I expect result1 and result2 below to be identical, but they aren’t. The discrepancy must be a bug, right? I’m working in R2024b, but the Run output below shows the issue exists as well in whatever the Matlab online engine is now running.
n = 5; m = 3;

S = sparse(n, m);
v=nan(n, 1);
D=sparse(diag(v));

result1=full(v.*S) %correct

result2=full(D*S) %incorrectI expect result1 and result2 below to be identical, but they aren’t. The discrepancy must be a bug, right? I’m working in R2024b, but the Run output below shows the issue exists as well in whatever the Matlab online engine is now running.
n = 5; m = 3;

S = sparse(n, m);
v=nan(n, 1);
D=sparse(diag(v));

result1=full(v.*S) %correct

result2=full(D*S) %incorrect I expect result1 and result2 below to be identical, but they aren’t. The discrepancy must be a bug, right? I’m working in R2024b, but the Run output below shows the issue exists as well in whatever the Matlab online engine is now running.
n = 5; m = 3;

S = sparse(n, m);
v=nan(n, 1);
D=sparse(diag(v));

result1=full(v.*S) %correct

result2=full(D*S) %incorrect sparse, nan, bug?, numeric MATLAB Answers — New Questions

​

Hello everyone,
I’m currently modeling a refrigerant test loop in Simscape. The flow path is:
High-pressure reservoir → condenser → short pipe → valve (TXV or EXV) → short pipe → evaporator → low-pressure reservoir. All parameters are set to realistic values (pressures, diameters, temperatures, etc.).
The issue I’m facing is that the evaporator appears to be removing heat from the refrigerant instead of adding it, the temperature at port B is higher than at port A, even though the evaporator’s ambient is hotter. This causes downstream effects such as superheat decreasing when the valve closes, which is the opposite of physical behavior. Meanwhile, the condenser behaves correctly (temperature drops from port A → B as it rejects heat). Has anyone seen this before or know why the evaporator heat direction might flip in Simscape?Hello everyone,
I’m currently modeling a refrigerant test loop in Simscape. The flow path is:
High-pressure reservoir → condenser → short pipe → valve (TXV or EXV) → short pipe → evaporator → low-pressure reservoir. All parameters are set to realistic values (pressures, diameters, temperatures, etc.).
The issue I’m facing is that the evaporator appears to be removing heat from the refrigerant instead of adding it, the temperature at port B is higher than at port A, even though the evaporator’s ambient is hotter. This causes downstream effects such as superheat decreasing when the valve closes, which is the opposite of physical behavior. Meanwhile, the condenser behaves correctly (temperature drops from port A → B as it rejects heat). Has anyone seen this before or know why the evaporator heat direction might flip in Simscape? Hello everyone,
I’m currently modeling a refrigerant test loop in Simscape. The flow path is:
High-pressure reservoir → condenser → short pipe → valve (TXV or EXV) → short pipe → evaporator → low-pressure reservoir. All parameters are set to realistic values (pressures, diameters, temperatures, etc.).
The issue I’m facing is that the evaporator appears to be removing heat from the refrigerant instead of adding it, the temperature at port B is higher than at port A, even though the evaporator’s ambient is hotter. This causes downstream effects such as superheat decreasing when the valve closes, which is the opposite of physical behavior. Meanwhile, the condenser behaves correctly (temperature drops from port A → B as it rejects heat). Has anyone seen this before or know why the evaporator heat direction might flip in Simscape? simscape MATLAB Answers — New Questions

​

Consider this code:

clear; close all;
n = 1e6;
m = 1e3;
s = 1e-3;
p = round(s^2 * n * m);
I = randperm(n, p);
J = randperm(m, p);
phi = sparse(I, J, 1, n, m);
Gx = randn(n, 1);
for k = 1:3
tic
L = Gx .* phi;
toc
end
Gx(1) = NaN;
for k = 1:3
tic
L = Gx .* phi;
toc
end

Now consider the output on my macbook pro M1 running Matlab 2024a (so through rosetta).

Elapsed time is 0.001591 seconds.
Elapsed time is 0.000291 seconds.
Elapsed time is 0.000282 seconds.
Elapsed time is 0.573762 seconds.
Elapsed time is 0.548177 seconds.
Elapsed time is 0.560880 seconds.

Notice the very large difference in computation times between the no-NaN version of Gx and the version that contains a single NaN value.

