I am writing a Convolutional Neural Network for regression in MATLAB R2021b. I’m using the trainNetwork function in Deep Learning Toolbox and in options I have ‘Plots’,’training-progress’.
I have understood that in the Training Progress, for each iteration, I have the MSE value computed on the mini-batch.
I wuold like to know whether I can plot in the training progress the MSE value on the overall Training Set and Validation Set respectively and not on the mini-batch.
Thanks in advance.I am writing a Convolutional Neural Network for regression in MATLAB R2021b. I’m using the trainNetwork function in Deep Learning Toolbox and in options I have ‘Plots’,’training-progress’.
I have understood that in the Training Progress, for each iteration, I have the MSE value computed on the mini-batch.
I wuold like to know whether I can plot in the training progress the MSE value on the overall Training Set and Validation Set respectively and not on the mini-batch.
Thanks in advance. I am writing a Convolutional Neural Network for regression in MATLAB R2021b. I’m using the trainNetwork function in Deep Learning Toolbox and in options I have ‘Plots’,’training-progress’.
I have understood that in the Training Progress, for each iteration, I have the MSE value computed on the mini-batch.
I wuold like to know whether I can plot in the training progress the MSE value on the overall Training Set and Validation Set respectively and not on the mini-batch.
Thanks in advance. training progress, trainnetwork MATLAB Answers — New Questions

​

Hello, I’m trying to connect to a custom target computer via a LAN cable at MATLAB/Simulink R2017b, but when I run the srttest an error comes out:.

## Simulink Real-Time Test Suite
### Host-Target interface is: TcpIp
### Test 1, Ping target computer ‘TargetPC1’ using system ping: … OK
### Test 2, Ping target computer ‘TargetPC1’ using SLRTPINGTARGET: … OK
### Test 3, Software reboot the target computer ‘TargetPC1’: error: slrttest (line 181)

message: "The input string format is not correct."
source: "mscorlib"

181 error( booterror.message );Hello, I’m trying to connect to a custom target computer via a LAN cable at MATLAB/Simulink R2017b, but when I run the srttest an error comes out:.

## Simulink Real-Time Test Suite
### Host-Target interface is: TcpIp
### Test 1, Ping target computer ‘TargetPC1’ using system ping: … OK
### Test 2, Ping target computer ‘TargetPC1’ using SLRTPINGTARGET: … OK
### Test 3, Software reboot the target computer ‘TargetPC1’: error: slrttest (line 181)

message: "The input string format is not correct."
source: "mscorlib"

181 error( booterror.message ); Hello, I’m trying to connect to a custom target computer via a LAN cable at MATLAB/Simulink R2017b, but when I run the srttest an error comes out:.

## Simulink Real-Time Test Suite
### Host-Target interface is: TcpIp
### Test 1, Ping target computer ‘TargetPC1’ using system ping: … OK
### Test 2, Ping target computer ‘TargetPC1’ using SLRTPINGTARGET: … OK
### Test 3, Software reboot the target computer ‘TargetPC1’: error: slrttest (line 181)

message: "The input string format is not correct."
source: "mscorlib"

181 error( booterror.message ); srttest MATLAB Answers — New Questions

​

%a
cockrach=importdata(‘cockroach_population.xlsx’);
num_cockrach=cockrach.data;
days=num_cockrach(:,1);
cockraches=num_cockrach(:,2);
%b
ylin=cockraches;
xlin=log(days);
[a0,a1,r2] = linreg(xlin,ylin);
p=polyfit(days,cockraches,2);
alpha=(exp(1)).^a0;
Beta=a1;
text1=sprintf(‘parabola-cockraches rate: cockraches= %.2fx^2+%.2fx+%.2f’,p);
%c
exponential_curve= alpha.*exp(1).^(Beta.*days);
text2=sprintf(‘exponential-cockraches rate: cockraches= %.2fe^%.2fxn’,alpha, Beta);
figure
plot(days,cockraches,’b.’);
hold on
plot(days,polyval(p,days),’r-‘,’LineWidth’,3);
plot(days,exponential_curve,’g-‘,’LineWidth’,3)
title(‘numbers of cockraches vs times(days)’);
xlabel(‘cockraches’);
ylabel(‘days’);
legend(‘unlinearised’,text1,text2)%a
cockrach=importdata(‘cockroach_population.xlsx’);
num_cockrach=cockrach.data;
days=num_cockrach(:,1);
cockraches=num_cockrach(:,2);
%b
ylin=cockraches;
xlin=log(days);
[a0,a1,r2] = linreg(xlin,ylin);
p=polyfit(days,cockraches,2);
alpha=(exp(1)).^a0;
Beta=a1;
text1=sprintf(‘parabola-cockraches rate: cockraches= %.2fx^2+%.2fx+%.2f’,p);
%c
exponential_curve= alpha.*exp(1).^(Beta.*days);
text2=sprintf(‘exponential-cockraches rate: cockraches= %.2fe^%.2fxn’,alpha, Beta);
figure
plot(days,cockraches,’b.’);
hold on
plot(days,polyval(p,days),’r-‘,’LineWidth’,3);
plot(days,exponential_curve,’g-‘,’LineWidth’,3)
title(‘numbers of cockraches vs times(days)’);
xlabel(‘cockraches’);
ylabel(‘days’);
legend(‘unlinearised’,text1,text2) %a
cockrach=importdata(‘cockroach_population.xlsx’);
num_cockrach=cockrach.data;
days=num_cockrach(:,1);
cockraches=num_cockrach(:,2);
%b
ylin=cockraches;
xlin=log(days);
[a0,a1,r2] = linreg(xlin,ylin);
p=polyfit(days,cockraches,2);
alpha=(exp(1)).^a0;
Beta=a1;
text1=sprintf(‘parabola-cockraches rate: cockraches= %.2fx^2+%.2fx+%.2f’,p);
%c
exponential_curve= alpha.*exp(1).^(Beta.*days);
text2=sprintf(‘exponential-cockraches rate: cockraches= %.2fe^%.2fxn’,alpha, Beta);
figure
plot(days,cockraches,’b.’);
hold on
plot(days,polyval(p,days),’r-‘,’LineWidth’,3);
plot(days,exponential_curve,’g-‘,’LineWidth’,3)
title(‘numbers of cockraches vs times(days)’);
xlabel(‘cockraches’);
ylabel(‘days’);
legend(‘unlinearised’,text1,text2) exponential curve, nonlinear, linear MATLAB Answers — New Questions

