I have problem when play my sound to Motu. There is "nuzzling sound" during the silence part of your audio, especially in the end of audio.
original audio
after_played_to_16ch

Am I correctly use this code to play in using Motu?

I don’t have any issue when I play directly from Audacity or Reaper.

Thanks

% Define the file and volume settings
audioFile = ‘chapel2m_0_order3_02_eight_new_shoes.wav’; % Path to your audio file
volumeLevel = 0.01; % Desired volume level (0 to 1)

% Get audio file information
info = audioinfo(audioFile);
fs = info.TotalSamples / info.Duration; % Sample rate of the audio file
numChannels = info.NumChannels; % Number of channels in the audio file

% Create an audio file reader
fileReader = dsp.AudioFileReader(audioFile, ‘PlayCount’, Inf);
fileReader.SamplesPerFrame = 8192; % Larger buffer size to avoid underflow/overflow

% Create an audio device writer with ASIO driver and MOTU Pro Audio device
deviceWriter = audioDeviceWriter(‘Driver’, ‘ASIO’, ‘Device’, ‘MOTU Pro Audio’, …
‘SampleRate’, fs, ‘ChannelMappingSource’, ‘Auto’);

% Setup the audio device writer with the correct sample rate and channel mapping
setup(deviceWriter, zeros(fileReader.SamplesPerFrame, numChannels));

% Process and play the audio
while ~isDone(fileReader)
% Read audio data
audioData = fileReader();

% Adjust the audio data by the volume level
audioData = audioData * volumeLevel;

% Write the adjusted audio data to the device
deviceWriter(audioData);
end

% Release resources
release(fileReader);
release(deviceWriter);I have problem when play my sound to Motu. There is "nuzzling sound" during the silence part of your audio, especially in the end of audio.
original audio
after_played_to_16ch

Am I correctly use this code to play in using Motu?

I don’t have any issue when I play directly from Audacity or Reaper.

Thanks

% Define the file and volume settings
audioFile = ‘chapel2m_0_order3_02_eight_new_shoes.wav’; % Path to your audio file
volumeLevel = 0.01; % Desired volume level (0 to 1)

% Get audio file information
info = audioinfo(audioFile);
fs = info.TotalSamples / info.Duration; % Sample rate of the audio file
numChannels = info.NumChannels; % Number of channels in the audio file

% Create an audio file reader
fileReader = dsp.AudioFileReader(audioFile, ‘PlayCount’, Inf);
fileReader.SamplesPerFrame = 8192; % Larger buffer size to avoid underflow/overflow

% Create an audio device writer with ASIO driver and MOTU Pro Audio device
deviceWriter = audioDeviceWriter(‘Driver’, ‘ASIO’, ‘Device’, ‘MOTU Pro Audio’, …
‘SampleRate’, fs, ‘ChannelMappingSource’, ‘Auto’);

% Setup the audio device writer with the correct sample rate and channel mapping
setup(deviceWriter, zeros(fileReader.SamplesPerFrame, numChannels));

% Process and play the audio
while ~isDone(fileReader)
% Read audio data
audioData = fileReader();

% Adjust the audio data by the volume level
audioData = audioData * volumeLevel;

% Write the adjusted audio data to the device
deviceWriter(audioData);
end

% Release resources
release(fileReader);
release(deviceWriter); I have problem when play my sound to Motu. There is "nuzzling sound" during the silence part of your audio, especially in the end of audio.
original audio
after_played_to_16ch

Am I correctly use this code to play in using Motu?

I don’t have any issue when I play directly from Audacity or Reaper.

Thanks

% Define the file and volume settings
audioFile = ‘chapel2m_0_order3_02_eight_new_shoes.wav’; % Path to your audio file
volumeLevel = 0.01; % Desired volume level (0 to 1)

% Get audio file information
info = audioinfo(audioFile);
fs = info.TotalSamples / info.Duration; % Sample rate of the audio file
numChannels = info.NumChannels; % Number of channels in the audio file

% Create an audio file reader
fileReader = dsp.AudioFileReader(audioFile, ‘PlayCount’, Inf);
fileReader.SamplesPerFrame = 8192; % Larger buffer size to avoid underflow/overflow

