Answer :
To find a cascaded realization for the transfer function
[tex]\[ H(z) = \frac{z^2 (6z - 2)}{(z-1) \left(z^2 - \frac{1}{6}z - \frac{1}{6}\right)} \][/tex]
we will factorize both the numerator and the denominator into their respective roots and then express \( H(z) \) in terms of its zeros and poles.
### Step 1: Factorize the Numerator
The numerator of the transfer function is given by:
[tex]\[ z^2 (6z - 2) \][/tex]
We can express this in factored form as:
[tex]\[ z^2 (6z - 2) = z^2 \cdot 2(3z - 1) = 2z^2(3z - 1) \][/tex]
The numerator has roots at \( z = 0 \) and \( z = \frac{1}{3} \). Thus, the numerator can be written as:
[tex]\[ 2z^2 (z - \frac{1}{3}) \][/tex]
### Step 2: Factorize the Denominator
The denominator of the transfer function is given by:
[tex]\[ (z-1) \left(z^2 - \frac{1}{6}z - \frac{1}{6}\right) \][/tex]
We need to solve the quadratic equation:
[tex]\[ z^2 - \frac{1}{6}z - \frac{1}{6} = 0 \][/tex]
The roots of this quadratic equation are obtained using the quadratic formula:
[tex]\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where \( a = 1 \), \( b = -\frac{1}{6} \), and \( c = -\frac{1}{6} \).
Solving for the roots, we get:
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\left(\frac{1}{6}\right)^2 + 4 \cdot \frac{1}{6}}}{2} \][/tex]
Simplifying the expression inside the square root and solving for \( z \):
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\frac{1}{36} + \frac{2}{3}}}{2} \][/tex]
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\frac{1}{36} + \frac{12}{36}}}{2} \][/tex]
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\frac{13}{36}}}{2} \][/tex]
[tex]\[ z = \frac{\frac{1}{6} \pm \frac{\sqrt{13}}{6}}{2} \][/tex]
[tex]\[ z = \frac{1 \pm \sqrt{13}}{12} \][/tex]
Thus, the roots of the denominator are:
[tex]\[ z_1 = \frac{1 + \sqrt{13}}{12} \approx 0.5 \][/tex]
[tex]\[ z_2 = \frac{1 - \sqrt{13}}{12} \approx -0.333 \][/tex]
And we have one more root at \( z = 1 \) from the linear factor.
So, we can write the denominator as:
[tex]\[ (z-1)(z-0.5)(z+0.333) \][/tex]
### Step 3: Construct the Transfer Function in Factored Form
Using the roots, the transfer function can be written as:
[tex]\[ H(z) = \frac{2z^2 (z - \frac{1}{3})}{(z - 1)(z - 0.5)(z + 0.333)} \][/tex]
### Step 4: Cascaded Realization
The factored form of \( H(z) \) can be realized as a series of second-order sections.
1. The first stage is:
[tex]\[ H_1(z) = \frac{z - 0}{z - 1} = \frac{z}{z - 1} \][/tex]
2. The second stage is:
[tex]\[ H_2(z) = \frac{z - 0}{z - 0.5} = \frac{z}{z - 0.5} \][/tex]
3. The third stage is:
[tex]\[ H_3(z) = \frac{2(z - \frac{1}{3})}{z + 0.333} = \frac{2(z - \frac{1}{3})}{z + 0.333} \][/tex]
So the cascaded realization of \( H(z) \) is:
[tex]\[ H(z) = H_1(z) \cdot H_2(z) \cdot H_3(z) \][/tex]
where
[tex]\[ H_1(z) = \frac{z}{z - 1} \][/tex]
[tex]\[ H_2(z) = \frac{z}{z - 0.5} \][/tex]
[tex]\[ H_3(z) = \frac{2(z - \frac{1}{3})}{z + 0.333} \][/tex]
Each of these stages can be implemented using digital filters in a cascade system.
