Signals & Systems Questions and Answers – Properties of Z-Transforms – 1

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This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Z-Transforms – 1”.

1. The z-transform of δ[n-k]>0 is __________
a) Zk, Z>0
b) Z-k, Z>0
c) Zk, Z≠0
d) Z-k, Z≠0
View Answer

Answer: d
Explanation: Performing z-transform on δ[n-k], we get,
X (z) = \(∑_{n=0}^∞ x[n]z^{-n}\)
= Z-k, Z≠0.
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2. The z-transform of δ[n+k]>0 is __________
a) Z-k, Z≠0
b) Zk, Z≠0
c) Z-k, all Z
d) Zk, all Z
View Answer

Answer: d
Explanation: Performing z-transform on δ[n+k], we get,
X (z) = \(∑_{n=0}^∞ x[n]z^{-n}\)
= Zk, all Z

3. The z-transform of u[n] is _________
a) \(\frac{1}{1-z^{-1}}\), |Z|>1
b) \(\frac{1}{1-z^{-1}}\), |Z|<1
c) \(\frac{z}{1-z^{-1}}\), |Z|<1
d) \(\frac{z}{1-z^{-1}}\), |Z|>1
View Answer

Answer: a
Explanation: Performing z-transform on u[n], we get,
X (z) = \(∑_{n=0}^∞ x[n]z^{-n}\)
= \(\frac{1}{1-z^{-1}}\), |Z|>1.

4. The z-transform of \((\frac{1}{4})^n\) (u[n] – u[n-5]) is __________
a) \(\frac{z^5 – 0.25^5}{z^4 (z-0.25)}\), z>0.25
b) \(\frac{z^5 – 0.25^5}{z^4 (z-0.25)}\), z>0
c) \(\frac{z^5 – 0.25^5}{z^3 (z-0.25)}\), z<0.25
d) \(\frac{z^5 – 0.25^5}{z^4 (z-0.25)}\), all z
View Answer

Answer: d
Explanation: X (z) = \(∑_{n=0}^4 (\frac{1}{4} z^{-1})^4 \)
= \(\frac{1-(\frac{1}{4} z^{-1})^5}{1-(\frac{1}{4} z^{-1})^1}\)
= \(\frac{z^5 – 0.25^5}{z^4 (z-0.25)}\), all z.

5. The z-transform of \((\frac{1}{4})^4\) u[-n] is ___________
a) \(\frac{4z}{4z-1}\), |Z|>\(\frac{1}{4}\)
b) \(\frac{4z}{4z-1}\), |Z|<\(\frac{1}{4}\)
c) \(\frac{1}{1-4z}\), |Z|>\(\frac{1}{4}\)
d) \(\frac{1}{1-4z}\), |Z|<\(\frac{1}{4}\)
View Answer

Answer: d
Explanation: X (z) = \(∑_{n=-∞}^0 (\frac{1}{4} z^{-1})^n\)
= \(∑_{n=0}^∞ (4z)^n \)
= \(\frac{1}{1-4z}\), |Z|<\(\frac{1}{4}\).
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6. The z-transform of 3n u[-n-1] is ___________
a) \(\frac{z}{3-z}\), |Z|>3
b) \(\frac{z}{3-z}\), |Z|<3
c) \(\frac{3}{3-z}\), |Z|>3
d) \(\frac{3}{3-z}\), |Z|<3
View Answer

Answer: b
Explanation: X (z) = \(∑_{n=-∞}^{-1} (3z^{-1})^n \)
= \(∑_{n=1}^∞ (z \frac{1}{3})^n \)
= \(\frac{\frac{1}{3} z}{1-\frac{1}{3}z}\), |z|<3
= \(\frac{z}{3-z}\).

7. The z-transform of \((\frac{2}{3})^{[n]}\) is ____________
a) \(\frac{-5z}{(2z-3)(3z-2)}\), –\(\frac{3}{2} \) < z < –\(\frac{2}{3}\)
b) \(\frac{-5z}{(2z-3)(3z-2)}\), \(\frac{2}{3}\) < |z| < \(\frac{3}{2} \)
c) \(\frac{5z}{(2z-3)(3z-2)}\), \(\frac{2}{3}\) < |z|
d) \(\frac{5z}{(2z-3)(3z-2)}\), –\(\frac{3}{2} \) < z< –\(\frac{2}{3}\)
View Answer

Answer: b
Explanation: X(z) = \(∑_{-∞}^{-1} (\frac{3}{2} z^{-1})^n + ∑_{n=0}^∞ (\frac{2}{3} z^{-1})^n \)
= \(\frac{-1}{(1-\frac{3}{2} z^{-1})} + \frac{1}{(1-\frac{2}{3} z^{-1})}\)
= \(\frac{-5z}{(2z-3)(3z-2)}\), \(\frac{2}{3}\) < |z| < \(\frac{3}{2} \).

8. The z-transform of cos(\(\frac{π}{3}\) n) u[n] is __________
a) \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), 0<|z|<1
b) \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1
c) \(\frac{z}{2} \frac{(1-2z)}{(z^2-z+1)}\), 0<|z|<1
d) \(\frac{z}{2} \frac{(1-2z)}{(z^2-z+1)}\), |z|>1
View Answer

Answer: b
Explanation: Performing z-transform on anu[n], we get \(\frac{z}{z-a}\)
∴ x[n] = \(\frac{1}{2} e^{j(\frac{π}{3})n} u[n] + \frac{1}{2} e^{-j(\frac{π}{3})n}u [n]\)
Hence, X (z) =\( 0.5 \left(\frac{1}{1-e^{j(\frac{π}{3})} z^{-1}} + \frac{1}{1-e^{j(\frac{π}{3})} z^{-1}}\right)\)
Hence, X (z) = \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1.

