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The z-transform of a signal is $$ X(z)=\frac{1}{z^2+z+1}$$ I attempted to solve for the the inverse z-transform by decomposing the denominator into complex roots, $\alpha$ and $\alpha^\ast$, to get

$$\frac{1}{z^2+z+1} = \frac{A}{z-\alpha}+\frac{B}{z-\alpha^\ast}=\frac{\frac{-i\sqrt{3}}{3}}{z-\alpha}+\frac{\frac{i\sqrt{3}}{3}}{z-\alpha^\ast}$$ for $$\alpha = e^{2\pi i /3} \ \ \text{and} \ \ \alpha^\ast = e^{-2\pi i /3}$$

And ultimately this leads to a long process of manipulating a series to relate it to the definition of the z-tranform. I was wondering if there was a faster or easier way to solve this problem, perhaps with a more direct series expansion

Xander Henderson
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John
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2 Answers2

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What you may do next is write $$\frac{A}{z-\alpha} + \frac{B}{z-\alpha^*} = \frac{A}{-\alpha}\frac{1}{1-\frac{z}{\alpha}} + \frac{B}{-\alpha^*}\frac{1}{1-\frac{z}{\alpha^*}} = \frac{A}{-\alpha}\sum_{n=0}^\infty \left( \frac{1}{\alpha^n} z^n \right) + \frac{B}{-\alpha^*}\sum_{n=0}^\infty \left( \frac{1}{\alpha^{*n}} z^n \right)$$

$$= \sum_{n=0}^\infty\left[ \frac{A}{-\alpha}\left( \frac{1}{\alpha^n} \right) +\frac{B}{-\alpha^*} \left( \frac{1}{\alpha^{*n}}\right)\right]z^n = \sum_{n=0}^\infty x_n z^n$$

Thus $$x_n = \left[ \frac{A}{-\alpha}\left( \frac{1}{\alpha^n} \right) +\frac{B}{-\alpha^*} \left( \frac{1}{\alpha^{*n}}\right)\right]$$

Joel
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  • Actually, I suppose my assumption was that you wanted an explicit formula for $x_n$... here it is if you wanted it! – Joel Jun 06 '14 at 21:03
  • Thanks for your answer. I did the same thing but with the actual values which made things more tedious. The z-transform that I am referring to is defined by $z^{-n}$ which means that the final answer needs to be multiplied by $-u[-n]$, right? – John Jun 06 '14 at 21:07
  • Yes, I believe so. – Joel Jun 06 '14 at 21:08
  • Is there another signal that has the same transform but with $u[n-1]$ and of course a different ROC? Because looking at examples there are multiple signals for a single transform but with different ROC. I was wondering how one can find the others? – John Jun 06 '14 at 21:10
  • You might shift the signal by multiplying by $z^{N}$. The signal itself will be a delayed version of the original. Other than that, I am not sure. Honestly, most of my intuition comes from using generating functions in combinatorics. – Joel Jun 06 '14 at 21:21
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Notice that $(1-z)(1+z+z^2)=1-z^3$. (That's a trick worth noting whenever you're working with $1+z+z^2+\ldots z^n$.)

So $1/(z^2+z+1)$ is actually $(1-z)/(1-z^3)$.

$1/(1-z^3)$ is the sum of a geometric progression ie. $1+z^3+z^6+\ldots$.

So the final result is $1+z^3+z^6+\ldots-z(1+z^3+z^6+\ldots)$

So if we write $X(z)=\sum_{n=0}^\infty a_n z^n$ we have

$a_n=\cases{1&$n=0$ mod $3$\cr -1&$n=1$ mod $3$\cr0&$n=2$ mod 3}$

It's a surprisingly simple result!

Dan Piponi
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    Good point. I should have thought of that. You can obtain the same result with my method by recognizing that $\alpha$ and $\alpha^*$ are both 3rd roots of unity. Very slick. – Joel Jun 06 '14 at 22:04
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    Combining our answers gives an interesting way to write $(1,-1,0,1,-1,0,\ldots)$. Occasionally these things are useful when people want to write sequences like that without using conditional expressions. – Dan Piponi Jun 06 '14 at 22:58