[tex]\[ H(z) = \frac{z^2 (6z - 2)}{(z-1) \left(z^2 - \frac{1}{6}z - \frac{1}{6}\right)} \][/tex]
we will factorize both the numerator and the denominator into their respective roots and then express \( H(z) \) in terms of its zeros and poles.
### Step 1: Factorize the Numerator
The numerator of the transfer function is given by:
[tex]\[ z^2 (6z - 2) \][/tex]
We can express this in factored form as:
[tex]\[ z^2 (6z - 2) = z^2 \cdot 2(3z - 1) = 2z^2(3z - 1) \][/tex]
The numerator has roots at \( z = 0 \) and \( z = \frac{1}{3} \). Thus, the numerator can be written as:
[tex]\[ 2z^2 (z - \frac{1}{3}) \][/tex]
### Step 2: Factorize the Denominator
The denominator of the transfer function is given by:
[tex]\[ (z-1) \left(z^2 - \frac{1}{6}z - \frac{1}{6}\right) \][/tex]
We need to solve the quadratic equation:
[tex]\[ z^2 - \frac{1}{6}z - \frac{1}{6} = 0 \][/tex]
The roots of this quadratic equation are obtained using the quadratic formula:
[tex]\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where \( a = 1 \), \( b = -\frac{1}{6} \), and \( c = -\frac{1}{6} \).
Solving for the roots, we get:
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\left(\frac{1}{6}\right)^2 + 4 \cdot \frac{1}{6}}}{2} \][/tex]
Simplifying the expression inside the square root and solving for \( z \):
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\frac{1}{36} + \frac{2}{3}}}{2} \][/tex]
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\frac{1}{36} + \frac{12}{36}}}{2} \][/tex]
[tex]\[ z = \frac{\frac{1}{6} \pm \sqrt{\frac{13}{36}}}{2} \][/tex]
[tex]\[ z = \frac{\frac{1}{6} \pm \frac{\sqrt{13}}{6}}{2} \][/tex]
[tex]\[ z = \frac{1 \pm \sqrt{13}}{12} \][/tex]
Thus, the roots of the denominator are:
[tex]\[ z_1 = \frac{1 + \sqrt{13}}{12} \approx 0.5 \][/tex]
[tex]\[ z_2 = \frac{1 - \sqrt{13}}{12} \approx -0.333 \][/tex]
And we have one more root at \( z = 1 \) from the linear factor.
So, we can write the denominator as:
[tex]\[ (z-1)(z-0.5)(z+0.333) \][/tex]
### Step 3: Construct the Transfer Function in Factored Form
Using the roots, the transfer function can be written as:
[tex]\[ H(z) = \frac{2z^2 (z - \frac{1}{3})}{(z - 1)(z - 0.5)(z + 0.333)} \][/tex]
### Step 4: Cascaded Realization
The factored form of \( H(z) \) can be realized as a series of second-order sections.
1. The first stage is:
[tex]\[ H_1(z) = \frac{z - 0}{z - 1} = \frac{z}{z - 1} \][/tex]
2. The second stage is:
[tex]\[ H_2(z) = \frac{z - 0}{z - 0.5} = \frac{z}{z - 0.5} \][/tex]
3. The third stage is:
[tex]\[ H_3(z) = \frac{2(z - \frac{1}{3})}{z + 0.333} = \frac{2(z - \frac{1}{3})}{z + 0.333} \][/tex]
So the cascaded realization of \( H(z) \) is:
[tex]\[ H(z) = H_1(z) \cdot H_2(z) \cdot H_3(z) \][/tex]
where
[tex]\[ H_1(z) = \frac{z}{z - 1} \][/tex]
[tex]\[ H_2(z) = \frac{z}{z - 0.5} \][/tex]
[tex]\[ H_3(z) = \frac{2(z - \frac{1}{3})}{z + 0.333} \][/tex]
Each of these stages can be implemented using digital filters in a cascade system.