% Create an audio device writer with ASIO driver and MOTU Pro Audio device
deviceWriter = audioDeviceWriter(‘Driver’, ‘ASIO’, ‘Device’, ‘MOTU Pro Audio’, …
‘SampleRate’, fs, ‘ChannelMappingSource’, ‘Auto’);

% Setup the audio device writer with the correct sample rate and channel mapping
setup(deviceWriter, zeros(fileReader.SamplesPerFrame, numChannels));

% Process and play the audio
while ~isDone(fileReader)
% Read audio data
audioData = fileReader();

% Adjust the audio data by the volume level
audioData = audioData * volumeLevel;

% Write the adjusted audio data to the device
deviceWriter(audioData);
end

% Release resources
release(fileReader);
release(deviceWriter); audio, motu, 16-channel MATLAB Answers — New Questions

​

Greetings
I’m working on deploying deep learning on raspberry pi 4 with matlab version 2024a. I’m facing timeout error when output of entry-point function with PIL configuration is vector. however when configuration is mex, the code has successfully implemtned. Simple code below can demonstrate the issue.

Kindly, how the code with PIL configuration can be fixed so that the output from entry-point function is vector or matrix ?
the driver for this request is described in more details in my case study below.

% raspberry connection and configuration
r = raspi(‘raspberrypi’,’pi’,’raspberry’);

cfg = coder.config(‘lib’,’ecoder’,true);
cfg.VerificationMode = ‘PIL’;
cfg.TargetLang = ‘C++’;

dlcfg = coder.DeepLearningConfig(‘arm-compute’);
dlcfg.ArmComputeVersion = ‘20.02.1’;
dlcfg.ArmArchitecture = ‘armv7’;
cfg.DeepLearningConfig = dlcfg;
cfg.MATLABSourceComments = 1;

hw = coder.hardware(‘Raspberry Pi’);
cfg.Hardware = hw;
cfg.CodeExecutionProfiling = true;

Case 1: with mex configuration, successfuly implemented
type dummy

t = ones(1,3);
codegen dummy -args {t} -report -config:mex

% testing deployement
x=[1,2,3];
testDummy= dummy_mex(x);

Case 2: with pil configuraiton, timeout error generated
type dummy

cfg.Hardware.BuildDir = ‘~/dummy’;
t = ones(1,3);
codegen dummy -args {t} -report -config cfg

% testing deployement
x=[1,2,3];
testDummy= dummy_pil(x);
entry-point function
function out = dummy(in)
%#codegen
out = in;
end

Case Study
my case study has raw data consists of 8 channels where I need to extract features from wavelet scattering network then pass them into trained lstm network. My code is similar to the below example, but the difference is my input raw data is 8 channels, while the example is one channel. openExample(‘wavelet/CodeGenerationForFaultDetectionUsingWaveletAndRNNExample’).
I’m getting the below error:
Number of features of input passed to the predict method must be a code generation time constant

So, I decided to breakdown entry-point function into two functions for easier troubleshooting. First featureFunction is to extract wavelet features from raw data (batch: 500samples x 8 channels), the second predictFunction is to take the generated wavelet features (32 C x 102 T) into trained lstm network.
prdictFunction_pil worked find which accept input size (32 x 102), and get predicted labels.
featureFunction_pil gives the below error
Error using rtw.pil.SILPILInterface.throwMException (line 1774)
The timeout of 300 seconds for receiving data from the rtiostream interface has been exceeded. There might be multiple reasons for this communications failure.

You should:
(a) Check that the target hardware configuration is correct, for example, check that the byte ordering is correct.
(b) Confirm that the target application is running on the target hardware.
(c) Consider the possibility of application run-time failures (e.g. divide by zero exceptions, incorrect custom code integration, etc.).
Note (c): To identify possible reasons for the run-time failure, consider using SIL, which supports signal handlers and debugging.
If you cannot find a solution, consider using the method setTimeoutRecvSecs of rtw.connectivity.RtIOStreamHostCommunicator to increase the timeout value.Greetings
I’m working on deploying deep learning on raspberry pi 4 with matlab version 2024a. I’m facing timeout error when output of entry-point function with PIL configuration is vector. however when configuration is mex, the code has successfully implemtned. Simple code below can demonstrate the issue.

Kindly, how the code with PIL configuration can be fixed so that the output from entry-point function is vector or matrix ?
the driver for this request is described in more details in my case study below.

