Question

Please answer fast. With full step by step solution.​

Answer 1

Let f(z) = (4z ² + 2z) / (2z ² - 3z + 1).

First, carry out the division:

f(z) = 2 + (8z - 2) / (2z ² - 3z + 1)

Observe that

2z ² - 3z + 1 = (2z - 1) (z - 1)

so you can separate the rational part of f(z) into partial fractions. We have

(8z - 2) / (2z ² - 3z + 1) = a / (2z - 1) + b / (z - 1)

8z - 2 = a (z - 1) + b (2z - 1)

8z - 2 = (a + 2b) z - (a + b)

so that a + 2b = 8 and a + b = 2, yielding a = -4 and b = 6.

So we have

f(z) = 2 - 4 / (2z - 1) + 6 / (z - 1)

or

f(z) = 2 - (2/z) (1 / (1 - 1/(2z))) + (6/z) (1 / (1 - 1/z))

Recall that for |z| < 1, we have

`\displaystyle\frac1{1-z}=\sum_(n=0)^\infty z^n`

Replace z with 1/z to get

`\displaystyle\frac1{1-\frac1z}=\sum_(n=0)^\infty z^(-n)`

so that by substitution, we can write

`\displaystyle f(z) = 2 - \frac2z \sum_(n=0)^\infty (2z)^(-n) + \frac6z \sum_(n=0)^\infty z^(-n)`

Now condense f(z) into one series:

`\displaystyle f(z) = 2 - \sum_(n=0)^\infty 2^(-n+1) z^(-(n+1)) + 6 \sum_(n=0)^\infty z^(-n-1)`

`\displaystyle f(z) = 2 - \sum_(n=0)^\infty \left(6+2^(-n+1)\right) z^(-(n+1))`

`\displaystyle f(z) = 2 - \sum_(n=1)^\infty \left(6+2^(-(n-1)+1)\right) z^(-n)`

`\displaystyle f(z) = 2 - \sum_(n=1)^\infty \left(6+2^(2-n)\right) z^(-n)`

So, the inverse Z transform of f(z) is
`\boxed{6+2^(2-n)}`.