Inverse Z-transform with a complex root

Question

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

Answer 1

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!

Answer 2

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