2:S.N-1)+dh(1:S.N-2));
u_flux=NaN(S.N,1);
u_flux(2:S.N-1)=a_1*P_L(S,nu_half(2:S.N-1))+(a_2*zeta(nu_half(2:S.N-1))./nu_half(2:S.N-1)).*u_grad;
% The system of equationsY
dnu_dt = nu_flux(2:S.N)-nu_flux(1:S.N-1);
du_dt =u_flux(2:S.N)-u_flux(1:S.N-1);

% Return the derivatives
dydt = [dnu_dt; du_dt];
end
function y=P_L(S,nu)
y=S.K./nu;
end
function y=zeta(nu)
y=(2/3)*(nu).^(-2)./(nu-1);
end anonymous, function MATLAB Answers — New Questions

​

I created this bandpass filter and plotted it’s impulse response:
% Bandpass filter parameters
f_nyquist = sampling_frequency / 2;
low_cutoff_frequency = 6; % Low cutoff frequency in Hz
high_cutoff_frequency = 400; % High cutoff frequency in Hz
% Design the bandpass filter
[b, a_filter] = butter(2, [low_cutoff_frequency, high_cutoff_frequency] / (f_nyquist), ‘bandpass’);
% Create an impulse signal (Dirac delta function)
impulse_signal = zeros(1, 1000); % Change 1000 to the desired length
impulse_signal(1) = 1; % Set the first sample to 1
% Apply bandpass filter to the impulse signal
impulse_response = filtfilt(b, a_filter, impulse_signal);
% Time vector for plotting
time = (0:length(impulse_response)-1) / sampling_frequency;
% Plot the impulse response
figure(‘Name’, ‘Impulse Response of the Bandpass Filter’);
plot(time, impulse_response);
title(‘Impulse Response of the Bandpass Filter’);
xlabel(‘Time (s)’);
ylabel(‘Amplitude’);
grid on;
[h, f] = freqz(b, a_filter, 1024, ‘whole’); % Frequency response
figure;
plot(f/pi * (sampling_frequency/2), abs(h)); % Plot magnitude response
xlabel(‘Frequency (Hz)’);
ylabel(‘Magnitude’);
title(‘Frequency Response of the Bandpass Filter’);
grid on;

Why is there a spike from zero to -0.1?
That’s the bandpass filtered sEMG signal. How can I remove the information that may be the transient response around 0 and the info around 4 seconds where I attenuate my signal in order to not include the switch (stop of an experiment)?

That’s the impulse response of the bandpass filter.I created this bandpass filter and plotted it’s impulse response:
% Bandpass filter parameters
f_nyquist = sampling_frequency / 2;
low_cutoff_frequency = 6; % Low cutoff frequency in Hz
high_cutoff_frequency = 400; % High cutoff frequency in Hz
% Design the bandpass filter
[b, a_filter] = butter(2, [low_cutoff_frequency, high_cutoff_frequency] / (f_nyquist), ‘bandpass’);
% Create an impulse signal (Dirac delta function)
impulse_signal = zeros(1, 1000); % Change 1000 to the desired length
impulse_signal(1) = 1; % Set the first sample to 1
% Apply bandpass filter to the impulse signal
impulse_response = filtfilt(b, a_filter, impulse_signal);
% Time vector for plotting
time = (0:length(impulse_response)-1) / sampling_frequency;
% Plot the impulse response
figure(‘Name’, ‘Impulse Response of the Bandpass Filter’);
plot(time, impulse_response);
title(‘Impulse Response of the Bandpass Filter’);
xlabel(‘Time (s)’);
ylabel(‘Amplitude’);
grid on;
[h, f] = freqz(b, a_filter, 1024, ‘whole’); % Frequency response
figure;
plot(f/pi * (sampling_frequency/2), abs(h)); % Plot magnitude response
xlabel(‘Frequency (Hz)’);
ylabel(‘Magnitude’);
title(‘Frequency Response of the Bandpass Filter’);
grid on;

Why is there a spike from zero to -0.1?
That’s the bandpass filtered sEMG signal. How can I remove the information that may be the transient response around 0 and the info around 4 seconds where I attenuate my signal in order to not include the switch (stop of an experiment)?