​

close all
iter=10;
A=im2double(imread(‘1.JPG’));
B=im2double(imread(‘2.JPG’));
A=imresize(A, [512 512]);
B=imresize(B, [512 512]);
D=B-A;
D=D./iter;
for i=1:iter
A=A+D;
imshow(A);
pause (0.1);
endclose all
iter=10;
A=im2double(imread(‘1.JPG’));
B=im2double(imread(‘2.JPG’));
A=imresize(A, [512 512]);
B=imresize(B, [512 512]);
D=B-A;
D=D./iter;
for i=1:iter
A=A+D;
imshow(A);
pause (0.1);
end close all
iter=10;
A=im2double(imread(‘1.JPG’));
B=im2double(imread(‘2.JPG’));
A=imresize(A, [512 512]);
B=imresize(B, [512 512]);
D=B-A;
D=D./iter;
for i=1:iter
A=A+D;
imshow(A);
pause (0.1);
end image processing, morphing, morph images MATLAB Answers — New Questions

​

clear all
close
%% 参数定义parameter definition
P = 42;
c = 800;
E = 15000000;
K = 1.8;
P_ya = 18000;
F = 2;
y = 26.8;
R = 20; % radius
syms H B
%H=100;
% 计算破裂角 a Calculate the rupture angle
if K <= 0.5
a = 90;
elseif K <= 1
a = -90 * K + 135;
elseif K <= 3
a = -22.5 * K + 67.5;
else
a = 0;
end

% 显示计算得到的 a 的值
disp([‘当 K = ‘, num2str(K), ‘ 时，破裂角 a = ‘, num2str(a), ‘°’]);

%% 求解初始破裂角相关量 Solving the initial rupture angle related quantities
L = H + R * (1 – sind(a));
G_1 = (y * L^2) / (2 * tand(B)); % 三角形块体的自重
p = atan2(tand(P), F); % 折减后的内摩擦角
p = rad2deg(p); % 将结果转换为度,便于后面统一计算
C = c * L; % 竖直面上粘聚力合力
C_s = c * L / (F * sind(B)); % 破裂面上粘聚力合力
G_0 = 2 * y * H * cosd(a);
z = 0.9 * P; % 按照围岩等级取值，三级围岩取0.9

%% 定义目标函数 E(B) Define the objective function E (B)
%E_func = @(B) (y ./ (2 .* tand(B))) .* sind(B + p) ./ cosd(B + p – z);
E_func=@(B) (cosd(B+p).*sind(B)).*cosd(B).*cosd(B+p-z)+sind(B+p).*sind(B).*(sind(B+p-z).*cosd(B)+cosd(B+p-z).*sind(B));
%% 数值求导函数 Numerical derivative function
% 使用中心差分法计算导数
dE_func = @(B) (E_func(B + 1e-6) – E_func(B – 1e-6)) / (2e-6);
%% 数值寻找导数为零的 B 值
% 只寻找一个接近的 B 值
B_range = [0, 90]; % B 的取值范围
B_init = 45; % 初始猜测值，设置为 45 度
% 使用 fzero 寻找导数为零的 B 值
try
B_zero = fzero(dE_func, B_init);
% 检查找到的 B 值是否满足条件
if abs(dE_func(B_zero)) < 1e-6
disp([‘找到满足条件的 B 值为：’, num2str(B_zero)]);
else
disp(‘没有找到导数接近零的 B 值’);
end
catch
disp(‘fzero 计算失败，未找到满足条件的 B 值’);
end
B=B_zero
%% 计算埋深 Calculate burial depth
f1 = ((G_1 – C) .* sind(p + B_zero) + C_s .* cosd(p)) ./ cosd(B_zero + p – z);
f2 = (P_ya – G_0 – 2 .* C) / (2 * sind(z));
% 定义控制方程，解出 H
eqn = f1 – f2 == 0;
% 使用 solve 反解出 H
sol_H_sym = solve(eqn, H);
% 将符号解转换为具体的数值
sol_H_num = double(subs(sol_H_sym));
% 显示结果
disp([‘解出的 H 的值为：’, num2str(sol_H_num)]);clear all
close
%% 参数定义parameter definition
P = 42;
c = 800;
E = 15000000;
K = 1.8;
P_ya = 18000;
F = 2;
y = 26.8;
R = 20; % radius
syms H B
%H=100;
% 计算破裂角 a Calculate the rupture angle
if K <= 0.5
a = 90;
elseif K <= 1
a = -90 * K + 135;
elseif K <= 3
a = -22.5 * K + 67.5;
else
a = 0;
end

% 显示计算得到的 a 的值
disp([‘当 K = ‘, num2str(K), ‘ 时，破裂角 a = ‘, num2str(a), ‘°’]);

%% 求解初始破裂角相关量 Solving the initial rupture angle related quantities
L = H + R * (1 – sind(a));
G_1 = (y * L^2) / (2 * tand(B)); % 三角形块体的自重
p = atan2(tand(P), F); % 折减后的内摩擦角
p = rad2deg(p); % 将结果转换为度,便于后面统一计算
C = c * L; % 竖直面上粘聚力合力
C_s = c * L / (F * sind(B)); % 破裂面上粘聚力合力
G_0 = 2 * y * H * cosd(a);
z = 0.9 * P; % 按照围岩等级取值，三级围岩取0.9

%% 定义目标函数 E(B) Define the objective function E (B)
%E_func = @(B) (y ./ (2 .* tand(B))) .* sind(B + p) ./ cosd(B + p – z);
E_func=@(B) (cosd(B+p).*sind(B)).*cosd(B).*cosd(B+p-z)+sind(B+p).*sind(B).*(sind(B+p-z).*cosd(B)+cosd(B+p-z).*sind(B));
%% 数值求导函数 Numerical derivative function
% 使用中心差分法计算导数
dE_func = @(B) (E_func(B + 1e-6) – E_func(B – 1e-6)) / (2e-6);
%% 数值寻找导数为零的 B 值
% 只寻找一个接近的 B 值
B_range = [0, 90]; % B 的取值范围
B_init = 45; % 初始猜测值，设置为 45 度
% 使用 fzero 寻找导数为零的 B 值
try
B_zero = fzero(dE_func, B_init);
% 检查找到的 B 值是否满足条件
if abs(dE_func(B_zero)) < 1e-6
disp([‘找到满足条件的 B 值为：’, num2str(B_zero)]);
else
disp(‘没有找到导数接近零的 B 值’);
end
catch
disp(‘fzero 计算失败，未找到满足条件的 B 值’);
end
B=B_zero
%% 计算埋深 Calculate burial depth
f1 = ((G_1 – C) .* sind(p + B_zero) + C_s .* cosd(p)) ./ cosd(B_zero + p – z);
f2 = (P_ya – G_0 – 2 .* C) / (2 * sind(z));
% 定义控制方程，解出 H
eqn = f1 – f2 == 0;
% 使用 solve 反解出 H
sol_H_sym = solve(eqn, H);
% 将符号解转换为具体的数值
sol_H_num = double(subs(sol_H_sym));
% 显示结果
disp([‘解出的 H 的值为：’, num2str(sol_H_num)]); clear all
close
%% 参数定义parameter definition
P = 42;
c = 800;
E = 15000000;
K = 1.8;
P_ya = 18000;
F = 2;
y = 26.8;
R = 20; % radius
syms H B
%H=100;
% 计算破裂角 a Calculate the rupture angle
if K <= 0.5
a = 90;
elseif K <= 1
a = -90 * K + 135;
elseif K <= 3
a = -22.5 * K + 67.5;
else
a = 0;
end

% 显示计算得到的 a 的值
disp([‘当 K = ‘, num2str(K), ‘ 时，破裂角 a = ‘, num2str(a), ‘°’]);

%% 求解初始破裂角相关量 Solving the initial rupture angle related quantities
L = H + R * (1 – sind(a));
G_1 = (y * L^2) / (2 * tand(B)); % 三角形块体的自重
p = atan2(tand(P), F); % 折减后的内摩擦角
p = rad2deg(p); % 将结果转换为度,便于后面统一计算
C = c * L; % 竖直面上粘聚力合力
C_s = c * L / (F * sind(B)); % 破裂面上粘聚力合力
G_0 = 2 * y * H * cosd(a);
z = 0.9 * P; % 按照围岩等级取值，三级围岩取0.9

%% 定义目标函数 E(B) Define the objective function E (B)
%E_func = @(B) (y ./ (2 .* tand(B))) .* sind(B + p) ./ cosd(B + p – z);
E_func=@(B) (cosd(B+p).*sind(B)).*cosd(B).*cosd(B+p-z)+sind(B+p).*sind(B).*(sind(B+p-z).*cosd(B)+cosd(B+p-z).*sind(B));
%% 数值求导函数 Numerical derivative function
% 使用中心差分法计算导数
dE_func = @(B) (E_func(B + 1e-6) – E_func(B – 1e-6)) / (2e-6);
%% 数值寻找导数为零的 B 值
% 只寻找一个接近的 B 值
B_range = [0, 90]; % B 的取值范围
B_init = 45; % 初始猜测值，设置为 45 度
% 使用 fzero 寻找导数为零的 B 值
try
B_zero = fzero(dE_func, B_init);
% 检查找到的 B 值是否满足条件
if abs(dE_func(B_zero)) < 1e-6
disp([‘找到满足条件的 B 值为：’, num2str(B_zero)]);
else
disp(‘没有找到导数接近零的 B 值’);
end
catch
disp(‘fzero 计算失败，未找到满足条件的 B 值’);
end
B=B_zero
%% 计算埋深 Calculate burial depth
f1 = ((G_1 – C) .* sind(p + B_zero) + C_s .* cosd(p)) ./ cosd(B_zero + p – z);
f2 = (P_ya – G_0 – 2 .* C) / (2 * sind(z));
% 定义控制方程，解出 H
eqn = f1 – f2 == 0;
% 使用 solve 反解出 H
sol_H_sym = solve(eqn, H);
% 将符号解转换为具体的数值
sol_H_num = double(subs(sol_H_sym));
% 显示结果
disp([‘解出的 H 的值为：’, num2str(sol_H_num)]); transferred, syms, symbolic MATLAB Answers — New Questions

​

How to sort dlarrays like in
sort(a)
, where a is a numeric array? Is there a way of sorting next(mbq) by specifying some options in the following code?
mbq = minibatchqueue(ads,MiniBatchSize=miniBatchSize,MiniBatchFormat="BC");
next(mbq)How to sort dlarrays like in
sort(a)
, where a is a numeric array? Is there a way of sorting next(mbq) by specifying some options in the following code?
mbq = minibatchqueue(ads,MiniBatchSize=miniBatchSize,MiniBatchFormat="BC");
next(mbq) How to sort dlarrays like in
sort(a)
, where a is a numeric array? Is there a way of sorting next(mbq) by specifying some options in the following code?
mbq = minibatchqueue(ads,MiniBatchSize=miniBatchSize,MiniBatchFormat="BC");
next(mbq) deep learning MATLAB Answers — New Questions

​

When using spectrogram to calculate power spectral density for a large dataset, the plot produced by directly running the command with no output arguments is slightly different from what I get when I run it with an output argument and plot that output myself.

For example, the following plots spectrograms for a subset of my data and compares the direct result of spectrogram with the output raster:
load(‘exampledata.mat’)
window = 10000;
noverlap = round(0.1*window);
fs = 10000;

figure(1);clf
t = tiledlayout(3,1, ‘TileIndexing’, ‘columnmajor’);

% Plot spectrogram directly from spectrogram function
s(1) = nexttile;
spectrogram(exampledata,window,noverlap,window,fs);
title(‘Original spectrogram plot (real data)’)
PSDdB_orig = get(get(gca,’Children’),’CData’);
caxis([min(PSDdB_orig,[],’all’) max(PSDdB_orig,[],’all’)])

% Calculate power spectral density and plot as decibels
[~,F,T,PSD] = spectrogram(exampledata,window,noverlap,window,fs);
PSDdB = (10.*log10(PSD))’;

s(2) = nexttile;
imagesc(F./1000,T./60,PSDdB)
title(‘Spectrogram output raster (real data)’)
c = colorbar;
caxis([min(PSDdB,[],’all’) max(PSDdB,[],’all’)])
set(get(c,’label’),’string’,’Power/frequency (dB/Hz)’,’Rotation’,90);

% Calculate ratio of the two approaches above and plot that
PSDdiff = PSDdB_orig./PSDdB;

s(3) = nexttile;
imagesc(F./1000,T./60,PSDdiff)
title(‘Original plot data divided by output raster data’)
c = colorbar;
caxis([min(PSDdiff,[],’all’) max(PSDdiff,[],’all’)])
set(get(c,’label’),’string’,’original/output’,’Rotation’,90);
set(s(2:3),’YDir’,’normal’)
xlabel(s(2:3), s(1).XLabel.String)
ylabel(s(2:3), s(1).YLabel.String)

From the figures above, it looks like they produce nearly the same result, except at high frequencies (the middle figure looks more different than the first mostly due to the different color scales).
Comparing the ranges and plotting a cross-section (example spectra at a single time) from each one shows that the difference seems to be that spectrogram’s direct plotting method is cutting off local minima:
disp([‘min/max of original spectrogram plot: ‘,num2str(min(PSDdB_orig,[],’all’)),’, ‘,num2str(max(PSDdB_orig,[],’all’))])
disp([‘min/max of spectrogram output raster: ‘,num2str(min(PSDdB,[],’all’)),’, ‘,num2str(max(PSDdB,[],’all’))])
% Plot example spectra from each approach
figure(2);clf
subplot(2,1,1);
plot(F./1000,PSDdB(10,:),F./1000,PSDdB_orig(10,:),’–‘);
legend(‘Spectrogram output raster’,’Original spectrogram’)
title(‘Example spectra’)
xlabel(‘Frequency (kHz)’)
ylabel(‘Power/frequency (dB/Hz)’)

% zoom in on high frequency range where values differ
subplot(2,1,2,copyobj(gca,gcf))
xlim([4.75 4.85])
title(‘high frequency close-up’)

But weirdly, I can’t reproduce this with randomly generated synthetic data:
x = randn(length(exampledata),1);

figure(3);clf

t = tiledlayout(3,1, ‘TileIndexing’, ‘columnmajor’);

s(1) = nexttile;
spectrogram(x,window,noverlap,window,fs);
title(‘Original spectrogram plot (synthetic data)’)
PSDdB_orig = get(get(gca,’Children’),’CData’);
caxis([min(PSDdB_orig,[],’all’) max(PSDdB_orig,[],’all’)])

[~,F,T,PSD] = spectrogram(x,window,noverlap,window,fs);
PSDdB = (10.*log10(PSD))’;
PSDdiff = PSDdB_orig./PSDdB;

s(2) = nexttile;
imagesc(F./1000,T./60,PSDdB)
title(‘Spectrogram output raster (synthetic data)’)
c = colorbar;
caxis([min(PSDdB,[],’all’) max(PSDdB,[],’all’)])
set(get(c,’label’),’string’,’Power/frequency (dB/Hz)’,’Rotation’,90);

s(3) = nexttile;
imagesc(F./1000,T./60,PSDdiff)
title(‘Original plot data divided by output raster data’)
c = colorbar;
caxis([min(PSDdiff,[],’all’) max(PSDdiff,[],’all’)])
set(get(c,’label’),’string’,’original/output’,’Rotation’,90);

set(s(2:3),’YDir’,’normal’)
xlabel(s(2:3), s(1).XLabel.String)
ylabel(s(2:3), s(1).YLabel.String)
disp([‘min/max of original spectrogram plot: ‘,num2str(min(PSDdB_orig,[],’all’)),’, ‘,num2str(max(PSDdB_orig,[],’all’))])
disp([‘min/max of spectrogram output raster: ‘,num2str(min(PSDdB,[],’all’)),’, ‘,num2str(max(PSDdB,[],’all’))])

