1.$H1(z) = 6 + z^(-1) +6z^{-2}$

2.$H2(z) = 1 - z^{-1} - 6z^{-2}$

3.$H3(z) = 1 - \frac{5}{2} z^{-1} - \frac{3}{2} z^{-2}$

4.$H4(z) = 1 - \frac{5}{2} z^{-1} - \frac{2}{3} z^{-2}$

i) $H(z)=6+z^{-1}+6z^{-2}$

$=6+\frac{1}{z}+\frac{6}{z^2}$

$=\frac{(6z^2+z+6)}{z^2}$

Poles: 0, 0

Zeros:-0.08±0.99j

$z_1=0.99 ∠ 85.38°$

$z_2=0.99 ∠ 85.38°$

Both the zeros lie inside the unit circle, poles are also at origin. Hence , the system is “Stable Minimum phase”

ii)$H_2 (z)=1-z^{-1}-6z^{-2}$

$=1-\frac{1}{z}-\frac{6}{z^2}$

$=\frac{(z^2-z-6)}{z^2}$

Poles: 0, 0 Zeros: 3, -2

Both the zeros lie …