That’s the impulse response of the bandpass filter. I created this bandpass filter and plotted it’s impulse response:
% Bandpass filter parameters
f_nyquist = sampling_frequency / 2;
low_cutoff_frequency = 6; % Low cutoff frequency in Hz
high_cutoff_frequency = 400; % High cutoff frequency in Hz
% Design the bandpass filter
[b, a_filter] = butter(2, [low_cutoff_frequency, high_cutoff_frequency] / (f_nyquist), ‘bandpass’);
% Create an impulse signal (Dirac delta function)
impulse_signal = zeros(1, 1000); % Change 1000 to the desired length
impulse_signal(1) = 1; % Set the first sample to 1
% Apply bandpass filter to the impulse signal
impulse_response = filtfilt(b, a_filter, impulse_signal);
% Time vector for plotting
time = (0:length(impulse_response)-1) / sampling_frequency;
% Plot the impulse response
figure(‘Name’, ‘Impulse Response of the Bandpass Filter’);
plot(time, impulse_response);
title(‘Impulse Response of the Bandpass Filter’);
xlabel(‘Time (s)’);
ylabel(‘Amplitude’);
grid on;
[h, f] = freqz(b, a_filter, 1024, ‘whole’); % Frequency response
figure;
plot(f/pi * (sampling_frequency/2), abs(h)); % Plot magnitude response
xlabel(‘Frequency (Hz)’);
ylabel(‘Magnitude’);
title(‘Frequency Response of the Bandpass Filter’);
grid on;

Why is there a spike from zero to -0.1?
That’s the bandpass filtered sEMG signal. How can I remove the information that may be the transient response around 0 and the info around 4 seconds where I attenuate my signal in order to not include the switch (stop of an experiment)?

That’s the impulse response of the bandpass filter. bandpass filter, impulse response, emg signals, transient response MATLAB Answers — New Questions

​

this is te first time i used MATLAB i have eyetrack record like this
eyetrackRecord =

struct with fields:

x: [1.3756e+03 1.3734e+03 1.3798e+03 1.3789e+03 … ] (1×5128 double)
y: [416.1000 415.5000 416.6000 410.8000 … ] (1×5128 double)
pa: [1416 1423 1423 1421 1420 1424 1431 1417 … ] (1×5128 double)
t: [281349 281395 281403 281409 281418 281423 … ] (1×5128 double)
missing: [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … ] (1×5128 double)
How can I plot the moving heatmap of gazing over the video file in MATLABthis is te first time i used MATLAB i have eyetrack record like this
eyetrackRecord =

struct with fields:

x: [1.3756e+03 1.3734e+03 1.3798e+03 1.3789e+03 … ] (1×5128 double)
y: [416.1000 415.5000 416.6000 410.8000 … ] (1×5128 double)
pa: [1416 1423 1423 1421 1420 1424 1431 1417 … ] (1×5128 double)
t: [281349 281395 281403 281409 281418 281423 … ] (1×5128 double)
missing: [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … ] (1×5128 double)
How can I plot the moving heatmap of gazing over the video file in MATLAB this is te first time i used MATLAB i have eyetrack record like this
eyetrackRecord =

struct with fields:

x: [1.3756e+03 1.3734e+03 1.3798e+03 1.3789e+03 … ] (1×5128 double)
y: [416.1000 415.5000 416.6000 410.8000 … ] (1×5128 double)
pa: [1416 1423 1423 1421 1420 1424 1431 1417 … ] (1×5128 double)
t: [281349 281395 281403 281409 281418 281423 … ] (1×5128 double)
missing: [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … ] (1×5128 double)
How can I plot the moving heatmap of gazing over the video file in MATLAB eyetracking, heatmap, video MATLAB Answers — New Questions

