Dear all,
How do I describe a permanent magnet in the PDE toolbox ?
I could find nothing in electromagneticProperties or electromagneticSourceDear all,
How do I describe a permanent magnet in the PDE toolbox ?
I could find nothing in electromagneticProperties or electromagneticSource Dear all,
How do I describe a permanent magnet in the PDE toolbox ?
I could find nothing in electromagneticProperties or electromagneticSource pde magnetostatics magnet MATLAB Answers — New Questions

​

Howdy,
I’m working with simscape multibody and I’ve seen using inertia sensor there’s chance to select a local group of connected blocks or whole mechanicsms. I was wondering if for a complex branched mechanism would be possible to take also a composite rigid body inertia measure (using actual configurations given by joints positions) for single branches in palce of whole mechanism of any workaround exist for.

Cheers,
Soldà EnricoHowdy,
I’m working with simscape multibody and I’ve seen using inertia sensor there’s chance to select a local group of connected blocks or whole mechanicsms. I was wondering if for a complex branched mechanism would be possible to take also a composite rigid body inertia measure (using actual configurations given by joints positions) for single branches in palce of whole mechanism of any workaround exist for.

Cheers,
Soldà Enrico Howdy,
I’m working with simscape multibody and I’ve seen using inertia sensor there’s chance to select a local group of connected blocks or whole mechanicsms. I was wondering if for a complex branched mechanism would be possible to take also a composite rigid body inertia measure (using actual configurations given by joints positions) for single branches in palce of whole mechanism of any workaround exist for.

Cheers,
Soldà Enrico simscape, inertia measures, composite rigid body MATLAB Answers — New Questions

​

Hi,
I encounter the error : " The ‘input port 1’ of ‘ddsblockset1/Outport’ requires a bus, but it receives a non bus signal from the ‘output port 1’ of ‘ddsblockset1/Write DDS Sample’."
The data type of the input and output ports has been selected as the "Dataype" defined by DDS. I don’t seem to have any difference from the official example settings. Why did I encounter an error?Hi,
I encounter the error : " The ‘input port 1’ of ‘ddsblockset1/Outport’ requires a bus, but it receives a non bus signal from the ‘output port 1’ of ‘ddsblockset1/Write DDS Sample’."
The data type of the input and output ports has been selected as the "Dataype" defined by DDS. I don’t seem to have any difference from the official example settings. Why did I encounter an error? Hi,
I encounter the error : " The ‘input port 1’ of ‘ddsblockset1/Outport’ requires a bus, but it receives a non bus signal from the ‘output port 1’ of ‘ddsblockset1/Write DDS Sample’."
The data type of the input and output ports has been selected as the "Dataype" defined by DDS. I don’t seem to have any difference from the official example settings. Why did I encounter an error? dds blockset, simulink, bus signal MATLAB Answers — New Questions

​

I want to read in a very large histological image file (file size: 2.05 GB, *.ndpi file, pixel size: 57856 x 1.111e+05) and don’t have enough RAM.
I tried blockedImage and thought it would fit great but still, I am unable to get my data in highest resolution.
I tried:

% p,f are path and filename for the 2.05 GB *.ndpi file

A = blockedImage(strcat(p,f));

levelSizeInfo = table((1:length(A.Size))’, …
A.Size(:,1), …
A.Size(:,2), …
A.Size(:,1)./A.Size(:,2), …
‘VariableNames’,["Resolution Level" "Image Width" "Image Height" "Aspect Ratio"])
and obtained
levelSizeInfo =
9×4 table

Resolution Level Image Width Image Height Aspect Ratio
________________ ___________ ____________ ____________

1 57856 1.111e+05 0.52074
2 28928 55552 0.52074
3 14464 27776 0.52074
4 7232 13888 0.52074
5 3616 6944 0.52074
6 1808 3472 0.52074
7 904 1736 0.52074
8 407 1188 0.34259
9 203 594 0.34175
So I want to assess the resolution level 1 information and therefore tried to split it up in smaller patches:
bls = selectBlockLocations(A,"Levels",1,"BlockSize",[4096, 4096]);
bimds = blockedImageDatastore(A, "BlockLocationSet", bls);
However, I cannot assess these 4096 x 4096 blocks.
I tried to work with:
>> while hasdata(bimds)
[data, info] = read(bimds);
disp(info);
end

