The table merely says that given a channel with a fixed bandwidth, if you transmit a signal through it (input can contain infinite harmonics), at the output only the first harmonics will get through, where a high baud rate (high fundamental frequency) will get less harmonics through than a low baud rate signal. In other words, a low baud rate signal will come out the other side relatively unaffected by the channel, while a high baud rate signal can be severely distorted (even completely filtered out).
This is important because if the signal gets too distorted, the original bits cannot be reliably recovered due to intersymbol interference and the increased relevance of noise perturbations.
So a low number of harmonics does not imply a low bit rate. Quite the opposite, a low bit rate implies a high number of harmonics going through the channel.
The general criterion for finding baud rate vs channel bandwidth limits are given by the Nyquist Criterion. Note that, although related, this is different than the sampling theorem mentioned in another answer (Shannon-Nyquist), because you are not trying to reconstruct an original "bandwidth limited signal" in the analog sense from a set of samples, it is a matter of determining the original symbols, which is related to threshold decisions, which can be based on sampling, transmitted power (areas), etc.
What are the sources of harmonics? Why are they generated?
Assume that we have a voltage waveform of the form:
\$v_s(t) = \cos(\omega t) \$
In words, the source voltage waveform is composed of a single sinusoid of (angular) frequency \$\omega\$.
If this waveform is the input to a linear circuit, the output will also be composed of a single sinusoid of the same frequency as the input.
For example, a linear voltage amplifier scales the input signal by some constant \$A_v \$:
\$v_o(t) = A_v v_s(t) = A_v \cos(\omega t)\$
Now, consider what happens when the amplifier is non-linear. For example:
\$v_o(t) = A_v v_s(t) + 2\alpha v^2_s(t)\$
This amplifier has a 2nd order non-linearity. By a simple trigonometry identity, we have:
\$v_o(t) = A_v \cos(\omega t) + \alpha[1 + \cos(2\omega t)]\$
See what happened? The output is no longer composed of a single frequency but, due to the non-linear term, now has a DC component as well as 2nd harmonic component.
If instead of a 2nd order non-linearity, the amplifier had a 3rd order non-linearity:
\$v_o(t) = A_v v_s(t) + 4\beta v^3_s(t) \$
you might guess that a 3rd harmonic will be generated. Let's see:
\$v_o(t) = (A_v + 3\beta)\cos(\omega t) + \beta \cos(3\omega t)\$
Note that the 3rd order non-linearity creates a 3rd harmonic as well as an additional 1st order term.
Essentially, even-order nonlinearities generate even harmonics while odd-order nonlinearities generate odd harmonics.
Now, a symmetric circuit, such as a complementary push-pull circuit, generates odd-order harmonics for the reason that the even-order nonlinearities cancel.
An example of a circuit that creates 2nd order harmonics is a single-ended FET (a square-law device) amplifier.
Best Answer
You know that a square wave with perfect symmetry has NO EVEN harmonics. (typo fixed)
You can imagine if a magnetic material had a "hard saturation" limit instead of a "Landau"? curve with hysteresis, you would get a square wave current for a sine wave voltage. Thus all the current is fundamental & odd harmonics. But with a symmetrical soft saturation curve the harmonics are attenuated. The harmonics increase as the current enters the soft saturation region, yet still remain ODD multiples as long as the non-linear soft saturation curve is symmetrical .
Thus the only time you would get even harmonics in a magnetic material is if there was remanence such as from DC bias. Then the voltage swing produces more asymmetrical current and no longer is a symmetrical soft square wave but asymmetrical even harmonic producing fourier components of the distorted waveform. Then the apparent inductance drops quickly and is usually rated for -10% at rated DC or ac current.
Normally large transformers experience remenance from abrupt disconnects and flux leakage resistance balances the current after several seconds which is apparent by the hum of large MVA transformers during reconnect. This is why "smart reclosures" remember the phase of disconnect and reclose at exactly the same phase to minimize Remenance and saturation currents that produce many forces inside transformers and even harmonics.
To visualize fourier components with a hand-drawn waveform or std. waveform try this java app. then choose boxes for mag/phase and log view and change the frequency spikes with mouse and see the effects on time domain signal. Note the absence of even harmonics in both a square wave and a triangle wave but the phase is different in each harmonic.