4

My book says that the input impedance of an infinitely long line is equal to the characteristic impedance as long as the attenuation constant is not equal to 0.

Is this true? I think it may be an error in the book because the formula for wave impedance of a lossless line, expressed in terms of the distance, is equal to the characteristic impedance when d-> infinity

JobHunter69
  • 915
  • 1
  • 15
  • 26
  • 1
    Yes. The same is true if the lossless line is terminated with a resistor of its characteristic impedance. Characteristic impedance is the impedance that the source "feels" until a reflection comes back from the termination at the end of the line. If the line is infinitely long, or if it is terminated in the characteristic impedance, no reflection ever comes back, and the impedance does not ever change. – user57037 Jan 31 '17 at 07:28
  • 2
    Is this true? It depends what sort of truth. It's mathematically true. Once you start bringing zero and infinity into equations that are supposed to describe physical reality, then you have automatically left behind the concept of physical truth. It might be better to use the mathematical tool of limits. If the attenuation constant is non zero, then the input impedance will tend to a limit of Z as the line length tends to infinity. The closer to zero, the slower it tends to that limit. – Neil_UK Jan 31 '17 at 07:30

2 Answers2

2

Yes its true, for a transmission line with a fixed length the characteristic impedance is:

\$Z_0 = \sqrt{\dfrac{R+jwL}{G+jwC}} \$

If we say the line is lossless with no attenuation then that means it has no resistance. So \$ R=0 \$ and \$ G=0 \$ and we get:

\$Z_0 = \sqrt{\dfrac{jwL}{jwC}} \$

It doesn't matter if the line is 1cm, 1m, 1km or infinitely long (*well technically short lines do matter in some cases). Why? because as long as the impedance is the same the wave will travel unimpeded. If you start a wave on one end it will continue to to travel indefinitely until it reaches a section with an unmatched impedance. With matched impedance you have maximum power transfer.

This is why the length doesn't matter: Without getting too much in the derivation look at this example: Lets say my transmission media impedance is 50Ω. The more sections I add, I still have 50Ω looking into each section and out of each section. The source still sees 50Ω's in each case. However one key difference is the length, if I send a wave down the transmission line, it takes longer and longer the more sections I add. So if I add an infinite number of sections the wave will keep traveling down the line, it never gets attenuated by resistance and never loses energy.

schematic

There is a medium that does this, its called free space. And its characteristic impedance is 377Ω, if you send a wave in free space it goes forever at the speed of light.

\$Z_0 = \sqrt{\dfrac{jwL}{jwC}} = \sqrt{\dfrac{\mu_0}{\epsilon_0}} \$

Voltage Spike
  • 82,181
  • 41
  • 84
  • 220
  • This sentence in your answer confuses me and may also confuse the OP: "It doesn't matter if the line is 1cm, 1m, 1km or infinitely long. If you start a wave on one end it will continue to to travel indefinitely." You may want to clarify. – user57037 Jan 31 '17 at 07:30
  • Yeah so the attenuation constant has to be 0 for the input impedance to be the characteristic for infinite line? – JobHunter69 Jan 31 '17 at 20:58
  • Yep, if you want the wave to travel infinitely far then you need 1) Matched impedance 2) No resistance. The amplitude of the wave attenuates over distance is resistance is added. If you get into non-ideal realworld transmission lines, you have to make the resistance sufficiently small for your wave to get from one end to the other. The equations for these are the telegraphers equations and include the relationship of the amplitude with the distance. By setting R and G to zero you can also see how the wave goes on forever. – Voltage Spike Jan 31 '17 at 23:35
1

To transfer power from A to B over a long distance, source A cannot know what impedance load B is so, the interconnecting cable is the only reference point and (say) 50 ohm coax presents a load equivalent to 50 ohms.

If A supplies 1 volt dc, 20 mA immediately starts to flow and if the cable is infinitely long then 20 mA will continue to flow. If the load happens to be 50 ohms then 1 volt and 20 mA meeting a 50 ohm load means there is no imbalance and there will be no reflection.

If the load does not equal 50 ohms there is an imbalance and the power that cannot be consumed is reflected back to the source.

My book says that the input impedance of an infinitely long line is equal to the characteristic impedance as long as the attenuation constant is not equal to 0.

I think it should say "as long as the attenuation constant is equal to 0"

Andy aka
  • 456,226
  • 28
  • 367
  • 807