Error using images.blocked.TIFF/getIOBlock (line 139)
Out of memory.
But get the above error message. Is there any possibility that I read in my large image and split it up until my memory is sufficient and then assess the data? I don’t want to use lower resolution, since I am really interested in some specific histological information only reliable visible as highest resolution.
I would be also happy, when I would provide voxel coordinate start and end position and split it up manually such that it fits my RAM. I can do it in python but I would like to work with this dataset in Matlab as well.
Similar to this (which was not working either):
>> img = bigimage(strcat(p,f));
%Starting coordinates
coordStart = [250 250];
%Ending Coordinates
coordEnd = [750,750];
%Extract the region between the aforementioned coordinates
blk1 = getRegion(img,1,coordStart, coordEnd);
%Display the image
bigimageshow(bigimage(blk1))

Error using imread (line 440)
Unable to read TIFF file "SSES-1_MAX.ndpi". File is corrupt or image does not contain any readable strips.
Thanks in advance!I want to read in a very large histological image file (file size: 2.05 GB, *.ndpi file, pixel size: 57856 x 1.111e+05) and don’t have enough RAM.
I tried blockedImage and thought it would fit great but still, I am unable to get my data in highest resolution.
I tried:

% p,f are path and filename for the 2.05 GB *.ndpi file

A = blockedImage(strcat(p,f));

levelSizeInfo = table((1:length(A.Size))’, …
A.Size(:,1), …
A.Size(:,2), …
A.Size(:,1)./A.Size(:,2), …
‘VariableNames’,["Resolution Level" "Image Width" "Image Height" "Aspect Ratio"])
and obtained
levelSizeInfo =
9×4 table

Resolution Level Image Width Image Height Aspect Ratio
________________ ___________ ____________ ____________

1 57856 1.111e+05 0.52074
2 28928 55552 0.52074
3 14464 27776 0.52074
4 7232 13888 0.52074
5 3616 6944 0.52074
6 1808 3472 0.52074
7 904 1736 0.52074
8 407 1188 0.34259
9 203 594 0.34175
So I want to assess the resolution level 1 information and therefore tried to split it up in smaller patches:
bls = selectBlockLocations(A,"Levels",1,"BlockSize",[4096, 4096]);
bimds = blockedImageDatastore(A, "BlockLocationSet", bls);
However, I cannot assess these 4096 x 4096 blocks.
I tried to work with:
>> while hasdata(bimds)
[data, info] = read(bimds);
disp(info);
end

Error using images.blocked.TIFF/getIOBlock (line 139)
Out of memory.
But get the above error message. Is there any possibility that I read in my large image and split it up until my memory is sufficient and then assess the data? I don’t want to use lower resolution, since I am really interested in some specific histological information only reliable visible as highest resolution.
I would be also happy, when I would provide voxel coordinate start and end position and split it up manually such that it fits my RAM. I can do it in python but I would like to work with this dataset in Matlab as well.
Similar to this (which was not working either):
>> img = bigimage(strcat(p,f));
%Starting coordinates
coordStart = [250 250];
%Ending Coordinates
coordEnd = [750,750];
%Extract the region between the aforementioned coordinates
blk1 = getRegion(img,1,coordStart, coordEnd);
%Display the image
bigimageshow(bigimage(blk1))

Error using imread (line 440)
Unable to read TIFF file "SSES-1_MAX.ndpi". File is corrupt or image does not contain any readable strips.
Thanks in advance! I want to read in a very large histological image file (file size: 2.05 GB, *.ndpi file, pixel size: 57856 x 1.111e+05) and don’t have enough RAM.
I tried blockedImage and thought it would fit great but still, I am unable to get my data in highest resolution.
I tried:

% p,f are path and filename for the 2.05 GB *.ndpi file

A = blockedImage(strcat(p,f));

levelSizeInfo = table((1:length(A.Size))’, …
A.Size(:,1), …
A.Size(:,2), …
A.Size(:,1)./A.Size(:,2), …
‘VariableNames’,["Resolution Level" "Image Width" "Image Height" "Aspect Ratio"])
and obtained
levelSizeInfo =
9×4 table

