Multiple Choice Questions & Answers (MCQs) on “Z Transform”.

1. The Z-Transform X(z) of a discrete time signal x(n) is defined as ____________ a) ∑∞n=−∞x(n)zn b) ∑∞n=−∞x(n)z−n c) ∑∞n=0x(n)zn d) None of the mentioned

Answer: b a) Radius of convergence b) Radius of divergence c) Feasible solution d) None of the mentioned

Explanation: The z-transform of a real discrete time sequence x(n) is defined as a power of ‘z’ which is equal to X(z)=∑∞n=−∞x(n)z−n, where ‘z’ is a complex variable.

2. What is the set of all values of z for which X(z) attains a finite value?

Answer: a

Explanation: Since X(z) is a infinite power series, it is defined only at few values of z. The set of all values of z where X(z) converges to a finite value is called as Radius of Convergence(ROC).

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Email: info@matlabassignmentexperts.com a) 2 + 4z + 5z2 + 7z3 + z4 b) 2 + 4z + 5z2 + 7z3 + z5 c) 2 + 4z-1 + 5z-2 + 7z-3 + z-5 d) 2z2 + 4z + 5 +7z-1 + z-3

3. What is the z-transform of the following finite duration signal?

Answer: d

Explanation: We know that, for a given signal x(n) the ztransform is defined as X(z)=∑∞n=−∞x(n)z−n

Substitute the values of n from -2 to 3 and the corresponding signal values in the above formula a) z=0 b) z=∞ c) Entire z-plane, except at z=0 d) Entire z-plane, except at z=∞

We get, X(z) = 2z2 + 4z + 5 + 7z-1 + z-3 .

4. What is the ROC of the signal x(n)=δ(n-k), k>0?

Answer: c

Explanation: We know that, the z-transform of a signal x(n) is X(z)=∑∞n=−∞x(n)z−n

Given x(n)=δ(n-k)=1 at n=k

=> X(z)=z-k

From the above equation, X(z) is defined at all values of z except at z=0 for k>0.

So ROC is defined as Entire z-plane, except at z=0.

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WhatsApp: +1(315)557-6473 a) 1/1−0.5z−1;ROC|z|>0.5 b) 1/1−0.5z−1;ROC|z|<0.5 c) 1/1+0.5z−1;ROC|z|>0.5 d) 1/1+0.5z−1;ROC|z|<0.5

5. What is the z-transform of the signal x(n)=(0.5)nu(n)?

Answer: a

Explanation: For a given signal x(n), its z-transform

X(z)=∑∞n=−∞x(n)z−n

Given x(n)=(0.5)nu(n)=(0.5)n for n≥0

So, X(z)=∑∞n=00.5nz−n=∑∞n=0(0.5z−1)n a) α-nu(n) b) αnu(n) c) α-nu(-n) d) αnu(n)

This is an infinite GP whose sum is given as X(z)=11−0.5z−1 under the condition that |0.5z-1|<1 => X(z)=11−0.5z−1 and ROC is |z|>0.5.

6. Which of the following series has an ROC as mentioned below?

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Answer: b

Explanation:

Let x(n)= αnu(n) a) 1/1−αz−1;ROC |z|<|α| b) −1/1+αz−1;ROC |z|<|α| c) −1/1−αz−1;ROC |z|>|α| d) −1/1−αz−1;ROC |z|<|α|

The z-transform of the signal x(n) is given as X(z)=X(z)=∑∞n=0αnz−n=∑∞n=0αz−1n => X(z)=11−αz−1 and ROC is |z|>α which is as given in the question.

7. What is the z-transform of the signal x(n) = -αnu(-n-1)?

Answer: d

Explanation: Given x(n)= -αnu(-n-1)=0 for n≥0 =-αn for n≤-1 a) |a|<|z|<|b| b) |a|>|z|>|b| c) |a|>|z|<|b| d) |a|<|z|>|b|

From the definition of z-transform, we have X(z)=∑−1n=−∞(−αn)z−n=−∑−1n=−∞(α−1z)−n=−11−αz−1 and |α-1z| <1 => |z|<|α|.

8. What is the ROC of the z-transform of the signal x(n)= anu(n) +bnu(-n-1)?

Answer: a a) z=0 b) z=∞ c) Entire z-plane, except at z=0 d) Entire z-plane, except at z=∞

Explanation: We know that, ROC of z-transform of anu(n) is |z|>|a|. ROC of z-transform of bnu(-n-1) is |z|<|b|.

By combining both the ROC’s we get the ROC of z-transform of the signal x(n) as |a|<|z|<|b|.

9. What is the ROC of z-transform of finite duration anti-causal sequence?

Answer: d a) |z|>r1 b) |z|<r1 c) r2<|z|<r1 d) None of the mentioned

Explanation: Let us an example of anti causal sequence whose ztransform will be in the form X(z)=1+z+z2 which has a finite value at all values of ‘z’ except at z=∞. So, ROC of an anti-causal sequence is entire z-plane except at z=∞.

10. What is the ROC of z-transform of an two sided infinite sequence?

Answer: c

Explanation: Let us plot the graph of z-transform of any two sided sequence which looks as follows.

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11. The z-transform of a sequence x(n) which is given as X(z)=∑∞n=−∞x(n)z−n is known as _____________ a) Uni-lateral Z-transform b) Bi-lateral Z-transform c) Tri-lateral Z-transform d) None of the mentioned

Answer: b a) Entire z-plane, except at z=0 b) Entire z-plane, except at z=∞ c) Contain unit circle d) None of the mentioned

Explanation: The entire timing sequence is divided into two parts n=0 to ∞ and n=-∞ to 0.

Since the z-transform of the signal given in the questions contains both the parts, it is called as Bi-lateral z-transform.

12. What is the ROC of the system function H(z) if the discrete time LTI system is BIBO stable?

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Answer: c

Explanation: A discrete time LTI is BIBO stable, if and only if its impulse response h(n) is absolutely summable. That is, ∑∞n=−∞|h(n)|<∞

We know that, H(z)= ∑∞n=−∞h(n)z−n

Let z=ejω so that |z|=|ejω|=1

Then, |H(ejω)|=|H(ω)|=| ∑∞n=−∞h(n)

=∑∞n=−∞|h(n)|<∞ a) True b) False

Hence, we see that if the system is stable, then H(z) converges for z=ejω . That is, for a stable discrete time LTI system, ROC of H(z) must contain the unit circle |z|=1.

13. The ROC of z-transform of any signal cannot contain poles.

Answer: a

Explanation: Since the value of z-transform tends to infinity, the ROC of the z-transform does not contain poles.

14. Is the discrete time LTI system with impulse response h(n)=an(n) (|a| < 1) BIBO stable? a) True b) False

Answer: a

Explanation: Given h(n)=an(n) (|a|<1)

The z-transform of h(n) is H(z)=zz−a, ROC is |z|>|a|

If |a|<1, then the ROC contains the unit circle. So, the system is BIBO stable. Visit: www.matlabassignmentexperts.com