I would like to prove that: $$ L = (\frac{1+z^{-1}}{1+0.5z^{-1}} ) \cdot (\frac{1}{1-z^{-1}}) = \frac{0.166}{z + 0.5} + \frac{1.33}{z - 1} + 1$$
How do I get from left to right? The same problem have other solution that I succeed to get $L = \frac43\frac{z}{z-1} - \frac13\frac{z}{z + 0.5}$(and Its good enough for solving the engineering problem) but I want to understand how do I get to the first
First, write the expression as a ratio of expanded polynomials in lowest terms, that is, $$ \frac{2z^2+2z}{2z^2-z-1}. $$ Next, use long division to find the quotient and remainder of $\frac{2z^2+2z}{2z^2-z-1}$ and then rewrite it as the quotient plus the remainder over the denominator, as follows $$ 1+\frac{3z+1}{2z^2-z-1}. $$ Hence, we only have to find the partial fraction decomposition of $\frac{3z+1}{2z^2-z-1}$. To do so, factor the denominator into linear and irreducible quadratic terms and equal it to the partial fraction expansion form $$ \frac{3z+1}{(z-1)(2z+1)}=\frac{\theta_1}{z-1}+\frac{\theta_2}{2z+1}. $$ Multiplying both sides by $(z-1)(2z+1)$ and rewriting the previous identity yields $$ 3z+1=\theta_1-\theta_2+z(2\theta_1+\theta_2). $$ Finally, equating the coefficients on both sides yields the following system $$ \begin{cases} 1=\theta_1-\theta_1\\ 3=2\theta_1+\theta_2 \end{cases} $$ which yields $\theta_1=\frac43$ and $\theta_2=\frac13$. Hence, $$ \left(\frac{1+z^{-1}}{1+0.5z^{-1}} \right)\left(\frac{1}{1-z^{-1}}\right)=1+\frac{4}{3(z-1)}+\frac{1}{3(2z+1)}=1+\frac{1.(3)}{z-1}+\frac{0.1(6)}{z+0.5}. $$ Note that $\frac43=1.333...=1.(3)$ and $\frac{1}{6}=0.1666...=0.1(6)$.