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Claim:

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?

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

$$x[\infty]=\lim_{k\rightarrow\infty}x[k]$$

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

$$x[k]=x[\infty]u[k]+y[k]\tag{1}$$

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

$$X(z)=x[\infty]\cdot\frac{z}{z-1}+Y(z)\tag{2}$$

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

$$(z-1)X(z)=x[\infty]z+(z-1)Y(z)\tag{3}$$

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.

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  • $\begingroup$ This is very well written. But how did you know that $x[k] = x[\infty]u[k] + y[k]$? $\endgroup$ – Carlos - the Mongoose - Danger Feb 1 '15 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 '15 at 17:55

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