​

Hallo everyone,
I’m following this example(in link) in Matlab to make ego location. In extended Kalman Filter, if I use fucntion-helpTuneinsEKF of example, matlab always reminds me that the matrix is singular in the accuracy of the calculation. I try to redefine the initial value of additive process noise for optimal tuning, but I habe no idea how can I define these values like additive process noise for orientation and…
apnoise(stateinfo(filter, "Orientation")) = [0.02014531; 100; 7.0076611; 1.308652];
apnoise(stateinfo(filter, "AngularVelocity")) = [0.0007698; 100; 100];

thank you
Best regards
https://de.mathworks.com/help/driving/ug/ego-vehicle-localization-using-gps-and-imu-fusion-for-scenario-generation.htmlHallo everyone,
I’m following this example(in link) in Matlab to make ego location. In extended Kalman Filter, if I use fucntion-helpTuneinsEKF of example, matlab always reminds me that the matrix is singular in the accuracy of the calculation. I try to redefine the initial value of additive process noise for optimal tuning, but I habe no idea how can I define these values like additive process noise for orientation and…
apnoise(stateinfo(filter, "Orientation")) = [0.02014531; 100; 7.0076611; 1.308652];
apnoise(stateinfo(filter, "AngularVelocity")) = [0.0007698; 100; 100];

thank you
Best regards
https://de.mathworks.com/help/driving/ug/ego-vehicle-localization-using-gps-and-imu-fusion-for-scenario-generation.html Hallo everyone,
I’m following this example(in link) in Matlab to make ego location. In extended Kalman Filter, if I use fucntion-helpTuneinsEKF of example, matlab always reminds me that the matrix is singular in the accuracy of the calculation. I try to redefine the initial value of additive process noise for optimal tuning, but I habe no idea how can I define these values like additive process noise for orientation and…
apnoise(stateinfo(filter, "Orientation")) = [0.02014531; 100; 7.0076611; 1.308652];
apnoise(stateinfo(filter, "AngularVelocity")) = [0.0007698; 100; 100];

thank you
Best regards
https://de.mathworks.com/help/driving/ug/ego-vehicle-localization-using-gps-and-imu-fusion-for-scenario-generation.html extended kalman filter, ego location MATLAB Answers — New Questions

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I am trying to estimate a second-order plus time delay (SOPTD) process model using MATLAB’s System Identification Toolbox. However, in some cases, the estimated model does not start from zero, even though my input data is normalized and clamped to ensure non-negative values. and somecases they start above zero
I have tried several steps to fix this:
Adding an offset to the normalized data to avoid negative values.
Setting the process gain (Kp) to 1 and fixing it.
Setting constraints on the time delay (Td) and ensuring it is non-negative
Despite these attempts, 5 out of my 20 plots still exhibit negative values at the start. I need help understanding why this happens and how to ensure the model never goes negative.I am trying to estimate a second-order plus time delay (SOPTD) process model using MATLAB’s System Identification Toolbox. However, in some cases, the estimated model does not start from zero, even though my input data is normalized and clamped to ensure non-negative values. and somecases they start above zero
I have tried several steps to fix this:
Adding an offset to the normalized data to avoid negative values.
Setting the process gain (Kp) to 1 and fixing it.
Setting constraints on the time delay (Td) and ensuring it is non-negative
Despite these attempts, 5 out of my 20 plots still exhibit negative values at the start. I need help understanding why this happens and how to ensure the model never goes negative. I am trying to estimate a second-order plus time delay (SOPTD) process model using MATLAB’s System Identification Toolbox. However, in some cases, the estimated model does not start from zero, even though my input data is normalized and clamped to ensure non-negative values. and somecases they start above zero
I have tried several steps to fix this:
Adding an offset to the normalized data to avoid negative values.
Setting the process gain (Kp) to 1 and fixing it.
Setting constraints on the time delay (Td) and ensuring it is non-negative
Despite these attempts, 5 out of my 20 plots still exhibit negative values at the start. I need help understanding why this happens and how to ensure the model never goes negative. idproc, procest, system identification toolbox, ma MATLAB Answers — New Questions

