I have been trying for a while to understand Coaxial Cable impedances for ham radio etc. use. I only know what it is not though.

  • Coaxial Cable does not measure the rated impedance from one end to the other

  • Coaxial Cable does not act like a load itself at the rated impedance, they may loose some power to heat, but they mainly carry the signal to the antenna (or whatever other load)

So, what is a coaxial cable impedance rating, and why can't I use a 75 ohm in place of a 50 ohm on the output of my transmitter?


3 Answers 3


Imagine, for a minute that electricity travels quite slowly.

When you turn your light switch on what happens? Current starts to flow - it starts working its way down the wire and so does the voltage. The current that flows is determined by two things: -

  • The voltage and
  • An "impression" of what the load resistance might be.

Will the current be too small or will the current be too much? It is the cable (and its properties) that dictate the amount of current flowing.

Voltage and current are traveling to the "unknown" load and because V and I are flowing there is power flowing (P = VI). When the current and voltage reach the bulb, if the bulb's resistance doesn't match the V/I relationship not all the power is consumed.

This means the excess (or deficit) of power has to be reflected back up the wire to the switch. It's got nowhere else to go.

In the real world of data comms or radio, this causes "reflections" and these can add or subtract to the forward power traveling down the wire and, in the case of data, it can become misshaped leading to possible data corruptions. In the case of an RF carrier, there will be points along the wire where it appears unmeasurable.

The cable dictates how much current initially flows based (mainly) on its inductance, capacitance and resistance. The formula is this: -

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

R is resistance per metre, L is inductance per metre, C is parallel capacitance per metre and G is parallel conductance per metre. At high frequencies (>1MHz) the impedance starts to largely become: -

\$\sqrt{\dfrac{L}{C}}\$ and if you look at some coax specs you'll see that 50 ohms is the result of this calculation.

why can't I use a 75 ohm in place of a 50 ohm on the output of my transmitter?

Hopefully, by now you should be able to answer this.

  • \$\begingroup\$ You don't mean just 50 ohms. 75 and 300 ohm transmission lines are widely available, for example. \$\endgroup\$
    – user128351
    Jan 31, 2017 at 18:11
  • \$\begingroup\$ @stretch of course not - the t-line impedance value I mentioned is just a common value. \$\endgroup\$
    – Andy aka
    Jan 31, 2017 at 18:17

Imagine a bunch of equal hanging weights connected together with equal springs. When it's just hanging there, every weight has the same amount of force trying to pull it up and down, so it will remain motionless. If an external stimulus jerks the top weight upward, then the next weight will have more force pulling it up than had been pulling it down, so it will move upward. This will in turn cause it to pull on the next weight. Each weight will receive energy from the weight above, and transfer that energy to the weight below, until the motion reaches the bottom weight.

If that weight is fixed so that it cannot move, then it will pull back twice as hard on the weight above as each of the previous weights had done. This in turn will make the weight above behave as though it had been jerked downward, and such motion will then propagate back up the bunch. Note that the original wave was a decompression wave, and the reflection will be as well, though it will represent downward rather than upward movement.

If the bottom weight has nothing affixed to it, so that it's pulling on nothing, then the lack of anything below it to absorb its energy will mean it moves up twice as far as it "should". This will in turn cause it to push the weight above it upward, and that weight to push the weight above it, etc. This will create a reflected compression wave, with the weights moving upward (same direction as they did before).

If the bottom weight is fastened to something that provides just the right amount of friction, it will precisely absorb all the energy from the weight above, resulting in no energy reflecting back.

If the weight provides some friction, but it's above or below the ideal amount, then some energy will be reflected upward while some is absorbed. The direction of motion represented by the reflected energy will be determined by whether there was too much or too little friction.


You can use cables and loads with mismatched impedances. There are impedance matching schemes using transformers, LC networks and resistors. A trivial (and wasteful) example would matching a 75 ohm antenna to a 50 ohm transmission line by connecting 150 ohms across the antenna leads.

If you'd used a search engine instead of asking the question, you'd have found many different 50 to 75 impedance matchers offered for sale.

  • \$\begingroup\$ I would have flipped that example around. Making the antenna the antenna 50 ohms and the line 75. Almost all broadcast systems are 50 ohm based(the antenna), whereas 75 ohm is used for video distribution(i.e. coax). It's valid either way, meaning your going to get a VSWR of about 1.5 no matter which is which, it just seems "better" with the antenna being 50 ohms. \$\endgroup\$
    – GB - AE7OO
    Jan 1, 2020 at 10:28

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