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I'm studying the trasmission of signals though coaxial cable; I'm using a coaxial cable (of course!), an oscilloscope, a generator an some different endings. This is the scheme: enter image description here

I was told that if I use a very long cable that has impedence \$Z_0= 50 \Omega\$ and put, at the end of it, a resistor of \$47\Omega \$, the signal will be reflected.

I have obtained this:

enter image description here

If I have correctly understood, the height of the step pointed by the green arrow is the half of the V given by the generator.

I can't understand why the step pointed by the red arrow is shorter than the one that is on the other side. And I can't understand what's the physical meaning of the step pointed by the red arrow.


Then, I have terminated the cable with a short circuit and I have obtained this: enter image description here

Could you explain me what's up at the points indicated by the two arrows and the yellow and red points?


Then, I have terminated the cable with a resistor of 100 ohm and I have obtained this: enter image description here

Could you explain me, why the step pointed by the green arrow is higher than the step pointed by the red arrow?

This is the last image: enter image description here

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    \$\begingroup\$ I don't think your cable was terminated by 47Ω - if it were 50Ω cable you'd barely see a reflection at all. \$\endgroup\$ – Andy aka Jul 2 '13 at 16:22
  • \$\begingroup\$ @Andyaka I was told that resistor was 47 ohm. Using a termination of 50 ohm, I didn't see any reflection. \$\endgroup\$ – sunrise Jul 2 '13 at 17:30
  • \$\begingroup\$ This would very nearly be the same with 47R \$\endgroup\$ – Andy aka Jul 2 '13 at 17:44
  • \$\begingroup\$ @Andyaka Could you explain me why in the short circuit there are three steps? Many thanks!! \$\endgroup\$ – sunrise Jul 2 '13 at 19:54
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With nothing connected to the end of the cable the reflection occurs in phase, so that cancellation occurs in those regions where the positive half cycle of one wave overlaps with the negative half cycle of the other. The duration of this overlap is the round-trip time of the signal through the cable. This is shown in the photograph.

enter image description here

It looks like your transmission line is open circuit!

After the exchange of comments I think I now know what is happening. The coaxial cable is indeed 50 ohm. The resistor used wasn't a 47 ohm (yellow, purple, BLACK) but a 470 ohm resistor (yellow, purple, BROWN). This explains why the reflection looks more like an open circuit cable because it was being terminated by a resistance 10x higher than the impedance of the cable. The 100 ohm was much closer and shows less distortion but it was still twice the impedance. If you look at the 100 ohm trace it has a similar but less pronounced castle shape with an asymmetry between the rising and falling edge of the pulse.

You get the step because this waveform is a superposition (addition) of the pulse and the reflections which are shifted in time by the round circuit trip. When the cable is terminated with 50 ohms these reflections disappear.

It would be interesting to see this happen by using a variable (non inductive) resistance starting high (say 500 ohms) and ending with a short circuit. Measuring the time shift should also give you an estimate of the cables length knowing that the signal travels at about 2/3 speed of light in vacuum. (about 10m per 100nS)

http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/76.18.html

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  • \$\begingroup\$ Thanks for your answer. I have tried to understand, but I haven't understood... :( In order to explain better my doubts, I have edited the question. If you might answer, I'll be very grateful to you! \$\endgroup\$ – sunrise Jul 2 '13 at 15:31
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    \$\begingroup\$ I think you need to mention that the pictures in your answer were at the sending end after the transmit resistor. \$\endgroup\$ – Andy aka Jul 2 '13 at 16:20
  • \$\begingroup\$ @ Andy - I think you just did (+1) \$\endgroup\$ – JIm Dearden Jul 2 '13 at 16:52
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    \$\begingroup\$ @Sunrise I'm beginning to wonder If the cable used is actually 50 ohm. The 100 ohm seems a better match but still has a bit of mismatch (hence the slightly odd looking reflection.) Try it with 75 ohms and see if it improves the signal. \$\endgroup\$ – JIm Dearden Jul 2 '13 at 16:58
  • \$\begingroup\$ @JImDearden It was 50 ohm, because using a termination of 50 ohm, I didn't see any reflection. Considering the 100 ohm termination, could you help me to understand why the step pointed by the green arrow is higher than the step pointed by the red arrow? \$\endgroup\$ – sunrise Jul 2 '13 at 17:34
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Here is a picture of what happens when the cable is unterminated. I hope to complete this with a picture of what happens when the cable is terminated in a short so bear with me for that: -

enter image description here

Note this is an incomplete answer so bear with me while I finish it off.

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  • \$\begingroup\$ Thanks a lot for your answer, even if it is incomplete, it helps me to understand many aspects! By for now :) \$\endgroup\$ – sunrise Jul 8 '13 at 20:44

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