Why antennas have usually 50 ohm resistance and ? what the source of this 50 ohm resistance

 

Welcome.
It is not resistance; it is impedance and it is not 50 ohm, it is near 50 ohm as a compromise standard. There is another standard of 75 ohm as that figure nears the transition from emission into air/space in less lossy performance. The 50 Ω impedance happens at one frequency and not a band of frequencies.
I will look for a reference link that explains it better than me and add later.

 

----> https://electronics.stackexchange.c...put-impedance-of-antennas-whereas-the-free-sp

 

The short answer is that it's derived from a compromise between the impedance that causes the minimum loss in a coax cable (77 Ohms) and the impedance that allows for maximum power handling (30 Ohms) in the same coax cable.
Here's a solid article on the topic.

 

Edit to Add. You cannot use a multimeter, which measures DC resistance in Ohms, on a coaxial cable. The 50Ω on a cable is the magnitude of a complex number, which you may remember is computed as:

\( Z\;=\;\sqrt{Re^2+Im^2} \)

AND this value is frequency dependent.

 

Edit to Add. You cannot use a multimeter, which measures DC resistance in Ohms, on a coaxial cable. The 50Ω on a cable is the magnitude of a complex number, which you may remember is computed as:

\( Z\;=\;\sqrt{Re^2+Im^2} \)

AND this value is frequency dependent.

What is frequency dependent?

It's a ratio of EM electric field and magnetic field energy components (voltage and current) as they propagate thru the coax dielectric.
Wave impedance: https://www.electricity-magnetism.org/wave-impedance-formula/

 

Edit to Add. You cannot use a multimeter, which measures DC resistance in Ohms, on a coaxial cable. The 50Ω on a cable is the magnitude of a complex number, which you may remember is computed as:

\( Z\;=\;\sqrt{Re^2+Im^2} \)

AND this value is frequency dependent.

No, it is the characteristic impedance of a transmission line, which is well-modeled by the Telegraphers Equation.

\(
Z_0 \; = \; \sqrt{\frac{R \; + \; j \omega L}{G \; + \; j \omega C}}
\)

While, in general, it is a complex value, in well-made transmission lines R and G are very small compared to L and C and so this is closely approximated by

\(
Z_0 \; \approx \; \sqrt{\frac{L}{C}}
\)

which is a real number with very little frequency dependence over a very wide range of frequencies.

 

Do not deviate from the original question; is was about antennas and see comments oriented toward transmission lines. An antenna exhibits 50Ω at only one frequency.

 

Do not deviate from the original question; is was about antennas and see comments oriented toward transmission lines. An antenna exhibits 50Ω at only one frequency.

Answered here already as you know. https://forum.allaboutcircuits.com/...usually-50-ohm-resistance.199208/post-1889444

They are very related in terms of why 50 ohms was selected and why antennas were designed to match that.

A simple dipole variant might meet your criteria of 50Ω at one frequency but there are many more that have wideband impedance matching properties related to transmission line properties in the antenna design.

https://en.wikipedia.org/wiki/Discone_antenna
https://antena.fe.uni-lj.si/literat...ut_the_Discone_Antenna,_QEX,_Jan-Feb-2007.pdf