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April 27

Pink noise and differintegrals

In Colors of noise we state that pure red noise (power density decreasing 6 dB per octave) can be generated directly from pure white noise (equal power per bandwidth) by integrating it (instead of choosing a number at random for each value, we choose a number at random and add it to the previous value).

In Pink noise, though, we say that there is no way to generate pink noise (power density decreasing 3 dB per octave) directly; there are only approximations with obvious "bumps"[1] in the spectrum. This also seems to be the consensus among the newsgroups,[2][3] though some of it is a little over my head. I know that in electronics, the 6 dB per octave fall-off is the simplest you can build, and all filters are multiples of 6.

Now, it seems to me that since the spectrum of pink noise is exactly between the spectrums of white and red noise, it should be possible to generate directly from white noise by doing something in-between unity and integration, something like a fractional integral or differintegral (I know nothing about fractional calculus). Is this true? Could ideal white noise be passed through an ideal fractional order integrator to produce ideal pink noise?

I think it could, though it looks like this just shifts the blame, since the ideal fractional order integrator itself can not be built; but only approximated (by, for instance, ladders of 6 dB per octave RC filters in the analog domain, approximating the ideal 3 dB per octave response). Am I getting this right?

However, in Pink noise and Colors of noise it's pointed out that the ideal pink spectrum occurs often in natural systems. For instance, Flicker noise has a pink spectrum. If white noise can be generated by natural random processes, and we can simulate it, and pink noise can be created by natural random processes, why can't it be perfectly simulated? — Omegatron (talk) 01:16, 27 April 2008 (UTC)[reply]

3dB per octave pink noise is not possible to implement as an exact solution using linear analog electronic components. This is equivalent to saying (sticking my neck out here on the maths page being as I've wandered in off the science page) that it is not possible to implement with linear differential equations. The reason for this is as follows; the components available to build a linear circuit (using the Laplace transform notation) are R, sL and 1/sC. The mathematical operations you can get from these in the time domain are multiply by a constant, differentiate and integrate. You can have an amplifier as well but that is still just multiply by a constant >1. Consequently the transfer function of any linear electronic device built out of these discrete components must have a transfer function of the form,

A(s)={\frac {P(s)}{Q(s)}}

P(s),Q(s)\,\!

polynomial in s

Polynomial in s in the s-domain means that the transfer function must be polynomial in iω in the frequency (Fourier transform) domain and therefore also polynomial in f. Note that transfer functions are expressed as voltage or current ratios not power ratios. 3dB/8ve is indeed 1/f as a power ratio but as a voltage ratio it is;

{\frac {1}{\sqrt {f}}}

This cannot be expressed as a polynomial in f. Of course, there are many ways of generating a close approximation, but an exact mathematical expression, and hence circuit implementation, is impossible.

Addressing the question of how this could occur in nature, well building a circuit is done using discrete components. In a system where the quantities are distributed limits have to be taken and the solution for the transfer function may well not be polynomial. For instance, the well known solution for the propagation coefficient of a transmission line is,

\gamma ={\sqrt {ZY}}

where

Z=R+sL\,

and

Y=G+sC\,

The square root here clearly preventing the expression from being polynomial in the general case.

SpinningSpark 20:31, 27 April 2008 (UTC)[reply]

I have restated the maths part of this question in a new question here for clarity SpinningSpark 16:17, 1 May 2008 (UTC)[reply]

Powers of 2 and Powers of 3

Any positive integer can be expressed by adding (integer) powers of 2. That's obvious.

Any positive integer can be expressed by adding or subtracting powers of 3 (without repeating them). For example:

1 = 3^0
2 = 3^1 - 3^0
3 = 3^1
4 = 3^1 + 3^0
5 = 3^2 - 3^1 - 3^0
6 = 3^2 - 3^1
...

I don't know where to continue from here on. What can you do with powers of 4? Are there any articles on this? Thanks.

