# 6.6 Gibbs phenomena  (Page 2/2)

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Below we will use the square wave, along with its Fourier Series representation, and show several figures that revealthis phenomenon more mathematically.

## Square wave

The Fourier series representation of a square signal below says that the left and right sides are "equal." In order tounderstand Gibbs Phenomenon we will need to redefine the way we look at equality.

$s(t)={a}_{0}+\sum_{k=1}$ a k 2 k t T k 1 b k 2 k t T

Figure 1 shows several Fourier series approximations of the square wave using a varied number of terms, denoted by $K$ :

When comparing the square wave to its Fourier series representation in [link] , it is not clear that the two are equal. The fact that thesquare wave's Fourier series requires more terms for a given representation accuracy is not important. However, closeinspection of [link] does reveal a potential issue: Does the Fourier series reallyequal the square wave at all values of $t$ ? In particular, at each step-change in the square wave, theFourier series exhibits a peak followed by rapid oscillations. As more terms are added to the series, theoscillations seem to become more rapid and smaller, but the peaks are not decreasing. Consider this mathematicalquestion intuitively: Can a discontinuous function, like the square wave, be expressed as a sum, even an infinite one, ofcontinuous ones? One should at least be suspicious, and in fact, it can't be thus expressed. This issue brought Fourier much criticism from the French Academy of Science (Laplace, Legendre, and Lagrange comprised thereview committee) for several years after its presentation on 1807. It was not resolved for also a century, and itsresolution is interesting and important to understand from a practical viewpoint.

The extraneous peaks in the square wave's Fourier series never disappear; they are termed Gibb's phenomenon after the American physicist Josiah Willard Gibbs. They occur whenever the signal isdiscontinuous, and will always be present whenever the signal has jumps.

## Redefine equality

Let's return to the question of equality; how can the equal sign in the definition of the Fourier series be justified? The partial answer is that pointwise--each and every value of $t$ --equality is not guaranteed. What mathematicians later in the nineteenth century showed was that the rmserror of the Fourier series was always zero.

$\lim_{K\to }K\to$ rms ε K 0
What this means is that the difference between an actual signaland its Fourier series representation may not be zero, but the square of this quantity has zero integral! It is through the eyes of the rms value that we define equality:Two signals ${s}_{1}(t)$ , ${s}_{2}(t)$ are said to be equal in the mean square if $\mathrm{rms}({s}_{1}-{s}_{2})=0$ . These signals are said to be equal pointwise if ${s}_{1}(t)={s}_{2}(t)$ for all values of $t$ . For Fourier series, Gibb's phenomenon peaks have finite height and zero width: Theerror differs from zero only at isolated points--whenever the periodic signal contains discontinuities--and equalsabout 9% of the size of the discontinuity. The value of a function at a finite set of points does not affect itsintegral. This effect underlies the reason why defining the value of a discontinuous function at its discontinuity ismeaningless. Whatever you pick for a value has no practical relevance for either the signal's spectrum or for how asystem responds to the signal. The Fourier series value "at" the discontinuity is the average of the values oneither side of the jump.

## Visualizing gibb's phenomena

The following VI demonstrates the occurrence of Gibb's Phenomena. Note how the wiggles near the square pulse to the left remain even if you drastically increase the order of the approximation, even though they do become narrower. Also notice how the approximation of the smooth region in the middle is much better than that of the discontinuous region, especially at lower orders.

## Conclusion

We can approximate a function by re-synthesizing using only some of the Fourier coefficients(truncating the F.S.)

$\frac{d {f}_{N}(t)}{d }}=\sum {c}_{n}e^{i{\omega }_{0}nt}$
This approximation works well where $f(t)$ is continuous, but not so well where $f(t)$ is discontinuous. In the regions of discontinuity, we will always find Gibb's Phenomena, which never decrease below 9% of the height of the discontinuity, but become narrower and narrower as we add more terms.

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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Commplementary angles
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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