I do notice that the minimum PSD value for my data (-196 dB) is significantly lower than the minimum of the synthetic data. Is there a lower limit where spectrogram starts to censor spectra (say, around -156.5351 dB)? Or is something else going on here? Any help would be much appreciated!! Thanks!When using spectrogram to calculate power spectral density for a large dataset, the plot produced by directly running the command with no output arguments is slightly different from what I get when I run it with an output argument and plot that output myself.

For example, the following plots spectrograms for a subset of my data and compares the direct result of spectrogram with the output raster:
load(‘exampledata.mat’)
window = 10000;
noverlap = round(0.1*window);
fs = 10000;

figure(1);clf
t = tiledlayout(3,1, ‘TileIndexing’, ‘columnmajor’);

% Plot spectrogram directly from spectrogram function
s(1) = nexttile;
spectrogram(exampledata,window,noverlap,window,fs);
title(‘Original spectrogram plot (real data)’)
PSDdB_orig = get(get(gca,’Children’),’CData’);
caxis([min(PSDdB_orig,[],’all’) max(PSDdB_orig,[],’all’)])

% Calculate power spectral density and plot as decibels
[~,F,T,PSD] = spectrogram(exampledata,window,noverlap,window,fs);
PSDdB = (10.*log10(PSD))’;

s(2) = nexttile;
imagesc(F./1000,T./60,PSDdB)
title(‘Spectrogram output raster (real data)’)
c = colorbar;
caxis([min(PSDdB,[],’all’) max(PSDdB,[],’all’)])
set(get(c,’label’),’string’,’Power/frequency (dB/Hz)’,’Rotation’,90);

% Calculate ratio of the two approaches above and plot that
PSDdiff = PSDdB_orig./PSDdB;

s(3) = nexttile;
imagesc(F./1000,T./60,PSDdiff)
title(‘Original plot data divided by output raster data’)
c = colorbar;
caxis([min(PSDdiff,[],’all’) max(PSDdiff,[],’all’)])
set(get(c,’label’),’string’,’original/output’,’Rotation’,90);
set(s(2:3),’YDir’,’normal’)
xlabel(s(2:3), s(1).XLabel.String)
ylabel(s(2:3), s(1).YLabel.String)

From the figures above, it looks like they produce nearly the same result, except at high frequencies (the middle figure looks more different than the first mostly due to the different color scales).
Comparing the ranges and plotting a cross-section (example spectra at a single time) from each one shows that the difference seems to be that spectrogram’s direct plotting method is cutting off local minima:
disp([‘min/max of original spectrogram plot: ‘,num2str(min(PSDdB_orig,[],’all’)),’, ‘,num2str(max(PSDdB_orig,[],’all’))])
disp([‘min/max of spectrogram output raster: ‘,num2str(min(PSDdB,[],’all’)),’, ‘,num2str(max(PSDdB,[],’all’))])
% Plot example spectra from each approach
figure(2);clf
subplot(2,1,1);
plot(F./1000,PSDdB(10,:),F./1000,PSDdB_orig(10,:),’–‘);
legend(‘Spectrogram output raster’,’Original spectrogram’)
title(‘Example spectra’)
xlabel(‘Frequency (kHz)’)
ylabel(‘Power/frequency (dB/Hz)’)

% zoom in on high frequency range where values differ
subplot(2,1,2,copyobj(gca,gcf))
xlim([4.75 4.85])
title(‘high frequency close-up’)

But weirdly, I can’t reproduce this with randomly generated synthetic data:
x = randn(length(exampledata),1);

figure(3);clf

t = tiledlayout(3,1, ‘TileIndexing’, ‘columnmajor’);

s(1) = nexttile;
spectrogram(x,window,noverlap,window,fs);
title(‘Original spectrogram plot (synthetic data)’)
PSDdB_orig = get(get(gca,’Children’),’CData’);
caxis([min(PSDdB_orig,[],’all’) max(PSDdB_orig,[],’all’)])

[~,F,T,PSD] = spectrogram(x,window,noverlap,window,fs);
PSDdB = (10.*log10(PSD))’;
PSDdiff = PSDdB_orig./PSDdB;

s(2) = nexttile;
imagesc(F./1000,T./60,PSDdB)
title(‘Spectrogram output raster (synthetic data)’)
c = colorbar;
caxis([min(PSDdB,[],’all’) max(PSDdB,[],’all’)])
set(get(c,’label’),’string’,’Power/frequency (dB/Hz)’,’Rotation’,90);

s(3) = nexttile;
imagesc(F./1000,T./60,PSDdiff)
title(‘Original plot data divided by output raster data’)
c = colorbar;
caxis([min(PSDdiff,[],’all’) max(PSDdiff,[],’all’)])
set(get(c,’label’),’string’,’original/output’,’Rotation’,90);

set(s(2:3),’YDir’,’normal’)
xlabel(s(2:3), s(1).XLabel.String)
ylabel(s(2:3), s(1).YLabel.String)
disp([‘min/max of original spectrogram plot: ‘,num2str(min(PSDdB_orig,[],’all’)),’, ‘,num2str(max(PSDdB_orig,[],’all’))])
disp([‘min/max of spectrogram output raster: ‘,num2str(min(PSDdB,[],’all’)),’, ‘,num2str(max(PSDdB,[],’all’))])

I do notice that the minimum PSD value for my data (-196 dB) is significantly lower than the minimum of the synthetic data. Is there a lower limit where spectrogram starts to censor spectra (say, around -156.5351 dB)? Or is something else going on here? Any help would be much appreciated!! Thanks! When using spectrogram to calculate power spectral density for a large dataset, the plot produced by directly running the command with no output arguments is slightly different from what I get when I run it with an output argument and plot that output myself.

