We want to judge whether the system is stable or not.
Given the below transfer function.
$$ H\left( z \right) =\frac{\left( 1+2 z^{-1} \right) }{\left( 2+z^{-1} \right) } $$
$$ H\left( z \right) =2-\frac{3}{2+z^{-1} } $$
$$ 2+\frac{-3}{2+z^{-1} } $$
$$S:=\frac{-3}{2+z^{-1} }$$
$$ S ~~\text{is the sum of each element of the geometric sequence.} $$
$$ -3 ~~ \leftarrow~~ \text{initial term} $$
$$ z^{-1} ~~ \leftarrow~~ \text{common ratio} $$
$$ i \geq1 \rightarrow \text{ith term} = -3 \cdot \left( z^{-1} \right) ^{i-1} $$
$$ = -3 \cdot z^{1-i} =-3 \cdot z^{-0}~,~-3z^{-1}~,~-3 z^{-2} ~,~ \cdot\cdot\cdot $$
$$ H\left( z^{} \right) =\sum_{ n=-\infty }^{ \infty } h\left[ n \right] z^{-n} $$
$$ \sum_{ n=-\infty }^{ \infty } \left( 2 \delta\left[n \right] +\left( -3 \right) u\left[ n \right] \right)z^{-n} $$
$$\displaystyle \therefore ~~ h\left[ n \right] =2 \delta\left[n \right] +\left( -3 \right) u\left[ n \right] $$
$$ \sum_{ n=-\infty }^{ \infty } \left| h\left[ n \right] \right| $$
$$ = \sum_{ n=-\infty }^{ \infty } \left| 2 \delta\left[n \right] +\left( -3 \right) u\left[ n \right]\right| = \infty $$
$$ \therefore ~~ ~~\text{The system is unstable.}~~ $$
However from the another approach,
$$ H\left( z \right) =2-\frac{3}{2+z^{-1} } $$
$$ = \frac{2 \left( 2+ z^{-1} \right) -3}{2+z^{-1} } $$
$$2+z^{-1} = 0 $$
$$ 2+\frac{1}{z^{} } =0 $$
$$ \frac{1}{z^{} } =-2 $$
$$ z=-\frac{1}{2} =-0.5 ~~ \leftarrow~~ ~~\text{pole}~~ $$
$$ ~~\text{Since }~~ \left| \text{pole} \right| =\left| -0.5 \right| =\left| 0.5 \right| <1 ~~\text{is held, the system is stable.}~~ $$
Why the different concolusions were gained?
What I've been missing?