The solution is quite obvious, just remove the NaN values with your favorite method (isnan, isfinite, etc.). So this is not my question.

What intregues me is:
why the computation time changes so drastically?Consider this code:

clear; close all;
n = 1e6;
m = 1e3;
s = 1e-3;
p = round(s^2 * n * m);
I = randperm(n, p);
J = randperm(m, p);
phi = sparse(I, J, 1, n, m);
Gx = randn(n, 1);
for k = 1:3
tic
L = Gx .* phi;
toc
end
Gx(1) = NaN;
for k = 1:3
tic
L = Gx .* phi;
toc
end

Now consider the output on my macbook pro M1 running Matlab 2024a (so through rosetta).

Elapsed time is 0.001591 seconds.
Elapsed time is 0.000291 seconds.
Elapsed time is 0.000282 seconds.
Elapsed time is 0.573762 seconds.
Elapsed time is 0.548177 seconds.
Elapsed time is 0.560880 seconds.

Notice the very large difference in computation times between the no-NaN version of Gx and the version that contains a single NaN value.

The solution is quite obvious, just remove the NaN values with your favorite method (isnan, isfinite, etc.). So this is not my question.

What intregues me is:
why the computation time changes so drastically? Consider this code:

clear; close all;
n = 1e6;
m = 1e3;
s = 1e-3;
p = round(s^2 * n * m);
I = randperm(n, p);
J = randperm(m, p);
phi = sparse(I, J, 1, n, m);
Gx = randn(n, 1);
for k = 1:3
tic
L = Gx .* phi;
toc
end
Gx(1) = NaN;
for k = 1:3
tic
L = Gx .* phi;
toc
end

Now consider the output on my macbook pro M1 running Matlab 2024a (so through rosetta).

Elapsed time is 0.001591 seconds.
Elapsed time is 0.000291 seconds.
Elapsed time is 0.000282 seconds.
Elapsed time is 0.573762 seconds.
Elapsed time is 0.548177 seconds.
Elapsed time is 0.560880 seconds.

Notice the very large difference in computation times between the no-NaN version of Gx and the version that contains a single NaN value.

The solution is quite obvious, just remove the NaN values with your favorite method (isnan, isfinite, etc.). So this is not my question.

What intregues me is:
why the computation time changes so drastically? sparse, mtimes MATLAB Answers — New Questions

