This is probably a really simple question, but I can't seem to find a definite answer anywhere. I'm guessing 50Ω cable means 50Ω per unit length.

What unit length is this? If this is not how its defined, how is it?

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    \$\begingroup\$ If I remember it correctly from my microwave course lectures, it was the impedance of the infinite length cable; assuming that its core charge carrier is a perfect conductor. The value of the impedance comes from the capacitance between two conductors (core and shield), and the inductance per unit length. The cable is not a lumped material, so this impedance value is calculated by solving a very complex multi-dimensional wave equation. \$\endgroup\$ Commented Dec 12, 2013 at 18:05

4 Answers 4


I see you have some accrurate but probably difficult to understand answers. I'll try to give you a better intuitive feel.

Consider what happens when you first apply a voltage to the end of a long cable. The cable has some capacitance, so it will draw some current. If that were all there was to it, you get a big current spike, then nothing.

However, it also has some series inductance. You can approximate it with a little series inductance, followed by a little capacitance to ground, followed by another series inductance, etc. Each one of these inductors and capacitors models a little length of the cable. If you make that length smaller, the inductance and the capacitance goes down and there are more of them in the same length. However, the ratio of the inductance to the capacitance stays the same.

Now imagine your initial applied voltage propagating down the cable. Each step of the way, it charges up a little capacitance. But, this charging up is slowed down by the inductances. The net result is that the voltage you applied to the end of the cable propagates slower than the speed of light, and it charges the capacitance along the length of the cable in a way to require a constant current. If you had applied twice the voltage, the capacitors would get charged to twice that voltage, therefore would require twice the charge, which would take twice the current to supply. What you have is the current that the cable draws being proportional to the voltage you applied. Gee, that's what a resistor does.

Therefore, while the signal is propagating down the cable, the cable looks resistive to the source. This resistance is only a function of the parallel capacitance and series inductance of the cable, and has nothing to do with what it connected to the other end. This is the characteristic impedance of the cable.

If you have a coil of cable on your bench that is short enough so that you can ignore the DC resistance of the conductors, then this all works as described until the signal propagates to the end of the cable and back. Until then, it looks like a infinite cable to whatever is driving it. In fact, it looks like a resistor at the characteristic impedance. If the cable is short enough and you short the end, for example, then eventually your signal source will see the short. But, at least for the time it takes the signal to propagate to the end of the cable and back, it will look like the characteristic impedance.

Now imagine that I put a resistor of the characteristic impedance accross the other end of the cable. Now the input end of the cable will look like a resistor forever. This is called terminating the cable, and has the nice property of making the impedance be consistant over time and preventing the signal from reflecting when it gets to the end of the cable. After all, to the end of the cable another length of cable would look the same as a resistor at the characteristic impedance.

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    \$\begingroup\$ This is the first time anyone's successfully explained cable impedance to me, thanks \$\endgroup\$
    – tom r.
    Commented Nov 28, 2018 at 5:09
  • \$\begingroup\$ This is a great explanation. It becomes immediately obvious why attaching a 50-ohm resistor to the inner and outer conductors of a 50-ohm characteristic impedence cable makes any signals that go down it dissappear; electrically, it is the same as adding an infinite length of the same cable. \$\endgroup\$
    – JoseOrtiz3
    Commented Mar 22 at 19:09

When we talk about a 50-Ohm cable, we are talking about characteristic impedance which is not quite the same as a lumped impedance.

When there is a signal propagating in the cable, there will be a voltage waveform and a current waveform associated with that signal. Because of the balance between capacitive and inductive characteristics of the cable, the ratio of these waveforms will be fixed.

When a cable has a 50 Ohm characteristic impedance, it means that if power is propagating in only one direction then at any point along the line the ratio of the voltage waveform and current waveform is 50 Ohms. This ratio is characteristic of the cable geometry and isn't something that increases or decreases if the length of the cable changes.

If we try to apply a signal where the voltage and current aren't in the appropriate ratio for that cable, then we will necessarily cause signals to propagate in both directions. This is essentially what happens when the terminating load doesn't match the cable characteristic impedance. The load can't support the same ratio of voltage to current without creating a reverse propagating signal to make things add up, and you have a reflection.

  • \$\begingroup\$ Why can't we say that the cable is like a previous load with and impedance Z which is equal to the cable's characteristic impedance? \$\endgroup\$ Commented Dec 10, 2013 at 18:12
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    \$\begingroup\$ @Felipe_Ribas, If you are looking in to one end of the cable, and if the other end is terminated with a matching load, then the cable would behave (as far as you can tell from the input end) like a fixed load with impedance Z. But that doesn't tell you what happens with other terminations, and it doesn't explain why it behaves that way. \$\endgroup\$
    – The Photon
    Commented Dec 10, 2013 at 18:15
  • \$\begingroup\$ Is the frequency of the signal a parameter as well, or is the characteristic impedance good for any frequency singal? \$\endgroup\$
    – deadude
    Commented Dec 10, 2013 at 18:16
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    \$\begingroup\$ @cagrigurleyuk A well-designed cable will have very close to the same characteristic impedance over a wide frequency range. Typically if the frequency goes too high either the cable loss increases unacceptable (see skin effect) or the cable becomes a multimode transmission line and can no longer be described with a single \$Z_0\$ parameter. \$\endgroup\$
    – The Photon
    Commented Dec 10, 2013 at 18:18
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    \$\begingroup\$ @Felipe_Ribas, no you cannot do that. For one thing, if the load is not matched, the overall reflection will depend not just on the Z0 of the cable but also the length. \$\endgroup\$
    – The Photon
    Commented Dec 10, 2013 at 18:32

In theory, if the cable in your example is infinitely long, then you will measure a 50Ω impedance between the two leads.