Resolution Level Image Width Image Height Aspect Ratio
________________ ___________ ____________ ____________

1 57856 1.111e+05 0.52074
2 28928 55552 0.52074
3 14464 27776 0.52074
4 7232 13888 0.52074
5 3616 6944 0.52074
6 1808 3472 0.52074
7 904 1736 0.52074
8 407 1188 0.34259
9 203 594 0.34175
So I want to assess the resolution level 1 information and therefore tried to split it up in smaller patches:
bls = selectBlockLocations(A,"Levels",1,"BlockSize",[4096, 4096]);
bimds = blockedImageDatastore(A, "BlockLocationSet", bls);
However, I cannot assess these 4096 x 4096 blocks.
I tried to work with:
>> while hasdata(bimds)
[data, info] = read(bimds);
disp(info);
end

Error using images.blocked.TIFF/getIOBlock (line 139)
Out of memory.
But get the above error message. Is there any possibility that I read in my large image and split it up until my memory is sufficient and then assess the data? I don’t want to use lower resolution, since I am really interested in some specific histological information only reliable visible as highest resolution.
I would be also happy, when I would provide voxel coordinate start and end position and split it up manually such that it fits my RAM. I can do it in python but I would like to work with this dataset in Matlab as well.
Similar to this (which was not working either):
>> img = bigimage(strcat(p,f));
%Starting coordinates
coordStart = [250 250];
%Ending Coordinates
coordEnd = [750,750];
%Extract the region between the aforementioned coordinates
blk1 = getRegion(img,1,coordStart, coordEnd);
%Display the image
bigimageshow(bigimage(blk1))

Error using imread (line 440)
Unable to read TIFF file "SSES-1_MAX.ndpi". File is corrupt or image does not contain any readable strips.
Thanks in advance! histological data, ndpi, blockedimage MATLAB Answers — New Questions

​

I am trying to add a second x-axis to each plot in a tiled layout. Using this tutorial, I was able to add a second x-axis to the first tile. However, as soon as I transition to the next tile, the previous plot disappears and I am just left with a single blank plot. Adding "hold on" to any line in the code didn’t help, neither did plotting the same data a second time. The same problem arises when using subplot().
I would greatly appreciate any help! My code looks like this (I only need the second axis in order to display different x-ticks w.r.t. to the same data, which is why I’m not adding any plots to the second axis):
% create tiledlayout
tiledlayout(2, 2)
nexttile

% plot toy example
hold on
plot(0:10, 0:10)
axis([0 10 0 10])
xticks([2 7])

% add second axis
ax1 = gca;
ax2 = axes(‘Position’, ax1.Position, ‘XAxisLocation’, ‘top’, ‘YAxisLocation’,’right’, ‘Color’,’none’);
axis([0 10 0 10])
xticks([1 8])

% after this line, the plot disappears
nexttileI am trying to add a second x-axis to each plot in a tiled layout. Using this tutorial, I was able to add a second x-axis to the first tile. However, as soon as I transition to the next tile, the previous plot disappears and I am just left with a single blank plot. Adding "hold on" to any line in the code didn’t help, neither did plotting the same data a second time. The same problem arises when using subplot().
I would greatly appreciate any help! My code looks like this (I only need the second axis in order to display different x-ticks w.r.t. to the same data, which is why I’m not adding any plots to the second axis):
% create tiledlayout
tiledlayout(2, 2)
nexttile

% plot toy example
hold on
plot(0:10, 0:10)
axis([0 10 0 10])
xticks([2 7])

% add second axis
ax1 = gca;
ax2 = axes(‘Position’, ax1.Position, ‘XAxisLocation’, ‘top’, ‘YAxisLocation’,’right’, ‘Color’,’none’);
axis([0 10 0 10])
xticks([1 8])

% after this line, the plot disappears
nexttile I am trying to add a second x-axis to each plot in a tiled layout. Using this tutorial, I was able to add a second x-axis to the first tile. However, as soon as I transition to the next tile, the previous plot disappears and I am just left with a single blank plot. Adding "hold on" to any line in the code didn’t help, neither did plotting the same data a second time. The same problem arises when using subplot().
I would greatly appreciate any help! My code looks like this (I only need the second axis in order to display different x-ticks w.r.t. to the same data, which is why I’m not adding any plots to the second axis):
% create tiledlayout
tiledlayout(2, 2)
nexttile

