Considering this coax cable open end reflection. What are the correct explanations for the flatter rise time, the rounding of the first reflection, and the much smaller what I assume to be secondary reflection? enter image description here


Cable: ~100m 50ohm "m17/028-rg-58"

Scaling is 500ns/div.

Signal: 100kHz square wave.


Suggestion: enter image description here


enter image description here enter image description here

  • \$\begingroup\$ What's the horizontal scale (it's not readable in the photo)? What kind of coax are you using? \$\endgroup\$ – The Photon Feb 22 '17 at 19:27
  • \$\begingroup\$ What is the length of the coax and scaling of the display - then you can match the display with the physical coax. \$\endgroup\$ – Kevin White Feb 22 '17 at 19:31
  • 1
    \$\begingroup\$ please see edit. \$\endgroup\$ – bretddog Feb 22 '17 at 19:35
  • \$\begingroup\$ The second step should be "almost" the same size as the first with more rolloff. However if you are looking at the total voltage it seems to be clamped at 15V which may be output voltage protection on your source or input protection on your detector. I would try the same experiment with a smaller amplitude step to see if you are able to get better step sizes. \$\endgroup\$ – KalleMP Feb 22 '17 at 20:13
  • 1
    \$\begingroup\$ Sorry, the second step is also expected to be somewhat smaller as some of the signal will be absorbed by the better terminated driver end. Down the page this has a very similar scope display to what you show. - allaboutcircuits.com/projects/… \$\endgroup\$ – KalleMP Feb 22 '17 at 20:17

The slower edge in the reflection indicates that high frequency signal components are in the reflection are attenuated compared to the low-frequency components.

The main reason for this is likely that the coax is lossier for the high frequency components than for the 100 kHz fundamental. A chart I found online indicates typical RG-58 has 6.6 dB/100 m loss at 30 MHz, and 16 dB/100 m loss at 100 MHz, for example. Remember that your reflected signal passes through the cable twice (so you need to consider 200 m worth of loss) when using these figures.

| improve this answer | |
  • \$\begingroup\$ @JackCreasy, rejected your edit because we don't expect the reflected signal to be proportional to the phase of the outgoing signal. I reworded to try to make what I said more clear. \$\endgroup\$ – The Photon Feb 22 '17 at 20:22
  • \$\begingroup\$ The other indication that the transmission line has losses is that the "shoulder" does not stay flat but slowly creeps up. Also it should be noted that the secondary reflection occurs because the driver internal impedance seems to be about 20% higher than the characteristic impedance of the cable. \$\endgroup\$ – Ale..chenski Feb 22 '17 at 21:23
  • \$\begingroup\$ @Ali Chen; How do you estimate the 20%? \$\endgroup\$ – bretddog Feb 22 '17 at 21:49
  • \$\begingroup\$ @bretddog, when the driver drives, a voltage divider is formed between R(out) and Z. The amplitude of initial step is about 3.3 units, while the settled amplitude is about 7.7 vertical units. The ratio between 3.3 and (7.7-3.3=4.4) indicates the ratio between Rout and cable characteristic impedance. \$\endgroup\$ – Ale..chenski Feb 22 '17 at 21:56
  • \$\begingroup\$ @Ali Chen: So I tried to understand this, and could only find that it would be most accurate the way I shown now in "edit 2". May you see if that would be correct or mistaken? Considering that the underlying signals would keep rising I figured that looked logical to measure this way (?) \$\endgroup\$ – bretddog Feb 22 '17 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.