# Signals & Systems Questions and Answers – Inverse Z-Transform

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Inverse Z-Transform”.

1. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to X(2z) is ___________
a) n23n u[2n]
b) $$(-\frac{3}{2})^n n^2 u[n]$$
c) $$(\frac{3}{2})^n n^2 u[n]$$
d) 6nn2u[n]

Explanation: Y (z) = X (2z) ↔ y[n] = $$\frac{1}{2^n}$$ x[n]
Or, y[n] = $$\frac{1}{2^n}$$ n2 3n u[n]
Or, y[n] = $$(\frac{3}{2})^n n^2 u[n]$$.

2. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to X(z-1) is ___________
a) n23-nu[n]
b) n23-nu[-n]
c) $$\frac{1}{n^2} 3^{\frac{1}{n}} u[n]$$
d) $$\frac{1}{n^2} 3^{\frac{1}{n}} u[-n]$$

Explanation: Y (z) = X ($$\frac{1}{z}$$) ↔ y[n] = X [-n]
Or, y[n] = n23-nu[-n].

3. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to $$\frac{dX(z)}{dz}$$ is ___________
a) (n-1)33n-1u[n-1]
b) n33nu[n-1]
c) (1-n)33n-1u[n-1]
d) (n-1)33n-1u[n]

Explanation: Y (z) = $$\frac{dX(z)}{dz}$$
= $$-z^{-1}\Big[-z\frac{dX(z)}{dz}\Big]$$
Now, Y (z) ↔ y (n) = – (n-1) X [-n-1] Or, y (n) = – (n-1)33n-1u[n-1].

4. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to $$\frac{z^2-z^{-2}}{2}$$ X(z) is ___________
a) $$\frac{1}{2}$$(x[n+2]-x[n-2])
b) (x[n+2]-x[n-2])
c) $$\frac{1}{2}$$(x[n-2]-x[n+2])
d) (x[n-2]-x[n+2])

Explanation: Y (z) = $$\frac{z^2-z^{-2}}{2}$$ X (z) ↔ y[n] = $$\frac{1}{2}$$(x[n+2]-x[n-2]).

5. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to {X(z)}2 is ___________
a) {x[n]}2
b) x[n]*x[n]
c) x[n]*x[-n]
d) x[-n]*x[-n]

Explanation: Y (z) = X (z) H (z)
Y (z) = X (z) X (z) ↔ y[n] Or, y [n] = x[n]*x[n]
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?
a) 2
b) 1
c) 0
d) -1

Explanation: Given equation = y (n) – 2y (n-1) + y (n-2) = X (n) – X (n-1) has y (n) = 0
For n = 0, y (0)2y (-1) + y (-2) = x (0) – x (-1)
∴ y(0) = x(0) – x(-1)
Or, y (n) = 0 for n<0
For n=1, y (1) = -2y (0) + y (-1) = x (1) – x (0)
Or, y (1) = x (1) – x (0) + 2x (0) – 2x (-1)
Or, y (1) = x (1) + x (0) – 2x (-1)
For n=2, y (2) = x (2) – x (1) + 2y (1) – y (0)
Or, y(2) = x(2) – x(1) + 2x(1) + 2x(0) – 4x(-1) – x(0) + x(-1)
∴y (2) = d (2) + d (1) + d (0) – 3d (-1).

7. The value of $$Z^{-1}\Big\{\frac{z^2}{(z-a)(z-b)}\Big\}$$ is ____________
a) $$\frac{a^{n+1} – b^{n+1}}{a+b}$$
b) $$\frac{a^{n+1} – b^{n+1}}{a-b}$$
c) $$\frac{a^{n+1} + b^{n+1}}{a-b}$$
d) $$\frac{a^{n+1} + b^{n+1}}{a+b}$$

Explanation: We know that, $$Z^{-1}{\frac{z}{z-a}} = a^n$$ and$$Z^{-1}{\frac{z}{z-b}} = b^n$$
∴$$Z^{-1}\Big\{\frac{z^2}{(z-a)(z-b)}\Big\} = Z^{-1}{\frac{z}{z-a} . \frac{z}{z-b}} = a^n * b^n$$
= $$∑_{m=0}^n a^m.b^{n-m}$$
= $$b^n ∑_{m=0}^n \frac{a^m}{b}$$
= $$b^n . \frac{\frac{a^{n+1}}{b}-1}{\frac{a}{b}-1}$$
= $$\frac{a^{n+1} – b^{n+1}}{a-b}$$.

8. Given the z-transform pair
$$X[n] \leftrightarrow \frac{32}{z^2-16}$$, |z|<4
The z-transform of the signal y[n] = nx[n] is _________
a) $$\frac{32z^2}{(z^2-16)^2}$$
b) $$\frac{-32z^2}{(z^2-16)^2}$$
c) $$\frac{32}{(z^2-16)^2}$$
d) $$\frac{-32z}{(z^2-16)^2}$$

Explanation: y[n] = n x[n] n x[n] ↔ Y (z) = $$-z\frac{dX(z)}{dz}$$
Now, $$-z\frac{dX(z)}{dz} = \frac{32z^2}{(z^2-16)^2}$$.

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

Explanation: x (n-n0) ↔ z-n0 X (z)
Now, y[n] = x [n+1] + x [n-1] ↔ Y (z)
Y (z) = (z+z-1) X (z)
∴ Y (z) = $$\frac{z^2 (1+z)}{z^2-16}$$.

10. The value of inverse Z-transform of log($$\frac{z}{z+1}$$) is _______________
a) (-1)n/n for n = 0; 0 otherwise
b) (-1)n/n
c) 0, for n = 0; (-1)n/n, otherwise
d) 0

Explanation: Putting z = $$\frac{1}{t}$$, U (z) = log $$\left(\frac{\frac{1}{y}}{\frac{1}{y}+1}\right)$$
= – log (1+y) = -y + $$\frac{1}{2}$$ y2 – $$\frac{1}{3}$$ y3 + …..
= -z-1 + $$\frac{1}{2}$$ z-2 – $$\frac{1}{3}$$ z-3 + …..
Thus, un = 0, for n = 0; (-1)n/n otherwise.

11. The inverse Z-transform of z/(z+1)2 is ______________
a) (-1)n+1
b) (-1)n-1 n
c) (-1)n-1
d) (-1)n+1 n

Explanation: U (z) = $$\frac{z}{z^2+2z+1}$$
= $$z^{-1} – \frac{2+z^{-1}}{z^2+2z+1}$$
= $$z^{-1} – 2z^{-2} + \frac{2z^{-2}+3z^{-1}}{z^2+2z+1}$$
= $$z^{-1} – 2z^{-2} + 3z^{-3} – \frac{4z^{-2}+3z^{-3}}{z^2+2z+1}$$
So, U (z) = $$∑_{n=0}^∞ (-1)^{n-1} nz^{-n}$$
Hence, un = (-1)n-1 n.

Sanfoundry Global Education & Learning Series – Signals & Systems.

To practice all areas of Signals & Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.