This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses 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

View Answer

Explanation: 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)=1/(1-1.5z^{-1}+0.5z^{-2} ) if ROC is |z|>1?

a) {1,3/2,7/4,15/8,31/16,….}

a) {1,2/3,4/7,8/15,16/31,….}

a) {1/2,3/4,7/8,15/16,31/32,….}

d) None of the mentioned

View Answer

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

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

3. What is the inverse z-transform of X(z)=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…..}

View Answer

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:

Thus

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|?

d)None of the mentioned

View Answer

X(z)= (1+3z

^{-1}+11/6 z

^{-2}+1/3 z

^{-3})/(1+5/6 z

^{-1}+1/6 z

^{-2})?

a) 1+2z -1+(1/6 z

^{-1})/(1+5/6 z

^{-1}+1/6 z

^{-2})

b) 1-2z -1+(1/6 z

^{-1})/(1+5/6 z

^{-1}+1/6 z

^{-2})

c) 1+2z -1+(1/3 z

^{-1})/(1+5/6 z

^{-1}+1/6 z

^{-2})

d) 1+2z -1-(1/6 z

^{-1})/(1+5/6 z

^{-1}+1/6 z

^{-2})

View Answer

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

X(z)= 1+2z -1+(1/6 z

^{-1})/(1+5/6 z

^{-1}+1/6 z

^{-2})

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

a) 2z/(z-1)-z/(z+0.5)

b) 2z/(z-1)+z/(z-0.5)

c) 2z/(z-1)+z/(z+0.5)

d) 2z/(z-1)-z/(z-0.5)

View Answer

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

Thus we obtain X(z)= 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

(X(z))/z = z/((z-1)(z-0.5)) = 2/((z-1) ) – 1/((z-0.5) )( obtained by applying partial fractions)

=>X(z)= 2z/(z-1)-z/(z-0.5)

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

a) (z(0.5-1.5j))/(z-0.5-0.5j) – (z(0.5+1.5j))/(z-0.5+0.5j)

b) (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j)

c) (z(0.5+1.5j))/(z-0.5-0.5j) – (z(0.5-1.5j))/(z-0.5+0.5j)

d) (z(0.5+1.5j))/(z-0.5-0.5j) + (z(0.5-1.5j))/(z-0.5+0.5j)

View Answer

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

X(z)=(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)= (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j)

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

a) z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z+1)〗^{2} )

b) z/(4(z+1)) + 3z/(4(z-1)) – z/(2〖(z+1)〗^{2} )

c) z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗^{2} )

d) z/(4(z+1)) + z/(4(z-1)) + z/(2〖(z+1)〗^{2} )

View Answer

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

(X(z))/z = z

^{2}/((z+1)〖(z-1)〗

^{2}) =A/(z+1) + B/(z-1) + 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 X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗

^{2}) .

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

a) (2-0.5^{n})u(n)

b) (2+0.5^{n})u(n)

c) (2^{n}-0.5^{n})u(n)

d) None of the mentioned

View Answer

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

X(z)= 2z/(z-1)-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)

^{n}u(n)-(0.5)

^{n}u(n)=(2-0.5

^{n})u(n)

^{-1}+0.5z

^{-2}) if ROC is |z|<0.5? a) [-2-0.5

^{n}]u(n)

b) [-2+0.5

^{n}]u(n)

c) [-2+0.5

^{n}]u(-n-1)

d) [-2-0.5

^{n}]u(-n-1)

View Answer

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

X(z)= 2z/(z-1)-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+0.5

^{n}]u(-n-1)

11. What is the inverse z-transform of X(z)= 1/(1-1.5z^{-1}+0.5z^{-2} ) if ROC is 0.5<|z|<1?
a) -2u(-n-1)+(0.5)^{n}u(n)

b) -2u(-n-1)-(0.5)^{n}u(n)

c) -2u(-n-1)+(0.5)^{n}u(-n-1)

d) 2u(n)+(0.5)^{n}u(-n-1)

View Answer

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

X(z)= 2z/(z-1)-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)

^{n}u(n)

12.What is the causal signal x(n) having the z-transform X(z)= 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)

View Answer

Explanation: The partial fraction expansion of X(z) is X(z)= z/(4(z+1)) + 3z/(4(z-1)) + 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)

**Sanfoundry Global Education & Learning Series – Digital Signal Processing.**

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