For example, the following plots spectrograms for a subset of my data and compares the direct result of spectrogram with the output raster:
load(‘exampledata.mat’)
window = 10000;
noverlap = round(0.1*window);
fs = 10000;

figure(1);clf
t = tiledlayout(3,1, ‘TileIndexing’, ‘columnmajor’);

% Plot spectrogram directly from spectrogram function
s(1) = nexttile;
spectrogram(exampledata,window,noverlap,window,fs);
title(‘Original spectrogram plot (real data)’)
PSDdB_orig = get(get(gca,’Children’),’CData’);
caxis([min(PSDdB_orig,[],’all’) max(PSDdB_orig,[],’all’)])

% Calculate power spectral density and plot as decibels
[~,F,T,PSD] = spectrogram(exampledata,window,noverlap,window,fs);
PSDdB = (10.*log10(PSD))’;

s(2) = nexttile;
imagesc(F./1000,T./60,PSDdB)
title(‘Spectrogram output raster (real data)’)
c = colorbar;
caxis([min(PSDdB,[],’all’) max(PSDdB,[],’all’)])
set(get(c,’label’),’string’,’Power/frequency (dB/Hz)’,’Rotation’,90);

% Calculate ratio of the two approaches above and plot that
PSDdiff = PSDdB_orig./PSDdB;

s(3) = nexttile;
imagesc(F./1000,T./60,PSDdiff)
title(‘Original plot data divided by output raster data’)
c = colorbar;
caxis([min(PSDdiff,[],’all’) max(PSDdiff,[],’all’)])
set(get(c,’label’),’string’,’original/output’,’Rotation’,90);
set(s(2:3),’YDir’,’normal’)
xlabel(s(2:3), s(1).XLabel.String)
ylabel(s(2:3), s(1).YLabel.String)

From the figures above, it looks like they produce nearly the same result, except at high frequencies (the middle figure looks more different than the first mostly due to the different color scales).
Comparing the ranges and plotting a cross-section (example spectra at a single time) from each one shows that the difference seems to be that spectrogram’s direct plotting method is cutting off local minima:
disp([‘min/max of original spectrogram plot: ‘,num2str(min(PSDdB_orig,[],’all’)),’, ‘,num2str(max(PSDdB_orig,[],’all’))])
disp([‘min/max of spectrogram output raster: ‘,num2str(min(PSDdB,[],’all’)),’, ‘,num2str(max(PSDdB,[],’all’))])
% Plot example spectra from each approach
figure(2);clf
subplot(2,1,1);
plot(F./1000,PSDdB(10,:),F./1000,PSDdB_orig(10,:),’–‘);
legend(‘Spectrogram output raster’,’Original spectrogram’)
title(‘Example spectra’)
xlabel(‘Frequency (kHz)’)
ylabel(‘Power/frequency (dB/Hz)’)

% zoom in on high frequency range where values differ
subplot(2,1,2,copyobj(gca,gcf))
xlim([4.75 4.85])
title(‘high frequency close-up’)

But weirdly, I can’t reproduce this with randomly generated synthetic data:
x = randn(length(exampledata),1);

figure(3);clf

t = tiledlayout(3,1, ‘TileIndexing’, ‘columnmajor’);

s(1) = nexttile;
spectrogram(x,window,noverlap,window,fs);
title(‘Original spectrogram plot (synthetic data)’)
PSDdB_orig = get(get(gca,’Children’),’CData’);
caxis([min(PSDdB_orig,[],’all’) max(PSDdB_orig,[],’all’)])

[~,F,T,PSD] = spectrogram(x,window,noverlap,window,fs);
PSDdB = (10.*log10(PSD))’;
PSDdiff = PSDdB_orig./PSDdB;

s(2) = nexttile;
imagesc(F./1000,T./60,PSDdB)
title(‘Spectrogram output raster (synthetic data)’)
c = colorbar;
caxis([min(PSDdB,[],’all’) max(PSDdB,[],’all’)])
set(get(c,’label’),’string’,’Power/frequency (dB/Hz)’,’Rotation’,90);

s(3) = nexttile;
imagesc(F./1000,T./60,PSDdiff)
title(‘Original plot data divided by output raster data’)
c = colorbar;
caxis([min(PSDdiff,[],’all’) max(PSDdiff,[],’all’)])
set(get(c,’label’),’string’,’original/output’,’Rotation’,90);

set(s(2:3),’YDir’,’normal’)
xlabel(s(2:3), s(1).XLabel.String)
ylabel(s(2:3), s(1).YLabel.String)
disp([‘min/max of original spectrogram plot: ‘,num2str(min(PSDdB_orig,[],’all’)),’, ‘,num2str(max(PSDdB_orig,[],’all’))])
disp([‘min/max of spectrogram output raster: ‘,num2str(min(PSDdB,[],’all’)),’, ‘,num2str(max(PSDdB,[],’all’))])

