From the definition I know:

$$|\sum_{n=-\infty}^{+\infty}x[n] \cdot z^{-n}| < \infty\space\space\space(1)$$

Is there any sequence of x[n] which can not be written as

$$x[n] = y[n] \cdot u[an-b]\space,\space\space\space a,b \in \mathbb{Z}$$ (where the dot means a multiplication and $u[k]$ means the discrete step function. In other words, x[n] is an infinity sequence)

but do satisfy the definition in (1)?

  • $\begingroup$ What does the dot mean in you equations? Multiplication, or convolution as well? Specifically, between $y$ and $u$. $\endgroup$
    – Phonon
    May 24 '14 at 1:08
  • $\begingroup$ Sorry about that. It's a multiplication and u[k] is the discrete step function (I've made the correction in the original question). $\endgroup$ May 24 '14 at 1:13
  • $\begingroup$ The step function changes for each $n$?? $\endgroup$ May 24 '14 at 2:01
  • $\begingroup$ What do you mean by change? Step function u[n] is defined to be equal to 1 when n>=0 and equals to zero when n < 0. In this case, the 'a' and 'b' constants may represent a time shift and/or an inversion $\endgroup$ May 24 '14 at 2:50
  • $\begingroup$ oops, was interpreting it wrong. too little sleep. :P $\endgroup$ May 24 '14 at 12:57

The answer is definitely yes. From your question you seem to think that the first equation can only be satisfied for sequences that vanish for $n>N$ or $n<N$ for some finite (positive or negative) $N$, i.e. for shifted versions of causal or anti-causal sequences. First of all note - as already pointed out in nidhin's answer - that your first equation only makes sense if you add "for $z$ inside the region of convergence (ROC)". For two-sided sequences the ROC is a ring, so $r_1<|z|<r_2$. Outside the ROC, the $\mathcal{Z}$-transform has isolated singularities, which are the poles of the transfer function (for which $X(z)\rightarrow\infty$).

As a counter-example, i.e. a sequence which cannot be written as suggested in your question, but for which the $\mathcal{Z}$-transform exists, consider the sequence

$$x[n]=\begin{cases}\frac{a^{n+1}}{a-b},&n\ge 0\\ \frac{b^{n+1}}{a-b},&n<0\end{cases}$$

with $0<|a|<1<|b|$. Note that this is a two-sided sequence, which is neither causal nor anti-causal, and for which $x[n]\neq 0$ is satisfied for all $n$. Its $\mathcal{Z}$-transform is given by


and its region of convergence is $|a|<|z|<|b|$.

  • $\begingroup$ Makes sense. What about a sequence like $x[n]=(0.5)^n$?. I'm failing to see its ROC. $\endgroup$ May 24 '14 at 17:41
  • $\begingroup$ @FELIPE_RIBAS: If it is a causal sequence (i.e. if your equation defines $x[n]$ only for $n\ge 0$), then its ROC is $|z|>0.5$. If $x[n]$ is defined for $-\infty<n<\infty$, then the (two-sided) Z-transform does not exist, i.e. the ROC is empty. $\endgroup$
    – Matt L.
    May 24 '14 at 19:22

Nop. For an infinite sequence, the convergence of Z transform depends on the value of $z$ also. So as far as $n\rightarrow\pm\infty$, equation (1) will not be satisfied for all values of $z$.

Consider a simple case, $x(n) = u(n)$ $$X(z) = \sum_{n=0}^{\infty}z^{-n}$$ $$X(z) = 1 + z^{-1}+z^{-2}+z^{-3}+ \cdots$$ for $z=0.5$, $$X(z) = 1 + 2+4+8+ \cdots = \infty$$ for this sequence to converge, $|z|>1$.


As indicated in nidhin and MattL's answers, the requirement for the existence of the $\mathcal{Z}$-transform must be restricted to a region of convergence. However, it should be noted that this region can be made arbitrarily large with functions such as:

$$ \begin{align} x[n] &= c^{-n^2} \end{align} $$ For any real constant $c$ such that $|c|<1$.

The region of convergence for the $\mathcal{Z}$-transform of this sequence is $|z|<\infty$, so definition (1) in your post is satisfied (almost everywhere). However, this sequence is strictly positive for all $n$ and as such cannot be expressed in the provided form. So, in short yes such sequences do exist.


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