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Using the time shifting, time reversal, and scaling, I want to derive the form of the Z Transform of

$$x[n]=-a^n u[-(n+1)]$$

$u[n]$ is the discrete-time unit step function:

$$ u[n] \triangleq \begin{cases} 0 \qquad & n < 0 \\ 1 \qquad & n \ge 0 \\ \end{cases}$$

(disregarding the ROC)

given the Z transform of the unit step:

$$ U(z) = \mathcal{Z} \Big\{u[n] \Big\} = \frac{1}{1-z^{-1}} $$


Let $X(z)$ be informally, the expected form of the Z Transform that $x[n]$ will be taking:

Using the time reversal property

$$ X(z) = U(z^{-1}) $$

Applying the time shifting property:

$$ X(z) = z^{1}U(z^{-1}) $$

Then, by scaling

$$ X(z) = zU(a^{-1}z^{-1}) $$

Returning to the Z Transform of $x[n]$

$$\begin{align} \mathcal{Z} \Big\{-a^nu[-(n+1)] \Big\} &= -zU(a^{-1}z^{-1}) \\ & \Longrightarrow \frac{-z}{1-az} \\ \end{align}$$

which is not the correct answer

$$ \frac{1}{1-az^{-1}} $$

Please note that I momentarily disregarded the concern for ROC in order to know if I can stack the three properties together to solve for the form of the Z Transform (similar to how I apply multiple properties of the Laplace and Fourier Transform Simultaneously)

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    $\begingroup$ When you apply scaling, $a^n x[n] \leftrightarrow X(z/a)$, but you didn't include the $a^{-1}$ outside G. $\endgroup$
    – Juancho
    Oct 8, 2021 at 15:50
  • $\begingroup$ @Juancho I'm sorry but I didn't exactly got what you meant by $a^{-1}$ outside $G(z)$. Are you referring to the $a^n$ of $f(n)$? It vanished after applying the scaling property $\endgroup$
    – wd violet
    Oct 8, 2021 at 16:29
  • $\begingroup$ they are actually the same answer. except for a minus sign. $\endgroup$ Oct 8, 2021 at 18:14
  • $\begingroup$ @robertbristow-johnson look again -- his answer has a pole at $z = 1/a$, the "correct" answer has a pole at $z = a$. $\endgroup$
    – TimWescott
    Oct 9, 2021 at 2:53
  • $\begingroup$ Thank You for all your help. I finally understood what you meant by $a^{-1}$ outside $G(z)$ $\endgroup$
    – wd violet
    Oct 9, 2021 at 3:41

1 Answer 1

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I propose the following answer for $$ X(z)=\mathcal{Z} \Big\{-a^nu[-(n+1)] \Big\} $$ given $$ U(z) = \mathcal{Z} \Big\{u[n] \Big\} = \frac{1}{1-z^{-1}} $$


Let

$$ \{X_1, X_2, X_3\}(z) $$ denote the transformation stages under which each property applies

Using the time reversal property

$$ X_1(z) = U(z^{-1}) $$

Applying the time shifting property:

$$ X_2(z) =z^1X_1= z^{1}U(z^{-1}) $$

Then, by scaling

$$ X_3(z) = X_2(a^{-1}z)=a^{-1}zU(az^{-1}) $$

Returning to the Z Transform of $x[n]$

$$\begin{align} \mathcal{Z} \Big\{-a^nu[-(n+1)] \Big\} &= -X_3(z) \\ & \Longrightarrow \frac{-za^{-1}}{1-a^{-1}z} \\ \end{align}$$

which after simplification finally agrees with the correct answer

$$ \frac{1}{1-az^{-1}} $$


It loosely concludes that I can use multiple properties of the Z Transform as long that each transformation stages is consistent (e.g. $X_1, X_2, X_3, ...$).

It is now left to determine if the properties also applies simultaneously with the ROC like how the form of the Z Transform did.

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