why 50-ohm impedance?

I wond how it defined that the typical Impedance in PCB is 50Ω.

from where they got this value?
why not 80Ω? or 20Ω?

This goes back to the days prior to 1939 and coaxial cable. For a fixed outer diameter, 77 ohms produced the least attenuation and 30 ohms had the highest power capability. So 50 ohms was a compromise. Then in modern times RF modules are made to interface to 50 ohm cables and then surface mount RF modules got the 50 ohm designed into them.

check the follwing question and discussion--
============================
Why 50 Ohms?

(Originally published in EDN Magazine, September 2000)

Q: Why do most engineers use 50W pc-board transmission lines (sometimes to the extent of this value becoming a default for pc-board layout)? Why not 60 or 70W ?—Tim Canales

A: Given a fixed trace width, three factors heavily influence pc-board-trace impedance decisions. First, the near-field EMI from a pc-board trace is proportional to the height of the trace above the nearest reference plane; less height means less radiation. Second, crosstalk varies dramatically with trace height; cutting the height in half reduces crosstalk by a factor of almost four. Third, lower heights generate lower impedances, which are less susceptible to capacitive loading.

All three factors reward designers who place their traces as close as possible to the nearest reference plane. What stops you from pressing the trace height all the way down to zero is the fact that most chips cannot comfortably drive impedances less than about 50W. (Exceptions to this rule include Rambus, which drives 27W, and the old National BTL family, which drives 17W).

It is not always best to use 50W. For example, an old NMOS 8080 processor operating at 100 kHz doesn't have EMI, crosstalk, or capacitive-loading problems, and it can't drive 50W anyway. For this processor, because very high-impedance lines minimize the operating power, you should use the thinnest, highest-impedance lines you can make.

Purely mechanical considerations also apply. For example, in dense, multilayer boards with highly compressed interlayer spaces, the tiny lithography that 70W traces require becomes difficult to fabricate. In such cases, you might have to go with 50W traces, which permit a wider trace width, to get a manufacturable board.

What about coaxial-cable impedances? In the RF world, the considerations are unlike the pc-board problem, yet the RF industry has converged on a similar range of impedances for coaxial cables. According to IEC publication 78 (1967), 75W is a popular coaxial impedance standard because you can easily match it to several popular antenna configurations. It also defines a solid polyethylene-based 50W cable because, given a fixed outer-shield diameter and a fixed dielectric constant of about 2.2 (the value for solid polyethylene), 50W minimizes the skin-effect losses.

You can prove the optimality of 50W coaxial cable from basic physics. The skin-effect loss, L , (in decibels per unit length) of the cable is proportional to the total skin-effect resistance, R , (per unit length) divided by the characteristic impedance, Z 0 , of the cable. The total skin-effect resistance, R , is the sum of the shield resistance and center conductor resistances. The series skin-effect resistance of the coaxial shield, at high frequencies, varies inversely with its diameter d 2 . The series skin-effect resistance of the coaxial inner conductor, at high frequencies, varies inversely with its diameter d 1 . The total series resistance, R , therefore varies proportionally to (1/d 2 +1/d 1 ). Combining these facts and given fixed values of d 2 and the relative electric permittivity of the dielectric insulation, E R , you can minimize the skin- effect loss, L , starting with the following equation:

In any elementary textbook on electromagnetic fields and waves, you can find the following formula for Z 0 as a function of d 2 , d 1 , and E R :

Substituting Equation 2 into Equation 1 , multiplying numerator and denominator by d 2 , and rearranging terms:

Equation 3 separates out the constant terms /60)´(1/d 2 )) from the operative terms ((1+d 2 /d 1 )/ln(d 2 /d 1 )) that control the position of the minimum. Close examination of Equation 3 reveals that the position of the minima is a function only of the ratio d 2 /d 1 and not of either E R or the absolute diameter d 2 .

A plot of the operative terms from L , as a function of the argument d 2 /d 1 , shows a minimum at d 2 /d 1 =3.5911. Assuming a solid polyethylene insulation with a dielectric constant of 2.25 corresponding to a relative speed of 66% of the speed of light, the value d 2 /d 1 =3.5911 used in Equation 2 gives you a characteristic impedance of 51.1W. A long time ago, radio engineers decided to simply round off this optimal value of coaxial-cable impedance to a more convenient value of 50W. It turns out that the minimum in L is fairly broad and flat, so as long as you stay near 50W, it doesn't much matter which impedance value you use. For example, if you produce a 75W cable with the same outer-shield diameter and dielectric, the skin-effect loss increases by only about 12%. Different dielectrics used with the optimal d 2 /d 1 ratio generate slightly different optimal impedances.
================================

I think it has answered ur question........