# Signals & Systems Questions and Answers – Properties of Z-Transforms – 2

This set of Tough Signals & Systems Questions and Answers focuses on “Properties of Z-Transforms – 2”.

1.Find the Z-transform of the causal sequence x(n) = {1,0,-2,3,5,4}. (1 as the reference variable)
a) 1 – 2z-2 + 3z-3 + 5z-4 + 4z-5
b) 1 – 2z2 + 3z3 + 5z4 + 4z5
c) z-1 – 2z2 + 3z3 + 5z4 + 4z5
d) z – 2z3 + 3z4 + 5z5 + 4z6

Explanation: Given sequence values are :
x(0)=1, x(1)=0, x(2)=-2, x(3)=3, x(4)=5, x(5)=4.
We know that
$$X(Z) = \sum\limits_{n=-∞}^{∞} x(n) z^{-n}$$
X(Z) = x(0) + x(1) z-1 + x(2) z-2 + x(3) z-3 + x(4) z-4 + x(5) z-5
X(Z) = 1 – 2z-2 + 3z-3 + 5z-4 + 4z-5.

2. Find the Z-transform of the anticausal sequence x(n) = {4,2,3,-1,-2,1}. (1 as the reference variable)
a) 4z5 + 2z4 + 3z3 – z2 – 2z + 1
b) 4z-5 + 2z-4 + 3z-3 -z-2 – 2z-1 + 1
c) -4z5 – 2z4 – 3z3 + z2 + 2z – 1
d) -4z-5 – 2z-4 – 3z-3 + z-2 + 2z-1 – 1

Explanation: Given sequence values are :
x(-5)=4, x(-4)=2, x(-3)=3, x(-2)=-1, x(-1)=-2, x(0)=1
We know that
$$X(Z) = \sum\limits_{n=-∞}^{∞} x(n) z^{-n}$$
X(Z) = x(-5) z5 + x(-4) z4 + x(-3) z3 + x(-2) z2 + x(-1)z + x(0)
X(Z) = 4z5 + 2z4 + 3z3 – z2 – 2z + 1.

3. Find the Z-transform of x(n) = u(-n).
a) $$\frac{1}{z-1}$$
b) $$\frac{1}{z+1}$$
c) $$\frac{1}{1-z}$$
d) $$\frac{-1}{z+1}$$

Explanation: Given x(n) = u(-n)
Time reversal property of Z-transform states that
If x(n) ↔ X(z), then x(-n) ↔ X($$\frac{1}{z}$$)
Z[u(-n)] = $$\left(\frac{1}{z-1}\right)_{z=(1/z)} = \frac{1/z}{(1/z)-1} = \frac{1}{1-z}$$.

4. Find the Z-transform of x(n) = u(-n-2).
a) $$\frac{z^2}{z-1}$$
b) $$\frac{z^2}{1-z}$$
c) $$\frac{z^2}{1+z}$$
d) $$\frac{z^2}{2-z}$$

Explanation: Given x(n) = u(-n-2)
Time shifting property of Z-transform states that
If x(n) ↔ X(z), then x(n-m) ↔ z-m X(z)
Z[u(-n-2)] = Z{u[-(n+2)]}=z2 Z[u(-n)] = $$\frac{z^2}{1-z}$$.

5. Find the Z-transform of x(n) = n2 u(n).
a) $$\frac{z(z-1)}{(z-1)^3}$$
b) $$\frac{z(z+1)}{(z-1)^3}$$
c) $$\frac{z(z+1)}{(z+1)^3}$$
d) $$\frac{z(z-1)}{(z+1)^3}$$

Explanation: Given x(n) = n2 u(n)
We know that X(z) = Z[x(n)] = Z[u(n)] = $$\frac{z}{1-z}$$
The multiplication of n or differentiation in z-domain property of Z-transform states that
If x(n) ↔ X(z), then nk x(n) ↔ (-1)k zk $$\frac{d^k X(z)}{dz^k}$$
Z[n2 u(n)] = z2 $$\frac{d^2 X(z)}{dz^2} = z^2 \frac{d^2}{dz^2}[\frac{z}{1-z}] = \frac{z(z+1)}{(z-1)^3}$$.

