Figure 5 has S-parameters for ADF4113 RF input. You can calculate input impedance from these.
\$R=\dfrac{Z_0(1-\rm{MAG}^2)}{1+\rm{MAG}^2-2\cdot\rm{MAG}\cdot\cos(\frac{\rm{ANG}}{360}\cdot 2\pi)}\$
\$X=\dfrac{2\cdot\rm{MAG}\cdot\sin(\frac{\rm{ANG}}{360}\cdot 2\pi)\cdot 50}{1+\rm{MAG}^2-2\cdot\rm{MAG}\cdot\cos(\frac{\rm{ANG}}{360}\cdot 2\pi)}\$
\$Z = \sqrt{R^2 + X^2}\$
Where \$Z_0 = 50\$
I hope I latexed the formulas correctly, just in case they are from this page.
Any kind of electrical power transfer has a typical ratio of voltage to current. For example, you can deliver 100 watts by 1 amp at 100 volts, or 10 amps at 10 volts, or any other product that comes to 100. It's convenient to express the ratio of volts to amps as a number of ohms (since dimensionally that's all ohms are anyway). Power sources, loads, and even transmission lines all have characteristic impedance, which tells you something about what will happen when things are connected together.
Impedance matching is the selection of components with identical impedance, or the addition of impedance transforming components to cause a component with an undesired impedance to appear as though it has a more desirable one.
As Brian Carlton pointed out, when you have matched impedance, you achieve maximal power transfer. This is often desirable, but not always. The thing to remember is that maximal power transfer is achieved at the cost of dissipating equal power in the source and the load.
So for example, a case for not matching impedance is when you want to efficiently use the energy from a source. A 0.1 ohm load would get optimal power out of a battery with a 0.1 ohm internal resistance, but half the energy would be dissipated in the battery itself, which would be rather wasteful of the stored energy. (Not to mention that the terminal voltage would fall to half!) By purposely using a load much with much higher resistance than the battery, most of the stored energy ends up doing work in the load.
On the other hand, you DO want to match impedance when, for example, you have an audio amplifier stage that ideally wants to drive a 600 ohm load, but you only have a 3.2 ohm speaker. An ordinary transformer, having a 1:N voltage ratio, will conveniently give you a 1:N^2 impedance ratio. Another common case is in RF work, where as volting pointed out, you need to minimize reflections, because reflected energy can cause excessive power dissipation in your source.
Best Answer
if you are designing a pre-amp: then there are a couple of things you need to take in consideration
Usually (but not always)
(I just find it handy to have some DI boxes ready, knowing your outputs/inputs, and wire accordingly when the situation requires... much easier than catering for every possible combination)
Generally speaking Guitar Outputs are "Line Level" "Unbalanced" "Hi-Z" If the Guitar pickup/electronics are "active" (e.g. there is a battery or so)
If it is a Guitar without active components, then it is likely to be "Mic Level" "Unbalanced" and can be either "Hi-Z" or "Lo-Z" depending on pickup/built-in-mic.
Some "effects" used with guitar may increase even over Line Level, which means you need to attenuate, which can be done either by DI or on the Mixing Desk by "pads" or "attenuators" (terminology differs from brand to brand)
... my 2 cents ...
[EDIT] As this question is migrated from sound.stackexchange.com to electronics.stackexchange.com, this answer is given from a "sound production perspective" without going into further electronics design. While I do have some electronic skills, I would not edit this answer further as I feel I would not be skilled enough to advise on electronic subjects like designing an audio-pre-amp to use with Pro-Audio applications. Please take that into account while you consider voting or commenting on this answer.