% raspberry connection and configuration
r = raspi(‘raspberrypi’,’pi’,’raspberry’);

cfg = coder.config(‘lib’,’ecoder’,true);
cfg.VerificationMode = ‘PIL’;
cfg.TargetLang = ‘C++’;

dlcfg = coder.DeepLearningConfig(‘arm-compute’);
dlcfg.ArmComputeVersion = ‘20.02.1’;
dlcfg.ArmArchitecture = ‘armv7’;
cfg.DeepLearningConfig = dlcfg;
cfg.MATLABSourceComments = 1;

hw = coder.hardware(‘Raspberry Pi’);
cfg.Hardware = hw;
cfg.CodeExecutionProfiling = true;

Case 1: with mex configuration, successfuly implemented
type dummy

t = ones(1,3);
codegen dummy -args {t} -report -config:mex

% testing deployement
x=[1,2,3];
testDummy= dummy_mex(x);

Case 2: with pil configuraiton, timeout error generated
type dummy

cfg.Hardware.BuildDir = ‘~/dummy’;
t = ones(1,3);
codegen dummy -args {t} -report -config cfg

% testing deployement
x=[1,2,3];
testDummy= dummy_pil(x);
entry-point function
function out = dummy(in)
%#codegen
out = in;
end

Case Study
my case study has raw data consists of 8 channels where I need to extract features from wavelet scattering network then pass them into trained lstm network. My code is similar to the below example, but the difference is my input raw data is 8 channels, while the example is one channel. openExample(‘wavelet/CodeGenerationForFaultDetectionUsingWaveletAndRNNExample’).
I’m getting the below error:
Number of features of input passed to the predict method must be a code generation time constant

So, I decided to breakdown entry-point function into two functions for easier troubleshooting. First featureFunction is to extract wavelet features from raw data (batch: 500samples x 8 channels), the second predictFunction is to take the generated wavelet features (32 C x 102 T) into trained lstm network.
prdictFunction_pil worked find which accept input size (32 x 102), and get predicted labels.
featureFunction_pil gives the below error
Error using rtw.pil.SILPILInterface.throwMException (line 1774)
The timeout of 300 seconds for receiving data from the rtiostream interface has been exceeded. There might be multiple reasons for this communications failure.

You should:
(a) Check that the target hardware configuration is correct, for example, check that the byte ordering is correct.
(b) Confirm that the target application is running on the target hardware.
(c) Consider the possibility of application run-time failures (e.g. divide by zero exceptions, incorrect custom code integration, etc.).
Note (c): To identify possible reasons for the run-time failure, consider using SIL, which supports signal handlers and debugging.
If you cannot find a solution, consider using the method setTimeoutRecvSecs of rtw.connectivity.RtIOStreamHostCommunicator to increase the timeout value. Greetings
I’m working on deploying deep learning on raspberry pi 4 with matlab version 2024a. I’m facing timeout error when output of entry-point function with PIL configuration is vector. however when configuration is mex, the code has successfully implemtned. Simple code below can demonstrate the issue.

Kindly, how the code with PIL configuration can be fixed so that the output from entry-point function is vector or matrix ?
the driver for this request is described in more details in my case study below.

% raspberry connection and configuration
r = raspi(‘raspberrypi’,’pi’,’raspberry’);

cfg = coder.config(‘lib’,’ecoder’,true);
cfg.VerificationMode = ‘PIL’;
cfg.TargetLang = ‘C++’;

dlcfg = coder.DeepLearningConfig(‘arm-compute’);
dlcfg.ArmComputeVersion = ‘20.02.1’;
dlcfg.ArmArchitecture = ‘armv7’;
cfg.DeepLearningConfig = dlcfg;
cfg.MATLABSourceComments = 1;

hw = coder.hardware(‘Raspberry Pi’);
cfg.Hardware = hw;
cfg.CodeExecutionProfiling = true;

Case 1: with mex configuration, successfuly implemented
type dummy

t = ones(1,3);
codegen dummy -args {t} -report -config:mex

% testing deployement
x=[1,2,3];
testDummy= dummy_mex(x);

Case 2: with pil configuraiton, timeout error generated
type dummy

cfg.Hardware.BuildDir = ‘~/dummy’;
t = ones(1,3);
codegen dummy -args {t} -report -config cfg

% testing deployement
x=[1,2,3];
testDummy= dummy_pil(x);
entry-point function
function out = dummy(in)
%#codegen
out = in;
end

