tag:blogger.com,1999:blog-9571647.post3166183076364488492..comments2024-02-01T02:16:29.656+03:00Comments on Devil's Mind: A Mathematical Challenge: Integrating A Sinusoidal SignalDevil's Mindhttp://www.blogger.com/profile/10541884626112839842noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-9571647.post-20139105410414075302007-01-17T18:39:00.000+02:002007-01-17T18:39:00.000+02:00The answer to the challenge has been posted.The answer to the challenge has been posted.Devil's Mindhttps://www.blogger.com/profile/10541884626112839842noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-45495113901816337712007-01-16T23:37:00.000+02:002007-01-16T23:37:00.000+02:00Yes, Ghaith is right, the frequency response reach...Yes, Ghaith is right, the frequency response reaches its maximum when the two frequencies are the same. <br /><br />Also, this challenge is purely math, it has nothing to do with physics. Not yet, to say the least.Devil's Mindhttps://www.blogger.com/profile/10541884626112839842noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-62538525813043877502007-01-16T17:08:00.000+02:002007-01-16T17:08:00.000+02:00I think that you might have mixed things up. The i...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.<br /><br />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"?<br /><br />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! :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9571647.post-67002014008741203572007-01-16T14:12:00.000+02:002007-01-16T14:12:00.000+02:00well, you could be right, i wont argue about it bu...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..Talahttps://www.blogger.com/profile/17308820896939744575noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-42483990833402270492007-01-16T13:43:00.000+02:002007-01-16T13:43:00.000+02:00Eh, what is it with you people? I think that the p...Eh, what is it with you people? I think that the point Zeid is making is straight forward :)<br /><br />Try a=1, a=2, a=3<br /><br />Try with a=-1, a=-2, a=-3<br /><br />You'll always get the same answer. Zeid is asking: would a=0 be any different?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9571647.post-36999911159261943872007-01-16T03:09:00.000+02:002007-01-16T03:09:00.000+02:00infinityinfinityTalahttps://www.blogger.com/profile/17308820896939744575noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-56369445768668050622007-01-13T03:22:00.000+02:002007-01-13T03:22:00.000+02:00Yeah!! Kind of...Yeah!! Kind of...Devil's Mindhttps://www.blogger.com/profile/10541884626112839842noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-42543463914931474952007-01-13T02:27:00.000+02:002007-01-13T02:27:00.000+02:00we have till next wednesday right?we have till next wednesday right?Talahttps://www.blogger.com/profile/17308820896939744575noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-35331492665808552452007-01-11T00:15:00.000+02:002007-01-11T00:15:00.000+02:00Cool Ghaith, we reach an understanding! Now I want...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!! :(Devil's Mindhttps://www.blogger.com/profile/10541884626112839842noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-11857018966722745372007-01-10T23:57:00.000+02:002007-01-10T23:57:00.000+02:00It does make sense actually. I didn’t come to ment...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!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9571647.post-1093034052666888352007-01-10T23:18:00.000+02:002007-01-10T23:18:00.000+02:00Exactly... This is the result that I got. But what...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:<br />X = X1 + X2<br />X1 = 6 * SQRT(2) * SIN (5000t)<br />X2 = 8 * SQRT(2) * SIN (5000t)<br /><br />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 ].<br /><br />The simplest reasoning I can give you is that, since:<br />Y = Y1 + Y2<br />Y1 = 6 * SQRT(2) * SIN ((5000+a)t)<br />Y2 = 8 * SQRT(2) * SIN (5000t)<br /><br />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)!!!<br /><br />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!<br /><br />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!Devil's Mindhttps://www.blogger.com/profile/10541884626112839842noreply@blogger.comtag:blogger.com,1999:blog-9571647.post-68662220487442648162007-01-10T22:16:00.000+02:002007-01-10T22:16:00.000+02:00This comment is not an answer to your challenge. T...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 :)<br /><br />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 <b>over the period of the signal</b>. <br /><br />Let assume we have the signals X, X1, X2 where<br /><br />X = 14 * SQRT(2) * SIN (5000t)<br />X1 = 6 * SQRT(2) * SIN (5000t)<br />X2 = 8 * SQRT(2) * SIN (5000t)<br /><br />Obviously, (X = X1 + X2) is physically and mathematically a sound equation. However, consider the following set of signals …<br /><br />Y = 14 * SQRT(2) * SIN (5000t)<br />Y1 = 6 * SQRT(2) * SIN (5001t)<br />Y2 = 8 * SQRT(2) * SIN (5000t)<br /><br />Stating that (Y) is approximately equal to (Y1 + Y2) is unacceptable both mathematically AND physically. In fact, <b>it is not even close!</b><br /><br />The reason for this is because by changing the frequency of either Y1 or Y2 (even with a minimal amount), <b>the period of the function (Y1+Y2) changes!</b> 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.<br /><br />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.<br /><br />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 <b>the largest possible change</b> that could be done to the signal, in terms of signal period.<br /><br /><a href="http://www.geocities.com/gtarawneh/RMS.zip">This program</a> demonstrates the point numerically, have fun :DAnonymousnoreply@blogger.com