​

Hello,
I am currently running a script in matlab that sends an email out when it is through with a .pdf attachment. This was working flawlessly two days ago and now I can’t seem to get it to work.
I am recieving the following error message:
Error using sendmail (line 175)
530 5.7.57 Client not authenticated to send mail. [BLAP220CA0019.NAMP220.PROD.OUTLOOK.COM]
Here is my code:
mail = ’email@address.com’; % false email for forum
password = ‘password’; % false pass for forum
mailingList = {‘people@email.com’}; % false email for forum
server = ‘smtp-mail.outlook.com’;
props = java.lang.System.getProperties;
props.setProperty(‘mail.smtp.port’,’587′);
props.setProperty(‘mail.smtp.starttls.enable’,’true’);
setpref(‘Internet’,’E_mail’,mail);
setpref(‘Internet’,’SMTP_Server’,server);
setpref(‘Internet’,’SMTP_Username’,mail);
setpref(‘Internet’,’SMTP_Password’,password);
messageBody = sprintf(‘ Good morning team!’);
messageBody = sprintf(‘%snn Here is the data analysis from yesterdays data. Have a great day!’, messageBody);
messageBody = sprintf(‘%snn – Brian Gregory’, messageBody);
messageBody = sprintf(‘%snnn *This email was generated and sent automatically via MATLAB’, messageBody);
sendmail(mailingList, …
"Data Analysis – "+date, …
messageBody,fileDir7);Hello,
I am currently running a script in matlab that sends an email out when it is through with a .pdf attachment. This was working flawlessly two days ago and now I can’t seem to get it to work.
I am recieving the following error message:
Error using sendmail (line 175)
530 5.7.57 Client not authenticated to send mail. [BLAP220CA0019.NAMP220.PROD.OUTLOOK.COM]
Here is my code:
mail = ’email@address.com’; % false email for forum
password = ‘password’; % false pass for forum
mailingList = {‘people@email.com’}; % false email for forum
server = ‘smtp-mail.outlook.com’;
props = java.lang.System.getProperties;
props.setProperty(‘mail.smtp.port’,’587′);
props.setProperty(‘mail.smtp.starttls.enable’,’true’);
setpref(‘Internet’,’E_mail’,mail);
setpref(‘Internet’,’SMTP_Server’,server);
setpref(‘Internet’,’SMTP_Username’,mail);
setpref(‘Internet’,’SMTP_Password’,password);
messageBody = sprintf(‘ Good morning team!’);
messageBody = sprintf(‘%snn Here is the data analysis from yesterdays data. Have a great day!’, messageBody);
messageBody = sprintf(‘%snn – Brian Gregory’, messageBody);
messageBody = sprintf(‘%snnn *This email was generated and sent automatically via MATLAB’, messageBody);
sendmail(mailingList, …
"Data Analysis – "+date, …
messageBody,fileDir7); Hello,
I am currently running a script in matlab that sends an email out when it is through with a .pdf attachment. This was working flawlessly two days ago and now I can’t seem to get it to work.
I am recieving the following error message:
Error using sendmail (line 175)
530 5.7.57 Client not authenticated to send mail. [BLAP220CA0019.NAMP220.PROD.OUTLOOK.COM]
Here is my code:
mail = ’email@address.com’; % false email for forum
password = ‘password’; % false pass for forum
mailingList = {‘people@email.com’}; % false email for forum
server = ‘smtp-mail.outlook.com’;
props = java.lang.System.getProperties;
props.setProperty(‘mail.smtp.port’,’587′);
props.setProperty(‘mail.smtp.starttls.enable’,’true’);
setpref(‘Internet’,’E_mail’,mail);
setpref(‘Internet’,’SMTP_Server’,server);
setpref(‘Internet’,’SMTP_Username’,mail);
setpref(‘Internet’,’SMTP_Password’,password);
messageBody = sprintf(‘ Good morning team!’);
messageBody = sprintf(‘%snn Here is the data analysis from yesterdays data. Have a great day!’, messageBody);
messageBody = sprintf(‘%snn – Brian Gregory’, messageBody);
messageBody = sprintf(‘%snnn *This email was generated and sent automatically via MATLAB’, messageBody);
sendmail(mailingList, …
"Data Analysis – "+date, …
messageBody,fileDir7); email, error, matlab MATLAB Answers — New Questions

​

Hi. MATLAB Community. I am trying to get the following code to work. I’m working on a toy example that I hope to get running correctly before applying it to a larger numerical project. The issue I’m facing is that the values of k and rho_0 are not updating properly when I Area_Integral from the command window for different values of k and rho_0 (I am selecting k and rho_0 values so it stays in the if loop of the Area_Integral file. I uploaded the code for the various scripts I am using. When I write this in the command window, Area_Integral (1,0.4) and then Area_Integral (1.5,0.4) the value of k is not updated for the later case.
function out = Area_Integral(k, rho_0)
global k rho_0
if k>=1-rho_0^2
theta_array = linspace(0,pi/2,101);
root = zeros(size(theta_array));
for ii = 1:length(theta_array)
theta = theta_array(ii);
root(ii) = mnhf_secant(@poly4,[0.4 0.8],1e-8,0); %% Note: values of k, r0 and theta must be made accessible to poly4
end
out = 2*trapz(theta_array,root);
else

r_upperlim = sqrt(1+rho_0^2+sqrt(k^2+4*rho_0^2));
r_lowerlim = sqrt(1+rho_0^2-sqrt(k^2+4*rho_0^2));

fun_A2 = @(r) acos((r.^4+(1-rho_0^2)^2-k^2)./2./r.^2-rho_0^2).*r;

out = integral(fun_A2,r_lowerlim,r_upperlim);
end
if rho_0 == 1 && k <= 1e-12
out = pi;
end

end

function f = poly4(x)
global k rho_0 theta
disp([‘Values: k=’ num2str(k) ‘, rho_0=’ num2str(rho_0) ‘, theta=’ num2str(theta) ‘, x=’ num2str(x)])
disp([‘size(x): ‘ num2str(size(x))])
disp([‘size(k): ‘ num2str(size(k))])
disp([‘size(rho_0): ‘ num2str(size(rho_0))])
disp([‘size(theta): ‘ num2str(size(theta))])
k
%Equation B3 from Shuo and Li (LF20B)
f = (x.^2-2.*x.*cos(theta)+1-rho_0^2).*(x.^2-2.*x.*cos(theta+pi)+1-rho_0^2)-k.^2;
disp([‘f = ‘ num2str(f)])
end

function r=mnhf_secant(Fun,x,tol,trace)
%MNHF_SECANT finds the root of "Fun" using secant scheme.
%
% Fun – name of the external function
% x – vector of length 2, (initial guesses)
% tol – tolerance
% trace – print intermediate results
%
% Usage mnhf_secant(@poly1,[-0.5 2.0],1e-8,1)
% poly1 is the name of the external function.
% [-0.5 2.0] are the initial guesses for the root.

