[互助]射频同轴为什么是50Ω或者75Ω？

微波技术基础都有推到的

这个应该是先由于制造商的实际可制造性决定的，开始有许多不同的阻抗（如30，50，75欧等），而后才由于需要变成行业标准及国家/国际标准的。

选择50Ω基准阻抗是对空气同轴电缆的最小损耗和最大功率容量尺寸之间折中的结果。对于空气介质同轴电缆，为使损耗最小，要求外导体内径与内导体外径的最佳比值为3.6，对应于阻抗Zo为77Ω。虽然从损耗的角度看，这时的性能最佳，但这一尺寸比不能提供出现介质击穿前允许的最大峰值功率容量。最大功率容量时，外导体内径与内径导体之比应为1.65，对应的阻抗Zo为30Ω。77Ω和30Ω的几何平均近似为50Ω。这样，50Ω标准即是同轴电缆最小损耗和最大功率容量之间的折中值。 XAB/S8e  

谢谢您的回答！我大概明白了！

谢谢各位的回答，我在书上找到了理论解释，特拿出来分享。

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. k=bv!T_o  
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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). jO:<"l^+u  
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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. qm]ljut  
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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. 7qXgHrr0|U  
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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. 4b3p,$BWS  
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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: 5'V'~Q%  
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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 : B{1+0k  
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Substituting Equation 2 into Equation 1 , multiplying numerator and denominator by d 2 , and rearranging terms: N'
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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 . /'VCJjzZ
 
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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. [
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谢谢楼上的回答，不过英文让我看着头疼，不过有空了一定好好看看！