9. The z-transform of {3,0,0,0,0,6,1,-4} (1 as the reference variable) is ___________
a) 3z5 + 6 + z-1 – 4z-2, 0≤|z|<∞
b) 3z5 + 6 + z-1 – 4z-2, 0<|z|<∞
c) 3z5 + 6 + z – 4z-2 0<|z|<∞
d) 3z5 + 6 + z-1 – 4z-2, 0≤|z|<∞
View Answer

Answer: b
Explanation: Performing z-transform on x (n+n0), we get zn0 X (z)
Now, x[n] = 3δ[n+5] + 6δ[n] + δ[n-1] – 4δ[n-2]
So, X (z) = 3z5 + 6 + z-1 – 4z-2, 0<|z|<∞.
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10. The z-transform of x[n]= {2,4,5,7,0,1} (5 as the reference variable) is ___________
a) 2z2 + 4z + 5 +7z + z3, z≠∞
b) 2z-2 + 4z-1 + 5 + 7z + z3, z≠∞
c) 2z-2 + 4z-1 + 5 + 7z + z3, 0<|z|<∞
d) 2z2 + 4z + 5 + 7z-1 + z3, 0<|z|<∞
View Answer

Answer: d
Explanation: Performing z-transform on x (n+n0), we get zn0 X (z)
Now, x[n] = 2δ[n+2] + 4δ[n+1] + 5δ[n] + 7δ[n-1] + δ[n-3]
So, X (z) = 2z2 + 4z + 5 + 7z-1 + z3, 0<|z|<∞.

11. The z-transform of x[n]= {1,0,-1,0,1,-1} (1st 1 as the reference variable) is __________
a) 1+2z-2 -4 z-4 + 5z-5
b) 1-z-2 + z-4 – z-5
c) 1-2z2 + 4z4 – 5z5
d) 1-z2 + z4 – z5
View Answer

Answer: b
Explanation: Performing z-transform on x (n-n0), we get z-n0 X (z)
Now, x[n] = δ[n] – δ[n-2] + δ[n-4] – δ[n-5]
So, X (z) = 1-z-2 + z-4 – z-5, z≠0.

12. Given the z-transform pair
\(X[n] \leftrightarrow \frac{32}{z^2-16}\), |z|<4
The z-transform of the signal x [n-2] is _________
a) \(\frac{z^4}{z^2-16}\)
b) \(\frac{(z+2)^2}{(z+2)^2-16}\)
c) \(\frac{1}{z^2-16}\)
d) \(\frac{(z-2)^2}{(z-2)^2-16}\)
View Answer

Answer: c
Explanation: Performing z-transform on x (n-n0), we get zn0 X (z)
Now, z-transform of y[n] = x [n-2] is given by,
Y (z) = z-2 X (z)
= \(\frac{1}{z^2-16}\).

13. Given the z-transform pair
\(X[n] \leftrightarrow \frac{32}{z^2-16}\), |z|<4
The z-transform of the signal y[n] = \(\frac{1}{2^n}\) x[n] is _________
a) \(\frac{(z+2)^2}{(z+2)^2-16}\)
b) \(\frac{z^2}{z^2-4}\)
c) \(\frac{(z-2)^2}{(z-2)^2-16}\)
d) \(\frac{z^2}{z^2-64}\)
View Answer

Answer: b
Explanation: y[n] = \(\frac{1}{2^n}\) x[n] Performing z-transform on y[n], we get, Y (z) = X (2z)
∴ Y(z) = \(\frac{z^2}{z^2-4}\).
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14. Given the z-transform pair
\(X[n] \leftrightarrow \frac{32}{z^2-16}\), |z|<4
The z-transform of the signal x [-n]*x[n] is ____________
a) \(\frac{z^2}{16z^2-257z^4-16}\)
b) \(\frac{-16z^2}{(z^2-16)^2}\)
c) \(\frac{z^2}{(257z^2-16z^4-16)}\)
d) \(\frac{16z^2}{(z^2-16)^2}\)
View Answer

Answer: c
Explanation: y[n] = x [-n]*x[n] Performing z-transform on y[n], we get, Y (z) = X (\(\frac{1}{z}\)) X (z)
∴ X(\(\frac{1}{z}\)) ↔ x [-n].

15. Given the z-transform pair
\(X[n] \leftrightarrow \frac{32}{z^2-16}\), |z|<4
The z-transform of the signal x[n]*x [n-3] is __________
a) \(\frac{z^{-3}}{(z^2-16)^2}\)
b) \(\frac{z^7}{(z^2-16)^2}\)
c) \(\frac{z^5}{(z^2-16)^2}\)
d) \(\frac{z}{(z^2-16)^2}\)
View Answer

Answer: d
Explanation: y[n] = x[n]*x [n-3] Performing z-transform on y[n], we get, Y (z) = X (z) z-3X (z)
Or, Y (z) = \(\frac{z}{(z^2-16)^2}\).

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn