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I have kind of an algebra problem. The original question is a Bilinear transformation for analogue to digital filters. (This is not a homework question)

In my lecture notes, he goes to the answer like 1 step, enter image description here

I'm trying to work it out but I'm getting stuck at one place and I don't know how to proceed to get the same way he has it. enter image description here

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  • $\begingroup$ Multiply the numerator and the denominator by $(1+z^{-1})^3$ to get the cancellation that you are looking for and multiply it out. Often lecturers will shave some time off the presentation by skipping the tedious steps expecting that students can do the multiplication on their own time. It's likely that what was presented as one step took a few more steps but just weren't shown. $\endgroup$ – Mason Dec 11 '18 at 23:46
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First, multiply (and divide) by $8(1+z^{-1})^3$: you get $$ \frac{8(1+z^{-1})^3}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}. $$ Next, expand the numerator to get $$ \frac{8(1+3z^{-1}+3z^{-2}+z^{-3})}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}. $$ Now, to expand the denominator. Let's do each term: $$ 8(1+z^{-1})^3=8+24z^{-1}+24z^{-2}+8z^{-3}; $$ $$ 8(1+z^{-1})^2(1-z^{-1})=8+8z^{-1}-8z^{-2}-8z^{-3}; $$ $$ 4(1+z^{-1})(1-z^{-1})^2=4-4z^{-1}-4z^{-2}+4z^{-3}; $$ and $$ (1-z^{-1})^3=1-3z^{-1}+3z^{-2}-z^{-3}. $$ Collecting terms, the denominator becomes $$ (8+8+4+1)+(24+8-4-3)z^{-1}+(24-8-4+3)z^{-2}+(8-8+4-1)z^{-3} =21+25z^{-1}+15z^{-2}+3z^{-3}. $$

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  • $\begingroup$ Fantastic. Thank you so much!!! $\endgroup$ – AlfroJang80 Dec 12 '18 at 0:21
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Defining $m:=1-z^{-1}$ and $p:=1+z^{-1}$, we have $$\begin{align} \frac{1}{1+2\cdot\dfrac12\dfrac{m}{p}+2\cdot\dfrac14\dfrac{m^2}{p^2}+\dfrac18\dfrac{m^3}{p^3}}\cdot\frac{8p^3}{8p^3} &= \frac{8p^3}{8p^3+8p^2m+4pm^2+m^3} \\[2pt] &=\frac{8p^3}{\left(2p+m\right)\left(4p^2+2pm+m^2\right)} \end{align}$$ From here, expanding the various pieces is straightforward.

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I would set $t=z^{-1}$, to get for the denominator: \begin{align} {}&\phantom{={}}\;1+\frac{1-t}{1+t}+\frac{(1-t)^2}{2(1+t)^2}+\frac{(1-t)^3}{8(1+t)^3}\\ &= \frac{8(1+t)^3+8(1-t)(1+t)^2+4(1-t)^2(1+t)+(1-t)^3}{8(1+t)^3} \\ &= \frac{8(1+t)^3+8(1-t^2)(1+t)+4(1-t)(1-t^2)+(1-t)^3}{8(1+t)^3} \\ &=\dotsm \end{align}

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Your problem looks very similar to this one.$$\frac{1}{t^3+2t^2+2t+1} = \frac{1}{t^3+3t^2+3t+1-t^2-t}=\frac{1}{(t+1)^3-t(t+1)}=\frac{1}{(t+1)(t^2+t+1)}$$

Can you take it from here?

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