Wednesday, January 10, 2007

A Mathematical Challenge: Integrating A Sinusoidal Signal

For the past two days my head has been processing at maximum load!! Thanks goes to an anonymous poster. Mr. Anonymous has brought my attention to an older post, titled "Inconsistent Circuits Formula". His comments triggered my thoughts, and as some do notice, our thoughts aren't clear all the time. At the time I posted my old entry, I had a limited vision - Today, I come with new understanding: An understanding of the nature of the phenomena of resonance.

I will not post my newly-reached understanding in this post. I have a clear vision of the phenomena of resonance, but I want to support it with some mathematical justification, and more importantly, numerical analysis data. The results should be ready within two weeks depending on my free time - No promises though! You can read the results here.

Mr. Anonymous has argued about my bold statement "We cannot prove two physical (even mathematical) quantities to be equal". This post should indirectly justify my position. I am still considering to post a more direct argument can be found here. Just keep in mind, my position is strictly regarding inclusion of equalities for continuous quantity spectrum.

Finally, this mathematical challenge question was tailored by me in order to clarify a point I want to make, so try to solve it, and tell me your results.

The Mathematical Challenge:

The Final Answer:
ZERO (assuming that no noise exists in the signal)
ANY VALUE is the general solution

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Explained Answer:
The interval [0, 2*PI/a] represents one period of the sinusoidal signal cos(ax). When integrating a sinusoidal signal over one period the answer is ZERO. In short, the answer is ZERO regardless of the value of a.

A source of confusion might arise due to the use of the value ZERO for 'a'. The easiest approach is to devise an answer which is independent of 'a', as has been explained above. Note that the sinusoidal signal has a constant value of ONE over the finite region. The interval of the integration extends to ( 2*PI/a ), which also extends to the infinity. For this reason, the transformation of cos(ax) to ONE is NOT valid, because the sinusoidal signal is equal to 1 ONLY in the finite region, but the integration extends to the infinite region which implies that this transformation leads to incorrect results.

This solution applies only if we consider that there is NO NOISE in the sinusoidal signal. If we consider the possibility of existence of noise, then the integration will take the value of the integral of noise over the whole interval.

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Update 1: The answer to the challenge has been posted.
Update 2: The paper about resonance has been posted here.


Ghaith said...

This comment is not an answer to your challenge. This is simply that way I see the claimed effective current formula problem, physically. Let’s see if we got to the same conclusion :)

The effective current formula calculates the RMS value for a sinusoidal signal by evaluating the root of its mean squared values. The sum of the squared images of the function (signal) is obtained by evaluating an integral over the period of the signal.

Let assume we have the signals X, X1, X2 where

X = 14 * SQRT(2) * SIN (5000t)
X1 = 6 * SQRT(2) * SIN (5000t)
X2 = 8 * SQRT(2) * SIN (5000t)

Obviously, (X = X1 + X2) is physically and mathematically a sound equation. However, consider the following set of signals …

Y = 14 * SQRT(2) * SIN (5000t)
Y1 = 6 * SQRT(2) * SIN (5001t)
Y2 = 8 * SQRT(2) * SIN (5000t)

Stating that (Y) is approximately equal to (Y1 + Y2) is unacceptable both mathematically AND physically. In fact, it is not even close!

The reason for this is because by changing the frequency of either Y1 or Y2 (even with a minimal amount), the period of the function (Y1+Y2) changes! The amount of match between the images of the individual sinusoidal functions now depends on t. For relatively small values of t, the two sinusoidals will approximately be in phase, but as t increases their peaks will move apart slowly till the two signals look like they’re 180 degrees out of phase. The integral of (Y1+Y2) over the period of the (Y1+Y2) signal will contain certain values where Y1 was 180 degrees out of phase with Y2.

As the change between the frequencies of any two signals S1, S2 becomes smaller; the period of the combined signal (S1+S2) will increase. As the change approaches but does not reach zero, the period of (S1+S2) approaches infinity.

Adding a tiny value to one of the equal frequencies of two sinusoidals composing a certain signal may look like a small step while in fact it is the largest possible change that could be done to the signal, in terms of signal period.

This program demonstrates the point numerically, have fun :D

Devil's Mind said...

Exactly... This is the result that I got. But what might surprise you is that I came with a different conclusion, that for the two signal of equal frequencies, like:
X = X1 + X2
X1 = 6 * SQRT(2) * SIN (5000t)
X2 = 8 * SQRT(2) * SIN (5000t)

Here you consider that X has period [ T= 2*PI/5000 ] - But I believe that in order to make more sense, we should consider [ T=infinity ].

The simplest reasoning I can give you is that, since:
Y = Y1 + Y2
Y1 = 6 * SQRT(2) * SIN ((5000+a)t)
Y2 = 8 * SQRT(2) * SIN (5000t)

Now, the period [T] of Y goes to infinity as (a) approaches zero. So the reasonable conclusion is that, when (a) reaches zero, T is infinity, NOT some small value like (2*PI/5000)!!!

If the above made sense to you, then we are all good. If not then wait until I post a detailed explanation why I consider the view that [T=infinity] is more reasonable!

This mathematical challenge I put here will also help to explain my point regarding why [T] should be regarded as infinity! But keep that until its time, lets not hurry up and lose track!

Ghaith said...

It does make sense actually. I didn’t come to mention the case when a=0 because I got no mathematical interpretation. I’ll wait for yours!

Devil's Mind said...

Cool Ghaith, we reach an understanding! Now I want to see some people try at the challenge I have put here - If no one posts an answer this challenge loses its meaning!! :(

Tala said...

we have till next wednesday right?

Devil's Mind said...

Yeah!! Kind of...

Tala said...


Ghaith said...

Eh, what is it with you people? I think that the point Zeid is making is straight forward :)

Try a=1, a=2, a=3

Try with a=-1, a=-2, a=-3

You'll always get the same answer. Zeid is asking: would a=0 be any different?

Tala said...

well, you could be right, i wont argue about it but what i understand is that if you have a system and you apply a frequency which is equal to its natural frequency the response would go to infinity..

Ghaith said...

I think that you might have mixed things up. The integral itself does not appear to resemble the response of a system due to an input frequency.

Also, an input with a frequency equal to the natural frequency of the system will result in the maximum amount of power transfer to the output as far as I can tell. What did you mean by "the response would go to infinity"?

Anyways, Zeid is apparently working out some mathematics to support his physical claims (which I totally agree with). You don't have to map the integral to physics, the challenge is purely mathematics! :)

Devil's Mind said...

Yes, Ghaith is right, the frequency response reaches its maximum when the two frequencies are the same.

Also, this challenge is purely math, it has nothing to do with physics. Not yet, to say the least.

Devil's Mind said...

The answer to the challenge has been posted.