$(i) H_1 (z) = 6 + z^{-1} + 6z^{-1} \ (ii) H_2 (z) = 1 – z^{-1} – 6z^{-1} \ (iii) H_3 (z) = 1 - \dfrac 52 z^{-1}-\dfrac 32z^{-1} \ (iv) H_4 (z) = 1-\dfrac 52 z^{-1}-\dfrac 23z^{-1}$

Comment on stability of minimum and maximum phase system.

Mumbai University > EXTC > Sem 6 > Discrete Time Signal Processing

Marks : 10

Year : DEC 2015

$$H_1 (z) = 6 + z^{-1} – z^{-2} $$

$H_1(z)=\dfrac {6z^2+z-1}{z^2}=\dfrac {6(z+1/2)(z-1/3)}{z^2} \\ zeros \space \space z_1=\dfrac {-1}2,z_2=\dfrac 13$

As both zeros are inside unit circle, this system is minimum phase system.

$1 – 2^{-1} -6z^{-2} \\ =\dfrac {z^2-z-6}{z^2}=\dfrac {(z+2)(z-2)}{z^2}\\ zeros \space \space z_1=-2,z_2=-3 $

As both zeros are outside unit circle, this system is maximum phase system.

$H_3(z)=1- (\dfrac 52)z^{-1}-(\dfrac 32)z^{-2} \\ H_3(z)=\dfrac {z^2-2.5z-1.5z}{z^2}=\dfrac {(z+0.5)(z-3)}{z^2}\\ zeros \space \space z_1=-0.5 ,z_2=3 $

For this system one zero is inside unit circle and one is outside the unit circle, Hence the system is mixed phase system.

$H_4(z)=1+ (\dfrac 53)z^{-1}-(\dfrac 23)z^{-2} =\dfrac {(z+0.5)(z-3)}{z^2} \\ =\dfrac {(z-\dfrac 13)(z+2)}{z^2} \\ zeros \space z_1=\dfrac 13, z_2=-2$

Hence, one zero lies inside the unit circle and one zero lies outside unit circle, Hence the system is Mixed phase system.

The first five points of DFT are $(0.25, 0.215-j0.3018, 0, 0.125 – j0.0518, 0) $

Given DFT is

$\{0.25, 0.125 – j0.3018, 0, 0.125 – j0.3018, 0\} \\ X [0] = 0.25, x [1] = 0.125 – j0.3018, x [2] = 0, x [3] = 0.125 – j0.0518, x [4] = 0$

Given sequence is real valued sequence .

By symmetry property

$X* [k] = x [N - k] \\ \text {Or} x [k] = x*[N - k] $

Here $, N = 8 \\ x [k] = x*[8 - k]$

Now

$x [5] = x*[8 - 5]= x* [3] \\ = 0.125 + j0.005818 \\ x [6] = x*[8 - 6]= x* [2] \\ = 0 \\ x [7] = x*[8 - 7]= x* [1] \\ = 0.125 + j0.3018$