% plot toy example
hold on
plot(0:10, 0:10)
axis([0 10 0 10])
xticks([2 7])

% add second axis
ax1 = gca;
ax2 = axes(‘Position’, ax1.Position, ‘XAxisLocation’, ‘top’, ‘YAxisLocation’,’right’, ‘Color’,’none’);
axis([0 10 0 10])
xticks([1 8])

% after this line, the plot disappears
nexttile subplot, tiledlayout, axis MATLAB Answers — New Questions

​

Please refer to this documentation: robgain
as well as this system:

In the example "Robust Performance of Closed-Loop System", it says that "Create a model of the controller, and build the closed-loop sensitivity function, S. The sensitivity measures the closed-loop response at the plant output to a disturbance at the plant input."
The corresponding codes are:
k = ureal(‘k’,10,’Percent’,40);
delta = ultidyn(‘delta’,[1 1]);
G = tf(18,[1 1.8 k]) * (1 + 0.5*delta);
C = pid(2.3,3,0.38,0.001);
S = feedback(1,G*C);
tf(S)
However, when I refer to the documentation of feedback, it seems that the code feedback(1,G*C) actually refers to the closed-loop response at the plant input, instead of that at the plant output. I use the function connect to prove this:
G.InputName = ‘e2’;
G.OutputName = ‘y’;
C.InputName = ‘e1’;
C.OutputName = ‘u’;
S1 = sumblk("e1 = r – y");
S2 = sumblk("e2 = u + d");
inputs = {‘d’};
outputs = {‘e2’};
S = connect(G,C,S1,S2,inputs,outputs);
tf(S)
The preceding two results are the same, which prove my assumption.
To measure the closed-loop response at the plant output to a disturbance at the plant input, the codes should be:
S = feedback(G,C);
tf(S)
or in the "connect" version
G.InputName = ‘e2’;
G.OutputName = ‘y’;
C.InputName = ‘e1’;
C.OutputName = ‘u’;
S1 = sumblk("e1 = r – y");
S2 = sumblk("e2 = u + d");
inputs = {‘d’};
outputs = {‘y’};
S = connect(G,C,S1,S2,inputs,outputs);
tf(S)
Please check other documentatins which also use this sensitivity function.

Best regards.Please refer to this documentation: robgain
as well as this system:

In the example "Robust Performance of Closed-Loop System", it says that "Create a model of the controller, and build the closed-loop sensitivity function, S. The sensitivity measures the closed-loop response at the plant output to a disturbance at the plant input."
The corresponding codes are:
k = ureal(‘k’,10,’Percent’,40);
delta = ultidyn(‘delta’,[1 1]);
G = tf(18,[1 1.8 k]) * (1 + 0.5*delta);
C = pid(2.3,3,0.38,0.001);
S = feedback(1,G*C);
tf(S)
However, when I refer to the documentation of feedback, it seems that the code feedback(1,G*C) actually refers to the closed-loop response at the plant input, instead of that at the plant output. I use the function connect to prove this:
G.InputName = ‘e2’;
G.OutputName = ‘y’;
C.InputName = ‘e1’;
C.OutputName = ‘u’;
S1 = sumblk("e1 = r – y");
S2 = sumblk("e2 = u + d");
inputs = {‘d’};
outputs = {‘e2’};
S = connect(G,C,S1,S2,inputs,outputs);
tf(S)
The preceding two results are the same, which prove my assumption.
To measure the closed-loop response at the plant output to a disturbance at the plant input, the codes should be:
S = feedback(G,C);
tf(S)
or in the "connect" version
G.InputName = ‘e2’;
G.OutputName = ‘y’;
C.InputName = ‘e1’;
C.OutputName = ‘u’;
S1 = sumblk("e1 = r – y");
S2 = sumblk("e2 = u + d");
inputs = {‘d’};
outputs = {‘y’};
S = connect(G,C,S1,S2,inputs,outputs);
tf(S)
Please check other documentatins which also use this sensitivity function.

