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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

arn
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  • @Rohan you should not edit posts than only contain a linked image, this is not encouraging people to use LateX, especially when it is a question without any personal effort at finding the solution. – zwim Sep 13 '19 at 14:54
  • @zwim I did tried and get what I linked in second image – arn Sep 13 '19 at 14:55
  • Have you tried the multiplication and division of $z^2$ on L.H.S, then resolve into partial fractions. – Vedant Chourey Sep 13 '19 at 15:03
  • @VedantChourey Yes and that how I got the second equation – arn Sep 13 '19 at 15:05
  • Ok you get that, now on R.H.S for first term write $z$ as $z+1-1$ and separate them similarly for second term write $z$ as $z + \frac{1}{2} - \frac {1}{2}$ you have the resolved ones. May this will help – Vedant Chourey Sep 13 '19 at 15:09
  • @arn: Is the confusion here that you have written "1.333" instead of $4/3$, and "0.166" where you mean "1/6"? To get rid of those z's in "picture number 1", notice for instance that $\frac{z}{z-1}=\frac{(z-1) + 1}{z-1}$... whoops! – Ben Sep 13 '19 at 15:09
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    Now note that $\frac{4}{3}\cdot\frac{z}{z-1} = \frac{4}{3 (z-1)}+\frac{4}{3}$ and that $\frac{1}{3}\frac{z}{z+0.5} = \frac{1}{3}-\frac{1}{6 \left(z+\frac{1}{2}\right)}$. – rogerl Sep 13 '19 at 15:09
  • Also, you must not have done partial fractions correctly; if you had, you would not have any $z$s in the numerators. – rogerl Sep 13 '19 at 15:10
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    I have clarified the problem – arn Sep 13 '19 at 15:20

1 Answers1

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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 $\frac{1}{6}=0.1666...=0.1(6)$.

sam wolfe
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