I do notice that the minimum PSD value for my data (-196 dB) is significantly lower than the minimum of the synthetic data. Is there a lower limit where spectrogram starts to censor spectra (say, around -156.5351 dB)? Or is something else going on here? Any help would be much appreciated!! Thanks! spectrogram, spectral analysis, signal processing MATLAB Answers — New Questions

​

hello community

please help me to understand why data does not show when I run a loop:
I have 2 files: @Data01.mat and @Data05.mat I want to extrac data points when conditions are met’
if I run my script with Data1 all the parameters are there:
figure;plot (Epm_nEngallc0_4, IKCtl_au16_0allc, ‘.’);

but if I run my script with Data1&2 the output is different and I get less data points

I would expect to have all data poins from @Data01.mat plus @Data05.mat
please help me to see what am I doing wrong.

clear all; %
clc; %

fprintf(‘select folder containing data filesn’);

currentpath=pwd; %%

datapath=uigetdir(”,’Select Data Folder’); %%

if (datapath == 0)
fprintf(‘Data directory was not selected…script will be terminatednn’);
return
end

prompt = ‘Do you want to export data as .xlsx?’;
save_data_excel = questdlg(prompt,’Save’,’Yes’,’No’,’Yes’);

cd(datapath); %%
files=dir(‘@*.mat’); %%
num_files=length(files(:,1)); %%

for i=1:num_files
fprintf(‘Progress: %d/%dnn’,i, num_files)
load(files(i,1).name)
idx = [false; diff(xzsen__ls_0_rs_) < 0];
idx1 = [false; diff(xzsen__ls_1_rs_) < 0];
idx2 = [false; diff(xzsen__ls_2_rs_) < 0];
idx3 = [false; diff(xzsen__ls_0_rs_) < 0];
idx4 = [false; diff(xzsen__ls_1_rs_) < 0];
idx5 = [false; diff(xzsen__ls_2_rs_) < 0];

for j=1:length(time)

if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j) > 0.165
Epm_nEngallc1_3(j,1) = Epm_nEng(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j) > 0.165
Epm_nEngallc2_5(j,1) = Epm_nEng(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j) > 0.165
Epm_nEngallc0_4(j,1) = Epm_nEng(j);
end

if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j) > 0.165
IKCtl_au16_1allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j);
IKCtl_au16_3allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j) > 0.165
IKCtl_au16_2allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j);
IKCtl_au16_5allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j) > 0.165
IKCtl_au16_0allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j);
IKCtl_au16_4allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j);
end

end

endhello community

please help me to understand why data does not show when I run a loop:
I have 2 files: @Data01.mat and @Data05.mat I want to extrac data points when conditions are met’
if I run my script with Data1 all the parameters are there:
figure;plot (Epm_nEngallc0_4, IKCtl_au16_0allc, ‘.’);

but if I run my script with Data1&2 the output is different and I get less data points

I would expect to have all data poins from @Data01.mat plus @Data05.mat
please help me to see what am I doing wrong.

clear all; %
clc; %

fprintf(‘select folder containing data filesn’);

currentpath=pwd; %%

datapath=uigetdir(”,’Select Data Folder’); %%

if (datapath == 0)
fprintf(‘Data directory was not selected…script will be terminatednn’);
return
end

prompt = ‘Do you want to export data as .xlsx?’;
save_data_excel = questdlg(prompt,’Save’,’Yes’,’No’,’Yes’);

cd(datapath); %%
files=dir(‘@*.mat’); %%
num_files=length(files(:,1)); %%

for i=1:num_files
fprintf(‘Progress: %d/%dnn’,i, num_files)
load(files(i,1).name)
idx = [false; diff(xzsen__ls_0_rs_) < 0];
idx1 = [false; diff(xzsen__ls_1_rs_) < 0];
idx2 = [false; diff(xzsen__ls_2_rs_) < 0];
idx3 = [false; diff(xzsen__ls_0_rs_) < 0];
idx4 = [false; diff(xzsen__ls_1_rs_) < 0];
idx5 = [false; diff(xzsen__ls_2_rs_) < 0];

for j=1:length(time)

if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j) > 0.165
Epm_nEngallc1_3(j,1) = Epm_nEng(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j) > 0.165
Epm_nEngallc2_5(j,1) = Epm_nEng(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j) > 0.165
Epm_nEngallc0_4(j,1) = Epm_nEng(j);
end

if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j) > 0.165
IKCtl_au16_1allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j);
IKCtl_au16_3allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j) > 0.165
IKCtl_au16_2allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j);
IKCtl_au16_5allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j) > 0.165
IKCtl_au16_0allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j);
IKCtl_au16_4allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j);
end

end

end hello community

please help me to understand why data does not show when I run a loop:
I have 2 files: @Data01.mat and @Data05.mat I want to extrac data points when conditions are met’
if I run my script with Data1 all the parameters are there:
figure;plot (Epm_nEngallc0_4, IKCtl_au16_0allc, ‘.’);

but if I run my script with Data1&2 the output is different and I get less data points

I would expect to have all data poins from @Data01.mat plus @Data05.mat
please help me to see what am I doing wrong.

