Digital Signal Processing

Inversion of Z Transform

Question 1
Marks : +2 | -2
Pass Ratio : 100%
What is the inverse z-transform of X(z)=\\(\\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\\) if ROC is |z| < 0.5?
{….62,30,14,6,2}
{…..62,30,14,6,2,0,0}
{0,0,2,6,14,30,62…..}
{2,6,14,30,62…..}
Explanation:
In this case the ROC is the interior of a circle. Consequently, the signal x(n) is anti causal. To obtain a power series expansion in positive powers of z, we perform the long division in the following way:
Question 2
Marks : +2 | -2
Pass Ratio : 100%
What is the partial fraction expansion of the proper function X(z)=\\(\\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\\)?
\\(\\frac{2z}{z-1}-\\frac{z}{z+0.5}\\)
\\(\\frac{2z}{z-1}+\\frac{z}{z-0.5}\\)
\\(\\frac{2z}{z-1}+\\frac{z}{z+0.5}\\)
\\(\\frac{2z}{z-1}-\\frac{z}{z-0.5}\\)
Explanation:
First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
What is the inverse z-transform of X(z)=log(1+az-1) |z|>|a|?
x(n)=(-1)n+1 \\(\\frac{a^{-n}}{n}\\), n≥1; x(n)=0, n≤0
x(n)=(-1)n-1 \\(\\frac{a^{-n}}{n}\\), n≥1; x(n)=0, n≤0
x(n)=(-1)n+1 \\(\\frac{a^{-n}}{n}\\), n≥1; x(n)=0, n≤0
None of the mentioned
Explanation:
Using the power series expansion for log(1+x), with |x|<1, we have
Question 4
Marks : +2 | -2
Pass Ratio : 100%
What is the partial fraction expansion of X(z)=\\(\\frac{1}{(1+z^{-1})(1-z^{-1})^2}\\)?
\\(\\frac{z}{4(z+1)} + \\frac{3z}{4(z-1)} + \\frac{z}{2(z+1)^2}\\)
\\(\\frac{z}{4(z+1)} + \\frac{3z}{4(z-1)} – \\frac{z}{2(z+1)^2}\\)
\\(\\frac{z}{4(z+1)} + \\frac{3z}{4(z-1)} + \\frac{z}{2(z-1)^2}\\)
\\(\\frac{z}{4(z+1)} + \\frac{z}{4(z-1)} + \\frac{z}{2(z+1)^2}\\)
Explanation:
First we express X(z) in terms of positive powers of z, in the form X(z)=\\(\\frac{z^3}{(z+1)(z-1)^2}\\)
Question 5
Marks : +2 | -2
Pass Ratio : 100%
What is the inverse z-transform of X(z)=\\(\\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\\) if ROC is |z|>1?
(2-0.5n)u(n)
(2+0.5n)u(n)
(2n-0.5n)u(n)
None of the mentioned
Explanation:
The partial fraction expansion for the given X(z) is
Question 6
Marks : +2 | -2
Pass Ratio : 100%
What is the inverse z-transform of X(z)=\\(\\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\\) if ROC is |z|>1?
{1,3/2,7/4,15/8,31/16,….}
{1,2/3,4/7,8/15,16/31,….}
{1/2,3/4,7/8,15/16,31/32,….}
None of the mentioned
Explanation:
Since the ROC is the exterior circle, we expect x(n) to be a causal signal. Thus we seek a power series expansion in negative powers of ‘z’. By dividing the numerator of X(z) by its denominator, we obtain the power series
Question 7
Marks : +2 | -2
Pass Ratio : 100%
Which of the following method is used to find the inverse z-transform of a signal?
Counter integration
Expansion into a series of terms
Partial fraction expansion
All of the mentioned
Explanation:
All the methods mentioned above can be used to calculate the inverse z-transform of the given signal.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
What is the inverse z-transform of X(z)=\\(\\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\\) if ROC is |z|<0.5?
[-2-0.5n]u(n)
[-2+0.5n]u(n)
[-2+0.5n]u(-n-1)
[-2-0.5n]u(-n-1)
Explanation:
The partial fraction expansion for the given X(z) is
Question 9
Marks : +2 | -2
Pass Ratio : 100%
What is the proper fraction and polynomial form of the improper rational transform
1+2z-1+\\(\\frac{\\frac{1}{6}z^{-1}}{1+\\frac{5}{6} z^{-1}+\\frac{1}{6} z^{-2}}\\)
1-2z-1+\\(\\frac{\\frac{1}{6} z^{-1}}{1+\\frac{5}{6} z^{-1}+\\frac{1}{6} z^{-2}}\\)
1+2z-1+\\(\\frac{\\frac{1}{3} z^{-1}}{1+\\frac{5}{6} z^{-1}+\\frac{1}{6} z^{-2}}\\)
1+2z-1–\\(\\frac{\\frac{1}{6} z^{-1}}{1+\\frac{5}{6} z^{-1}+\\frac{1}{6} z^{-2}}\\)
Explanation:
First, we note that we should reduce the numerator so that the terms z-2 and z-3 are eliminated. Thus we should carry out the long division with these two polynomials written in the reverse order. We stop the division when the order of the remainder becomes z-1. Then we obtain
Question 10
Marks : +2 | -2
Pass Ratio : 100%
What is the partial fraction expansion of X(z)=\\(\\frac{1+z^{-1}}{1-z^{-1}+0.5z^{-2}}\\)?
\\(\\frac{z(0.5-1.5j)}{z-0.5-0.5j} – \\frac{z(0.5+1.5j)}{z-0.5+0.5j}\\)
\\(\\frac{z(0.5-1.5j)}{z-0.5-0.5j} + \\frac{z(0.5+1.5j)}{z-0.5+0.5j}\\)
\\(\\frac{z(0.5+1.5j)}{z-0.5-0.5j} – \\frac{z(0.5-1.5j)}{z-0.5+0.5j}\\)
\\(\\frac{z(0.5+1.5j)}{z-0.5-0.5j} + \\frac{z(0.5-1.5j)}{z-0.5+0.5j}\\)
Explanation:
To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,