$\displaystyle T(z) = \frac{z}{2-z}$

$\displaystyle S^{-1}(z) = 4\frac{z-1}{z+4}$

Solve:

$\displaystyle L(z) = S^{-1}(T(z))$

Can someone show me the steps? I know to plug in for z but I can't seem to do it the right way...

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- Nov 7th 2008, 09:05 AMProof_of_lifeFactor this please quick!
$\displaystyle T(z) = \frac{z}{2-z}$

$\displaystyle S^{-1}(z) = 4\frac{z-1}{z+4}$

Solve:

$\displaystyle L(z) = S^{-1}(T(z))$

Can someone show me the steps? I know to plug in for z but I can't seem to do it the right way... - Nov 7th 2008, 09:18 AMProof_of_life
For Linear Functional Transformations. Thanks Moo, you just put a negative for the 4 instead of a positive when factoring out tho. James has the correct answer..

- Nov 7th 2008, 09:38 AMjames_bond
Repost, sorry I thought mine was wrong and Moo wrote it more nicely by the way.

$\displaystyle L(z) = S^{-1}(T(z))=S^{-1}\left(\frac{z}{2-z}\right)=4\frac{\frac{z}{2-z}-1}{\frac{z}{2-z}+4}=-\frac{8 (-1+z)}{-8+3 z}$

$\displaystyle z\ne 2\,\text{and}\, \frac 83$