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I took this from the Matlab help section:
text(size(I,2),size(I,1)+15, ...
'Image courtesy of Massachusetts Institute of Technology', ...
'FontSize',7,'HorizontalAlignment','right');
I don't understand how it works, could anyone explain it to me?
https://www.mathworks.com/help/matlab/ref/text.html
In summary:
x = size(I,2)
y = size(I,1)+15
"..." is just the statmentment line continuation character in matlab (it is basically saying, do not end this statement here, but rather continue reading next line)
actual text = 'Image courtesy of Massachusetts Institute of Technology'
The next 4 arguments are name-value pairs (as described in the link above). Basically, it allows you to grab a particular setting and apply it a value based on its name.
font size is set to 7 and horizontal alignment is set to right.
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When I substitue the function in to the integration with infinity as upper limit, maple fails.
https://i.stack.imgur.com/VAkeO.jpg
https://i.stack.imgur.com/ujfGS.jpg
I can substitue successfully the function in to the integration with another variable as upper limit as shown in the figure. I hope to keep the infinity symbol in the substituted result.
The following is the plain text code:
f := u -> theta*exp(-theta*u);
subs(f(u__b) = f(u), int(h(Q, u__b)*f(u__b), u__b = q__a .. infinity));
subs(f(u__b) = f(u), int(h(Q, u__b)*f(u__b), u__b = q__a .. IN));
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Is it possible to plot bars with filled pattern in Matlab, like this figure?
I attempted to use applyhatch_pluscolor link however, I couldn't make it work (attempt below). Is there an easy / feasible alternative?
Test code:
bar(rand(3,4));
[im_hatch,colorlist] = applyhatch_pluscolor(gcf,'\-x.',1,[],[],150);
Then I changed the source code, from bits = hardcopy(h,'-dzbuffer',['-r' num2str(dpi)]); to bits = print(h,'-RGBImage',['-r' num2str(dpi)]);.
I got the figure below. However, this is still far away form the desired result.
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I have been trying for a while to get this to work but I don't seem to be getting any closer to a solution.
I'm not sure of what the pattern is and what to write for the "d" value in a:d:b
Clearly you want to start from 0 (a) and go to 2*pi (b).
Now the question comes what is your step size (d)?
From your example you can see that you are changing from 0 to pi/n.
And from pi/n to 2*pi/n.
This means your step size is d=pi/n
Once you defined your n, e.g:
n=10;
you can do the rest like this:
x=0:(pi/n):(2*pi)
y=sin(x);
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Could someone explain me what I must change in my model?
Model
The error messages are pretty clear and self-explanatory. The reason you get the error is because B is of dimension 4x2 and you are trying to do B * Xr where Xr is of dimension 1. According to your equation, you need to do B*U where U = [dXr/dt; Xr];. However, using the derivative block is never a good idea in Simulink if you can avoid it, especially with a step input. Think about how you want to formulate the inputs to your state-space.
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I have 2 data set point cloud and I want to crop a part of them together.
Because of volume of them is too too large I couldn't crop them with below codes.
Can you help me to how can I crop them?
Used codes are:
selectedl=[];%% last pulse
for i=1:size(indexl)
selectl=lr(indexl(i),:);
selectedl=[selectedl;selectl];
end
selectedf=[];%% first pulse
for i=1:size(indexf)
selectf=fr(indexf(i),:);
selectedf=[selectedf;selectf];
end
Thank U all.
It is a bit difficult to understand what you want to do, as lr, fr, indexl and indexf are missing.
But assuming something like
lr = rand(5,3) ;
indexl = [2 5] ;
I would advise to allocate selectedl above the loop
selectedl = NaN(length(indexl),size(lr,2)) ;
for i = 1:length(indexl)
selectedl(i,:) = lr(indexl(i),:) ;
end
This might not be needed for this example, but if the data size becomes larger this will speed up the loop.