--wj32 ^t/c 08:50, 27 April 2008 (UTC)[reply]

Radix and Non-standard positional numeral systems would be good places to start. Basically, any positive integer can be expressed as a weighted sum of powers of 2, where the weights are 0 or 1 (2 possibilities). Any integer can be expressed as a weighted sum of powers of 3, where the weights are 1, 0 or -1 (3 possibilities). Likewise, any integer can be expressed as a weighted sum of powers of 4, where the weights are 2, 1, 0 or -1 (4 possibilities). The details can change depending on the base and choice of weights, but the general idea is that you need n different weights to express numbers with powers of n. -- Meni Rosenfeld (talk) 09:39, 27 April 2008 (UTC)[reply]

(edit conflict) The proper generalization, I think is: given a base

b>1

, if you choose a set of integers

a_{1}

through

a_{b-1}

such that

a_{k}

is congruent to

k

modulo

b

, then any integer can be expressed as a sum whose terms are of the form

a_{m}b^{n}

. Your first example

b=2,a_{1}=1

. Your second example is

b=3,a_{1}=1,a_{2}=-1

. You're basically writing the number in balanced ternary notation, which uses the digits (0,1,-1) instead of (0,1,2).

In base 4, you could do something similar by using the digits (0,1,-1,2) or (0,1,-1,-2). It can't be described as compactly as you did the base 3 case because we don't have a single-word description for "adding something multiplied by 2", like we do for "adding something multiplied by -1" (subtracting).

1 = 4^0
2 = 2*4^0
3 = 4^1 - 4^0
4 = 4^1
5 = 4^1 + 4^0
6 = 4^1 + 2*4^0
...

I don't know if the generalization as I stated it has a name. It seems like an obvious extension of "any integer can be written in any base". --tcsetattr (talk / contribs) 09:44, 27 April 2008 (UTC)[reply]

I don't think it's true as stated. Take b=2, a₁=3 for an example where it fails. An additional condition is that the a_m should be (collectionwise) coprime, but that's still not enough (you can't write 1 as a sum of integers of the form 17·3^k and 42·3^k, say). Bikasuishin (talk) 19:39, 27 April 2008 (UTC)[reply]

Well, it is certainly possible to represent an integer z in some base n (an integer) using the digits 0, 1, ... , n-1, but really all you need is that you have n integers that are distinct modulo n. Then you can represent it as the sum of

a_{n}b^{n}

with the

a_{i}

one of your n integers. The question is really only whether or not such a representation is a finite sum. In your case, representing base 2, the series certainly does not terminate (for, if it does, as you stated, it would be 0 modulo 3, which 1 is not). Uniqueness also does not have to hold. Someletters<Talk> 03:18, 28 April 2008 (UTC)[reply]

Might also be some convergence issues too actually. Been working in realms where convergence is free for too long...Someletters<Talk> 03:28, 28 April 2008 (UTC)[reply]

Thanks! --wj32 ^t/c 10:26, 27 April 2008 (UTC)[reply]

Well, not any base. I don’t think a base using any root of unity would be able to give you any integer, when using a finite character set. I wonder if every other base capable of representing all integers? GromXXVII (talk) 12:40, 27 April 2008 (UTC)[reply]

You sure can use a root of unity as a base; it's just stupid because it amounts to using unary. For example, in base

-1

, the integers are 1, 101, 10101, 1010101... —Keenan Pepper 18:11, 27 April 2008 (UTC)[reply]

The problem is, you completely lose uniqueness (normal radix systems aren't completely unique - see 0.999..., but they're not far off). The integers are also 100, 10100, 1010100, 101010100..., for example. --Tango (talk) 19:15, 27 April 2008 (UTC)[reply]

soda

The Sparkling Soda Company has asked you to design a new can for its soda that holds the same amount but increases its height by 25%. By what percent must the radius be decreased to accomplish this? The original can is 5 in. high and 2 in. in diameter.

Charlotte —Preceding unsigned comment added by 96.248.177.61 (talk) 20:34, 27 April 2008 (UTC)[reply]

Charlotte, the refdesk isn't allowed to do your homework for you. Please attempt it and the skilled mathematicians who frequent here will help you if you get stuck.

As a start, realise that "amount" is the same as volume and a soda can is a cylinder.

Zain Ebrahim (talk) 20:38, 27 April 2008 (UTC)[reply]