Z Transform Inversion MCQ Quiz – Objective Question with Answer for Z Transform Inversion

All the methods mentioned above can be used to calculate the inverse z-transform of the given signal.

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

X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}=1+\frac{3}{2}z^{-1}+\frac{7}{4}z^{-2}+\frac{15}{8}z^{-3}+\frac{31}{16}z^{-4}+…\)

So, we obtain x(n)= {1,3/2,7/4,15/8,31/16,….}.

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:

Thus

X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}=2z^2+6z^3+14z^4+30z^5+62z^6+…\)

In this case x(n)=0 for n≥0.Thus we obtain x(n)= {…..62,30,14,6,2,0,0}

Using the power series expansion for log(1+x), with |x|<1, we have

X(z)=\(\sum_{n=1}^∞ \frac{(-1)^{n+1} a^n z^{-n}}{n}\)

Thus

x(n)=(-1)n+1 \(\frac{a^n}{n}\), n≥1

=0, n≤0.

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

X(z)=1+2z-1+\(\frac{\frac{1}{6} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}\).

First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.

Thus we obtain X(z)=\(\frac{z^2}{z^2-1.5z+0.5}\)
The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be

\(\frac{X(z)}{z} = \frac{z}{(z-1)(z-0.5)} = \frac{2}{(z-1)} – \frac{1}{(z-0.5)}\)

(obtained by applying partial fractions)

=>X(z)=\(\frac{2z}{(z-1)}-\frac{z}{(z-0.5)}\).

To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,

X(z)=\(\frac{z(z+1)}{z^{-2}-z+0.5}\)

The poles of X(z) are complex conjugates p1=0.5+0.5j and p2=0.5-0.5j
Consequently the expansion will be

X(z)= \(\frac{z(0.5-1.5j)}{z-0.5-0.5j} + \frac{z(0.5+1.5j)}{z-0.5+0.5j}\).

First we express X(z) in terms of positive powers of z, in the form

X(z)=\(\frac{z^3}{(z+1)(z-1)^2}\)

X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial fraction expansion is

\(\frac{X(z)}{z} = \frac{z^2}{(z+1)(z-1)^2} = \frac{A}{z+1} + \frac{B}{z-1} + \frac{C}{(z-1)^2}\)

On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.

Therefore, we get \(\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z-1)^2}\).

The partial fraction expansion for the given X(z) is

\(X(z)= \frac{2z}{z-1}-\frac{z}{z-0.5}\)

In case when ROC is |z|>1, the signal x(n) is causal and both the terms in the above equation are causal terms. Thus, when we apply inverse z-transform to the above equation, we get

x(n)=2(1)nu(n)-(0.5)nu(n)=(2-0.5n)u(n).

The partial fraction expansion for the given X(z) is

\(X(z)= \frac{2z}{z-1}-\frac{z}{z-0.5}\)

In case when ROC is |z|<0.5, the signal is anti causal. Thus both the terms in the above equation are anti causal terms. So, if we apply inverse z-transform to the above equation we get

x(n)= [-2+0.5n]u(-n-1).