Z Transform MCQ [Free PDF] – Objective Question Answer for Z Transform Quiz

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).

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

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

In this case, ROC is 0.5<|z|<1 is a ring, which implies that the signal is two-sided. Thus one of the signals corresponds to a causal signal and the other corresponds to an anti-causal signal.

Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole p2 = 0.5 provides the causal part and the pole p1 = 1 provides the anti causal part. SO, if we apply the inverse z-transform we get

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

The partial fraction expansion of X(z) is

\(X(z) = \frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}2{(z-1)^2}\)

When we apply the inverse z-transform for the above equation, we get

x(n) = [1/4(-1)n+3/4+n/2]u(n).