Case Study
my case study has raw data consists of 8 channels where I need to extract features from wavelet scattering network then pass them into trained lstm network. My code is similar to the below example, but the difference is my input raw data is 8 channels, while the example is one channel. openExample(‘wavelet/CodeGenerationForFaultDetectionUsingWaveletAndRNNExample’).
I’m getting the below error:
Number of features of input passed to the predict method must be a code generation time constant

So, I decided to breakdown entry-point function into two functions for easier troubleshooting. First featureFunction is to extract wavelet features from raw data (batch: 500samples x 8 channels), the second predictFunction is to take the generated wavelet features (32 C x 102 T) into trained lstm network.
prdictFunction_pil worked find which accept input size (32 x 102), and get predicted labels.
featureFunction_pil gives the below error
Error using rtw.pil.SILPILInterface.throwMException (line 1774)
The timeout of 300 seconds for receiving data from the rtiostream interface has been exceeded. There might be multiple reasons for this communications failure.

You should:
(a) Check that the target hardware configuration is correct, for example, check that the byte ordering is correct.
(b) Confirm that the target application is running on the target hardware.
(c) Consider the possibility of application run-time failures (e.g. divide by zero exceptions, incorrect custom code integration, etc.).
Note (c): To identify possible reasons for the run-time failure, consider using SIL, which supports signal handlers and debugging.
If you cannot find a solution, consider using the method setTimeoutRecvSecs of rtw.connectivity.RtIOStreamHostCommunicator to increase the timeout value. code generation, matlab coder, raspberry, deployment MATLAB Answers — New Questions

​

I am trying to find the width of a curve taken from the intensity vs x-index of an image (see below) at intensity = 1/e^2, essentially finding the beam diameter of a laser.

The resulting plot is:

The value of 1/e^2 is -3.55e4. Right now its calculating the width from the red points on the plot, but I’d like it to calculate the points from the green points. I’ve tried making it so it finds the points where the slope is negative, but that doesn’t seem to work. Even if you have an idea of how to do this that would be helpful, I can try to figure out the code from there. Still pretty new to matlab.
My code is:
clear
im=imread("D:THORS Imaging8.06.24Result of on_20000-off_20000.tif")
im=imrotate(im, -96,"bilinear")
im=imcrop(im, [200, 450, 700, 150]);
result=im
imshow(result)
%Sums pixel intensity in each verticle column
vertsum = sum(result, 1);

%Subtracts maximum intensity
vertsum =vertsum-max(vertsum);

%Sets threshold for baseline intensity
thresholdhigh = -100; %Threshold value chosen base on intensity of image

%Remove points above the threshold
vertsum = vertsum(vertsum <= thresholdhigh);

%Performs moving average on data
vertsum = movmean(vertsum,100);

%Sets data points fo create the Gaussian fit
[xData, yData] = prepareCurveData( [], vertsum );

%Sets conversion factor between pixels and cm
conversion = 3.5/1024;

%Convert pixels to cm
xcm = xData * conversion;

%Generates plot
plot(xcm, yData );

%% Find width of barrier
%Set the height of the fitted curve equal to the distance between 0 and the
%Height of intensity value
min = min(yData)
max=max(yData)
height=max-min
%Sets variable for 1/e^2, chosen due to gaussian shape of the beam
e = (1/((2.718281828459045)^2))
%Sets y-value on the curve at 1/e^2
pointY = max-(height * e)
%Finds nearest left side x-value on curve at 1/e^2
leftIndexWidth = find(yData <= pointY, 1, ‘first’)
%Finds nearest right side y-value on curve at 1/e^2
rightIndexWidth = find(yData <= pointY, 1, ‘last’)
%Converts pixel coordinate to cm
leftIndexWidth = leftIndexWidth * conversion
rightIndexWidth = rightIndexWidth * conversion

%Calculates width of curve at 1/e^2
widthpx = rightIndexWidth – leftIndexWidthI am trying to find the width of a curve taken from the intensity vs x-index of an image (see below) at intensity = 1/e^2, essentially finding the beam diameter of a laser.

