# I tried two approaches and gained the different conclusions of judging the stability of the transfer function of the system

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?

\begin{align}-\frac{3}{2+z^{-1}}&=-\frac32\frac{1}{1+\frac12z^{-1}}\\&=-\frac32\sum_{n=0}^{\infty}\left(-\frac12\right)^nz^{-n}\end{align}
$$h[n]=2\delta[n]-\frac32\left(-\frac12\right)^nu[n]$$