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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

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

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  • $\begingroup$ Have you considered solving the problem yourself to see who is right? $\endgroup$
    – Matt L.
    Apr 22 at 8:41
  • $\begingroup$ @MattL. Yes. I personally agree with wolfram. But I am new to z-transform. I want confirmation from senior people. $\endgroup$
    – kile
    Apr 22 at 8:49
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    $\begingroup$ The important part is how you arrive at your answer, not how any one "expert" answers. Please edit your question to show how you came up with Wolfram's answer (which is correct assuming a large-enough region of convergence, BTW). $\endgroup$
    – TimWescott
    Apr 22 at 14:16

2 Answers 2

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To uniquely determine the inverse Z-transform, the key is the region of convergence (ROC). Note that

$$ {\cal{Z}}^{-1}\{\frac{1}{z-a}\} = {\cal{Z}}^{-1}\{\frac{z^{-1}}{1-az^{-1}}\} = \left\{\begin{array}{ll}a^{n-1}u[n-1] & \text{ROC:}~|z| > |a| \\ -a^{n-1}u[-n] & \text{ROC:}~|z| < |a|\end{array}\right . $$

Note that

$$ \left\{ \begin{array}{ll} {\cal{Z}}^{-1}\{\frac{1}{1-az^{-1}}\} = {\cal{Z}}^{-1}\{\frac{z}{z-a}\} = a^{n}u[n] & \text{ROC:} |z| > |a| \\ {\cal{Z}}^{-1}\{\frac{1}{1-az^{-1}}\} = {\cal{Z}}^{-1}\{\frac{z}{z-a}\} = -a^{n}u[-n-1] & \text{ROC:} |z| < |a| \end{array} \right. $$

Given the values $a = 2,3$, there are three inverse Z transforms

  1. Right-sided sequence $|z| > 3$ (left)

$$ {\cal{Z}}^{-1}\left\{5\left(\frac{1}{z-2}-\frac{1}{z-3}\right)\right\} = 5\left(2^{n-1} u[n-1] - 3^{n-1} u[n-1]\right) $$

  1. Left-right sided sequence $2 < |z| < 3$ (center)

$$ {\cal{Z}}^{-1}\left\{5\left(\frac{1}{z-2}-\frac{1}{z-3}\right)\right\} = 5\left(2^{n-1}u[n-1] + 3^{n-1}u[-n]\right) $$

  1. Left-sided sequence $|z| < 2$ (right)

$$ {\cal{Z}}^{-1}\left\{5\left(\frac{1}{z-2}-\frac{1}{z-3}\right)\right\} = 5\left(-2^{n-1}u[-n] + 3^{n-1}u[-n]\right) $$

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Based on the applications, each of these sequences are valid inverse Z transforms.

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    $\begingroup$ It may be useful to pay attention to the fact that the $z$ transform is often taught without much regard for the region of convergence, and that the OP is looking to see if they got their arithmetic right. Adding a section that says "so if we assume that the ROC is $|z| > 3$ is ...", with the specific answer, would be more helpful than introducing concepts which, at this point, will just confuse things. $\endgroup$
    – TimWescott
    Apr 22 at 14:18
  • $\begingroup$ @TimWescott What's the inverse transform in other regions? $\endgroup$
    – kile
    Apr 24 at 9:49
  • $\begingroup$ @TimWescott What's the inverse transform in other regions? Is it the same? $\endgroup$
    – kile
    Apr 24 at 9:54
  • $\begingroup$ Look carefully at @AHT's results. Their first equation gives the difference -- if the ROC is less than some magnitude, then the inverse z transform is "in reverse", being nonzero from $-\infty < n <0$. If the ROC is greater than some magnitude then the inverse z transform is "forward", being nonzero from $0 \le n < \infty$. $\endgroup$
    – TimWescott
    Apr 25 at 16:15
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The z-transform of $f_n$ is simply $$ F(z) = \sum_{n=-\infty}^\infty f_nz^{-n} $$ For nice, stable sequences $f_n$, this can often be reduced to a rational transfer function, often simplifying the geometric sum as $$ \sum_{n=0}^\infty a^n = {1\over 1-a} $$ (Intuitively, this comes from long division of $1-a$ into $1+0*a+0*a^2+...$)

To answer your question, start with the possible answers and plug into the general, infinite sum for the $F(z)$ transform, then use the general geometric sum formula, substituting for $a$ to get the rational transfer function. Is either one correct? (Hint: if the $f_n$ values grow exponentially for a causal response, it more likely needs to be anti-causal, and neither is right.)

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  • $\begingroup$ Can you discuss each of ROC? $\endgroup$
    – kile
    Apr 24 at 20:59
  • $\begingroup$ ROC talks to the values of $z$ for which the z-transform sum is summable. For example, if $f_n = 2^nu(n)$ and $F(z) = \sum_{n=0}^\infty (2/z)^n$, what range of $z$ will make this sum less than infinite magnitude? Answer: $|z|>2$. However, if we allow $f_n$ to be anti-causal, meaning it is non-zero for negative $n$, we can get ROC that includes $|z|=1$ for a similar $F(z)$. For illustration, consider $f_n=2^nu(-n)$ and compute $F(z)=\sum_{-\infty}^{n=0}f_nz^{-n}$. $\endgroup$
    – vml
    Apr 25 at 4:36
  • $\begingroup$ Can you take a look at answers by @AHT? Do you think it's correct in each region? $\endgroup$
    – kile
    Apr 25 at 7:48
  • $\begingroup$ I don't disagree regarding various ROCs, though I consider ROCs aside from the unit circle to be more academic gymnastics than useful, personally. Underneath it all is the understanding of what we're doing here and why, more than a set of transform mapping equations and rules. It's about where the sequence is $\ell_2$ finite in practice, but in the abstracted case, we can talk about ROC with $|z|$ away from unity, though I haven't found this useful beyond solving problems in an assignment some decades ago. $\endgroup$
    – vml
    Apr 26 at 2:38

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