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Hello everyone!
I have a question about create class
I want to have classname like this in variables window:
val = (classname)
nch: 4
N: 8192
F: 100
df: 0.012207
T: 81.92
dt: 0.01
quantity: force,acc
sifactor: 1
labels: 1F,1A
How can I do that? Thank you all.Hello everyone!
I have a question about create class
I want to have classname like this in variables window:
val = (classname)
nch: 4
N: 8192
F: 100
df: 0.012207
T: 81.92
dt: 0.01
quantity: force,acc
sifactor: 1
labels: 1F,1A
How can I do that? Thank you all. Hello everyone!
I have a question about create class
I want to have classname like this in variables window:
val = (classname)
nch: 4
N: 8192
F: 100
df: 0.012207
T: 81.92
dt: 0.01
quantity: force,acc
sifactor: 1
labels: 1F,1A
How can I do that? Thank you all. signal processing MATLAB Answers — New Questions

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%here are some of the codes.am i using the right codes?
%are there any other codes i can use?
clear all;
close all;
Ireal = imread(‘C:UsersNatiDesktopnati projectNew folderNew folderprogress natibirrIMG_20180513_10260.jpg’); % Real
Iscaned = imread(‘C:UsersNatiDesktopnati projectNew folderNew folderprogress natibirrIMG_20180513_102606.jpg’); % scaned

%%//Pre-analysis
hsvImageReal = rgb2hsv(Ireal);
hsvImagescaned = rgb2hsv(Iscaned);
figure;
imshow([hsvImageReal(:,:,1) hsvImageReal(:,:,2) hsvImageReal(:,:,3)]);
title(‘Real pre process’);
figure;
imshow([hsvImagescaned(:,:,1) hsvImagescaned(:,:,2) hsvImagescaned(:,:,3)]);
title(‘scaned’);

%%//Initial segmentation
croppedImageReal = hsvImageReal(:,90:95,:);
croppedImagescaned = hsvImagescaned(:,93:98,:);
satThresh = 0.4;
valThresh = 0.3;
BWImageReal = (croppedImageReal(:,:,2) > satThresh & croppedImageReal(:,:,3) < valThresh);
figure;
subplot(1,2,1);
imshow(BWImageReal);
title(‘Real initial seg’);
BWImagescaned = (croppedImagescaned(:,:,2) > satThresh & croppedImagescaned(:,:,3) < valThresh);
subplot(1,2,2);
imshow(BWImagescaned);
title(‘scaned’);

%%//Post-process
se = strel(‘line’, 6, 90);
BWImageCloseReal = imclose(BWImageReal, se);
BWImageClosescaned = imclose(BWImagescaned, se);
figure;
subplot(1,2,1);
imshow(BWImageCloseReal);
title(‘Real post process’);
subplot(1,2,2);
imshow(BWImageClosescaned);
title(‘scaned’);

