This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Rational Z Transform”.

1. What are the values of z for which the value of X(z)=0?

a) Poles

b) Zeros

c) Solutions

d) None of the mentioned

View Answer

Explanation: For a rational z-transform X(z) to be zero, the numerator of X(z) is zero and the solutions of the numerator are called as ‘zeros’ of X(z).

2. What are the values of z for which the value of X(z)=∞?

a) Poles

b) Zeros

c) Solutions

d) None of the mentioned

View Answer

Explanation: For a rational z-transform X(z) to be infinity, the denominator of X(z) is zero and the solutions of the denominator are called as ‘poles’ of X(z).

3. If X(z) has M finite zeros and N finite poles, then which of the following condition is true?

a) |N-M| poles at origin(if N>M)

b) |N+M| zeros at origin(if N>M)

c) |N+M| poles at origin(if N>M)

d) |N-M| zeros at origin(if N>M)

View Answer

Explanation: If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z)=z -M+N.X'(z).

So, if N>M then z has a positive power. So, it has |N-M| zeros at origin.

4. If X(z) has M finite zeros and N finite poles, then which of the following condition is true?

a) |N-M| poles at origin(if N < M)

b) |N+M| zeros at origin(if N < M)

c) |N+M| poles at origin(if N < M)

d) |N-M| zeros at origin(if N < M)

View Answer

Explanation: If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z)=z

^{-M+N}.X'(z).

So, if N < M then z has a negative power. So, it has |N-M| poles at origin.

5. Which of the following signals have a pole-zero plot as shown below?

a) a.u(n)

b) u(an)

c) a^{n}u(n)

d) None of the mentioned

View Answer

Explanation: From the given pole-zero plot, the z-transform of the signal has one zero at z=0 and one pole at z=a.

So, we obtain X(z)=z/(z-a)

By applying inverse z-transform for X(z), we get

x(n)= a

^{n}u(n)

6. Which of the following signals have a pole-zero plot as shown below?(Let M=8 in the figure)

View Answer

Explanation: From the figure given, the z-transform of the signal has 8 zeros on circle of radius ‘a’ and 7 poles at origin.

7. The z-transform X(z) of the signal x(n)=a^{n}u(n) has:

a) One pole at z=0 and one zero at z=a

b) One pole at z=0 and one zero at z=0

c) One pole at z=a and one zero at z=a

d) One pole at z=a and one zero at z=0

View Answer

Explanation: The z-transform of the given signal is X(z)= z/(z-a)

So, it has one pole at z=a and one zero at z=0.

8. What is the nature of the signal whose pole-zero plot is as shown?

a) Rising signal

b) Constant signal

c) Decaying signal

d) None of the mentioned

View Answer

Explanation: From the pole-zero plot, it is shown that r < 1, so the signal is a decaying signal.

9. What are the values of z for which the value of X(z)=0?

a) Poles

b) Zeros

c) Solutions

d) None of the mentioned

View Answer

Explanation: For a rational z-transform X(z) to be zero, the numerator of X(z) is zero and the solutions of the numerator are called as ‘zeros’ of X(z).

10. If Y(z) is the z-transform of the output function, X(z) is the z-transform of the input function and H(z) is the z-transform of system function of the LTI system, then H(z)=?

a) (Y(z))/(X(z))

b) (X(z))/(Y(z))

c) Y(z).X(z)

d) None of the mentioned

View Answer

Explanation: We know that for an LTI system, y(n)=h(n)*x(n)

On applying z-transform on both sides we get, Y(z)=H(z).X(z)=>H(z)= ( Y(z))/(X(z) ).

11. What is the system function of the system described by the difference equation y(n)=0.5y(n-1)+2x(n)?

View Answer

Explanation: Given difference equation of the system is y(n)=0.5y(n-1)+2x(n)

On applying z-transform on both sides we get, Y(z)=0.5z

^{-1}Y(z)+2X(z)

12. What is the unit sample response of the system described by the difference equation y(n)=0.5y(n-1)+2x(n)?

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

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

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

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

View Answer

Explanation: By applying the z-transform on both sides of the difference equation given in the question we obtain,

By applying the inverse z-transform we get h(n)= 2(0.5)

^{n}u(n).

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