Best regards. Please refer to this documentation: robgain
as well as this system:

In the example "Robust Performance of Closed-Loop System", it says that "Create a model of the controller, and build the closed-loop sensitivity function, S. The sensitivity measures the closed-loop response at the plant output to a disturbance at the plant input."
The corresponding codes are:
k = ureal(‘k’,10,’Percent’,40);
delta = ultidyn(‘delta’,[1 1]);
G = tf(18,[1 1.8 k]) * (1 + 0.5*delta);
C = pid(2.3,3,0.38,0.001);
S = feedback(1,G*C);
tf(S)
However, when I refer to the documentation of feedback, it seems that the code feedback(1,G*C) actually refers to the closed-loop response at the plant input, instead of that at the plant output. I use the function connect to prove this:
G.InputName = ‘e2’;
G.OutputName = ‘y’;
C.InputName = ‘e1’;
C.OutputName = ‘u’;
S1 = sumblk("e1 = r – y");
S2 = sumblk("e2 = u + d");
inputs = {‘d’};
outputs = {‘e2’};
S = connect(G,C,S1,S2,inputs,outputs);
tf(S)
The preceding two results are the same, which prove my assumption.
To measure the closed-loop response at the plant output to a disturbance at the plant input, the codes should be:
S = feedback(G,C);
tf(S)
or in the "connect" version
G.InputName = ‘e2’;
G.OutputName = ‘y’;
C.InputName = ‘e1’;
C.OutputName = ‘u’;
S1 = sumblk("e1 = r – y");
S2 = sumblk("e2 = u + d");
inputs = {‘d’};
outputs = {‘y’};
S = connect(G,C,S1,S2,inputs,outputs);
tf(S)
Please check other documentatins which also use this sensitivity function.

Best regards. robust control toolbox, control, help, robgain MATLAB Answers — New Questions

​

How can we make a boxplot like the one below for both x and y data (the box widths of each group that are x-data related are different). I tried boxplot(X,Y, group) but it did not work.How can we make a boxplot like the one below for both x and y data (the box widths of each group that are x-data related are different). I tried boxplot(X,Y, group) but it did not work. How can we make a boxplot like the one below for both x and y data (the box widths of each group that are x-data related are different). I tried boxplot(X,Y, group) but it did not work. boxplot, different box width MATLAB Answers — New Questions

​

Hey all:
I am trying to accelerate the speed of calculation of high dimisional matrix expotential by using GPU, but I find that the speed of calculating them on CPU is faster than GPU, and I can’t find where the problem is. The code is:
dev = gpuDevice();
CPU_time = 0;
GPU_time = 0;
for i = 1:10
CPU_matrix = rand(4096, 4096);
GPU_matrix = gpuArray(complex(CPU_matrix));
tic;
Exp_CPU = expm(-1i * CPU_matrix);
CPU_time = CPU_time + toc;
tic;
Exp_GPU = expm(-1i * GPU_matrix);
GPU_time = GPU_time + toc;
end
disp("CPU time:" + string(CPU_time));
disp("GPU time:" + string(GPU_time));
I tested this code using my computer, and its CPU configuration is: Intel(R) Core(TM) i7-10750H CPU @ 2.60GHz 2.59 GHz, RAM 16 GB. Its GPU configuration is: NVIDIA GeForce GTX 1650. The final result is:
CPU time:452.1338
GPU time:915.5892
Why the speed of GPU is slower than CPU?
ThanksHey all:
I am trying to accelerate the speed of calculation of high dimisional matrix expotential by using GPU, but I find that the speed of calculating them on CPU is faster than GPU, and I can’t find where the problem is. The code is:
dev = gpuDevice();
CPU_time = 0;
GPU_time = 0;
for i = 1:10
CPU_matrix = rand(4096, 4096);
GPU_matrix = gpuArray(complex(CPU_matrix));
tic;
Exp_CPU = expm(-1i * CPU_matrix);
CPU_time = CPU_time + toc;
tic;
Exp_GPU = expm(-1i * GPU_matrix);
GPU_time = GPU_time + toc;
end
disp("CPU time:" + string(CPU_time));
disp("GPU time:" + string(GPU_time));
I tested this code using my computer, and its CPU configuration is: Intel(R) Core(TM) i7-10750H CPU @ 2.60GHz 2.59 GHz, RAM 16 GB. Its GPU configuration is: NVIDIA GeForce GTX 1650. The final result is:
CPU time:452.1338
GPU time:915.5892
Why the speed of GPU is slower than CPU?
Thanks Hey all:
I am trying to accelerate the speed of calculation of high dimisional matrix expotential by using GPU, but I find that the speed of calculating them on CPU is faster than GPU, and I can’t find where the problem is. The code is:
dev = gpuDevice();
CPU_time = 0;
GPU_time = 0;
for i = 1:10
CPU_matrix = rand(4096, 4096);
GPU_matrix = gpuArray(complex(CPU_matrix));
tic;
Exp_CPU = expm(-1i * CPU_matrix);
CPU_time = CPU_time + toc;
tic;
Exp_GPU = expm(-1i * GPU_matrix);
GPU_time = GPU_time + toc;
end
disp("CPU time:" + string(CPU_time));
disp("GPU time:" + string(GPU_time));
I tested this code using my computer, and its CPU configuration is: Intel(R) Core(TM) i7-10750H CPU @ 2.60GHz 2.59 GHz, RAM 16 GB. Its GPU configuration is: NVIDIA GeForce GTX 1650. The final result is:
CPU time:452.1338
GPU time:915.5892
Why the speed of GPU is slower than CPU?
Thanks gpu, matrix, matlab, time, expotential MATLAB Answers — New Questions

