Voltages along on a transmission line...

Try to figure out the difference between "amplitude" and instantaneous value.

All continuous waves have both.

...... Phil

Yes, at any one instant it might look like that. A little later, it will look different. The generator is extruding a sine wave that travels along the line, towards the load, at a goodly fraction of the speed of light. At any intermediate point along the line, there's the same sine wave voltage, just shifted in phase.

If you measure any point along the line with an oscilloscope, you'll see the sine wave rise and fall. If you measure it with a voltmeter, it will look the same everywhere. It's a matter of the time scale over which you take the measurements.

If the line is unmatched, there will be waves traveling in both directions, in which case the voltmeter will see different voltages in different places, as the waves add in some locations and subtract in others.

John

Fantastic info John -- that really cleared up my confusion on this subject!

Thanks again for the great help!

-Bill

Although you seem to have cleared this up, here's a thought exercise:

You have a driver output that has an effective rise time of 300 picoseconds (not uncommon nowadays). It is driving a 6 inch track.

On FR4, the transmission velocity is approximately 160 picoseconds per inch. So at the initiation of the transmission, the line is 0V everywhere.

At 50% amplitude output (at the driver, and assuming 3.3V systems and a linear output) the voltage 1 inch away **is still zero**; i.e. the energy has not yet reached this point.

Now continue this until the entire system has stabilised.

Cheers

PeteS

A matched line will have uniform RF voltage along its length, since there are no standing waves. A (theoretical) time domain voltage probe would show the same signal, delayed in phase by (vf * distance from source)/wavelength in free space.

Frank Raffaeli