Re: factoring an infinite impulse response into SOS

On 24 Jun., 17:38, Robert Adams <robert.ad...@xxxxxxxxxx> wrote:
On Jun 24, 11:15 am, Andor <andor.bari...@xxxxxxxxx> wrote:





On 24 Jun., 15:14, Robert Adams <robert.ad...@xxxxxxxxxx> 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
.

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