The resulting plot is:

The value of 1/e^2 is -3.55e4. Right now its calculating the width from the red points on the plot, but I’d like it to calculate the points from the green points. I’ve tried making it so it finds the points where the slope is negative, but that doesn’t seem to work. Even if you have an idea of how to do this that would be helpful, I can try to figure out the code from there. Still pretty new to matlab.
My code is:
clear
im=imread("D:THORS Imaging8.06.24Result of on_20000-off_20000.tif")
im=imrotate(im, -96,"bilinear")
im=imcrop(im, [200, 450, 700, 150]);
result=im
imshow(result)
%Sums pixel intensity in each verticle column
vertsum = sum(result, 1);

%Subtracts maximum intensity
vertsum =vertsum-max(vertsum);

%Sets threshold for baseline intensity
thresholdhigh = -100; %Threshold value chosen base on intensity of image

%Remove points above the threshold
vertsum = vertsum(vertsum <= thresholdhigh);

%Performs moving average on data
vertsum = movmean(vertsum,100);

%Sets data points fo create the Gaussian fit
[xData, yData] = prepareCurveData( [], vertsum );

%Sets conversion factor between pixels and cm
conversion = 3.5/1024;

%Convert pixels to cm
xcm = xData * conversion;

%Generates plot
plot(xcm, yData );

%% Find width of barrier
%Set the height of the fitted curve equal to the distance between 0 and the
%Height of intensity value
min = min(yData)
max=max(yData)
height=max-min
%Sets variable for 1/e^2, chosen due to gaussian shape of the beam
e = (1/((2.718281828459045)^2))
%Sets y-value on the curve at 1/e^2
pointY = max-(height * e)
%Finds nearest left side x-value on curve at 1/e^2
leftIndexWidth = find(yData <= pointY, 1, ‘first’)
%Finds nearest right side y-value on curve at 1/e^2
rightIndexWidth = find(yData <= pointY, 1, ‘last’)
%Converts pixel coordinate to cm
leftIndexWidth = leftIndexWidth * conversion
rightIndexWidth = rightIndexWidth * conversion

%Calculates width of curve at 1/e^2
widthpx = rightIndexWidth – leftIndexWidth I am trying to find the width of a curve taken from the intensity vs x-index of an image (see below) at intensity = 1/e^2, essentially finding the beam diameter of a laser.

The resulting plot is:

The value of 1/e^2 is -3.55e4. Right now its calculating the width from the red points on the plot, but I’d like it to calculate the points from the green points. I’ve tried making it so it finds the points where the slope is negative, but that doesn’t seem to work. Even if you have an idea of how to do this that would be helpful, I can try to figure out the code from there. Still pretty new to matlab.
My code is:
clear
im=imread("D:THORS Imaging8.06.24Result of on_20000-off_20000.tif")
im=imrotate(im, -96,"bilinear")
im=imcrop(im, [200, 450, 700, 150]);
result=im
imshow(result)
%Sums pixel intensity in each verticle column
vertsum = sum(result, 1);

%Subtracts maximum intensity
vertsum =vertsum-max(vertsum);

%Sets threshold for baseline intensity
thresholdhigh = -100; %Threshold value chosen base on intensity of image

%Remove points above the threshold
vertsum = vertsum(vertsum <= thresholdhigh);

%Performs moving average on data
vertsum = movmean(vertsum,100);

%Sets data points fo create the Gaussian fit
[xData, yData] = prepareCurveData( [], vertsum );

%Sets conversion factor between pixels and cm
conversion = 3.5/1024;

%Convert pixels to cm
xcm = xData * conversion;

%Generates plot
plot(xcm, yData );

%% Find width of barrier
%Set the height of the fitted curve equal to the distance between 0 and the
%Height of intensity value
min = min(yData)
max=max(yData)
height=max-min
%Sets variable for 1/e^2, chosen due to gaussian shape of the beam
e = (1/((2.718281828459045)^2))
%Sets y-value on the curve at 1/e^2
pointY = max-(height * e)
%Finds nearest left side x-value on curve at 1/e^2
leftIndexWidth = find(yData <= pointY, 1, ‘first’)
%Finds nearest right side y-value on curve at 1/e^2
rightIndexWidth = find(yData <= pointY, 1, ‘last’)
%Converts pixel coordinate to cm
leftIndexWidth = leftIndexWidth * conversion
rightIndexWidth = rightIndexWidth * conversion

%Calculates width of curve at 1/e^2
widthpx = rightIndexWidth – leftIndexWidth fwhm, curve fitting MATLAB Answers — New Questions

​