% Check inputs.
if nargin < 4, trace = 1; end
if nargin < 3, tol = 1e-8; end
if (length(x) ~= 2)
error(‘Please provide two initial guesses’)
end

f = feval(Fun,x); % Fun is assumed to accept a vector

for i = 1:100
x3 = x(1)-f(1)*(x(2)-x(1))/(f(2)-f(1)); % Update the guess.
f3 = feval(Fun,x3); % Function evaluation.
if trace, fprintf(1,’%3i %12.5f %12.5fn’, i,x3,f3); end
if abs(f3) < tol % Check for convergenece.
r = x3;
return
else % Reset values for x(1), x(2), f(1) and f(2).
x(1) = x(2); f(1) = f(2); x(2) = x3; f(2) = f3;
end
end

r = NaN;Hi. MATLAB Community. I am trying to get the following code to work. I’m working on a toy example that I hope to get running correctly before applying it to a larger numerical project. The issue I’m facing is that the values of k and rho_0 are not updating properly when I Area_Integral from the command window for different values of k and rho_0 (I am selecting k and rho_0 values so it stays in the if loop of the Area_Integral file. I uploaded the code for the various scripts I am using. When I write this in the command window, Area_Integral (1,0.4) and then Area_Integral (1.5,0.4) the value of k is not updated for the later case.
function out = Area_Integral(k, rho_0)
global k rho_0
if k>=1-rho_0^2
theta_array = linspace(0,pi/2,101);
root = zeros(size(theta_array));
for ii = 1:length(theta_array)
theta = theta_array(ii);
root(ii) = mnhf_secant(@poly4,[0.4 0.8],1e-8,0); %% Note: values of k, r0 and theta must be made accessible to poly4
end
out = 2*trapz(theta_array,root);
else

r_upperlim = sqrt(1+rho_0^2+sqrt(k^2+4*rho_0^2));
r_lowerlim = sqrt(1+rho_0^2-sqrt(k^2+4*rho_0^2));

fun_A2 = @(r) acos((r.^4+(1-rho_0^2)^2-k^2)./2./r.^2-rho_0^2).*r;

out = integral(fun_A2,r_lowerlim,r_upperlim);
end
if rho_0 == 1 && k <= 1e-12
out = pi;
end

end

function f = poly4(x)
global k rho_0 theta
disp([‘Values: k=’ num2str(k) ‘, rho_0=’ num2str(rho_0) ‘, theta=’ num2str(theta) ‘, x=’ num2str(x)])
disp([‘size(x): ‘ num2str(size(x))])
disp([‘size(k): ‘ num2str(size(k))])
disp([‘size(rho_0): ‘ num2str(size(rho_0))])
disp([‘size(theta): ‘ num2str(size(theta))])
k
%Equation B3 from Shuo and Li (LF20B)
f = (x.^2-2.*x.*cos(theta)+1-rho_0^2).*(x.^2-2.*x.*cos(theta+pi)+1-rho_0^2)-k.^2;
disp([‘f = ‘ num2str(f)])
end

function r=mnhf_secant(Fun,x,tol,trace)
%MNHF_SECANT finds the root of "Fun" using secant scheme.
%
% Fun – name of the external function
% x – vector of length 2, (initial guesses)
% tol – tolerance
% trace – print intermediate results
%
% Usage mnhf_secant(@poly1,[-0.5 2.0],1e-8,1)
% poly1 is the name of the external function.
% [-0.5 2.0] are the initial guesses for the root.

