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
Explanation: Performing z-transform on δ[n-k], we get,
X (z) = \(∑_{n=0}^∞ x[n]z^{-n}\)
= Z-k, Z≠0.
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
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
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
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
Explanation: X (z) = \(∑_{n=-∞}^0 (\frac{1}{4} z^{-1})^n\)
= \(∑_{n=0}^∞ (4z)^n \)
= \(\frac{1}{1-4z}\), |Z|<\(\frac{1}{4}\).
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
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
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
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
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|<∞.
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
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
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
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
Explanation: y[n] = \(\frac{1}{2^n}\) x[n] Performing z-transform on y[n], we get, Y (z) = X (2z)
∴ Y(z) = \(\frac{z^2}{z^2-4}\).
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
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
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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