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Winograd

Name of this serie [SOLVED]

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Hi! Is there a name for the serie below? EDIT: sum(n = 0 to N-1, cn*zn^k), where cn and zn have n as a index and k is a positive integer. And can you point to any useful site containing information about series of that form? [Edited by - Winograd on December 1, 2004 1:18:44 PM]

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Not homework, although I admit that it might seem one ;)

MathWorld didn't help.. and even if my serie is in some dark corner of mathworld I would still need to know the name first or crawl through all series and sums it has (and pretty much this is what I did :) )

This serie or sum emerges from close-form solution for recurrence equation of following kind:

y[n]=sum(k=0 to N-1, ak*y[n-k] )

Those recursive equations produce the impulse response of the IIR-filter having unity nominator in the z-plane equation. for example: H(z)=1/(1+az^-1) ==> y[n]=-a*y[n-1]

So I would REALLY appreciate if someone knows a name or place for information for the serie I described in my first post!

And even if this would be homework (which it isn't.. trust me) I don't ask you guys to solve anything for me. And what kind of home assignment goes: "Find the name for following expression."

:D

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I am very familiar with that type of expression, but I don't have a better name for it than "sum of geometric sequences". What is it that you want to know about those sequences?

EDIT: Nevermind, I read your formula wrong.

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Sounds a bit odd for a homework question, so I'll give you the benifit of the doubt.
EDIT: Nevermind. There was only 1 reply when I started my post [smile]

It's called the Power Series.

[Edited by - joanusdmentia on December 1, 2004 6:22:58 AM]

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Quote:
Original post by Winograd
This serie or sum emerges from close-form solution for recurrence equation of following kind:

y[n]=sum(k=0 to N-1, ak*y[n-k] )

Those recursive equations produce the impulse response of the IIR-filter having unity nominator in the z-plane equation. for example: H(z)=1/(1+az^-1) ==> y[n]=-a*y[n-1]


Thought it might be something along those lines :)

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Joanusdmentia, you missed one crucial point.
This is my serie:
sum(n = 0 to N-1, cn*zn^n)

This is power serie:
sum(n = 0 to N-1, cn*z^n)

Notice the index in z.

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So there is.....but that doesn't seem right. I'm assuming that z is in the z-plane (and you didn't just happen to chose z for something else [smile]), in which case having a subscript doesn't make any sense.

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Whoopsy!! Made a HUGE typo.. sorry about that! I corrected it in my first post.

So the exponent should be independent positive integer variable k and not n as in my first post it used to be.

So this is kind of family of series. One for every k :)

Here it is again.. this time correctly:

sk = sum(n = 0 to N-1, cn*zn^k), where cn and zn have n as a index and k is a positive integer

EDIT: Now the question really should be that is there name for that family of series. I'm actually intrested in solving (or aproximating the solution) following group of equations for cn and zn:

sk = sum(n=0 to N-1, cn*zn^k), for k in [0, M-1]

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Quote:
Original post by Winograd
sk = sum(n = 0 to N-1, cn*zn^k), where cn and zn have n as a index and k is a positive integer

Still doesn't make any sense. By zn^k, do you really mean (z^-n)^k? How did you get your equation?

EDIT: Forgot a minus sign.

[Edited by - joanusdmentia on December 1, 2004 7:57:18 AM]

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Forgot to mention that this is not z-plane equation. ck and zn are calculated from the z-plane transfer function of iir-filter (or from characteristic equation of the recurrence equation (in time-plane)) and from pre-given initial-state (in my case unit-impulse). And I think I have already mentioned that it gives you closed-form solution for the impulse response, not recursive solution usually given. This does _not_ give any speed increase calculating the impulse response. I'm heading toward different angle on designing IIR-filters, thats all.

The actual relation of my serie to filters and z-plane are irrelevant. You can take the group of equations as is. I'm just intrested in more information on solving such equation groups and any other properties that arise from that particular family of series.

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Aahhhh...ok then. Damn you and you're choice of z!!! [smile]

As for the series, I doubt you'll find any information on it. Reason being that without knowing either cn or zn it could take on any form. Just to illustrate, if zn = 1/n then your dealing with a series similar to the harmonic series, but if zn = n the series is totally different.

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You're probably right joanusdmentia. It's rather general equation when one considers only one sum for some k.

But, the solving part is still open.. (perhaps I should start a new thread)

I want to solve system of equations for ck and zk. Something like this... (here N=2)

c0 + c1 = x0
c0*z0 + c1*z1 = x1
c0*z0^2 + c1*z1^2 = x2
c0*z0^3 + c1*z1^3 = x3
.
.
.
c0*z0^M + c1*z1^M = xM

I guess four equations is enough (M=3) to fully determine c0,c1,z0 and z1. Or is it... i have to think harder this one when I have more time.

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Quote:
Original post by Winograd
I guess four equations is enough (M=3) to fully determine c0,c1,z0 and z1. Or is it... i have to think harder this one when I have more time.

It's possible, it won't be fun but it's possible [smile]
You'll have to use numerical methods (read: Newtons method) to solve that generally.

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So it was a sum of geometric sequences (what I first thought), only that you wrote the wrong equation.


You can compute the z0, z1, z2, ... first and then compute c0, c1, c2, ..., which is just a linear system of equations.

Consider the polynomial P := (l-z0)*(l-z1)*...*(l-z{n-1}). Its roots are obviously z0, z1, ... Let's write the polynomial in expanded form: P = l^n + a{n-1}*l^(n-1) + ... + a2*l^2 + a1*l +a0

Your sequence (let's call it xn) satisfies the recurrence equation
x{k+n} = -a{n-1}*x{k+n-1} - ... - a2*x{k+2} - a1*x{k+1} - a0*xk

You can compute the values of a0, a1, ..., a{n-1} by solving a system of linear equations. Then compute all the roots of P (using Newton-Raphson, for instance), which will be z0, z1, ... z{n-1}. Then compute c0, c1, ..., c{n-1}, solving another system of linear equations.

If you need any more detail, ask.

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alvaro,
Maybe I did not understand you correctly, but it seems that what you're saying is that I need the recurrence equation to determine the values (ak and zk). Well, too bad, because I don't have the recurrence equation. And I already know how to do that. It is almost trivial task to obtain the values from recursive equation (4 lines of Matlab code).

The problem is roughly this: given a arbitrary real sequence of length N, derive a recursive equation that produces that sequence for the first N samples when feed unit-impulse.


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You can compute what the recursion formula is with a system of linear equations.

Example: Find the recursion formula of order 3 that the following sequence satisfies:
3 0 2 3 2 5 ...

A*xn+B*x{n+1}+C*x{n+2} = x{n+3}

A*3 + B*0 + C*2 = 3
A*0 + B*2 + C*3 = 2
A*2 + B*3 + C*2 = 5

Solve to find A=1, B=1, C=0. So the recurrence is x{n+3}=xn+x{n+1}.

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