3

I have a pair of GLC-LH-SMD devices which are 1000BASE-LX/LH

In the field they will be a couple of kilometers apart , but whilst I test the setup I only have 5m patch leads.

I looked at the specs and Cisco state they are

Transmit Power -3 to -9.5 dBm, Receive power -3 to -20 dBm

So this seems to me to mean that you can connect them with any length cable as the Transmitter maximum is the same as the receiver maximum (-3 dBm)

Yet if I connect them to each with a five meter patch lead to each other (in the same switch) - or if I connect one to itself I get a high RX power warning

Dec 19 22:44:44.946: %SFF8472-5-THRESHOLD_VIOLATION: Gi1/0/1: Rx power high warning; Operating value: -1.4 dBm, Threshold value: -3.0 dBm.t

C3750#sho int g1/0/1 tra
ITU Channel not available (Wavelength not available),
Transceiver is internally calibrated.
If device is externally calibrated, only calibrated values are printed.
++ : high alarm, +  : high warning, -  : low warning, -- : low alarm.
NA or N/A: not applicable, Tx: transmit, Rx: receive.
mA: milliamperes, dBm: decibels (milliwatts).

                                 Optical   Optical
           Temperature  Voltage  Tx Power  Rx Power
Port       (Celsius)    (Volts)  (dBm)     (dBm)
---------  -----------  -------  --------  --------
Gi1/0/1      27.4       3.30      -5.1      -1.4 +

C3750#sho int g1/0/2 tra
ITU Channel not available (Wavelength not available),
Transceiver is internally calibrated.
If device is externally calibrated, only calibrated values are printed.
++ : high alarm, +  : high warning, -  : low warning, -- : low alarm.
NA or N/A: not applicable, Tx: transmit, Rx: receive.
mA: milliamperes, dBm: decibels (milliwatts).

                                 Optical   Optical
           Temperature  Voltage  Tx Power  Rx Power
Port       (Celsius)    (Volts)  (dBm)     (dBm)
---------  -----------  -------  --------  --------
Gi1/0/2      33.3       3.30      -5.5      -1.9 +

C3750#

If I connect the transceiver XMT to its own RCV I get

                                 Optical   Optical
           Temperature  Voltage  Tx Power  Rx Power
Port       (Celsius)    (Volts)  (dBm)     (dBm)
---------  -----------  -------  --------  --------
Gi1/0/1      31.4       3.30      -5.5      -1.9 +

I don't see how a transmit power of -5.5 dBm is being received as -1.4 dBm so my question is , Is it real - how am I receiving more power than I transmit? Can I control the transmit power ? (other than inserting the lead but not clicking it into place - which actually seems to work as shown by the receive level below

                                 Optical   Optical
           Temperature  Voltage  Tx Power  Rx Power
Port       (Celsius)    (Volts)  (dBm)     (dBm)
---------  -----------  -------  --------  --------
Gi1/0/1      31.8       3.30      -5.1     -13.0

And more importantly is this a problem ? I can see that the 1000BASE-ZX has a much higher transmit power than a receive power - so this would need an attenuation cable

  • What Cisco router or switch are you using and what IOS version? – user36472 Dec 20 '17 at 7:24
  • I think you will need an attenuation cable for your test if you don't want to damage your SFPs, that's why they made DDM transceivers. The range is up to 10KM, so i think 5 meters is a bit too small. – user36472 Dec 20 '17 at 7:38
  • It's a 3750 not x and not + , the iOS is 12.0 – Ross Dec 20 '17 at 10:51
2

I don't see how a transmit power of -5.5 dBm is being received as -1.4 dBm so my question is , Is it real - how am I receiving more power than I transmit?

Possible? Yes under the right set of circumstances. To begin with, you would need to be on a short fiber run (per your description, check) and very few insertion points (again by your description, check). Beyond that, any interference created would have to be nearly perfectly constructive.

To understand the last, you need to know that part of the signal is reflected back at the end of the cable. When this reflected signal reaches the original source end of the cable, again part of it will be reflected back. If the wave forms align correctly, you can then have a form of constructive interference.

Let's put this in VERY simplistic and unscientific percentages to illustrate the effect. You transmit on side A of the cable at 100%. Some signal is lost along the way, but being a short cable, 98% reaches side B. A tenth of that, 9.8%, is reflected back toward A. 9.6% reaches side A, and a tenth of that, or 0.96% is reflected to constructively interfere perfectly with the new 100% source, becoming 100.96%. As this happens in a continuous cycle, the gain can be significant.

Again, that is simplistic and not reflective (pun intended) of the real life math/physics, but it was meant more as an illustration of the effect, not the full details. In real life, the numbers change depending on a wide range of factors and there are limits to the medium.

Of course, you could just have faulty hardware or a bug in the software was well.

Can I control the transmit power ?

As in manual adjust the transmit power of the transmitter up or down? No. Many will adjust to some degree automatically, but you won't have control over it.

However you can reduce the signal strength in a number of ways (or a combination of the below):

  • try a different cable; another 5m cable may be a slightly different length altering the effect
  • use attenuators to reduce the signal strength (best option and you can pick them up inexpensively)
  • use a longer cable (I have a 1000' SM fiber for temporary deployments that works in a pinch)
  • connect (couple) multiple cables together
  • damage the cable (generally wrapping it around a relatively narrow object)

This last has been alluded to by others in their answers, but this will likely permanently damage your cable. I believe the basis for this solution is in mandrel wrapping, which is a technique used in multi-mode fiber to limit or eliminate higher modes allowing for better signal loss measurements. This also results in the attenuation of the signal, and has often been used in a pinch for that purpose.