%%//Area open the image
figure;
areaopenReal = bwareaopen(BWImageCloseReal, 15);
subplot(1,2,1);
imshow(areaopenReal);
title(‘Real area open image’);
subplot(1,2,2);
areaopenscaned = bwareaopen(BWImageClosescaned, 15);
imshow(areaopenscaned);
title(‘scaned’);
%//Count how many objects there are
[~,countReal] = bwlabel(areaopenReal);
[~,countscaned] = bwlabel(areaopenscaned);
disp([‘The total number of black lines for the real note is: ‘ num2str(countReal)]);
disp([‘The total number of black lines for the scaned note is: ‘ num2str(countscaned)]);
%%
% convert to gray scale
grt = rgb2gray(Ireal);
grs = rgb2gray(Iscaned);
% contrast enhance the gray image to emphasize dark lines in lighter background
grt = imadjust(grt);
grs = imadjust(grs);
% close rgb. choose a larger k. idea is to remove the dark line
k = 7;
se = ones(k);
Irealcl = imclose(Ireal, se);
Iscanedcl = imclose(Iscaned, se);
% convert closed image to gray scale
grtcl = rgb2gray(Irealcl);
grscl = rgb2gray(Iscanedcl);
% take the difference (closed-gray-scale – contrast-enhanced-gray-scale)
difft = grtcl – grt;
diffs = grscl – grs;
% take the projection of the difference
pt = sum(difft’);
pf = sum(diffs’);
% smooth the projection
ptfilt = conv(pt, ones(1, k)/k, ‘same’);
pffilt = conv(pf, ones(1, k)/k, ‘same’);
% threshold (multiplication by max element is just for visualization)
tht = (pt > graythresh(pt))*max(pt);
thf = (pf > graythresh(pf))*max(pf);
% get the number of segments.
[lblt, nt] = bwlabel(tht);
[lblf, nf] = bwlabel(thf);

figure,
subplot(2, 1, 1), imshow(difft’), title(‘difference image for solid line’)
subplot(2, 1, 2), imshow(diffs’), title(‘difference image for broken line’)

figure,
subplot(2, 1, 1), plot(1:length(pt), pt, 1:length(pt), ptfilt, 1:length(pt), tht),
title(‘solid line image’),
legend(‘projection’, ‘smoothed’, ‘thresholded’, ‘location’, ‘eastoutside’)
subplot(2, 1, 2), plot(1:length(pf), pf, 1:length(pf), pffilt, 1:length(pf), thf),
title(‘broken line image’),
legend(‘projection’, ‘smoothed’, ‘thresholded’, ‘location’, ‘eastoutside’)

%%%
%% //Extract the black strips for each image
blackStripReal = Ireal(:,200:725,:);
blackStripscaned = Iscaned(:,200:725,:);

figure;
subplot(1,3,1);
imshow(blackStripReal);
title(‘Real black stripe’);
subplot(1,3,2);
imshow(blackStripscaned);
title(‘scaned’);

%%%
%% //Convert into grayscale then threshold
blackStripReal = rgb2gray(blackStripReal);
blackStripscaned = rgb2gray(blackStripscaned);

figure;
subplot(1,3,1);
imshow(blackStripReal);
title(‘Realblack strip’);
subplot(1,3,2);
imshow(blackStripscaned);
title(‘scaned’);%here are some of the codes.am i using the right codes?
%are there any other codes i can use?
clear all;
close all;
Ireal = imread(‘C:UsersNatiDesktopnati projectNew folderNew folderprogress natibirrIMG_20180513_10260.jpg’); % Real
Iscaned = imread(‘C:UsersNatiDesktopnati projectNew folderNew folderprogress natibirrIMG_20180513_102606.jpg’); % scaned

%%//Pre-analysis
hsvImageReal = rgb2hsv(Ireal);
hsvImagescaned = rgb2hsv(Iscaned);
figure;
imshow([hsvImageReal(:,:,1) hsvImageReal(:,:,2) hsvImageReal(:,:,3)]);
title(‘Real pre process’);
figure;
imshow([hsvImagescaned(:,:,1) hsvImagescaned(:,:,2) hsvImagescaned(:,:,3)]);
title(‘scaned’);

%%//Initial segmentation
croppedImageReal = hsvImageReal(:,90:95,:);
croppedImagescaned = hsvImagescaned(:,93:98,:);
satThresh = 0.4;
valThresh = 0.3;
BWImageReal = (croppedImageReal(:,:,2) > satThresh & croppedImageReal(:,:,3) < valThresh);
figure;
subplot(1,2,1);
imshow(BWImageReal);
title(‘Real initial seg’);
BWImagescaned = (croppedImagescaned(:,:,2) > satThresh & croppedImagescaned(:,:,3) < valThresh);
subplot(1,2,2);
imshow(BWImagescaned);
title(‘scaned’);