​

How to use stacked bar charts to draw multiple confidence intervals as follows

My sample data is as follows: the point estimated coefficient is
coef = [-0.0186
0.0057
-0.0067
-0.0007
0
-0.0295
-0.0517
-0.0651
-0.0689
-0.0862
-0.0866
];
The lower bounds for the point estimated coefficient at 1%, 5%, 10%, 15% and 20% levels are
Lower_Bound = [
-0.061944 -0.051528 -0.04632 -0.042792 -0.040104
-0.04203 -0.03056 -0.024825 -0.02094 -0.01798
-0.05314 -0.04198 -0.0364 -0.03262 -0.02974
-0.044302 -0.033824 -0.028585 -0.025036 -0.022332
0 0 0 0 0
-0.07723 -0.06576 -0.060025 -0.05614 -0.05318
-0.103042 -0.090704 -0.084535 -0.080356 -0.077172
-0.121602 -0.108024 -0.101235 -0.096636 -0.093132
-0.132368 -0.117116 -0.10949 -0.104324 -0.100388
-0.152506 -0.136572 -0.128605 -0.123208 -0.119096
-0.183092 -0.159904 -0.14831 -0.140456 -0.134472
];
The upper bounds for the point estimated coefficient at 1%, 5%, 10%, 15% and 20% levels are
Upper_Bound = [0.024744 0.014328 0.00912 0.005592 0.002904
0.05343 0.04196 0.036225 0.03234 0.02938
0.03974 0.02858 0.023 0.01922 0.01634
0.042902 0.032424 0.027185 0.023636 0.020932
0 0 0 0 0
0.01823 0.00676 0.001025 -0.00286 -0.00582
-0.000358 -0.012696 -0.018865 -0.023044 -0.026228
-0.008598 -0.022176 -0.028965 -0.033564 -0.037068
-0.005432 -0.020684 -0.02831 -0.033476 -0.037412
-0.019894 -0.035828 -0.043795 -0.049192 -0.053304
0.009892 -0.013296 -0.02489 -0.032744 -0.038728
];How to use stacked bar charts to draw multiple confidence intervals as follows

