A Quadrature Signals Tutorial: Complex, But Not Complicated

Understanding the 'Phasing Method' of Single Sideband Demodulation

Complex Digital Signal Processing in Telecommunications

Introduction to Sound Processing

Introduction of C Programming for DSP Applications

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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.

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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.
```

```
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
```

On Thu, 26 Jun 2008 02:52:20 -0500, "vasindagi" <v...@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.

```
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
```

```
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
```

"vasindagi" <v...@gmail.com> wrote in message news:E--d...@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