%%//Post-process
se = strel(‘line’, 6, 90);
BWImageCloseReal = imclose(BWImageReal, se);
BWImageClosescaned = imclose(BWImagescaned, se);
figure;
subplot(1,2,1);
imshow(BWImageCloseReal);
title(‘Real post process’);
subplot(1,2,2);
imshow(BWImageClosescaned);
title(‘scaned’);

%%//Area open the image
figure;
areaopenReal = bwareaopen(BWImageCloseReal, 15);
subplot(1,2,1);
imshow(areaopenReal);
title(‘Real area open image’);
subplot(1,2,2);
areaopenscaned = bwareaopen(BWImageClosescaned, 15);
imshow(areaopenscaned);
title(‘scaned’);
%//Count how many objects there are
[~,countReal] = bwlabel(areaopenReal);
[~,countscaned] = bwlabel(areaopenscaned);
disp([‘The total number of black lines for the real note is: ‘ num2str(countReal)]);
disp([‘The total number of black lines for the scaned note is: ‘ num2str(countscaned)]);
%%
% convert to gray scale
grt = rgb2gray(Ireal);
grs = rgb2gray(Iscaned);
% contrast enhance the gray image to emphasize dark lines in lighter background
grt = imadjust(grt);
grs = imadjust(grs);
% close rgb. choose a larger k. idea is to remove the dark line
k = 7;
se = ones(k);
Irealcl = imclose(Ireal, se);
Iscanedcl = imclose(Iscaned, se);
% convert closed image to gray scale
grtcl = rgb2gray(Irealcl);
grscl = rgb2gray(Iscanedcl);
% take the difference (closed-gray-scale – contrast-enhanced-gray-scale)
difft = grtcl – grt;
diffs = grscl – grs;
% take the projection of the difference
pt = sum(difft’);
pf = sum(diffs’);
% smooth the projection
ptfilt = conv(pt, ones(1, k)/k, ‘same’);
pffilt = conv(pf, ones(1, k)/k, ‘same’);
% threshold (multiplication by max element is just for visualization)
tht = (pt > graythresh(pt))*max(pt);
thf = (pf > graythresh(pf))*max(pf);
% get the number of segments.
[lblt, nt] = bwlabel(tht);
[lblf, nf] = bwlabel(thf);

figure,
subplot(2, 1, 1), imshow(difft’), title(‘difference image for solid line’)
subplot(2, 1, 2), imshow(diffs’), title(‘difference image for broken line’)

figure,
subplot(2, 1, 1), plot(1:length(pt), pt, 1:length(pt), ptfilt, 1:length(pt), tht),
title(‘solid line image’),
legend(‘projection’, ‘smoothed’, ‘thresholded’, ‘location’, ‘eastoutside’)
subplot(2, 1, 2), plot(1:length(pf), pf, 1:length(pf), pffilt, 1:length(pf), thf),
title(‘broken line image’),
legend(‘projection’, ‘smoothed’, ‘thresholded’, ‘location’, ‘eastoutside’)

%%%
%% //Extract the black strips for each image
blackStripReal = Ireal(:,200:725,:);
blackStripscaned = Iscaned(:,200:725,:);

figure;
subplot(1,3,1);
imshow(blackStripReal);
title(‘Real black stripe’);
subplot(1,3,2);
imshow(blackStripscaned);
title(‘scaned’);

%%%
%% //Convert into grayscale then threshold
blackStripReal = rgb2gray(blackStripReal);
blackStripscaned = rgb2gray(blackStripscaned);

figure;
subplot(1,3,1);
imshow(blackStripReal);
title(‘Realblack strip’);
subplot(1,3,2);
imshow(blackStripscaned);
title(‘scaned’); %here are some of the codes.am i using the right codes?
%are there any other codes i can use?
clear all;
close all;
Ireal = imread(‘C:UsersNatiDesktopnati projectNew folderNew folderprogress natibirrIMG_20180513_10260.jpg’); % Real
Iscaned = imread(‘C:UsersNatiDesktopnati projectNew folderNew folderprogress natibirrIMG_20180513_102606.jpg’); % scaned

