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Ch 5.17 - Equation 5.56 (p.133-134)
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November 21, 2016
12:12 am
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Sidora83
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It mention that the equation : 1 + α1z-1 + α2z-2
can be factored to: (1-Z1z-1)(1-Z2z-1)
with some algebra.

Would anyone have the algebra steps to arrive at that conclusion?
Im a bit confuse about how this was deduced.

Thanks!
Yan

November 21, 2016
5:55 pm
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Sidora83
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Never mind. I was looking too deep into it :P.
Just realized the conjugate pair was added in to replace the "pre-FOIL" coefficients, with the knowledge the imaginary part would cancel out, leaving real numbers.

July 25, 2017
3:50 pm
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I did have exactly the same question. I wanted to know the algebra steps involved. Digging a little bit, I found this link with the explanation behind the algebra. Very useful and interesting.

https://www.dsprelated.com/freebooks/filters/Pole_Zero_Analysis_I.html

Here is a little summary

"we can write the general transfer function for the recursive LTI digital filter as

$displaystyle H(z) = gfrac{1 + eta_1 z^{-1}+ cdots + eta_M z^{-M}}{1 + a_1 z^{-1}+ cdots + a_N z^{-N}} protect$Image Enlarger

(...)

In the same way that

$ z^2 + 3z + 2$

can be factored into

$ (z + 1)(z + 2)$

we can factor the numerator and denominator to obtain.

$displaystyle H(z) = gfrac{(1-q_1z^{-1})(1-q_2z^{-1})cdots(1-q_Mz^{-1})}{(1-p_1z^{-1})(1-p_2z^{-1})cdots(1-p_Nz^{-1})}. protect$Image Enlarger

Hope this helps!

Thanks.

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