Can anyone help me clarify how this rewriting is done?
$$\frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} = \frac{16}{15}\frac{1}{(1-\frac{1}{4}z^{-1})}-\frac{16}{15}\frac{1}{(1-4z^{-1})}$$
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the question is repeated :http://math.stackexchange.com/questions/402830/frac11-dfrac14z-11-dfrac14z-frac-4z-11-dfrac1 – mnsh May 26 '13 at 12:59
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No it is not. This is the next part of the same equation. – Attaque May 26 '13 at 13:00
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Then, may I ask, what have you tried? – user49685 May 26 '13 at 13:01
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Are you familiar with Partial Fraction Dissociation? – Maazul May 26 '13 at 13:09
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No, i'm afraid not. Sorry that I am not responding, but I am trying to figure our the result using your help. – Attaque May 26 '13 at 13:30
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As Maazuls answer clearly illustrates, this question is not a repost. So why the vote down? – Attaque May 26 '13 at 14:33
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Maybe because it didn't show much effort on your part. – Maazul May 26 '13 at 18:12
1 Answers
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Let $x=z^{-1}$
Then
\begin{align} \frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} =\frac{-4x}{(1-\frac{1}{4}x)(1-4x)} \end{align}
Suppose
\begin{align} &\frac{-4x}{(1-\frac{1}{4}x)(1-4x)}=\frac{A}{(1-\frac{1}{4}x)}+\frac{B}{(1-4x)}\tag{1}\\&\frac{-4x}{(1-\frac{1}{4}x)(1-4x)}=\frac{A(1-4x)+B(1-\frac{1}{4}x)}{(1-\frac{1}{4}x)(1-4x)}\\&-4x=A(1-4x)+B(1-\frac{1}{4}x)&\\&-4x=x\left(-4A-\frac{B}{4}\right)+A+B\end{align}
Comparing coefficients, you'll have $2$ equations involving $A$ and $B$.
$$-4A-\frac{B}{4}=-4 \tag{2}$$ $$A+B=0 \tag{3}$$
Solve $(2)$ and $(3)$ to get $A=\frac{16}{15}$ and $B=-\frac{16}{15}$ then replace them in $(1)$ and also revert $x$ back into $z^{-1}$ to get your final result.
Maazul
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That is clever. Thank you for your very thorough answer. I'm reading up on Partial Fraction Dissociation now.. (the same as partial fraction decomposition right?) – Attaque May 26 '13 at 14:03
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