example 3.1.2
ZIP with MATLAB scripts and note:
example 3.1.2 notes:
Why common coaxial characteristic impedance Z0 is 50W or 75W?
Among the many answers to this question, one available is, Ahaneku and Duru [DURU] https://www.researchgate.net/publication/305083350_Analyzing_the_power_handling_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 specific parameters for Cu sigma_Cu=5.183e7 S/m Rs=(pi*f0/sigma_Cu)^.5 shows the optimisation ratio.
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 diameter while [POZAR]'s b and a are radii.
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
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
Z0=1/( 2*pi)*(mu0*mur/(e0*er))^.5*z(nzmin)
if coaxial fill-up were PTFE (Teflon)
er=2.2
Z0=1/( 2*pi)*(mu0*mur/(e0*er))^.5*z(nzmin)
But for the mass production of cheap coaxial might be too expensive, even with Teflon.
Let's have a look at a common 75 ohm coaxial
a=1.016;b=4.512; % both mm
since specs read 'foam PE' what relative permittivity has the filling material?
Z0=75;
er=(etha0/Z0*1/(2*pi)*log(b/a))^2
matching the relative permittivity of Foam Polyethylene in table
[PRAD] 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;
directly from the 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')
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, and then test again in your facility before assembly, with heavy penalties agreed in advance for specs deviations.
Z0 = 76.662073914302525
Z0 = 51.685559689169658
er = 1.201557466938515
loglog(X1,Y1) contrary to plot, only takes 1 pair of reference and function vectors. When attempting loglog(X1,Y1,X2,Y2) only the last trace prevails, this is why log is applied to data before plotting.
Reference [DURU]: Duru Ahenku at UoN Analyzing the power handling capability of coaxial transmission line