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

1. Which of the following method is used to find the inverse z-transform of a signal?

A. Counter integration
B. Expansion into a series of terms
C. Partial fraction expansion
D. All of the mentioned

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

2. What is the inverse z-transform of
X(z)=$$\frac{1}{1-1.5z^{-1}+0.5z^{-2}}$$ if ROC is |z|>1?

A. {1,3/2,7/4,15/8,31/16,….}

B. {1,2/3,4/7,8/15,16/31,….}

C. {1/2,3/4,7/8,15/16,31/32,….}

D. None of the mentioned

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,….}.

3. What is the inverse z-transform of
X(z)=$$\frac{1}{1-1.5z^{-1}+0.5z^{-2}}$$ if ROC is |z| < 0.5?

A. {….62,30,14,6,2}
B. {…..62,30,14,6,2,0,0}
C. {0,0,2,6,14,30,62…..}
D. {2,6,14,30,62…..}

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}

4. What is the inverse z-transform of X(z)=log(1+az-1) |z|>|a|?

A. x(n)=(-1)n+1 $$\frac{a^{-n}}{n}$$, n≥1; x(n)=0, n≤0

B. x(n)=(-1)n-1 $$\frac{a^{-n}}{n}$$, n≥1; x(n)=0, n≤0

C. x(n)=(-1)n+1 $$\frac{a^{-n}}{n}$$, n≥1; x(n)=0, n≤0

D. None of the mentioned

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.

5. What is the proper fraction and polynomial form of the improper rational transform

X(z)=$$\frac{1+3z^{-1}+\frac{11}{6} z^{-2}+\frac{1}{3} z^{-3}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}$$?

A. 1+2z-1+$$\frac{\frac{1}{6}z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}$$

B. 1-2z-1+$$\frac{\frac{1}{6} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}$$

C. 1+2z-1+$$\frac{\frac{1}{3} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}$$

D. 1+2z-1–$$\frac{\frac{1}{6} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}$$

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+$$\frac{\frac{1}{6} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}$$.

6. What is the partial fraction expansion of the proper function
X(z)=$$\frac{1}{1-1.5z^{-1}+0.5z^{-2}}$$?

A. $$\frac{2z}{z-1}-\frac{z}{z+0.5}$$

B. $$\frac{2z}{z-1}+\frac{z}{z-0.5}$$

C. $$\frac{2z}{z-1}+\frac{z}{z+0.5}$$

D. $$\frac{2z}{z-1}-\frac{z}{z-0.5}$$

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

7. What is the partial fraction expansion of
X(z)=$$\frac{1+z^{-1}}{1-z^{-1}+0.5z^{-2}}$$?

A. $$\frac{z(0.5-1.5j)}{z-0.5-0.5j} – \frac{z(0.5+1.5j)}{z-0.5+0.5j}$$

B. $$\frac{z(0.5-1.5j)}{z-0.5-0.5j} + \frac{z(0.5+1.5j)}{z-0.5+0.5j}$$

C. $$\frac{z(0.5+1.5j)}{z-0.5-0.5j} – \frac{z(0.5-1.5j)}{z-0.5+0.5j}$$

D. $$\frac{z(0.5+1.5j)}{z-0.5-0.5j} + \frac{z(0.5-1.5j)}{z-0.5+0.5j}$$

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

8. What is the partial fraction expansion of X(z)=$$\frac{1}{(1+z^{-1})(1-z^{-1})^2}$$?

A. $$\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z+1)^2}$$

B. $$\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} – \frac{z}{2(z+1)^2}$$

C. $$\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z-1)^2}$$

D. $$\frac{z}{4(z+1)} + \frac{z}{4(z-1)} + \frac{z}{2(z+1)^2}$$

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 $$\frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}{2(z-1)^2}$$.

9. What is the inverse z-transform of
X(z)=$$\frac{1}{1-1.5z^{-1}+0.5z^{-2}}$$ if ROC is |z|>1?

A. (2-0.5n)u(n)
B. (2+0.5n)u(n)
C. (2n-0.5n)u(n)
D. None of the mentioned

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).

10. What is the inverse z-transform of
X(z)=$$\frac{1}{1-1.5z^{-1}+0.5z^{-2}}$$ if ROC is |z|<0.5?

A. [-2-0.5n]u(n)
B. [-2+0.5n]u(n)
C. [-2+0.5n]u(-n-1)
D. [-2-0.5n]u(-n-1)

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