clear all; %
clc; %

fprintf(‘select folder containing data filesn’);

currentpath=pwd; %%

datapath=uigetdir(”,’Select Data Folder’); %%

if (datapath == 0)
fprintf(‘Data directory was not selected…script will be terminatednn’);
return
end

prompt = ‘Do you want to export data as .xlsx?’;
save_data_excel = questdlg(prompt,’Save’,’Yes’,’No’,’Yes’);

cd(datapath); %%
files=dir(‘@*.mat’); %%
num_files=length(files(:,1)); %%

for i=1:num_files
fprintf(‘Progress: %d/%dnn’,i, num_files)
load(files(i,1).name)
idx = [false; diff(xzsen__ls_0_rs_) < 0];
idx1 = [false; diff(xzsen__ls_1_rs_) < 0];
idx2 = [false; diff(xzsen__ls_2_rs_) < 0];
idx3 = [false; diff(xzsen__ls_0_rs_) < 0];
idx4 = [false; diff(xzsen__ls_1_rs_) < 0];
idx5 = [false; diff(xzsen__ls_2_rs_) < 0];

for j=1:length(time)

if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j) > 0.165
Epm_nEngallc1_3(j,1) = Epm_nEng(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j) > 0.165
Epm_nEngallc2_5(j,1) = Epm_nEng(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j) > 0.165
Epm_nEngallc0_4(j,1) = Epm_nEng(j);
end

if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j) > 0.165
IKCtl_au16_1allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_1_rs_(j);
IKCtl_au16_3allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_3_rs_(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j) > 0.165
IKCtl_au16_2allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_2_rs_(j);
IKCtl_au16_5allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_5_rs_(j);
end
if Epm_nEng(j) > 0 && IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j) > 0.165 && IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j) > 0.165
IKCtl_au16_0allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_0_rs_(j);
IKCtl_au16_4allc(j,1)= IKCtl_DiagRefLvlMinErr_au16__ls_4_rs_(j);
end

end

end loops, matlab MATLAB Answers — New Questions

​

I’m having difficulty getting a function to plot correctly in Matlab. The function is:
f(x) = x^3 – sin(x) – e^x
I’ve tried using the below but the graph is not how it should appear:
X = -4:0.01:10;
Y = X.^3 – (sin(X)) – (exp(X));
plot(X,Y)I’m having difficulty getting a function to plot correctly in Matlab. The function is:
f(x) = x^3 – sin(x) – e^x
I’ve tried using the below but the graph is not how it should appear:
X = -4:0.01:10;
Y = X.^3 – (sin(X)) – (exp(X));
plot(X,Y) I’m having difficulty getting a function to plot correctly in Matlab. The function is:
f(x) = x^3 – sin(x) – e^x
I’ve tried using the below but the graph is not how it should appear:
X = -4:0.01:10;
Y = X.^3 – (sin(X)) – (exp(X));
plot(X,Y) plot, function MATLAB Answers — New Questions

​

Hi all,
This might be a bit of a fruitless question but I just want to confirm there is no other method. I want to know if there is any way of vectorizing the following code:
array = [1,0,1,1,0,0,1];
for i = 2:length(array)-1
if array(i-1) ~= array(i+1)
% Generate a random bit (0 or 1)
random_bit = randi([0, 1]);
% Update the array in place
array(i) = mod(array(i) + random_bit, 2);
end
end
The array can be of any length as long as it is only fillled with 0s and 1s. The key part of the code is that I want to update the array by doing nearest neighbor checks and in such a way that the number of walls (array(i) ~= array(i+1) for i from 1 to size(array)) is constant. And it is this constrained logic of my code that leads me to believe I cannot do away with using for-loops.
The reason I wish to vectorize the code is because I will be doing monte-carlo sampling where I have a large number of these arrays and wish to update them all at the same time using the above update rule, so in the ideal scenario I wouldn’t have to use a nested for-loop to cycle through each sample.Hi all,
This might be a bit of a fruitless question but I just want to confirm there is no other method. I want to know if there is any way of vectorizing the following code:
array = [1,0,1,1,0,0,1];
for i = 2:length(array)-1
if array(i-1) ~= array(i+1)
% Generate a random bit (0 or 1)
random_bit = randi([0, 1]);
% Update the array in place
array(i) = mod(array(i) + random_bit, 2);
end
end
The array can be of any length as long as it is only fillled with 0s and 1s. The key part of the code is that I want to update the array by doing nearest neighbor checks and in such a way that the number of walls (array(i) ~= array(i+1) for i from 1 to size(array)) is constant. And it is this constrained logic of my code that leads me to believe I cannot do away with using for-loops.
The reason I wish to vectorize the code is because I will be doing monte-carlo sampling where I have a large number of these arrays and wish to update them all at the same time using the above update rule, so in the ideal scenario I wouldn’t have to use a nested for-loop to cycle through each sample. Hi all,
This might be a bit of a fruitless question but I just want to confirm there is no other method. I want to know if there is any way of vectorizing the following code:
array = [1,0,1,1,0,0,1];
for i = 2:length(array)-1
if array(i-1) ~= array(i+1)
% Generate a random bit (0 or 1)
random_bit = randi([0, 1]);
% Update the array in place
array(i) = mod(array(i) + random_bit, 2);
end
end
The array can be of any length as long as it is only fillled with 0s and 1s. The key part of the code is that I want to update the array by doing nearest neighbor checks and in such a way that the number of walls (array(i) ~= array(i+1) for i from 1 to size(array)) is constant. And it is this constrained logic of my code that leads me to believe I cannot do away with using for-loops.
The reason I wish to vectorize the code is because I will be doing monte-carlo sampling where I have a large number of these arrays and wish to update them all at the same time using the above update rule, so in the ideal scenario I wouldn’t have to use a nested for-loop to cycle through each sample. vectorization MATLAB Answers — New Questions

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