If the z-transform of x[n] is X(z)

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If the z-transform of x[n] is X(z), then what is the z-transform of x[n+1] in terms of X(z) ?
 
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symsane said:
If the z-transform of x[n] is X(z), then what is the z-transform of x[n+1] in terms of X(z) ?
There is a shift theorem, similar to the other kinds of reciprocal transforms. To prove it, just change n to n' in the sum, replace n' by n+1, and then shift the sum back to n.
 


OK I know this Shift Theorem, but I have a problem with it. If the z-transform of x[n] is X(z), I can solve the z-transform of x[n-1] in terms of X(z) and what I found is X(z)z-1+x[-1]. However what is the solution if I want to solve the z-transform of x[n+1] in terms of X(z) ? Or in this case, I can ask this question: How can I solve the z-transform of x[n] if I know the z-transform x[n-1] in terms of X(z) ( As I found x[n-1] = X(z)z-1+x[-1])?
 


I searched the net and I found left shift of z-transform. In this question it is very useful.When we applied this definition we acquire the solution.
 


symsane said:
If the z-transform of x[n] is X(z), I can solve the z-transform of x[n-1] in terms of X(z) and what I found is X(z)z-1+x[-1].
OK, so you are using the UNILATERAL transform (i.e. for causal systems). So, you should see some similarities to the Laplace transform. If you've studied transient response of LTI circuit, the Laplace transform is what you (probably) use there.



symsane said:
... I want to solve the z-transform of x[n+1] in terms of X(z) ?
If you know how to prove the result that you have above for xn-1, then you should be able to do this. Did you read my previous post? Are you having trouble arranging terms in the summation?

EDIT: Sorry, I had a typo in a previous post. I will restate the procedure:

Change n to n' in the sum. Replace n' with n-1. Of course, if you do what I said in the previous post, that will not give you a wrong answer, but it will not help either.
 
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