I've tried looking on the net for my answer but can't find it.

I could have this totally wrong but here I go. Do you need a wave length of ladder line for it's impedance to be what it is described as? How can a length of ladder line a meter long have the same impedance as a length of 100 meters?

It just does, in the same way that a metre of 50 Ohm co-ax has the same impedance as 100 meters of 50 Ohm co-ax.

I think you are talking about the "Characteristic Impedance" of a line; if so it is completely independent of the length of the line.

Say that we had an unterminated 1m length of 450 Ohm ladderline, and that we had some way to instantly apply a 100v RF signal across it. If we could measure the current flowing into the line it would be 0.22A instantaneously, and for as long as it takes for the signal to reach the end of the line and be reflected back to the source. It wouldn't matter how long the line was, we would always measure 0.22A (100/450) until the wave was reflected back to the source; with a longer line the 0.22A initial current would simply last longer.

The ratio of voltage to initial current is known as the "surge impedance" or "characteristic impedance".

Hope that helps,
Steve G3TXQ

Thanks for the replies, that clears that up for me.

Just another question. Again I maybe way of the point. Say you have a off center fed dipole with a feed point of 200 ohms and feed it with 600 ohm ladder. Am I right in thinking you have a total impedance off 150 ohms?

No - the impedance you see will vary depending on the length of the line. If the line is a half-wave long (or a multiple) the impedance will be 200 Ohms; if it's a quarter-wave long (or an odd multiple) the impedance will be 1800 Ohms. Other lengths will give an infinite variety of impedances, but the SWR on the line will always by 3:1 no matter its length.

Don't confuse the input impedance of the line with its characteristic impedance - they're two different things!

Steve G3TXQ

To expand on that:

Say that the 600 Ohm line was 300m long and we applied a signal from a radio at 3MHz to it. Initially the impedance will appear to be 600 Ohms - the surge impedance of the line. 1 uSec later when the signal reaches the antenna feedpoint, because of the 600 vs 200 mismatch some of it will get reflected and travel back towards the radio. After another 1 uSec it will reach the radio and interfere with the signal from the radio, making the net impedance appear to be 200 Ohms.

Different lengths of line change the time taken for the signal to arrive back at the radio, and therefore change the phase difference between radio signal and the reflected signal; in turn that changes the observed net impedance.

That's the simplified version. Happy to expand if necessary.

Steve G3TXQ

I'm a little more confused, not to worry. I'll have a play tomorrow with my analyser and some ladder line.

''.....That's the simplified version''.

You've got to love him, haven't you?

Well, I knew that the explanation begged lots of further questions that might be confusing if I tried to address all in one go.

Looks like I failed

Steve G3TXQ

No, I understand it, I just wouldn't have DARED to quote actual figures like you did

Is there such a thing as a bal-to-bal ''un'', to transform Hi-Z twin feeder down for a 50 Ohm dipole at the feedpoint?

Hmmm. Wouldn't be an 'un' would it? I'll go and stand in the corner.

No you haven't failed I didn't see your second post.

I've got plenty more questions.

Steve, can you expand on the bit about the phase difference altering the impedance?

The impedance you see looking into the line is simply the ratio of the voltage to the current at that point.

The voltage and the current at the input to the line are made up of a number of component waves (in theory an infinite number) including the forward wave from the radio, plus the first reflected wave coming back from the mismatch at the antenna, plus that wave re-reflected at the input to the line, plus that wave reflected from the antenna etc etc ad infinitum.

Depending on the phase relationships of all those components at the input to the line, you could have for example some voltage cancellation giving you low voltage/high current (equivalent to a low input impedance), or you could have some voltage addition giving you high voltage/low current (equivalent to a high input impedance, or anything in between.

The phase relationships between all the different components - which determine the overall net voltage and current - depend on the "round trip" times for the waves, and therefore the line lengths.

If you could open up the line at some point along its length, the phase relationships between the various components - and therefore the net voltage and current - would be different than at the input, and therefore the impedance would be different. That's why the voltage, current and impedance vary all the way along a mis-matched transmission line.

Hope that helps some more. If not I'll try an example with some worked numbers.

Steve G3TXQ