answer with MATLAB to the question: Why common coaxial characteristic impedance Z0 is 50Ω or 75Ω?

Published by John_2016 in the blog blog title. Views: 64


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Among the many answers to this question, one available is, Ahaneku and Duru [DURU]

https://www.researchgate.net/public...dling_capability_of_coaxial_transmission_line

Following, with MATLAB, the part explaining how Z0(b/a) is optimised, that explains common values of coaxial Z0:

While

Z0=etha0/(2*pi)*(mu0*mur/(e0*er))^.5*log(b/a)

the alpha of the conductor, the attenuation caused by metal is

alpha_c=Rs/(a*etha*(mur/er)^.5)*(b/a+1)/log(b/a)


After plugging Cu conductivity

sigma_Cu=5.183e7 % S/m

the expression with the ratio to optimise is

Rs=(pi*f0/sigma_Cu)^.5



Ed is |E| to be reached in order to disrupt dielectric.

For case of air

Ed_air=3e6 % V/m

here ignored but also depending upon air humidity and air pressure.


Note [DURU]'s b and a are while [POZAR] Microwave Engineering 4th edition chapter 3 coaxial b and a are radii.

https://www.amazon.co.uk/Microwave-Engineering-David-M-Pozar/dp/0470631554


[z=[0:.0001:10];f2=10*log10((exp(z)+1)./z);
figure(20);plot(z,f2);
nzmin=find(f2==min(f2))
f2(nzmin)
z(nzmin)
b_over_a_optim=exp(z(nzmin))
hold on;plot(z(nzmin),f2(nzmin),'ro')
text(z(nzmin),20+f2(nzmin),['optim b/a = exp(zmin) =' num2str(b_over_a_optim)])
etha0=377;

mur=1;er=1
% no filling material
mu0=4*pi*10^-7; % H/m
e0=8.853*10^-12; % F/m





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no fill-up or fill-up with materials that have er close to 1, that tend to be cheaper

than those with er>>1 when fillip up transmission lines


er=1
Z0=1/( 2*pi)*(mu0*mur/(e0*er))^.5*z(nzmin)

Z0=
76.662073914302525


when coaxial fill-up is PTFE (Teflon)


er=2.2
Z0=1/( 2*pi)*(mu0*mur/(e0*er))^.5*z(nzmin)

Z0 =
51.685559689169658



Now let's have a look at real 75Ω coaxial specifications:

https://www.commscope.com/catalog/cables/pdf/part/46892/7451203_F677TSVV_XP.pdf

since these specs read 'foam PE' what is the relative permittivity of the filling material?

a=1.016;b=4.512; % both mm
Z0=75;
er=(etha0/Z0*1/(2*pi)*log(b/a))^2

er =
1.201557466938515



matching the relative permittivity of Foam Polyethylene shown in table [PRAD]

http://www.pulsedpower.net/Info/common_dielectrics.htm


Yet one has to be aware of the wide frequency span that supplier claims the coaxial is operative because the attenuation varies widely depending upon what frequencies used.


sigma_Cu=5.183e7 % S/m
etha0=377;
f1=5e6;f2=1.8e9;f0=(f1+f2)/2;
% chosen centre band in specs
df=(f2-f1)/25;f=[f1:df:f2];

Rs=(pi*f/sigma_Cu).^.5

a=1.016;b=4.512;
% both mm, Commscope F677TSW EXPRESS
mur=1;er=1.2 % cheap coax
alpha_c=Rs./(2*etha0*(mur/er)^.5)*(1/a+1/b)*1/log(b/a);
Np2dB=10*log10(exp(1)^2);
alpha_c_dB=alpha_c*Np2dB;
alpha_c_dB_100m=alpha_c_dB*100;




and directly from the previous manufacturer's datasheet:



Lcoax75=[1.9 5.25 6.4 6.46 9.35 9.84 10 10.82 11.64 12.63 13.61 14.43 15.29 16.08 16.73 18.54 20.01 21.49 23.66 24.71 25.71 26.68 27.63 28.54 29.39 29.44];
figure;semilogx(f,alpha_c_dB_100m);
hold on;
semilogx(f,Lcoax75);grid on
legend('theoretical','real measurements')




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We can start feeling the need to test, if not all, at least all the critical components of any expensive system. Theoretical models? datasheets?

Test the one you buy.

when systems are really expensive one of the strategies is to send someone to factory to test what you buy right at the factory just before shipping, and then test again just on arrival at your premises before any further assembly, with agreed heavy penalties on contract.

MATLAB scripts cannot be uploaded as attachment to forum posts yet, so I am sending copy of the .m script to anyone asking by email, check my profile for contact details.
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