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 outside the unit circle, poles are at origin. Hence, the system is “Stable Maximum Phase”

iii)$ H_3 (z)=1-\frac{5}{2 z^{-1}}-\frac{3}{2 z^{-2}}$

$=1-\frac{5}{2z}-\frac{3}{2z^2 }$

$=\frac{z^2-\frac{5}{2 z}-\frac{3}{2}}{z^2}$

Poles: 0, 0 Zero: 3, -0.5

One of the zero is inside the unit circle and other is outside the unit circle and poles are at origin. Hence the system is “Stable Mixed Phase”

iv) $H_4 (z)=1-\frac{5}{2 z^{-1}}-\frac{2}{3 z^{-2}}$

$=1-\frac{5}{2z}-\frac{2}{3z}$

$=\frac{z^2-\frac{5}{2z}-\frac{2}{3}}{z^2}$

Poles: 0, 0 Zeros: 2.74, -0.24

One of the zero is inside the unit circle and other one is outside the unit circle and poles are at origin. Hence, the system is “Stable Mixed Phase”.