Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am faced with the following question and would appreciate any help you may be able to offer: It is not homework, I know the first and second shift theorems and based on the other examples I have done, I know you start by taking the z-transform of the equation, then factor out X(z) and move the rest of the equation across the equals sign, then you take the inverse z-transform which usually involves partial fractions and it should yield the answer given. Cheers

Use z-transforms to solve:

$$2x[k+1]-x[k]=2^k \,,\quad x[0]=2$$

The answer is:


share|cite|improve this question
If it's homework use the appropriate tag. Moreover: What have you trief and do you know helpful rules for the z-tranform regarding shifts? – Dirk Apr 18 '12 at 9:57

1 Answer 1

up vote 1 down vote accepted

by definition we have \begin{align*} X(z) &= \sum_{k\ge 0} x[k]z^{-k}\\\ &= 2 + \sum_{k\ge 0} x[k+1]z^{-(k+1)}\\\ &= 2 + (2z)^{-1}\sum_{k \ge 0} \left(2^{k} + x[k]\right)z^{-k}\\\ &= 2 + (2z)^{-1}\sum_{k\ge 0} 2^{k}z^{-k} + (2z)^{-1}X(z)\\\ &= 2 + (2z)^{-1} \frac 1{1 - \frac 2z} + (2z)^{-1} X(z). \end{align*} Now oyu can solve for $X(z)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.