formula for the zeros of a parallel connection of filters?

[Music Engineer]

this time, i have a math-question. i'll first motivate it by some filter problem. lets assume, i have two filters specified in terms of their poles and zeros. now consider a parallel connection of these two filters. the resulting filter has the transfer function:

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   P                     R
 -----                 -----
  | |                   | |
  | | (z - z_p)         | | (z - z_r)
 
  p=1                   r=1

-----------------  +  -----------------

   Q                     S
 -----                 -----
  | |                   | |
  | | (z - z_q)         | | (z - z_s)
 
  q=1                   s=1

which, expressed over a common denominator, becomes:

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   P              S                   R              Q
 -----          -----               -----          -----
  | |            | |                 | |            | |
  | | (z - z_p)  | | (z - z_s)  +    | | (z - z_r)  | | (z - z_q)
 
  p=1            s=1                 r=1            q=1

------------------------------------------------------------------

                  Q                   S
                -----               -----
                 | |                 | |
                 | | (z - z_q)       | | (z - z_s)
 
                 q=1                 s=1

from which we can directly see, that the set of poles of the new filter is formed by union of the sets of poles the two original filters because the product of the two products in the new denominator could also be written as a single product running from 1 to Q+S. but how about the zeros? is there some easy way to figure out where the new zeros are? we can write the first summand (consisting of two products) in the numerator as one product (running up to N = P+S) and similarly write the second summand as product running up to M = R+Q, so the denominator becomes:

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   N                   M
 -----               -----
  | |                 | |
  | | (z - z_n)  +    | | (z - z_m)
 
  n=1                 m=1

now the question boils down to finding the roots of a sum of two polynomials which are each given in product form (that is, via their roots) - is anyone aware of a formula/theroem how these roots can be obtained from the roots of the two constituting polynomials?

thanks, robin

[Music Engineer]

...just found a nice list of theorems for polynomials myself:

http://homepage.smc.edu/kennedy_john/polytheorems.pdf

but it doesn't seem to address my problem. but i looked only briefly up to now.

[robrokken]

You guys must be aliens to understand this stuff lol

~Rob.

[l0calh05t]

Mathematicians, physicists and electrical engineers shouldn't have much trouble understanding these things. No need to wait for our alien overlords

[Dean Roddey]

The answer of course is 42.

[docdued]

I don't think there's a way to obtain the roots of a sum of two polynomials via the roots of each of the polynomials (unless they have a root in common, of course).

[l0calh05t]

I don't think there's a way to obtain the roots of a sum of two polynomials via the roots of each of the polynomials (unless they have a root in common, of course).

And even then only the common root would remain the same. All others would change.

[Music Engineer]

The answer of course is 42.

you mean, i accidentaly found the question of all questions?

i'm beginning to believe that there is indeed no such formula. i guess, that means then: multiply out the numerator-products, add them and throw a root-finder at the result. i hoped to avoid that.

[MackTuesday]

I think I might have an approach to this problem.

Let P(z) and Q(z) represent your two parallel feedforward filters and

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S(z) = P(z) + w∙Q(z)

where w is a free parameter. When w=0 the roots of S are the same as the roots of P. Imagine these roots moving around the z plane as w gradually increases. For the kth root r_k,

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dr_k/dw = Q(r_k) / S'(r_k)

This gives a differential equation of the form

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[P'(y)+w∙Q'(y)]∙y' - Q(y) = 0

Find y(w,C), which will be an expression in terms of w with some free parameter C. For each root r_k find C such that

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y(0,C) = r_k

Then the new root is

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r_new = y(1,C)

So the big trick is finding the general form for y. I don't see how to do this but someone with more knowledge of diffeqs might know.

Then again, I wouldn't be surprised if finding y required knowing the roots of S'.

[Music Engineer]

I think I might have an approach to this problem. Let P(z) and Q(z) represent your two parallel feedforward filters and

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S(z) = P(z) + w∙Q(z)

where w is a free parameter. When w=0 the roots of S are the same as the roots of P. Imagine these roots moving around the z plane as w gradually increases. For the kth root r_k

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dr_k/dw = Q(r_k) / S'(r_k)

interesting approach! errmm...maybe i'm overlooking something obvious, but could you explain how you arrived at that expression for the derivative?

[MackTuesday]

errmm...maybe i'm overlooking something obvious, but could you explain how you arrived at that expression for the derivative?

Yeah, let me look at it again here.

Consider dP/dz near a root and ask which direction the root will migrate due to an infinitesimal change in w. We're looking to attain

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dS = 0

with infinitesimal changes in both r and w. We have

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dS/dz = dP/dz + w∙dQ/dz
dS/dw = dP/dw + d[wQ]/dw = 0 + Q = Q

Now here's a little trick I used that might not have been kosher. Since we always think of r as a root of P, and since r is changing, P must be changing as well. We can say that, due to these infinitesimal increments of w, P continually changes by dQ and Q changes by -dQ. And w remains at 0. Therefore we say

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dS/dz = dP/dz + w∙dQ/dz
 = dP/dz + 0∙dQ/dz
 = dP/dz

and then

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dr/dw = dS/dw * dr/dS
 = Q / dP/dr

Ahh, but I think I see a problem. Since this formulation requires that P and Q change as the root migrates, that differential equation becomes a whole lot more complicated than I thought. Hm.

I dunno, this approach might still be made to work if the resulting diffeq isn't too cruel.

[MackTuesday]

So maybe what we need to do is keep P and Q constant and think of r as a root of S. Let's see, if dS=0,

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dS/dz∙dz + dS/dw∙dw = (dP/dz + w dQ/dz) dz + Q dw = 0

and

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dz/dw = -Q / (dP/dz + w dQ/dz) = -Q / dS/dz

Yikes, I had a sign wrong the first time around. Anyway, maybe we can say

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z = z0 + INTEGRAL {w=0..1} -Q dw / (dP/dz + w dQ/dz)

But like I feared, this requires knowing the roots of dS/dz unless someone has a fantastic way of finding the antiderivative when there's a polynomial in the denominator.

But even if you had the roots, the partial fraction expansion requires solving a system of equations -- just as expensive as finding roots.

[MackTuesday]

Oh but wait a second. The integral here

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z = z0 + INTEGRAL {w=0..1} -Q dw / (dP/dz + w dQ/dz)

is only first-degree in w on the bottom there. I was conflating w with z in that integral. So it turns out to be

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z = z0 - (Q / dQ/dz) [ln ((dP/dz + dQ/dz) / Q) - ln (dP/dz / Q)]
 = z0 - (Q / dQ/dz) ln (1 + dQ/dz / dP/dz)

So that appears to be the answer we sought. Now it seems a little too good to be true, considering this answer doesn't appear to be well known and doesn't seem to have any mechanism for roots increasing or decreasing in number as we know is possible when you add two polynomials together. Hm.

[vadarfone]

Good lord, this is above me.

I promised myself that I would have a simple VST plugin coded by the end of the year, so if anyone could point me in the direction of something that would give a beginner an idea of what has just been discussed in this thread, I would be very grateful.

Thanks!

[mystran]

Good lord, this is above me.

I promised myself that I would have a simple VST plugin coded by the end of the year, so if anyone could point me in the direction of something that would give a beginner an idea of what has just been discussed in this thread, I would be very grateful.

Actually I would put this thread into the "advanced" bin of the local topics. We have a nice number of people that actually like math, I guess (I do at least; though I'm not that great at it). What robin wants to solve is the total filter he gets when combining two filters in parallel, which probably happens to be approximately the deepest math you might encounter when working with (idealized) linear time-invariant filters. Why he actually wants to do that is beyond me as well... not that it isn't curious exercise anyway.

But for writing your first plugin, you most definitely don't need any of this (like said; I can't think of why he wants to do it that way anyway), so don't worry if you can't follow.