My sample data is as follows: the point estimated coefficient is
coef = [-0.0186
0.0057
-0.0067
-0.0007
0
-0.0295
-0.0517
-0.0651
-0.0689
-0.0862
-0.0866
];
The lower bounds for the point estimated coefficient at 1%, 5%, 10%, 15% and 20% levels are
Lower_Bound = [
-0.061944 -0.051528 -0.04632 -0.042792 -0.040104
-0.04203 -0.03056 -0.024825 -0.02094 -0.01798
-0.05314 -0.04198 -0.0364 -0.03262 -0.02974
-0.044302 -0.033824 -0.028585 -0.025036 -0.022332
0 0 0 0 0
-0.07723 -0.06576 -0.060025 -0.05614 -0.05318
-0.103042 -0.090704 -0.084535 -0.080356 -0.077172
-0.121602 -0.108024 -0.101235 -0.096636 -0.093132
-0.132368 -0.117116 -0.10949 -0.104324 -0.100388
-0.152506 -0.136572 -0.128605 -0.123208 -0.119096
-0.183092 -0.159904 -0.14831 -0.140456 -0.134472
];
The upper bounds for the point estimated coefficient at 1%, 5%, 10%, 15% and 20% levels are
Upper_Bound = [0.024744 0.014328 0.00912 0.005592 0.002904
0.05343 0.04196 0.036225 0.03234 0.02938
0.03974 0.02858 0.023 0.01922 0.01634
0.042902 0.032424 0.027185 0.023636 0.020932
0 0 0 0 0
0.01823 0.00676 0.001025 -0.00286 -0.00582
-0.000358 -0.012696 -0.018865 -0.023044 -0.026228
-0.008598 -0.022176 -0.028965 -0.033564 -0.037068
-0.005432 -0.020684 -0.02831 -0.033476 -0.037412
-0.019894 -0.035828 -0.043795 -0.049192 -0.053304
0.009892 -0.013296 -0.02489 -0.032744 -0.038728
]; How to use stacked bar charts to draw multiple confidence intervals as follows

My sample data is as follows: the point estimated coefficient is
coef = [-0.0186
0.0057
-0.0067
-0.0007
0
-0.0295
-0.0517
-0.0651
-0.0689
-0.0862
-0.0866
];
The lower bounds for the point estimated coefficient at 1%, 5%, 10%, 15% and 20% levels are
Lower_Bound = [
-0.061944 -0.051528 -0.04632 -0.042792 -0.040104
-0.04203 -0.03056 -0.024825 -0.02094 -0.01798
-0.05314 -0.04198 -0.0364 -0.03262 -0.02974
-0.044302 -0.033824 -0.028585 -0.025036 -0.022332
0 0 0 0 0
-0.07723 -0.06576 -0.060025 -0.05614 -0.05318
-0.103042 -0.090704 -0.084535 -0.080356 -0.077172
-0.121602 -0.108024 -0.101235 -0.096636 -0.093132
-0.132368 -0.117116 -0.10949 -0.104324 -0.100388
-0.152506 -0.136572 -0.128605 -0.123208 -0.119096
-0.183092 -0.159904 -0.14831 -0.140456 -0.134472
];
The upper bounds for the point estimated coefficient at 1%, 5%, 10%, 15% and 20% levels are
Upper_Bound = [0.024744 0.014328 0.00912 0.005592 0.002904
0.05343 0.04196 0.036225 0.03234 0.02938
0.03974 0.02858 0.023 0.01922 0.01634
0.042902 0.032424 0.027185 0.023636 0.020932
0 0 0 0 0
0.01823 0.00676 0.001025 -0.00286 -0.00582
-0.000358 -0.012696 -0.018865 -0.023044 -0.026228
-0.008598 -0.022176 -0.028965 -0.033564 -0.037068
-0.005432 -0.020684 -0.02831 -0.033476 -0.037412
-0.019894 -0.035828 -0.043795 -0.049192 -0.053304
0.009892 -0.013296 -0.02489 -0.032744 -0.038728
]; stack, bar, baseline MATLAB Answers — New Questions

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Hello,
I did what the task asked. The result of it is same as the answer. I cannot move on the new task since the question says you cannot meet the requirements although I did. Could you help me please?Hello,
I did what the task asked. The result of it is same as the answer. I cannot move on the new task since the question says you cannot meet the requirements although I did. Could you help me please? Hello,
I did what the task asked. The result of it is same as the answer. I cannot move on the new task since the question says you cannot meet the requirements although I did. Could you help me please? simulink fundamentals, online course, self-paced, masking subsystems, disp() MATLAB Answers — New Questions

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