in 50Ω or 75 Ω cable this mean that if i have 10 ampere go through this cable i will have voltage drop 500v or 750v in this cable
thanks
thanks
It always remains a transmission line, it will become so lossy it is not usable as a transmission line anymore.As I understand the physics, the characteristic impedance is not a function of frequency. It is a function of the geometry of the coaxial cable. As you increase the frequency the losses change and the electrical length of the transmission line gets shorter and shorter. As the frequency goes up the wavelength becomes shorter and shorter to the point where the dimensions of the cable are of the same order of magnitude as the wavelength of the signal and it stops being a transmission line and starts being a waveguide.
Have a look at this article:Actually, I meant as I increase the frequency for say 0.01Hz to hundreds of megahertz, where does the cable start act as it has some impedance? Or does it have 50ohms even at that 0.01Hz?
Actually, I meant as I increase the frequency for say 0.01Hz to hundreds of megahertz, where does the cable start act as it has some impedance? Or does it have 50ohms even at that 0.01Hz?
True, it's not the DC resistance. But an infinite length of ideal coax will show its characteristic impedance for DC or AC. Thus if you apply a step DC voltage to an infinite length cable, the current will be equal to the voltage divided by the cable characteristic impedance. For a finite length of cable it will look like its characteristic impedance for the time it takes the step to travel to the end of the cable and back to the start. At the point the reflected wave will add or subtract from the input current depending upon whether the other end is open or shorted.No. 50Ω or 75Ω cable is the AC impedance, not the DC resistance.
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