# Checking coax for signal integrity



## cmckeeman (Jan 13, 2015)

So for starters this is for a non video project but I am using coax to send a dc voltage signal from a load cell to be monitored and recorded elsewhere. Is there an easy way to test if there is any signal loss or interference in coax with just a multimeter?


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## Edrick (Jan 14, 2015)

Why are you using coax for this purpose?


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## cmckeeman (Jan 14, 2015)

Edrick said:


> Why are you using coax for this purpose?


It was recommended by the professor since there might be long runs to get the signal to my computer.


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## Edrick (Jan 14, 2015)

I misread it, thought you were just using it to run Voltage from a battery cell, as in using it to run power to a device. But looks like you're running an actual signal. I'm not sure that a VOM would do the job. They do make handheld units that detect signal leakage for RF. But what you're working on is out of my range! Hopefully someone will come along.


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## ruinexplorer (Jan 14, 2015)

cmckeeman said:


> It was recommended by the professor since there might be long runs to get the signal to my computer.


Coaxial cable is a two conductor cable consisting of a core and shield. If it is video cable (or digital audio), it will be 75ohm. Standard audio cable (often used for wireless antennae) will be 50ohm. The best way to check the stability of the signal is with a waveform monitor. This will tell you what kind of interferrence you are getting. 

If you are using the multimeter, will it record minimum and maximum voltage? Interferrence could be introduced along the cable depending on how close it is to other power sources. This was always something to be considered when running analog signals. To remove most of the interferrence on analog signals, we would use "humbuckers" (video ground loop isolation transformer).


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## cmckeeman (Jan 14, 2015)

ruinexplorer said:


> Coaxial cable is a two conductor cable consisting of a core and shield. If it is video cable (or digital audio), it will be 75ohm. Standard audio cable (often used for wireless antennae) will be 50ohm. The best way to check the stability of the signal is with a waveform monitor. This will tell you what kind of interferrence you are getting.
> 
> If you are using the multimeter, will it record minimum and maximum voltage? Interferrence could be introduced along the cable depending on how close it is to other power sources. This was always something to be considered when running analog signals. To remove most of the interferrence on analog signals, we would use "humbuckers" (video ground loop isolation transformer).


It is video, the cable i have is the same as you would find for cable in your house. And my MM will only give me a live reading and won't record a min and max.


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## FMEng (Jan 17, 2015)

ruinexplorer said:


> Coaxial cable is a two conductor cable consisting of a core and shield. If it is video cable (or digital audio), it will be 75ohm. Standard audio cable (often used for wireless antennae) will be 50ohm. The best way to check the stability of the signal is with a waveform monitor. This will tell you what kind of interferrence you are getting.
> 
> If you are using the multimeter, will it record minimum and maximum voltage? Interferrence could be introduced along the cable depending on how close it is to other power sources. This was always something to be considered when running analog signals. To remove most of the interferrence on analog signals, we would use "humbuckers" (video ground loop isolation transformer).



A transformer won't pass the DC voltage to be measured measure. If the input impedance of the computer interface is high, and I'd bet it is, the coax cable will not lose any voltage. The coax will reduce interference from outside sources to a degree, provided the shield is connected to an effective ground at the computer end. If the shield is floating, it won't prevent any interference.

The pros that do this sort of thing tend to use balanced, shielded, twisted pair cables driven from a balanced source and received by a differential amplifier, also known as an instrumentation amplifier. The noise is rejected because it is common to both wires, equal in value and polarity, so the amp subtracts it. The desired signal is equal in value and opposite in polarity on both wires, so it gets amplified. The technique is far more effective than just shielding. All microphones and mic preamps work this way because the signals are small and prone to interference.


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## themuzicman (Jan 18, 2015)

cmckeeman said:


> So for starters this is for a non video project but I am using coax to send a dc voltage signal from a load cell to be monitored and recorded elsewhere. Is there an easy way to test if there is any signal loss or interference in coax with just a multimeter?



You can figure out loss in dB with a simple Coax loss calculator -- http://www.arrg.us/pages/Loss-Calc.htm

Your most accurate way to figure out loss is to use an analyzer with a tracking generator feature -- a calibrated output that goes through the cable and an analyzer on the other end that. You want a spectrum analyzer as opposed to an oscilloscope because a spectrum analyzer figures out things with a frequency domain (X axis is freq, Y axis is power) and a scope is a time-domain device (X axis is time, Y axis is power) and from there you can figure out frequency dependent loss in the cable.


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## FMEng (Jan 18, 2015)

The OP is asking about running a DC voltage through coax, not an AC radio frequency signal. A coax loss calculator only applies to RF and it doesn't apply to DC. Instead, plain old Ohm's law with the resistance per foot of conductor is how this would be calculated. Because the input impedance of the measurement device is likely to be extremely high, it renders any current-resistance losses completely negligable, unless the cable is thousands of feet long.


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## themuzicman (Jan 18, 2015)

FMEng said:


> The OP is asking about running a DC voltage through coax, not an AC radio frequency signal. A coax loss calculator only applies to RF and it doesn't apply to DC. Instead, plain old Ohm's law with the resistance per foot of conductor is how this would be calculated. Because the input impedance of the measurement device is likely to be extremely high, it renders any current-resistance losses completely negligable, unless the cable is thousands of feet long.



Solid, thanks for clarifying that -- you can clearly see where my knowledge lays here!


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