However mandrel wrapping done correctly doesn't damage the cable as it follows specific standards as well. Your typical mandrel will be about 2cm in diameter, with some variance depending on the core size and jacket thickness of the MM fiber in question. Wrapping around a smaller diameter object can introduce cracks in the cable, allowing light to escape; the smaller the diameter, the more damage and the more signal loss.

Additionally, by it's nature, SM fiber doesn't have multiple modes, so any real loss is a result of damaging the cable.

And more importantly is this a problem ?

Yes. Optics can and will burn out. Just like your eyes, if the receiver is exposed to a light source that is too bright for too long, it will cause damage. The brighter the source, the shorter the exposure necessary.

You should take some action to attenuate the signal.

  • The reflection theory above isn't clear to me. The reflected power is taken out of the transmitted power. If it is reflected back with a perfect phase alignment it'll add back to the (later) transmitted wave. How can the sum the higher than the original amplitude? – Zac67 Dec 22 '17 at 8:39
  • When you have constructive interference, the sum of the amplitudes of the two are summed. If you are adding such interference at the source, any gain in amplitude will be higher than the original amplitude. Each iteration (approximately 30 million per second on a 5m cable) adding slightly more power. Also keep in mind the reflected light will make multiple passes back and forth. Using an OTDR you can see the spikes created by reflections at connectors and you can see these power levels as higher than original power (from the insertion point), especially on short runs. – YLearn Dec 22 '17 at 21:15
  • The fiber patch is not a laser. Every bit of power you add to the later signal has been taken from the earlier signal. Equally, the later signal suffers the same attenuation by reflection than the earlier signal. There can't be more power than at the first place (1st law of thermodynamics). If 1% of the power is reflected, 99% of the original power is delivered plus .01% of the power reflected twice plus .0001% of the power reflected four times and so on - adding up to 99.01010101..%. This implies constructive interference. Destructive interference subtracts_ the power (to 89.98989898..%). – Zac67 Dec 22 '17 at 21:25
  • You are correct, the power gain comes from earlier in the signal. However, we aren't talking percentages, we are talking the power level and with each iteration. So on your first iteration, using a fictional measure of 100 bloits of light at the source end of the cable, on the second iteration this may be 101 bloits (original source plus reflection). While the same % may be reflected, since the number of bloits has increased, the amount of bloits reflected increases. Small loss on a very short run of SM fiber can be exceeded by the power "trapped" in the cable without other sources of loss. – YLearn Dec 22 '17 at 22:26
  • I did find on the FOA thefoa.org/tech/FAQS/FAQ-TEST.HTM site a statement about this reflectance causing this problem on short SM patch leads – Ross Dec 22 '17 at 23:32
2

The readings you get from the SFPs are in my experience more a "best guess" at the signal level, and are usually only useful for telling you if you have a signal, especially on the TX side. For something close to a real measurement you'd need an optical power meter.

For testing purposes, you could wrap a patch cable tightly around a pencil and wrap it with electrical tape. The sharp bend will cause the cable to leak signal, and as such attenuate the signal. (You can test this with a test laser if you have one, the patch cable will start glowing red if you bend it enough.) Some modern patch cables have really tight minimum bend radiuses, so the cheaper the cable the better. I know it sounds like the dirtiest AD hoc trick you've ever heard of, but I know a lot of fiber technicians who use it in the field to fix issues temporarily when they don't have the right attenuation cable handy.

  • I think I read something about that I think red light was mm but I will try that – Ross Dec 20 '17 at 10:52
1

I do not have an answer as to why those Tx/Rx readings are as such, but I do know that connecting fiber SFPs without proper attenuation (within threshold) is a great way to burn out your optics. I was taught a similar Ad-hoc method as Stuggi...guy had me wrap the jumper around my screw driver a bunch of times and twist tie it after we got the Tx/Rx within spec.

  • Yes my question though is the specification says it's fine, the sfp cannot transmit more power than it can receive so why does. It claim it is receiving too much power , obviously the measurements don't match the spec. It pears from everyone's statements the specification is wrong. – Ross Dec 20 '17 at 20:18
0

Mostly agreeing with the answer above, you're completely right in that this should work.

My guess is that the receive level is a little too high and that the receiver fails to give a more meaningful reading in that domain. Physically, it's impossible to receive more power than has been transmitted, of course.

  • Actually, it is possible to receive more power than has been transmitted as the result of constructive interference. It takes a special set of circumstances for this to occur, but it can happen. – YLearn Dec 22 '17 at 2:55
  • This is single-mode fiber, please elaborate how interference should be possible. Additionally, interference can't raise effective receive power, only momentary. – Zac67 Dec 22 '17 at 8:23
  • Any waveform (light included) is subject to interference. Not sure what you are trying to get at with the second point, but if the two wave forms are both continuously travelling the same medium, at the same frequency, in the same direction, and in the same mode, they will continue to interfere (either constructively or deconstructively), it isn't a momentary effect. – YLearn Dec 22 '17 at 21:31
  • See my comment above. What you're saying defies the 1st law of thermodynamics. The power at the receiver must come from somewhere and borrowing from an earlier signal doesn't cut it when the current signal is subject to the same borrowing. – Zac67 Dec 22 '17 at 21:37
  • No. You miss that with each iteration (after the first) the same base source power level is applied and increased a small amount by reflected signal. This higher total signal will result in slightly higher reflected signal on the next iteration, resulting in a higher total signal for the next iteration which results in higher reflected signal, resulting in higher total signal which results in higher reflected signal....put another way: base*x=reflected1 -> (base+reflected1*x)*x=reflected2 -> (base+reflected2*x)*x=reflected3 - not accounting for multiple circuits by the first iteration light. – YLearn Dec 22 '17 at 22:35

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.