Partial Fraction Expansion (Inverse Z-Transform)

ElfenKiller
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Homework Statement



H(z) = \frac{6-z^{-1}}{1+0.5z^{-1}} + \frac{2}{1-0.4z^{-1}} = k + \frac{A}{1+0.5z^{-1}} + \frac{2}{1-0.4z^{-1}}

where A = (6-z^{-1}) is evaluated at z^{-1}=-2.

Homework Equations



Partial fraction expansion.

The Attempt at a Solution



Why is z^{-1} set equal to -2? I thought normally you use the terms in the denominators for z^{-1} ...
 
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What do you mean by "use the terms in the denominator for z^{-1}"? Use them how?
 
vela said:
What do you mean by "use the terms in the denominator for z^{-1}"? Use them how?

Like here: http://dspcan.homestead.com/files/Ztran/zinvpart.htm"
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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