Hi, I m trying to learn more about gibbs phenomenon. I found various sites explaining about what is gibbs phenomenon but none of them epxlained the reason behind the phenomenon. Can anyone please explain the reason for the gibbs phenomenon or send me links about the same. Thanks in advance.

# Gibbs Phenomenon

Started by ●June 26, 2008

Posted by ●June 26, 2008

On Jun 26, 7:52 pm, "vasindagi" <vish...@gmail.com> wrote:> Hi, > I m trying to learn more about gibbs phenomenon. I found various sites > explaining about what is gibbs phenomenon but none of them epxlained the > reason behind the phenomenon. > Can anyone please explain the reason for the gibbs phenomenon or send me > links about the same. > Thanks in advance.For a periodic waveform it can be made of of sine-waves of various harmonic frequencies. If you miss out the higher frequencies you end up with a "bad-fit" - loads on the net. K.

Posted by ●June 26, 2008

On 26 Jun, 09:52, "vasindagi" <vish...@gmail.com> wrote:> Hi, > I m trying to learn more about gibbs phenomenon. I found various sites > explaining about what is gibbs phenomenon but none of them epxlained the > reason behind the phenomenon.The Fourier transform is defined for *piecewise* continuous functions, which are continuous everywhere except in some discrete points where there are jumps. One example is a rectangular pulse, where the leading and trailing flanks are jump discontinuities. The FT, on the other hand, represents the function in terms of functions that are continuous. The Gibbs phenomenon is a consequence from trying to represent discontinuous functions in terms of continuous ones. Rune

Posted by ●June 26, 2008

On Thu, 26 Jun 2008 02:52:20 -0500, "vasindagi" <vishur6@gmail.com> wrote:>Hi, >I m trying to learn more about gibbs phenomenon. I found various sites >explaining about what is gibbs phenomenon but none of them epxlained the >reason behind the phenomenon.Gibbs phenomenon happens when one tries to represent a signal with discontinuities with Fourier series. If the number of components used in the expansion is finite, it can't represent the discontinuity as a finite sum continuous functions is by definition continuous so there will always be an error.

Posted by ●June 26, 2008

vasindagi schrieb:> Hi, > I m trying to learn more about gibbs phenomenon. I found various sites > explaining about what is gibbs phenomenon but none of them epxlained the > reason behind the phenomenon. > Can anyone please explain the reason for the gibbs phenomenon or send me > links about the same.Depends on "how deep" you want to dig to understand the reason. I would say, the reason for this phenomenon is the "uncertainty relation of signal processing", meaning that you cannot represent a signal that is sharp both in the frequency (Fourier) space and in the spatial (regular/time) domain. Specifically, spikes or edges (depending on whether that's audio or visual information) representing localized objects or phenomena relate to "unlocalized" data in the Fourier domain (e.g. the spectrum of a sharp spike covers more or less all frequencies), and also vice versa. That is, as soon as you represent data in the Fourier domain, and then introduce some kind of loss in this domain by, for example, dropping small coefficients or band-limiting the signal, you end up with "sharpness loss" and the typical Gibbs artifacts near edges; to describe them precisely, an infinite Fourier spectrum is required. Mathematically, one can show that the Fourier series (or the Fourier integral) converges for continuous functions, and converges to the average of the two values at the discontinuity, but the closer you get to the discontinuity, the slower the rate of convergence will become. Specifically, one can also see that for every finite number of terms in the Fourier series approximating a square wave, there is always a point near the edge where the series differs by a constant term from the source signal - mathematically, the convergence is only in l^2 sense (mean square error) and not in the pointwise (l^infinity) sense. This difference is exactly the Gibbs phenomenon. So long, Thomas

Posted by ●June 26, 2008

On Jun 26, 3:52 am, "vasindagi" <vish...@gmail.com> wrote:> Hi, > I m trying to learn more about gibbs phenomenon. I found various sites > explaining about what is gibbs phenomenon but none of them epxlained the > reason behind the phenomenon. > Can anyone please explain the reason for the gibbs phenomenon or send me > links about the same. > Thanks in advance.They explanation about the limitation of not using higher frequencies isn't really correct - even though it displays the same type of behaviour. Even if you used an infinite number of frequencies the result does not converge at that point i.e. the discontinuity. So what you're seeing is the limitation of sum of sinusoids representation - it does not form a complete basis. It is like trying to represent 3D space with only 2 vectors. A linear combination of 2 vectors can only map out a plane and so you can only get so close to representing any point in 3D space. Hope that helps. Cheers, Dave

Posted by ●June 26, 2008

"vasindagi" <vishur6@gmail.com> wrote in message news:E--dnXOJr4Yp1_7VnZ2dnUVZ_orinZ2d@giganews.com...> Hi, > I m trying to learn more about gibbs phenomenon. I found various sites > explaining about what is gibbs phenomenon but none of them epxlained the > reason behind the phenomenon. > Can anyone please explain the reason for the gibbs phenomenon or send me > links about the same. > Thanks in advance.http://en.wikipedia.org/wiki/Gibbs_phenomenon This makes it pretty clear along with the explanations given by others. My 2 cents: You are trying to represent a waveform of infinite frequency content - and the series doesn't converge as was pointed out. You may want to ask: "How do I get rid of it?" The answer is to *not* try to represent discontinuities that are as sharp as a pure theorectical square wave would be. In the real world they don't exist anyway - although you can certainly generate a waveform that has noticeable Gibbs phenomenon which is a bandlimited signal. Use a lowpass filter of some reasonable bandwidth to smooth out those edges and the ringing goes away. Fred

Posted by ●June 26, 2008

On 26 Jun., 14:43, Dave <dspg...@netscape.net> wrote:> On Jun 26, 3:52 am, "vasindagi" <vish...@gmail.com> wrote: > > > Hi, > > I m trying to learn more about gibbs phenomenon. I found various sites > > explaining about what is gibbs phenomenon but none of them epxlained the > > reason behind the phenomenon. > > Can anyone please explain the reason for the gibbs phenomenon or send me > > links about the same. > > Thanks in advance. > > They explanation about the limitation of not using higher frequencies > isn't really correct - even though it displays the same type of > behaviour. Even if you used an infinite number of frequencies the > result does not converge at �that point i.e. the discontinuity. > > So what you're seeing is the limitation of sum of sinusoids > representation - it does not form a complete basis.You are right in that those do not form a complete basis. However, finite sums of sinusoids approximation to periodic functions don't necessarily display Gibbs ringing at points of discontinuity. Truncating the Fourier series of a periodic function after the first N terms is the best N-bandlimited approximation in the L2 sense (and results in Gibb's). However, using windowing one can reduce or do away with Gibbs ringing altogether for N term sum of sinusoids approximation - one of the points in windowed FIR design is to reduce Gibbs ringing in the approximation of ideal frequency responses. The resulting windowed N term approximation won't be optimal in the L2 sense, but you can suppress the Gibbs ringing. Regards, Andor

Posted by ●June 26, 2008

vasindagi wrote:

Posted by ●June 26, 2008

On Jun 26, 3:52�am, "vasindagi" <vish...@gmail.com> wrote: