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I have a 3m long transmission line that transmits a 100ns bipolar square pulse every 200us. At the beginning of the coax cable is a pulsing circuit board, that generates the waveform, then it sends the pulse through a 50 ohm 3 m long coax line and at the other end of the coax line is connected to a load of 20 ohm.

I have noticed that by placing 6dB attenuators near at the output of the pulsing board then attaching the coax lines in series results in a greater power output from the load.

Is this due to the damping of reflections?

In addition would placing the attenuation close to the load result in a greater power transfer since the reflections are attenuated before travelling back a good portion of the cable? I can not try this scenario yet as I do not have the correct connectors to attempt this.

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    \$\begingroup\$ What is the load? How do you measure its power output? How many 6-dB attenuators did you put between the source board and the cable? \$\endgroup\$ – The Photon Sep 11 '14 at 20:36
  • \$\begingroup\$ Also, what are the rise and fall times of your pulse? \$\endgroup\$ – The Photon Sep 11 '14 at 20:37
  • \$\begingroup\$ @ThePhoton the load is a piezoelectric transducer I am measuring the strength of the pressure wave. The rise and fall times are roughly 2 - 4ns. \$\endgroup\$ – user1084113 Sep 11 '14 at 20:57
  • \$\begingroup\$ Adding a 6-dB attenuator should reduce the delivered power by about 4x. Even if there is a benefit from reducing reflections, I wouldn't expect a net increase in output power due to these effects. More likely there is something going on with a nonlinearity in the piezo device or related to the fact the piezo is not a resistive load. Piezo's aren't my area, but I'd guess that more details about the device could help someone give you a good answer. \$\endgroup\$ – The Photon Sep 11 '14 at 21:13
  • \$\begingroup\$ What is the bandwidth of your pressure-measurement instrument? If it isn't on the order of 10 MHz or more, the apparent increase in power could be nothing more than a shift of energy from frequencies outside its bandwith to those inside the bandwidth. This could occur even if the total power delivered to the transducer is less. \$\endgroup\$ – Dave Tweed Sep 11 '14 at 22:09
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Instead of attenuation, consider matching the load to the transmission line with a transformer.

Z1 = (turns ratio)^2 * Z2

Try a turns ratio of 11/7

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You can use a transformer as john suggested, if you can find or make one, or you can use a 30 ohm resistor in series with your load:

20 + 30 = 50, which matches the cable.

You might not be able to get exactly 30 ohms in one package, so you may have to use 27 or 33 or a series/parallel combination of other values to get closer to 30. But even 27 or 33 will be a lot better than what you have now.

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