how separate the first one (Xp) from (Xn).?

now when I compleat the first signal, But now I want to separate Xn from Xp in the following

rs=[Xn, Xp]

but the length of Xn and Xp is different, therefor we truck the length

XNsize = length(Xn);

rf =[XNsize, Xp, Xn];

Answer 1

[John D'Errico]

I have no idea why you are trying to do this using a vector. Far better would be to store your variables in a struct or a cell array. For example,

rs = {Xn,Xp};

Now you can recover Xn of Xp as simply rs{1} or rs{2}.

Or use

rs.Xn = Xn;
rs.Xp = Xp;

Here you extract Xn or Xp using rs.Xn or rs.Xp.

In either case, Xn and Xp are easily extracted. You need never pack the length of Xn into the vector, a kludge that serves no purpose, since in order to use those variables, you need to unpack them anyway.

Can you unpack the result from the vector? OF COURSE YOU CAN! But the point is, packing it all into a vector serves no purpose, especially since you have no idea how to unpack them in the first place. Yes, you are still out there, asking but if I really, really, really must do it the way I desperately wanted to do it, how would I have solved the problem using the kludge of packing it all into a vector?

Xn = rf(2:rf(1)+1);
Xp = rf(2 + rf(1):end);

Again, better would be just to learn to use MATLAB. Then you don't need to use a kludge solution in the first place.

[abdullah qasim]

thank u very mauch

I have another question

I want reshape (or damege ) to orginal signal

meaning

in same vector

[abdullah qasim]

in this method some thing is wrong in it

Xn = rs(2:rs(1)+1);

there is two bits lost

Xp = rs(2 + rs(1):end);

here two bits increase

[John D'Errico]

No. Sorry. But there is not.

As you stated to claim to create rs...

x = 1:3;
y = 2:2:10;
rf = [length(x),x,y]
rf =
     3     1     2     3     2     4     6     8    10

Now, recover the two vectors:

Xn = rf(2:rs(1)+1);
Xp = rf(2 + rs(1):end);

Xn
Xn =
     1     2     3

Xp
Xp =
     2     4     6     8    10

There is NOTHING lost.

Even so, if you use the method you seem to want to use, AND you try to reconstruct it as I show, AND you think that something was lost, then you either do not have what you explicitly stated you created, or you do not understand what the code does, and you never even tried it.

Regardless, this is a silly way to do things.

[John D'Errico]

And I have no idea what you are asking in your followup question, which seems totally unrelated to this question. If you have another question, then ask it separately. And if you want a useful answer, then be more clear about what you mean by the words "reshape" and "damage", as they have no meaning on their own. Only you know what they mean to you.

Answer 2

[abdullah qasim]

thank u very much

All the appreciation to you thank you very much