6. Find the Z-transform of x(n) = 2n u(n-2).
a) $$\frac{z}{z-2}$$
b) $$\frac{z}{z+2}$$
c) $$\frac{z}{z(z-2)}$$
d) $$\frac{4}{z(z-2)}$$

Explanation: Given x(n) = 2n u(n-2)
Time shifting property of Z-transform states that
If x(n) ↔ X(z), then x(n-m) ↔ z-m X(z)
Z[u(n-2)] = z-2 Z[u(n)] = $$z^{-2} \frac{z}{z-1} = \frac{1}{z(z-1)}$$
The multiplication by an exponential sequence property of Z-transform states that
If x(n) ↔ X(z), then an x(n) ↔ X(z/a)
Z[2n u(n-2)] = Z[u(n-2)]|z=(z/2) = $$\Big[\frac{1}{z(z-1)}\Big]_{z=(z/2)}$$
$$= \frac{1}{(z/2)[(z/2)-1]} = \frac{4}{z(z-2)}$$.

7. Find the Z-transform of x(n) = n[an u(n)].
a) $$\frac{z}{z(z-a)}$$
b) $$\frac{az}{z(z-a)}$$
c) $$\frac{az}{z(z+a)}$$
d) $$\frac{a}{z(z-a)^2}$$

Explanation: Given x(n) = n[an u(n)]
We know that an u(n) ↔ $$\frac{z}{z-a}$$
Time differentiation property states that
If x(n) ↔ X(z), then nx(n) ↔ -z $$\frac{dX(z)}{dz}$$
Z[x(n)] = Z{n[an u(n)]} = -z $$\frac{dX(z)}{dz} = -z \frac{d}{dz} \Big[\frac{z}{z-a}\Big] = \frac{a}{z(z-a)^2}$$.

8. Find the Z-transform of x(n) = ($$\frac{1}{2}$$)n u(n)*($$\frac{1}{4}$$)n u(n).
a) $$\frac{z}{z-(1/2)} \frac{z}{z-(1/4)}$$
b) $$\frac{z}{z-(1/2)} + \frac{z}{z-(1/4)}$$
c) $$\frac{z}{z+(1/2)} * \frac{z}{z-(1/4)}$$
d) $$\frac{z}{z-(1/2)} – \frac{z}{z+(1/4)}$$

Explanation: We know that an u(n) ↔ $$\frac{z}{z-a}$$
Let x1 (n)=($$\frac{1}{2}$$)n u(n) and x2 (n) = ($$\frac{1}{4}$$)n u(n)
∴X1 (z) = $$\frac{z}{z-(1/2)}$$ and X2 (z) = $$\frac{z}{z-(1/4)}$$
Given x(n) = x1 (n) * x2 (n)
The convolution property of Z-transform states that
x1 (n) * x2 (n) ↔ X1 (z) X2 (z)
∴Z[x(n)] = X(z) = Z[x1 (n)*x2 (n)] = X1 (z) X2 (z) = $$\frac{z}{z-(1/2)} \frac{z}{z-(1/4)}$$.

9. Find x(∞) if X(z) = $$\frac{Z+1}{(z-0.6)^2}$$.
a) 1
b) 0
c) ∞
d) 0.6

Explanation: Given X(z) = $$\frac{Z+1}{(z-0.6)^2}$$
The final value theorem of Z-transform states that
If x(n) ↔ X(z), then x(∞) = Ltz→1⁡ (z-1)X(z)
x(∞) = Ltz→1 (z-1)X(z) = Ltz→1 (z-1) $$\frac{Z+1}{(z-0.6)^2}$$ = 0.

10. Find x(∞) if X(z) = $$\frac{z+3}{(z+1)(z+2)}$$.
a) ∞
b) -1
c) 1
d) 0

Explanation: Given X(z) = $$\frac{z+3}{(z+1)(z+2)} = \frac{z[1+(3/z)]}{z^2 [1+(1/z)][1+(2/z)]} = \frac{1}{z} \frac{1+(3/z)}{[1+(1/z)][1+(2/z)]}$$
The initial value theorem of Z-transform states that
If x(n) ↔ X(z), then x(0) = Ltz→∞⁡ X(z)
x(0) = Ltz→∞⁡ X(z) = $$Lt_{z→∞} \frac{1}{z} \frac{[1+(3/z)]}{[1+(1/z)][1+(2/z)]}$$ = 0.

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