What will happen to a single ended (not differential) 1Vpp analog audio signal(4khz bandwidth, not modulated) if it runs through long cat5 cable (length can be 50 meters)? How much signal voltage can drop? Can there be any change in frequency? What will happen if the input signal is differential (say 0.5Vpp signal in each differential line)?

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    \$\begingroup\$ Do you know what a telephone sounds like? \$\endgroup\$ – Dave Tweed Sep 24 '18 at 12:10
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    \$\begingroup\$ A small consideration: the audio technically won’t run over “cat5”. It will run on the individual lines within he cable. Audio over twisted-pair is common in the AV industry. \$\endgroup\$ – user133290 Sep 24 '18 at 14:28
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    \$\begingroup\$ Professional audio is typically run over 600 ohm shielded twisted pairs (STP) as a differential signal. \$\endgroup\$ – Sparky256 Sep 25 '18 at 1:50
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    \$\begingroup\$ I've used Cat 5 and worse in the past for audio applications (speaker output of PC to power amplifier). Does it work? Yes. Will an audiophile scream at you for using the "wrong" cable? Probably. But for 20m it worked fine. \$\endgroup\$ – Mast Sep 25 '18 at 6:09

Your question is impossible to answer quantitatively since the level and type of interference along the cable is unknown, and you haven't told us how the end is terminated.

Look up the resistance per unit length of the cable you want to use. From that you can calculate the resistance in series with your signal. Don't forget that the signal makes a round trip, so the total wire length is twice the cable length.

That all said, sending single ended audio over a long unshielded cable sounds like a really bad idea to me.

My first reaction would be to send the data digitally. That's what cat 5 cable is intended for. You can easily send a few Mbit/s over one pair. To put this in persepective, 16 bit samples at 50 kHz rate (better than CD quality) is only 800 kbit/s.

Digital gives you a certain noise immunity directly. It also allows for easy common mode rejection at the receiving end, by using opto-isolators or pulse transformers.

My second choice would be to use one pair of the cat 5 cable to send differential audio. I would make the voltage higher than you suggest. That makes it larger relative to the inevitable noise along the way. A small audio transformer at the receiver is a easy way to eliminate most common mode noise, and that also removes the need to for a third wire to carry ground.

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    \$\begingroup\$ Whoever downvoted this, what do you think is wrong? \$\endgroup\$ – Olin Lathrop Sep 24 '18 at 15:44
  • \$\begingroup\$ OP could also reduce sensitivity to noise by using shielded Cat 5, though of course that's more expensive than plain Cat 5. And, if they don't have a transformer to use at the receiver, they may be able to tie some of the other pairs in the Cat 5 to ground and prevent ground loops that way – though the feasibility of this depends on the actual potential difference between source ground and receiver ground (Cat 5 is only rated to 60V and 360mA). \$\endgroup\$ – Jazz Sep 24 '18 at 18:02

What will happen if the input signal is differential (say 0.5Vpp signal in each differential line)?

When sending an analogue signal from one place to another, the biggest hurdle faced is usually interference pick-up. To minimize pick-up you should make the impedance of each differential wire the same. The result is that when interference comes along, the same interfering voltage is developed equally on both wires.

This is called balancing the impedances and it leads to the term "a balanced transmission". It does not mean that you have to send signals differentially although this helps because it doubles the signal amplitude resulting in a 6 dB improvement in signal-to-noise ratio.

To balance the impedance on each wire, the wires have to be very similar in construction and, of course, twisted pair is the usual solution but, the driving impedance also needs to be the same for both the hot lead and the return wire. For example, if you are driving the line with an op-amp, it is usual to put (say) 600 ohms in series with the hot-wire and the same impedance in series with the return wire.

This ensures that interference induces the same unwanted voltage on each wire.

