Find the causal signal x(n) for the ff. z transform
$\displaystyle X(z) = \frac{0.5z}{z^2-z+0.5}$
Dividing top and bottom by $\displaystyle z^{2}$ yields
$\displaystyle X(z)=\dfrac{0.5z^{-1}}{1-z^{-1}+0.5z^{-2}}.$
Comparing this expression to # 20 in this list, we see that we must have, if possible,
$\displaystyle 0.5=a\sin(\omega_{0}),$
$\displaystyle 1=2a\cos(\omega_{0}),$ and
$\displaystyle 0.5=a^{2}.$
You can make this work out. What do you get for $\displaystyle a$ and $\displaystyle \sin(\omega_{0})$ and $\displaystyle \cos(\omega_{0})?$ What is your radius of convergence?
The 'rexcursive relation' corresponding the the z-transform...
$\displaystyle \displaystyle X(z)= \frac{1}{2}\ \frac{z^{-1}}{1-z^{-1} + \frac{z^{-2}}{2}}$ (1)
... is...
$\displaystyle \displaystyle x(n)= x(n-1)- \frac{x(n-2)}{2} + \frac{\delta(n-1)}{2}\\,\\ x(-1)=0\\,\\ x(-2)=0$ (2)
... that has to be solved...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
the region of convergence is |z| > 0.707106
solving for a
$\displaystyle \sqrt(0.5) = a$
$\displaystyle a = 0.707106$
solving for $\displaystyle \omega_o$
$\displaystyle \sin^{-1}(\frac{1}{2*0.707106}}) = \omega_o$
$\displaystyle \omega_o = 45 degrees$
the inverse z transform is
$\displaystyle x[n] = (0.707107)^n \sin(\frac{\pi}{4}n)u[n]$
is this correct?