If $\lim_{k\rightarrow\infty} x[k]$ exists and is finite then $X(z)$, the Z-transform of $x[k]$, has no poles in the region $|z|>1$ and at most 1 pole at $z = 1$.

Attempt: \begin{align*} X(z) &= \sum_{k\ge 0} x[k]z^{-k}\\\ H(z)/G(z) &= \sum_{k\ge 0} x[k]z^{-k}\\\ \end{align*}

First prove that no pole can be in $|z|>1$ \begin{align*} H(z)/G(z) &= \sum_{k\ge 0} x[k]z^{-k}\\\ H(z)/[(z-a)G'(z)] &= \sum_{k\ge 0} x[k]z^{-k}, a>1\\\ H(z)/[(z-a)G'(z)] &= x[0] + x[1]z^{-1} +x[2]z^{-2}+x[3]z^{-3}...\\\ H(z)/G'(z) &= (z-a)(x[0] + x[1]z^{-1} +x[2]z^{-2}+x[3]z^{-3}...)\\\ \end{align*}

Can someone help carry this further?


1 Answer 1


Let me show you a simple way to see this property. Assume $x[k]$ is a causal sequence and let


be finite. Then the sequence $x[k]$ can be written as


where $u[k]$ is the unit step sequence, and $y[k]$ is a causal sequence that decays to zero as $k\rightarrow\infty$. Taking the $\mathcal{Z}$-transform of (1) gives


The poles of $X(z)$ are determined by the two terms on the right-hand side of (2). Since $y[k]$ is a causal decaying sequence, its $\mathcal{Z}$-transform $Y(z)$ must have all its poles inside the unit circle. Since the first term on the right-hand side of (2) can only contribute a single pole at $z=1$, $X(z)$ cannot have any poles outside the unit circle. If $x[\infty]\neq 0$, $X(z)$ has exactly one pole at $z=1$ from the first term on the right-hand side of (2). If $x[\infty]=0$, i.e. $x[k]$ decays to zero as $k\rightarrow\infty$, then that first term disappears and $X(z)$ has no pole at $z=1$. This verifies the claim: if $x[k]$ has a finite limit as $k\rightarrow\infty$, $X(z)$ cannot have any poles outside the unit circle, and, if $x[\infty]\neq 0$, it has a single pole at $z=1$ in addition to its poles inside the unit circle.

Also note that the final value theorem directly follows from (2). Multiplying both sides of (2) by $z-1$ gives


Taking the limit $z\rightarrow 1$ yields the final value theorem of the $\mathcal{Z}$-transform:

$$\lim_{z\rightarrow 1}(z-1)X(z)=x[\infty]\tag{4}$$

Note that the limit $\lim_{k\rightarrow\infty}x[k]$ must be guaranteed to exist in order for (4) to make any sense.

  • $\begingroup$ This is very well written. But how did you know that $x[k] = x[\infty]u[k] + y[k]$? $\endgroup$
    – Fraïssé
    Feb 1, 2015 at 17:21
  • $\begingroup$ @IllegalImmigrant: I can simply decompose any causal $x[k]$ with a finite limit $x[\infty]$ as $x[\infty]u[k]+y[k]$ with $\lim_{k\rightarrow\infty}y[k]=0$, because then $\lim_{k\rightarrow\infty}x[k]=x[\infty]\cdot \lim_{k\rightarrow\infty}u[k]+\lim_{k\rightarrow\infty}y[k]=x[\infty]$ is always guaranteed. $\endgroup$
    – Matt L.
    Feb 1, 2015 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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