Question:
The causal signal with z-transform $z^{2}(z-a)^{-2}$ is ($u[n]$ is the unit step signal)
The causal signal with z-transform $z^{2}(z-a)^{-2}$ is ($u[n]$ is the unit step signal)
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Remember: $(n+1)a^n u[n]$ always corresponds to $z^{2}/(z-a)^2$ in the Z-transform table.
Updated On: Dec 29, 2025
- $a^{2n}u[n]$
- $(n+1)a^{n}u[n]$
- $n^{-1} a^{n} u[n]$
- $n^{2} a^{n} u[n]$
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The Correct Option is B
Solution and Explanation
We start from the standard Z-transform identity:
\[
\mathcal{Z}\{ (n+1)a^{n}u[n] \} = \frac{z^{2}}{(z-a)^{2}}
\]
Given the z-transform in the question is \[ z^{2}(z-a)^{-2}, \] it exactly matches the Z-transform of the sequence \[ (n+1)a^{n}u[n]. \]
Thus, the corresponding causal time-domain signal is $(n+1)a^{n}u[n]$.
Final Answer: $(n+1)a^{n}u[n]$
Given the z-transform in the question is \[ z^{2}(z-a)^{-2}, \] it exactly matches the Z-transform of the sequence \[ (n+1)a^{n}u[n]. \]
Thus, the corresponding causal time-domain signal is $(n+1)a^{n}u[n]$.
Final Answer: $(n+1)a^{n}u[n]$
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