Question

The causal signal with z-transform z² (z - a)^-2 is

(u[n] is the unit step signal)

1. (n + 1) aⁿ u[n]
2. a²ⁿ u[n]
3. n^-1 aⁿ u[n]
4. n² aⁿ u[n]

Answer 1

Correct Answer - Option 1 : (n + 1) aⁿ u[n]

Concept:

Causal Signals:

• Causal Signals are signals that are zero for all negative time.
• Causality in a system determines whether a system relies on future information of a signal x [n+1].
• Talking about “causality” in signals, it means whether they are zero to the left of t = 0 or zero to the right of t = 0.
• A causal signal is zero for t < 0.
• A system is said to be causal if its output depends upon present and past inputs, and does not depend upon future input. 

 

Calculation:

Given a casual signal,

X (Z) = z2 (z - a)-2

\(X(Z) = \frac{{{z^2}}}{{{{\left( {a - z} \right)}^2}}}\)

From standard Z-Transform,

\(n{\left( a \right)^n}u\left( n \right) \leftrightarrow \frac{{az}}{{{{\left( {z - a} \right)}^2}}}\)

\(n{\left( a \right)^{n - 1}}u\left( n \right) \leftrightarrow \frac{z}{{{{\left( {z - a} \right)}^2}}}\)

From the time-shifting property,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

\(\left( {n + 1} \right){\left( a \right)^{n + 1 - 1}}u\left( {n + 1} \right) \leftrightarrow z\left[ {\frac{z}{{{{\left( {z - a} \right)}^2}}}} \right]\)

\(\left( {n + 1} \right){\left( a \right)^n}u\left( {n + 1} \right) \leftrightarrow \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}\)

Since,

\(X\left( z \right) = \frac{{{z^2}}}{{{{\left( {z - a} \right)}^2}}}\)

Therefore,

\(x\left( n \right) = \left( {n + 1} \right){\left( a \right)^n}u\left( {n + 1} \right)\)

The given signal is causal, therefore,

\(x\left( n \right) = \left( {n + 1} \right){\left( a \right)^n}u\left( n \right)\)

Z-Transform basic functions:

Sequence	Z-Transform
δ (n)	1
u (n)	\(\frac{z}{{z - 1}}\)
an	\(\frac{z}{{z - a}}\)
n (an) u (n)	\(\frac{{az}}{{{{\left( {a - z} \right)}^2}}}\)
n (an - 1) u (n)	\(\frac{z}{{{{\left( {a - z} \right)}^2}}}\)
n (an - 1) u (n - 1)	\(\frac{1}{{z - a}}\)