Lets look at the poles and zeros of a finite duration signal $v(n) = \delta(n-N)$.

\begin{align*}
\hat{V}(z) &= \sum \limits_{n=-\infty}^{\infty} \delta(n-N) z^{-n} \\
\hat{V}(z) &= z^{-N}
\end{align*}

Lets consider the cases for different values of $N$:



I) $N \gt 0$

$$
\hat{V}(z) = z^{-N} = \frac{1}{z^N}
$$

There are $N$ poles at zero, hence, the region of convergence must exclude $0$.


II) $N = 0$

$$
\hat{V}(z) = 1
$$

The $RoC$ includes all $z$ including $0$ and $\infty$.


III) $N \lt 0$

$$
\hat{V}(z) = z^{-N} = z^M \mbox { where } M \gt 0
$$


We have $M$ poles at $\infty$ which is excluded from the $RoC$.

\begin{example}

Find the poles for the Z-transform of the signal $r(n) = \delta(n-1) + 2\delta(n-2) + 3\delta(n-3)$.

%\img{images/deltas/examples/one.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(5,2)
   

% line
\psline{-*}(1,0)(1,0.5)
%  value
\rput(1.5,0.5){(1)}
%  position
\rput(1,-0.3){1}


% line
\psline{-*}(2,0)(2,1)
%  value
\rput(2.3,1){(2)}
%  position
\rput(2,-0.3){2}


% line
\psline{-*}(3,0)(3,1.5)
%  value
\rput(3.5,1.5){(3)}
%  position
\rput(3,-0.3){3}

% axes
\psline{->}(0,0)(4,0)
\rput(4,-0.3){ $n$ }
\rput(4,1.25){ $r(n)$ }

\end{pspicture}
\end{center}

\begin{align*}
\hat{R}(z) &= \sum \limits_{n = -\infty}^{\infty} r(n) z^{-n} \\
\hat{R}(z) &= \sum \limits_{n = -\infty}^{\infty}\lr{\delta(n-1) + 2\delta(n-2) + 3\delta(n-3)} z^{-n} \\
\hat{R}(z) &= z^{-1} + 2z^{-2} + 3z^{-3} \\
\hat{R}(z) &= \frac{z^2 + 2z + 3}{z^3} \\
\end{align*}

We have a triple pole at 0. $RoC(v): 0 \lt \abs{z}$.

\end{example}

\begin{example}

Find the poles for the anti-casual signal $g(n)$ represented in the figure below.

%\img{images/deltas/examples/two.ps}

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\begin{center}
\begin{pspicture}(-1,-1)(5,2)
    

% line
\psline{-*}(3,0)(3,0.5)
%  value
\rput(3.5,0.5){(1)}
%  position
\rput(3,-0.3){0}


% line
\psline{-*}(2,0)(2,1)
%  value
\rput(2.3,1){(2)}
%  position
\rput(2,-0.3){-1}


% line
\psline{-*}(1,0)(1,1.5)
%  value
\rput(1.5,1.5){(3)}
%  position
\rput(1,-0.3){-2}

% axes
\psline{->}(0,0)(4,0)
\rput(4,-0.3){ $n$ }
\rput(4,1.5){ $g(n)$ }

\end{pspicture}
\end{center}

\begin{align*}
\hat{G}(z) &= \sum \limits_{n = -\infty}^{\infty} g(n) z^{-n} \\
\hat{G}(z) &= 1 + 2z + 3z^2 \\
\end{align*}

The poles are at infinity, $RoC(g): \abs{z} \lt \infty$.
\end{example}

\begin{example}

Find the poles for the two-sided signal $h(n)$ represented in the figure below.

%\img{images/deltas/examples/three.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(5,2)
    

% line
\psline{-*}(3,0)(3,0.5)
%  value
\rput(3.5,0.5){(1)}
%  position
\rput(3,-0.3){1}


% line
\psline{-*}(2,0)(2,1)
%  value
\rput(2.3,1){(2)}
%  position
\rput(2,-0.3){0}


% line
\psline{-*}(1,0)(1,1.5)
%  value
\rput(1.5,1.5){(3)}
%  position
\rput(1,-0.3){-1}

% axes
\psline{->}(0,0)(4,0)
\rput(4,-0.3){ $n$ }
\rput(4,1.5){ $h(n)$ }

\end{pspicture}
\end{center}

\begin{align*}
\hat{H}(z) &= \sum \limits_{n = -\infty}^{\infty} h(n) z^{-n} \\
\hat{H}(z) &= z^{-1} + 2 + 3z \\
\hat{H}(z) &= \frac{1 + 2z + 3z^2}{z} \\
\end{align*}

Poles are at 0 and $\infty$, $RoC(h): 0 \lt \abs{z} \lt \infty$.

\end{example}

Poles and Zeros