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I'm trying to find the sequence ${a_n}$ given the Z-transform $A(z)=1/z$

I'm not sure though how to calculate the inverse transform of $A(z)$, All help is much appreciated

wertyle
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  • See whether this helps you: http://math.stackexchange.com/questions/1224797/how-can-i-find-the-inverse-z-transform-of-1-z-a –  Jan 06 '17 at 13:56

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The Z-transform of a sequence $a_n$ is defined as $A(z)=\sum_{n=-\infty}^{\infty} a_n z^{-n}$.

In your case, $A(z)=1/z=z^{-1}$, so this must mean $a_n=0$ for all $n\neq 1$, and $a_1=1$. We don't need any fancy computations in this example, we just read off the one nonzero coefficient directly from $A$.

MonadBoy
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  • ok I get it now, thank you – wertyle Jan 06 '17 at 14:36
  • Does this mean that $\mathscr{X}^{-1}{\frac{1}{x}}=\delta_n$? Wolframalpha concludes it is $\delta(n-1)$ – Luthier415Hz Mar 08 '22 at 11:57
  • @Luthier415Hz Assuming that $\delta$ denotes the Kronecker delta (i.e. $\delta(x) = 1$ for $x=0$ and $\delta(x) = 0$ for $x\neq 0$), WolframAlpha is correct, since here $a_n = 1$ for $n=1$, and $a_n = 0$ for all other $n$. Inserting $n=1$ just to be sure, $\delta(n-1) = \delta(0) = 1$. – MonadBoy Apr 17 '22 at 07:56