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For $Y(b) = \text{thing}_1$, after a transform to another domain $Y(a)=\text{thing}_2$ , is there $Y(a,b)= \text{thing}_3$?

where $\text{thing}_3$ is related to $\text{thing}_1$ and $\text{thing}_2$ ?

More clarification:-

Ok, if I have a function in time say $Y(n) = u(n)$; after moving to $Z$-domain it will be $Y(Z)= \frac{z}{(z-1)}$, is there some way I can make a new function $Y(n,Z)$ = something that changes to n and Z ?

Edit 2:

I want if $(n =0)$ in $Y(n,Z) = Y(Z)$ and vice versa

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    Please clarify; I understand $\epsilon$.2012-06-01
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    Ok , if I have a function in time say Y(n) = u(n) ; after moving to Z-domain it will be Y(Z)= z/(z-1) , is there some way I can make a new function Y(n,Z) = something that changes to n and Z ?2012-06-01
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    I'd ask the Cat in the Hat.2012-06-01
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    I'm looking for a mathematical theory for this2012-06-01
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    You can make any new function you want, for example, $Y(n,Z)=Zu(n)/(Z-1)$; the question is, are there any properties you want $Y(n,Z)$ to have, other than just being a function of $n$ and $Z$. When you figure out what properties you'd like $Y(n,Z)$ to have, please edit them into your question.2012-06-01
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    @GerryMyerson I want if (n =0) in Y(n,Z) = Y(Z) and vice versa2012-06-01

1 Answers 1

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$$Y(n,Z)=(2/\pi)[\arctan(n/Z)u(n)+(\pi/2-\arctan(n/Z))Y(Z)]$$ seems to have the properties $Y(0,Z)=Y(Z)$ and $Y(n,0)=u(n)$.

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    is'n there a general rule to deduce such a thing ?2012-06-02
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    How general a rule do you want? What kind of situation do you have in mind?2012-06-02
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    Isn't there a rule or a theory that for any function and the transform of function map them into a new function depends on the previous functions2012-06-03
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    I have given you a formula that will take any two functions, whether one is a transform of the other or not, and give you a new function that "depends on the previous functions." But perhaps I still don't understand exactly what you want. Do you have an example of two functions and a new function that depends on them, an example that will make clear what you are looking for?2012-06-03
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    MM I guess this what I need for now :) thanks a lot man2012-06-03