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
factoring an infinite impulse response into SOS
Started by ●June 24, 2008
Reply by ●June 24, 20082008-06-24
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. �����������������������������������������������������������������������
Reply by ●June 24, 20082008-06-24
On Jun 24, 10:12�am, Jerry Avins <j...@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
Reply by ●June 24, 20082008-06-24
On 24 Jun, 15:14, Robert Adams <robert.ad...@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^-4The 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
Reply by ●June 24, 20082008-06-24
On 24 Jun., 15:14, Robert Adams <robert.ad...@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
Reply by ●June 24, 20082008-06-24
On Jun 24, 11:15�am, Andor <andor.bari...@gmail.com> wrote:> On 24 Jun., 15:14, Robert Adams <robert.ad...@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, > AndorThanks; yes, I am aware of this solution, but this gives me all poles and no zeroes. I guess a bigger question is whether or not there ARE any zeroes. I think there is a convergence problem since this is an infinite energy signal. I was hoping for a product formula where each SOS contributed a single zero-pair. Clearly at frequencies of w= N*PI there is no growth in the sum, so perhaps if I re-cast the problem such that each term decays by 1/a^n where a is slighly less than 1 and n is the sample index, and then take the limit as a approaches 1.0, I might see zeroes ?? Bob
Reply by ●June 24, 20082008-06-24
On 24 Jun., 17:38, Robert Adams <robert.ad...@analog.com> wrote:> On Jun 24, 11:15�am, Andor <andor.bari...@gmail.com> wrote: > > > > > > > On 24 Jun., 15:14, Robert Adams <robert.ad...@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 > > Thanks; yes, I am aware of this solution, but this gives me all poles > and no zeroes.Not quite. The transfer function H(z) = 1/(1+z^-1) = z/(z+1) has a first-order zero at 0 and a first-order pole at -1.> I guess a bigger question is whether or not there ARE > any zeroes. I think there is a convergence problem since this is an > infinite energy signal.The power series converges for all |z| < 1, ie. |z^1| > 1.> I was hoping for a product formula where each > SOS contributed a single zero-pair. Clearly at frequencies of w= N*PI > there is no growth in the sum, so perhaps if I re-cast the problem > such that each term decays by 1/a^n where a is slighly less than 1 and > n is the sample index, and then take the limit as a approaches 1.0, I > might see zeroes ??Then you'll see H_a(z) = 1/(1+a z^-1) = z/(z+a), which has the same zero and a pole at -a. Taking the limit a->1 takes you back to the first operator. There are simply not more poles or zeros in there :-). I have the feeling you are still after the zeros of the Zeta function, Zeta(s) = sum_{k=1}^infinity k^s. (1) There is the following neat mathematical trick that let's you evaluate the above operator H(z) at its pole position. You only have to note that H(-1) = Zeta(0) = -1/2, (2) because H(-1) = 1 + 1 + 1 + .... and Zeta(0) = 1 + 1 + 1 + ... The last part of equation (2) follows from the analytic continuation of the defintion (1). So, the frequency response of the inverted integrator at Nyquist frequency is -1/2. :-) Regards, Andor
Reply by ●June 24, 20082008-06-24
Robert Adams wrote:> On Jun 24, 10:12 am, Jerry Avins <j...@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 FellowBob, I know you've been around comp.dsp a long time. Honestly, I thought you were taking a refresher course. This is high-school algebra if you look at it the right way: the sum of an infinite geometric series (obviously, with exponent less than unity). Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 24, 20082008-06-24
Robert Adams wrote:> On Jun 24, 11:15 am, Andor <andor.bari...@gmail.com> wrote: >> On 24 Jun., 15:14, Robert Adams <robert.ad...@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 > > Thanks; yes, I am aware of this solution, but this gives me all poles > and no zeroes. I guess a bigger question is whether or not there ARE > any zeroes. I think there is a convergence problem since this is an > infinite energy signal. I was hoping for a product formula where each > SOS contributed a single zero-pair. Clearly at frequencies of w= N*PI > there is no growth in the sum, so perhaps if I re-cast the problem > such that each term decays by 1/a^n where a is slighly less than 1 and > n is the sample index, and then take the limit as a approaches 1.0, I > might see zeroes ??My guess is that it is what it is, and that it won't be twisted into something else by looking at it from some special angle. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 24, 20082008-06-24