If your cable is shorter than infinite, but longer than approximately 10% of the signal's wavelength* \$\lambda = \dfrac{c}{f}\$ (where \$c\approx 3\cdot 10^8 \text{[m/s]}\$), then you enter the area of transmission lines. So for a frequency of 1MHz, the wavelength will be approximately 300m and one tenth will be 30m. So if you are working with 1MHz and a cable shorter than 30m, you don't have to worry about its impedance too much.

*) Actually the wavelength in a cable is shorter than in vacuum. To be on the safe side, for practical example, just multiply the wavelength by 2/3. So in practice your cable worry-threshold with 1MHz should be 30m*2/3 = 20m.

Other answers have written a more theoretical text, I'll try to give some high level practical information.

In practice this means that you want to terminate your cable at both ends with a resistor that equals the characteristic impedance you can transmit a reasonably clean signal. If you do not properly terminate your cable, you get reflections.


simulate this circuit – Schematic created using CircuitLab

Reflections may distort (or attenuate) your signal at receiver end.

As the name suggests, the reflection also travels back from the far end of the cable to the transmitter. Often RF transmitters cannot cope with large reflecting signals and you may blow up the power stage. This is the reason why it is often strongly advised to not power a transmitter if the antenna is not connected.


The characteristic impedance of a cable is nothing to do with its physical length. It's quite complex to visualize but if you consider a long length of cable with a 100 ohm load at one end and a 10 volt battery at the other end and ask yourself how much current will flow down the cable when the 10 volt battery is connected.

Eventually 100 mA will flow but, in that short space of time when current is flowing down the cable and hasn't yet reached the load, how much current will be flowing from the 10 volt battery? If the characteristic impedance of the cable is 50 ohms then 200mA will flow and this represents a power of 2 watts (10 V x 200 mA). But this power cannot all be "consumed" by the 100 ohm resistor because it wants 100 mA at 10V. The excess power is reflected back from the load and back up the cable. Eventually things settle down but in the short space of time after the battery is applied it's a different story.

Characteristic impedance of the cable is defined by the cable's size and shape. This results in four parameters that define it's characteristic impedance Z\$_0\$: -

\$Z_0 = \sqrt{\dfrac{R+j\omega L}{G+j\omega C}}\$


  • R is the series resistance per metre (or per unit length)
  • L is the series inductance per metre (or per unit length)
  • G is the parallel conductance per metre (or per unit length) and
  • C is the parallel capacitance per metre (or per unit length)

In audio/telephony spheres the cable characteristic impedance is usually approximated to: -

\$Z_0 = \sqrt{\dfrac{R}{j\omega C}}\$

This is reasonable up to about 100 kHz because series R is usually much bigger than \$j\omega L\$ and G is usually negligible.

At RF, usually 1MHz and higher, the cable is regarded as having a characterisitic impedance of: -

\$Z_0 = \sqrt{\dfrac{L}{C}}\$

Because \$j\omega L\$ dominates R and as mentioned previously, G is regarded as negligible, however, dielectric losses at frequencies above 100MHz start to increase and G is sometimes used in the formula.

  • \$\begingroup\$ I'm not sure about your last paragraph. It may apply to high-precision work in the 100-1000 MHz range (not my field). But in the 1 GHz and up world, R losses tend to dominate rather than G losses. This causes a "square-root-of-f" loss characteristic that is a very big deal in gigabit communication work. \$\endgroup\$
    – The Photon
    Commented Dec 10, 2013 at 19:34
  • \$\begingroup\$ @ThePhoton you've got me there - above 1GHz certainly isn't my field but I have had to contend with G losses in the 100MHz area. Regarding the skin losses (I think you might be referring to those because of the square root of F loss you mentioned), Won't jwL always rise much quicker that sqrt(F). Maybe it's something else? \$\endgroup\$
    – Andy aka
    Commented Dec 10, 2013 at 19:46
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    \$\begingroup\$ Did a little search and found this: sigcon.com/Pubs/edn/LossyLine.htm. For a given dielectric, G losses tend to dominate at higher frequencies. But what the article doesn't say is, we can usually spend more money to get a better dielectric, but we're pretty much stuck with copper and skin effect no matter what we spend (aside from the possibility of using Litz wire for some applications) \$\endgroup\$
    – The Photon
    Commented Dec 10, 2013 at 20:10

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