250+ TOP MCQs on Inversion of Z Transform & Answers

Digital Signal Processing Multiple Choice Questions on “Inversion of Z Transform”.

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
Answer: D
Clarification: 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
Answer: A
Clarification: 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…..}
Answer: B
Clarification: 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
Answer: C
Clarification: 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.

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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}})
Answer: A
Clarification: 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})
Answer: D
Clarification: 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})
Answer: B
Clarification: 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})

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Answer: C
Clarification: 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
Answer: A
Clarification: 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)
Answer: C
Clarification: 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).

11. What is the inverse z-transform of X(z)=(frac{1}{1-1.5z^{-1}+0.5z^{-2}}) if ROC is 0.5<|z|<1?
A. -2u(-n-1)+(0.5)ⁿu(n)
B. -2u(-n-1)-(0.5)ⁿu(n)
C. -2u(-n-1)+(0.5)ⁿu(-n-1)
D. 2u(n)+(0.5)ⁿu(-n-1)
Answer: B
Clarification: 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 signal 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).

12. What is the causal signal x(n) having the z-transform X(z)=(frac{1}{(1+z^{-1})(1-z^{-1})^2})?
A. [1/4(-1)ⁿ+3/4-n/2]u(n)
B. [1/4(-1)ⁿ+3/4-n/2]u(-n-1)
C. [1/4+3/4(-1)ⁿ-n/2]u(n)
D. [1/4(-1)ⁿ+3/4+n/2]u(n)
Answer: D
Clarification: 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).

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