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Examples from studysmarter said

$$F(z) = \frac{5} {z-2} - \frac{5} {z-3} $$

Inverse z transform

$$f(n) = 5 \times \Bigl(2^n - 3^n \Bigl) $$

And Google gemini and ChatGPT 3.5 also agrees with studysmarter.

However, https://www.wolframalpha.com/ will give the following

$$f(n) = 5 \times \Bigl(2^{n-1} - 3^{n-1} \Bigl) \times u[n] $$

However, according to formula table.

$$f(n) = 5 \times \Bigl(2^{n-1} - 3^{n-1} \Bigl) \times u[n-1] $$

Which is the right one?

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    \$\begingroup\$ Note that LLMs are not capable of solving these kind of problems, they just represent the most likely answer based on their training, which could be studysmarter. Wolfram Alpha on the other hand uses a very different approach and is build upon Mathematica, which has solvers for this. \$\endgroup\$
    – Arsenal
    Commented Apr 22 at 11:41

3 Answers 3

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Technically, you cannot really give a definite answer to that question, because the ROC (region of convergence) is not specified. The z transform

$$ F(z) = \frac{1}{1-z^{-1}} $$

can have two possible time domain signals depending on the ROC:

$$ f[n] = \mathcal Z^{-1}\left\lbrace F(z)\right\rbrace[n]= \begin{cases}u[n] & |z| > 1 \\ -u[-n-1] & |z| < 1\end{cases} $$ For a definite answer you have to specify the region of convergence of your z transform. Otherwise your z Transform can have multiple valid time domain signals. With two poles, you can theoretically have 3 valid regions of convergence. Thus, 3 time domain signals. Your poles are at 2 and 3. So you can get ROCs at

\$|z| < 2\$,

\$ 2 < |z| < 3\$,

and \$|z| > 3\$

For the inverse z transform of your signal, I'm assuming you are searching for a causal signal (a signal defined for \$n \ge 0\$). Therefore I'm assuming the ROC \$ |z| > 3 \$

First, I'm going to multiply your fractions with \$\frac{z^{-1}}{z^{-1}}\$ to make all exponents negative, to ease lookup in a transformation table. It is valid, as the ROC we're looking at is defined for z > 3 and the fraction works for these numbers. Multiplying the the fraction essentially adds poles at \$z=0\$ which is not a problem as our ROC is \$|z| > 3\$

$$ F(z) = 5\cdot \left( \frac{z^{-1}}{1-2z^{-1}} - \frac{z^{-1}}{1-3z^{-1}} \right) = 5\cdot z^{-1}\left( \frac{1}{1-2z^{-1}} - \frac{1}{1-3z^{-1}} \right) $$

The table on wikipedia lists: $$ a^nu[n] $$

transformed as $$ \frac{1}{1-az^{-1}}, |z| > |a| $$ Your two components are basically this signal but one time step delayed (multiplied with \$z^{-1}\$). Therefore:

$$ f[n] = 5 u[n-1] \left( 2^{n-1} - 3^{n-1} \right) $$ Remember the assumption of the ROC.

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  • \$\begingroup\$ Can you write z- transform in other regions as well? \$\endgroup\$
    – kile
    Commented Apr 24 at 10:01
  • \$\begingroup\$ What's the transform \$z^{-1}\$? Do you have to multiply \$\delta [ n-1]\$ for that? \$\endgroup\$
    – kile
    Commented Apr 24 at 11:30
  • \$\begingroup\$ @kile correct. The time domain signal of \$ z^{-n_0} \$ is \$ \delta [n-n_0]\$. However, The term is multiplied in the Z-space. Multiplication in the frequency space of the z-Transform equals convolution in the time domain. So you have to calculate the convolution with the dirac pulse which is located at n = 1. Either you calculate it manually or you already know, that a convolution with a dirac pulse that is shiftet results in the same but equally shifted signal. Therefore everything in my signal f is shifted by one cycle. \$\endgroup\$
    – GNA
    Commented Apr 25 at 7:59
  • \$\begingroup\$ Could you please explain in details how you use convolution for this time domain? I have no idea how you derive this $$ {\cal{Z}}^{-1}\left\{5\left(\frac{1}{z-2}-\frac{1}{z-3}\right)\right\} = 5 u[n-1]\left(2^{n-1} - 3^{n-1}\right) $$ \$\endgroup\$
    – kile
    Commented Apr 25 at 8:53
  • \$\begingroup\$ The z transform is equivalent to $$ 5 z^{-1}\left( \frac{1}{1-2z^{-1}} - \frac{1}{1-3z^{-1}} \right)$$ simply by extending the fraction with \$z^{-1}\$. The terms in the parentheses can be looked up in a table (e.g. on wikipedia). The whole z transform is multiplied by 5. So the time domain signal is multiplied by 5 as well. Last, the term in the parentheses is multiplied with \$z^{-1}\$ This is equivalent to shifting it by 1 time step. \$ \mathcal{Z}\left\lbrace x[n-n_0] \right\rbrace = z^{-n_0} X(z) \$ is a basic property of the z transform. \$\endgroup\$
    – GNA
    Commented Apr 26 at 11:56
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The ambiguity likely resides in the lack of defining of the region of convergence (ROC). For z approaching infinity (i.e. no ROC), the first answer seems to be correct. If ROC is causal, meaning x[n] contains u[n], then the second answer is correct. Lastly, if ROC is noncausal, meaning x[n] contains u[±n±k], then the last answer is correct.

Edit: @GNA has a good mathematical explanation of this.

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Before the calculation, I remember the two methods for multiple and simple poles:

enter image description here

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  • \$\begingroup\$ I don't quite get your math. You list lim x->0 X(z) as the final value theorem. But instead lim z->1 (z-1)*X(z) would be the limit theorem. However, this is only valid, if there are no poles outside the unit circle. Your lim of z->0 gives you a value of -5/6? Clearly the series (2^n - 3^n) does not converge to a final limit. \$\endgroup\$
    – GNA
    Commented Apr 23 at 7:44
  • \$\begingroup\$ @GNA Thanks for pointing out the error to me. I will correct it immediately. It must be: lim(x(k)) for k->∞ = lim(((z-1)/z)x(z)) for z->1 valid only if the final value exsists, as demonstrated by R Isermann in Digital Control Systems, pag 37. \$\endgroup\$
    – Franc
    Commented Apr 23 at 9:00

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