% Check inputs.
if nargin < 4, trace = 1; end
if nargin < 3, tol = 1e-8; end
if (length(x) ~= 2)
error(‘Please provide two initial guesses’)
end

f = feval(Fun,x); % Fun is assumed to accept a vector

for i = 1:100
x3 = x(1)-f(1)*(x(2)-x(1))/(f(2)-f(1)); % Update the guess.
f3 = feval(Fun,x3); % Function evaluation.
if trace, fprintf(1,’%3i %12.5f %12.5fn’, i,x3,f3); end
if abs(f3) < tol % Check for convergenece.
r = x3;
return
else % Reset values for x(1), x(2), f(1) and f(2).
x(1) = x(2); f(1) = f(2); x(2) = x3; f(2) = f3;
end
end

r = NaN; Hi. MATLAB Community. I am trying to get the following code to work. I’m working on a toy example that I hope to get running correctly before applying it to a larger numerical project. The issue I’m facing is that the values of k and rho_0 are not updating properly when I Area_Integral from the command window for different values of k and rho_0 (I am selecting k and rho_0 values so it stays in the if loop of the Area_Integral file. I uploaded the code for the various scripts I am using. When I write this in the command window, Area_Integral (1,0.4) and then Area_Integral (1.5,0.4) the value of k is not updated for the later case.
function out = Area_Integral(k, rho_0)
global k rho_0
if k>=1-rho_0^2
theta_array = linspace(0,pi/2,101);
root = zeros(size(theta_array));
for ii = 1:length(theta_array)
theta = theta_array(ii);
root(ii) = mnhf_secant(@poly4,[0.4 0.8],1e-8,0); %% Note: values of k, r0 and theta must be made accessible to poly4
end
out = 2*trapz(theta_array,root);
else

r_upperlim = sqrt(1+rho_0^2+sqrt(k^2+4*rho_0^2));
r_lowerlim = sqrt(1+rho_0^2-sqrt(k^2+4*rho_0^2));

fun_A2 = @(r) acos((r.^4+(1-rho_0^2)^2-k^2)./2./r.^2-rho_0^2).*r;

out = integral(fun_A2,r_lowerlim,r_upperlim);
end
if rho_0 == 1 && k <= 1e-12
out = pi;
end

end

function f = poly4(x)
global k rho_0 theta
disp([‘Values: k=’ num2str(k) ‘, rho_0=’ num2str(rho_0) ‘, theta=’ num2str(theta) ‘, x=’ num2str(x)])
disp([‘size(x): ‘ num2str(size(x))])
disp([‘size(k): ‘ num2str(size(k))])
disp([‘size(rho_0): ‘ num2str(size(rho_0))])
disp([‘size(theta): ‘ num2str(size(theta))])
k
%Equation B3 from Shuo and Li (LF20B)
f = (x.^2-2.*x.*cos(theta)+1-rho_0^2).*(x.^2-2.*x.*cos(theta+pi)+1-rho_0^2)-k.^2;
disp([‘f = ‘ num2str(f)])
end

function r=mnhf_secant(Fun,x,tol,trace)
%MNHF_SECANT finds the root of "Fun" using secant scheme.
%
% Fun – name of the external function
% x – vector of length 2, (initial guesses)
% tol – tolerance
% trace – print intermediate results
%
% Usage mnhf_secant(@poly1,[-0.5 2.0],1e-8,1)
% poly1 is the name of the external function.
% [-0.5 2.0] are the initial guesses for the root.

% Check inputs.
if nargin < 4, trace = 1; end
if nargin < 3, tol = 1e-8; end
if (length(x) ~= 2)
error(‘Please provide two initial guesses’)
end

f = feval(Fun,x); % Fun is assumed to accept a vector

for i = 1:100
x3 = x(1)-f(1)*(x(2)-x(1))/(f(2)-f(1)); % Update the guess.
f3 = feval(Fun,x3); % Function evaluation.
if trace, fprintf(1,’%3i %12.5f %12.5fn’, i,x3,f3); end
if abs(f3) < tol % Check for convergenece.
r = x3;
return
else % Reset values for x(1), x(2), f(1) and f(2).
x(1) = x(2); f(1) = f(2); x(2) = x3; f(2) = f3;
end
end

r = NaN; global variables MATLAB Answers — New Questions

​

I am getting this error while trying to run the example code at GitHub – mathworks/OpenTrafficLab: OpenTrafficLab is a MATLAB environment capable of simulating simple traffic scenarios with modular vehicle and junction controllers.

The lastb code block is the one returning the error
%Advance the Scenario
while advance(scenario)
plotOpenPaths(trafficLight)
drawnow limitrate
end

Unrecognized method, property, or field ‘ReferencePoint’ for class ‘DrivingStrategy’.