%%//Pre-analysis
hsvImageReal = rgb2hsv(Ireal);
hsvImagescaned = rgb2hsv(Iscaned);
figure;
imshow([hsvImageReal(:,:,1) hsvImageReal(:,:,2) hsvImageReal(:,:,3)]);
title(‘Real pre process’);
figure;
imshow([hsvImagescaned(:,:,1) hsvImagescaned(:,:,2) hsvImagescaned(:,:,3)]);
title(‘scaned’);

%%//Initial segmentation
croppedImageReal = hsvImageReal(:,90:95,:);
croppedImagescaned = hsvImagescaned(:,93:98,:);
satThresh = 0.4;
valThresh = 0.3;
BWImageReal = (croppedImageReal(:,:,2) > satThresh & croppedImageReal(:,:,3) < valThresh);
figure;
subplot(1,2,1);
imshow(BWImageReal);
title(‘Real initial seg’);
BWImagescaned = (croppedImagescaned(:,:,2) > satThresh & croppedImagescaned(:,:,3) < valThresh);
subplot(1,2,2);
imshow(BWImagescaned);
title(‘scaned’);

%%//Post-process
se = strel(‘line’, 6, 90);
BWImageCloseReal = imclose(BWImageReal, se);
BWImageClosescaned = imclose(BWImagescaned, se);
figure;
subplot(1,2,1);
imshow(BWImageCloseReal);
title(‘Real post process’);
subplot(1,2,2);
imshow(BWImageClosescaned);
title(‘scaned’);

%%//Area open the image
figure;
areaopenReal = bwareaopen(BWImageCloseReal, 15);
subplot(1,2,1);
imshow(areaopenReal);
title(‘Real area open image’);
subplot(1,2,2);
areaopenscaned = bwareaopen(BWImageClosescaned, 15);
imshow(areaopenscaned);
title(‘scaned’);
%//Count how many objects there are
[~,countReal] = bwlabel(areaopenReal);
[~,countscaned] = bwlabel(areaopenscaned);
disp([‘The total number of black lines for the real note is: ‘ num2str(countReal)]);
disp([‘The total number of black lines for the scaned note is: ‘ num2str(countscaned)]);
%%
% convert to gray scale
grt = rgb2gray(Ireal);
grs = rgb2gray(Iscaned);
% contrast enhance the gray image to emphasize dark lines in lighter background
grt = imadjust(grt);
grs = imadjust(grs);
% close rgb. choose a larger k. idea is to remove the dark line
k = 7;
se = ones(k);
Irealcl = imclose(Ireal, se);
Iscanedcl = imclose(Iscaned, se);
% convert closed image to gray scale
grtcl = rgb2gray(Irealcl);
grscl = rgb2gray(Iscanedcl);
% take the difference (closed-gray-scale – contrast-enhanced-gray-scale)
difft = grtcl – grt;
diffs = grscl – grs;
% take the projection of the difference
pt = sum(difft’);
pf = sum(diffs’);
% smooth the projection
ptfilt = conv(pt, ones(1, k)/k, ‘same’);
pffilt = conv(pf, ones(1, k)/k, ‘same’);
% threshold (multiplication by max element is just for visualization)
tht = (pt > graythresh(pt))*max(pt);
thf = (pf > graythresh(pf))*max(pf);
% get the number of segments.
[lblt, nt] = bwlabel(tht);
[lblf, nf] = bwlabel(thf);

figure,
subplot(2, 1, 1), imshow(difft’), title(‘difference image for solid line’)
subplot(2, 1, 2), imshow(diffs’), title(‘difference image for broken line’)

figure,
subplot(2, 1, 1), plot(1:length(pt), pt, 1:length(pt), ptfilt, 1:length(pt), tht),
title(‘solid line image’),
legend(‘projection’, ‘smoothed’, ‘thresholded’, ‘location’, ‘eastoutside’)
subplot(2, 1, 2), plot(1:length(pf), pf, 1:length(pf), pffilt, 1:length(pf), thf),
title(‘broken line image’),
legend(‘projection’, ‘smoothed’, ‘thresholded’, ‘location’, ‘eastoutside’)

