Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
