Find the inverse z transforms of these functions in series form by synthetic division.
(a) X(z) = z/(z – 1/2) |z| > 1/2
(b) X(z) = (z – 1)/(z ^ 7 – 2z + 1) |z| > 1
(c) X(z) = z/(z – 1/2) |z| < 1/2
(d) X(z) = (z + 2)/(4z ^ 2 – 2z + 3) |z| < (sqrt(3)) / 2
The Correct Answer and Explanation is :
To determine the inverse Z-transforms of the given functions using synthetic division, we will first express each function in a form amenable to this method. Synthetic division is particularly useful for rational functions where the degree of the numerator is less than that of the denominator. It involves dividing the numerator by the denominator to obtain a power series representation, from which the inverse Z-transform can be directly read.
(a) ( X(z) = \frac{z}{z – \frac{1}{2}}, \ |z| > \frac{1}{2} )
Here, the numerator’s degree is less than the denominator’s. To apply synthetic division, rewrite the function as:
[
X(z) = \frac{z}{z – \frac{1}{2}} = \frac{z – \frac{1}{2} + \frac{1}{2}}{z – \frac{1}{2}} = 1 + \frac{\frac{1}{2}}{z – \frac{1}{2}}
]
Now, divide ( \frac{\frac{1}{2}}{z – \frac{1}{2}} ) using synthetic division:
- First term: Divide ( z ) by ( z – \frac{1}{2} ):
- Quotient: 1
- Remainder: ( \frac{1}{2} )
- Second term: Divide ( \frac{1}{2} ) by ( z – \frac{1}{2} ):
- Quotient: ( \frac{1}{4} )
- Remainder: ( \frac{1}{8} )
Continuing this process yields the series:
[
X(z) = 1 + \frac{1}{2} \left( \frac{1}{z – \frac{1}{2}} \right) = 1 + \frac{1}{2} \sum_{n=0}^{\infty} \left( \frac{1}{2z} \right)^n
]
This is a geometric series with a ratio ( \frac{1}{2z} ), converging when ( |z| > \frac{1}{2} ). The inverse Z-transform corresponds to the sequence:
[
x[n] = \frac{1}{2^{n+1}} u[n]
]
where ( u[n] ) is the unit step function.
(b) ( X(z) = \frac{z – 1}{z^7 – 2z + 1}, \ |z| > 1 )
For this function, the numerator’s degree is less than the denominator’s. Perform synthetic division by dividing ( z – 1 ) by ( z^7 – 2z + 1 ):
- First term: Divide ( z ) by ( z^7 – 2z + 1 ):
- Quotient: 0
- Remainder: ( z – 1 )
Since the quotient is zero, the series does not develop further, indicating that the inverse Z-transform is not expressible in a simple series form using synthetic division. Alternative methods, such as partial fraction expansion or contour integration, may be more suitable for this case.
(c) ( X(z) = \frac{z}{z – \frac{1}{2}}, \ |z| < \frac{1}{2} )
This function is similar to part (a), but with a different region of convergence (ROC). To apply synthetic division, express ( X(z) ) as:
[
X(z) = \frac{z}{z – \frac{1}{2}} = 1 + \frac{\frac{1}{2}}{z – \frac{1}{2}}
]
Dividing ( \frac{\frac{1}{2}}{z – \frac{1}{2}} ) by synthetic division:
- First term: Divide ( z ) by ( z – \frac{1}{2} ):
- Quotient: 1
- Remainder: ( \frac{1}{2} )
- Second term: Divide ( \frac{1}{2} ) by ( z – \frac{1}{2} ):
- Quotient: ( \frac{1}{4} )
- Remainder: ( \frac{1}{8} )
This process yields the series:
[
X(z) = 1 + \frac{1}{2} \sum_{n=0}^{\infty} \left( \frac{1}{2z} \right)^n
]
This is a geometric series with a ratio ( \frac{1}{2z} ), converging when ( |z| < \frac{1}{2} ). The inverse Z-transform corresponds to the sequence:
[
x[n] = -\frac{1}{2^{n+1}} u[-n-1]
]
where ( u[n] ) is the unit step function.
(d) ( X(z) = \frac{z + 2}{4z^2 – 2z + 3}, \ |z| < \frac{\sqrt{3}}{2} )
For this function, the numerator’s degree is less than the denominator’s. Perform synthetic division by dividing ( z + 2 ) by ( 4z^2 –