%%%
%% //Extract the black strips for each image
blackStripReal = Ireal(:,200:725,:);
blackStripscaned = Iscaned(:,200:725,:);

figure;
subplot(1,3,1);
imshow(blackStripReal);
title(‘Real black stripe’);
subplot(1,3,2);
imshow(blackStripscaned);
title(‘scaned’);

%%%
%% //Convert into grayscale then threshold
blackStripReal = rgb2gray(blackStripReal);
blackStripscaned = rgb2gray(blackStripscaned);

figure;
subplot(1,3,1);
imshow(blackStripReal);
title(‘Realblack strip’);
subplot(1,3,2);
imshow(blackStripscaned);
title(‘scaned’); fake currency detection system., currency, fake currency MATLAB Answers — New Questions

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daily_savings=zeros(1505,1);%initial saving vectors
daily_earnings=zeros(1505,1);%initial earning vectors
% Loop through each day to calculate savings and earnings % total days is 1505 (Date from(06/04/20 to 19/05/24))
for i = 0:1504 % days1 starts from 0 so 0:1504
% Calculate daily savings
daily_savings(i + 1) = O_Consumption(i + 1).*Cost;
% Calculate daily earnings
if i <= 1212 % Up to July 31, 2023 (day index 1212)
daily_earnings(i + 1) = energy_out(i + 1).*Feed(i + 1);
else % From August 1, 2023 onwards
if e_out(i + 1) < 10
daily_earnings(i + 1) = e_out(i + 1).*Feed(i + 1);
else
daily_earnings(i + 1) = ((e_out(i + 1) – 10) / 100) + 0.88;
end
end
above is the codedaily_savings=zeros(1505,1);%initial saving vectors
daily_earnings=zeros(1505,1);%initial earning vectors
% Loop through each day to calculate savings and earnings % total days is 1505 (Date from(06/04/20 to 19/05/24))
for i = 0:1504 % days1 starts from 0 so 0:1504
% Calculate daily savings
daily_savings(i + 1) = O_Consumption(i + 1).*Cost;
% Calculate daily earnings
if i <= 1212 % Up to July 31, 2023 (day index 1212)
daily_earnings(i + 1) = energy_out(i + 1).*Feed(i + 1);
else % From August 1, 2023 onwards
if e_out(i + 1) < 10
daily_earnings(i + 1) = e_out(i + 1).*Feed(i + 1);
else
daily_earnings(i + 1) = ((e_out(i + 1) – 10) / 100) + 0.88;
end
end
above is the code daily_savings=zeros(1505,1);%initial saving vectors
daily_earnings=zeros(1505,1);%initial earning vectors
% Loop through each day to calculate savings and earnings % total days is 1505 (Date from(06/04/20 to 19/05/24))
for i = 0:1504 % days1 starts from 0 so 0:1504
% Calculate daily savings
daily_savings(i + 1) = O_Consumption(i + 1).*Cost;
% Calculate daily earnings
if i <= 1212 % Up to July 31, 2023 (day index 1212)
daily_earnings(i + 1) = energy_out(i + 1).*Feed(i + 1);
else % From August 1, 2023 onwards
if e_out(i + 1) < 10
daily_earnings(i + 1) = e_out(i + 1).*Feed(i + 1);
else
daily_earnings(i + 1) = ((e_out(i + 1) – 10) / 100) + 0.88;
end
end
above is the code array, error MATLAB Answers — New Questions

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