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

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

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

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

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

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|<$$\frac{1}{4}$$

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

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< –$$\frac{2}{3}$$

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

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|<∞

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|<∞

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

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) $$\frac{(z-2)^2}{(z-2)^2-16}$$

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) $$\frac{z^2}{z^2-64}$$

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) $$\frac{16z^2}{(z^2-16)^2}$$

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) $$\frac{z}{(z^2-16)^2}$$

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) = $$\frac{z}{(z^2-16)^2}$$.

Sanfoundry Global Education & Learning Series – Signals & Systems.

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