factoring an infinite impulse response into SOS
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From: Robert Adams on 24 Jun 2008 09:14 Guys I have the following problem; Factor the following infinite series into a product of second-order sections; H(z) = 1 - Z^-1 + Z^-2 - Z^-3 + Z^-4 ...................................... Thanks for any pointers! Bob Adams
From: Jerry Avins on 24 Jun 2008 10:12 Robert Adams wrote: > Guys > > I have the following problem; > > Factor the following infinite series into a product of second-order > sections; > > > H(z) = 1 - Z^-1 + Z^-2 - Z^-3 + > Z^-4 ...................................... > > Thanks for any pointers! What's the name of the course? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
From: Robert Adams on 24 Jun 2008 10:28 On Jun 24, 10:12 am, Jerry Avins <j...(a)ieee.org> wrote: > Robert Adams wrote: > > Guys > > > I have the following problem; > > > Factor the following infinite series into a product of second-order > > sections; > > > H(z) = 1 - Z^-1 + Z^-2 - Z^-3 + > > Z^-4 ...................................... > > > Thanks for any pointers! > > What's the name of the course? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ Jerry Believe me, I WISH I were a student!! But, sadly, I have been grinding out audio chip designs for 25+ years. This problem is related to a rather bizarre mathematical problem I have been trying to solve for the last few years, unrelated to my "day job". Regards Bob Adams Analog Devices Fellow
From: Rune Allnor on 24 Jun 2008 11:09 On 24 Jun, 15:14, Robert Adams <robert.ad...(a)analog.com> wrote: > Guys > > I have the following problem; > > Factor the following infinite series into a product of second-order > sections; > > H(z) = 1 - Z^-1 + Z^-2 - Z^-3 + Z^-4 The way to do this is as follows: - Find the roots of the polynomial H(z) (use some numerical for this) - Group one pair of complex conjugated roots into one SOS as SOS_n = (1-z_n)(1-conj(z_n)) - Repeat for all pairs of complex conjugated roots. If you find real roots, by all means group them in pairs as well, and take care of any-leftover real root as a First-Order-Section. If the coefficients of H(z) are all real and you find single complex-valued roots, you have a problem. Rune
From: Andor on 24 Jun 2008 11:15
On 24 Jun., 15:14, Robert Adams <robert.ad...(a)analog.com> wrote: > Guys > > I have the following problem; > > Factor the following infinite series into a product of second-order > sections; > > H(z) = 1 - Z^-1 + Z^-2 - Z^-3 + > Z^-4 ...................................... > > Thanks for any pointers! Hi Bob That's the impulse response of a first order integrator multiplied by [... 1 -1 1 -1 ...], ie. spectrally inverted. The rational transfer function is simply H(z) = 1/(1 + z^-1). Hint: geometric series :-). Regards, Andor |