Error in driving.scenario.DefaultScenario/actorPoses (line 23)
if isa(obj.Scenario.Actors(i),"driving.scenario.Vehicle") && ~isa(motionStrategy,"driving.scenario.Stationary") && isequal(motionStrategy.ReferencePoint, "front-axle")

Error in drivingScenario/actorPoses (line 51)
a = actorPoses(obj.Scenario);

Error in drivingScenario/setUpSensorSimulation (line 474)
obj.SensorSimulation.addActors(actorProfiles(obj), actorPoses(obj));

Error in driving.scenario.DefaultScenario/advance (line 8)
setUpSensorSimulation(obj.Scenario);

Error in drivingScenario/advance (line 42)
running = advance(obj.Scenario);I am getting this error while trying to run the example code at GitHub – mathworks/OpenTrafficLab: OpenTrafficLab is a MATLAB environment capable of simulating simple traffic scenarios with modular vehicle and junction controllers.

The lastb code block is the one returning the error
%Advance the Scenario
while advance(scenario)
plotOpenPaths(trafficLight)
drawnow limitrate
end

Unrecognized method, property, or field ‘ReferencePoint’ for class ‘DrivingStrategy’.

Error in driving.scenario.DefaultScenario/actorPoses (line 23)
if isa(obj.Scenario.Actors(i),"driving.scenario.Vehicle") && ~isa(motionStrategy,"driving.scenario.Stationary") && isequal(motionStrategy.ReferencePoint, "front-axle")

Error in drivingScenario/actorPoses (line 51)
a = actorPoses(obj.Scenario);

Error in drivingScenario/setUpSensorSimulation (line 474)
obj.SensorSimulation.addActors(actorProfiles(obj), actorPoses(obj));

Error in driving.scenario.DefaultScenario/advance (line 8)
setUpSensorSimulation(obj.Scenario);

Error in drivingScenario/advance (line 42)
running = advance(obj.Scenario); I am getting this error while trying to run the example code at GitHub – mathworks/OpenTrafficLab: OpenTrafficLab is a MATLAB environment capable of simulating simple traffic scenarios with modular vehicle and junction controllers.

The lastb code block is the one returning the error
%Advance the Scenario
while advance(scenario)
plotOpenPaths(trafficLight)
drawnow limitrate
end

Unrecognized method, property, or field ‘ReferencePoint’ for class ‘DrivingStrategy’.

Error in driving.scenario.DefaultScenario/actorPoses (line 23)
if isa(obj.Scenario.Actors(i),"driving.scenario.Vehicle") && ~isa(motionStrategy,"driving.scenario.Stationary") && isequal(motionStrategy.ReferencePoint, "front-axle")

Error in drivingScenario/actorPoses (line 51)
a = actorPoses(obj.Scenario);

Error in drivingScenario/setUpSensorSimulation (line 474)
obj.SensorSimulation.addActors(actorProfiles(obj), actorPoses(obj));

Error in driving.scenario.DefaultScenario/advance (line 8)
setUpSensorSimulation(obj.Scenario);

Error in drivingScenario/advance (line 42)
running = advance(obj.Scenario); automated driving toolbox MATLAB Answers — New Questions

​

I have a psychology experiment paradigm which asks participants to give a verbal response immediately after they hear a beep sound. Participants may or may not respond to the beep, and their response could be quick or slow. I need to extract the time length between the end of the beep sound and the start of their verbal response. Such time length should be measured in miliseconds as the total time allowed for each response was 3 seconds (3000 ms). There are hundreds of trials so I would like to find a way to do the extraction automatically. How should I achieve this? Carload thanks to any suggestions!I have a psychology experiment paradigm which asks participants to give a verbal response immediately after they hear a beep sound. Participants may or may not respond to the beep, and their response could be quick or slow. I need to extract the time length between the end of the beep sound and the start of their verbal response. Such time length should be measured in miliseconds as the total time allowed for each response was 3 seconds (3000 ms). There are hundreds of trials so I would like to find a way to do the extraction automatically. How should I achieve this? Carload thanks to any suggestions! I have a psychology experiment paradigm which asks participants to give a verbal response immediately after they hear a beep sound. Participants may or may not respond to the beep, and their response could be quick or slow. I need to extract the time length between the end of the beep sound and the start of their verbal response. Such time length should be measured in miliseconds as the total time allowed for each response was 3 seconds (3000 ms). There are hundreds of trials so I would like to find a way to do the extraction automatically. How should I achieve this? Carload thanks to any suggestions! signal processing, audio signal processing MATLAB Answers — New Questions

​