Electronic – Nyquist-Shannon sampling theorem and highest frequency

I heard somewhere that Fourier transform only works on a signal that contains parts that are harmonic to each other (i.e. f, 2f, 3f etc.).

No. What you say would imply that the signal must be periodic, and that is not true. The Fourier transform can be computed for any signal, periodic or not, with finite energy or even with finite power. In the latter case (of finite power), the transform will, in most cases, be unbounded.

What you say is correct for Fourier series, not for the Fourier transform. The Fourier series decomposition is defined only for periodic signals. The spectrum of a periodic signal will be able to have contents only at \$f_0\$, \$2f_0\$, \$3f_0\$, etc. That does not mean that it will always have contents at all those frequencies. It means that it cannot have contents outside that set of frequencies.

Can anyone explain me on proof of this?

That was not right, so it cannot be proved.

What is the sufficient condition for Fourier transform?

There is no "the sufficient condition". There are several sufficient conditions.

For instance, one sufficient condition is that it satisfies both Dirichlet conditions:

1) Over any time interval of finite length, the function w(t) is single valued with a finite number of maxima and minima, and the number of discontinuities (if any) is finite.

2) w(t) is absolutely integrable.

That condition is sufficient, but not necessary.

A weaker sufficient condition for the existence of the Fourier transform is that the signal has finite energy. All physically realizable waveforms are finite-energy, so that means that all physical waveforms encountered in engineering practice are Fourier transformable. However, that is to have a Fourier transform with bounded values. If you allow yourself to work with unbounded values, then you can also compute the Fourier transform of a signal that has finite power (which is a less strict requirement). For instance, a sin(\$2\pi f_0·t\$) that exists for all t has finite power, but not finite energy. You can compute the Fourier transform of such a signal, but it will have unbounded values (in fact, "deltas") at \$f_0\$ and \$-f_0\$.

edited for more depth & clarity to avoid artifacts, pun intended

You can control your artifacts with an improved "anti-alias" filter to eliminate all content of signals above the Nyquist rate. This is often more critical if the image has been digitized already by another system at some unknown pixel rate.

For example a high resolution scan of magazine photo without the appropriate anti-alias filter setting will result in herringbone artifacts that are much worse than the original. This is a similar effect to the stripes of a shirt on TV with harmonic pixel interference herringbone patterns or the fringing patterns of lenses on a digital camera or camcorder in the presence of infrared light and similar resolution subject matter above the pixel resolution of the camera.

How you implement the anti-alias filter depends on the original signal.

• In the more general sense, if the original signal is natural light to be captured for 2D imaging, you might use an AA optical "blur filter".
• If the original signal is light reflected from a digital printed object in a scanner, you might tune the alias digital blur filter to match the resolution of the original print and synchronously if possible. e.g. 100 dpi for fax content
• If the original signal is a 3D scan using magneto-resonant imaging then you need to determine is it from stop-band remnant signals or pass-band group delay distortion.

The quality factors of low artifacts depend on the relative levels of interference from each to avoid alias artifacts.

Over-sampling makes the job of ideal bandstop filtering easier but ends up with huge file sizes so this can be followed by undersampling to achieve easier brick wall filters with desired blur filter in pass band.

Wiki has a more examples of various anti-alias filters and also spacial filtering applications.