At the receiving end, the input amplifier must be differential and present a balanced impedance to ground else it could also imbalance the interference voltage on one wire with respect to the other and the differential amplifier wouldn't be able to adequately cancel it at its output: -

enter image description here

This method also overcomes the problem of a different earth potential between transmitter and receiver caused possibly by earth currents from localized plant equipment.

Can there be any change in frequency?

There will be no change of frequency but there can be a relative attenuation of some frequencies with respect to others due to the complex input impedance presented by any cable at audio especially on long wires such as in telephones.

How much signal voltage can drop?

There won't be much attenuation for audio base-band on a 50m cable but, if in doubt, the cable can be modeled and attenuation figures acquired via simulation.

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  • \$\begingroup\$ @user287001 the voltages at each op-amp input are the same (due to op-amp negative feedback action), therefore if the resistors are the same value then the impedance looking into those resistors from the line is the same. However, to avoid other folk getting confused about this I shall consider making the input stage appear less ambiguous but, I'll wait for your counter-comment. \$\endgroup\$ – Andy aka Sep 25 '18 at 9:49
  • \$\begingroup\$ I removed it Apologies. \$\endgroup\$ – user287001 Sep 25 '18 at 10:29

Forget single ended signals - 50m is so long distance that any ground connection simultaneously at both ends make an enormous interference collecting loop. Hum and buzzes spoil the signals totally if they are only few hundred millivolts.

See Cat5 spec: https://en.wikipedia.org/wiki/Category_5_cable

All wave phenomenas at audio frequencies and 50m cable length are neglible. You can well think one pair as RC lowpass filter which has 10 Ohm resistor and C=2500pF. It alone affects virtually nothing, but the source resistance at the signal transmitting end will affect radically. If there's several kOhms at the transmitting end, you will get treble loss. Only 1500 Ohms can be tolerated for full 20kHz band audio. 4kHz allow 7500 Ohm, but the higher is the resistance, the more sensitive the system is to collect hum and other noises.

How much there's interfering fields in your operating environment. I cannot say it. An AC power cord side by side with your cable can cause troubles if the terminating resistance at both ends is high. You must do some tests.

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The other questions point out the electrical effects, but I also want to mention that there are commercial products that do exactly this with good results. The keywords to search for are "cat5 audio balun". For example, here's one meant for professional audio applications that is rated to work up to 1900ft/500m: https://www.markertek.com/product/ets-pa202m/ets-pa202m-instasnake-adaptor-4-mxlr-to-rj45-jack-all-pins . If you're looking to DIY something similar, the designs of these things would be a good starting point.

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Here is a bit of math, to inject a voltage from a magnetic field. We'll assume the wire-wire spacing of your signal/ground within the CAT5 is 2mm. We'll assume a single power wire (not hot / return) located 10cm away. And we'll assume the power wire is driven by a PURE 60Hz power line, thus no spikes, no rectifier diode fast-on edges, just a pure 60Hz sin, at 1 amp. By the way, dI/dT is 377 amps/second.

What will be induced, for worst-case orientation of the CAT5 versus the power wire? Use the formula

Vinduce = [MU0 * MUr * Area / (2 * pi * Distance)] * dI/dT

which we can rewrite for MU0 = 4 * pi * 1e-7 Henry/meter and for MUr = 1 (for air, vacuum, copper, aluminum) as

Vinduce = 2e-7 * Area/Distance * dI/dT

Inserting numbers, we find

Vinduce = [2e-7 H/meter * (50 meters * 2mm)/ 100mm ] * 377

Vinduce = 2e-7 * 377 = 754e-7 volts

= 75 microVolts

= 0.075 milliVolts


If your power wire (again, this is just ONE WIRE, not the standard field-reducing HOT+RETURN) has any spikes on it, such as from a microwave-oven where the internal 2,000 volts have fast slewrate and thus rectifier diodes can turn on in 100nanosecond, thus dI/dT can be 10 amps peak / 100nanoSec or 10^8 amp/sec, your interference will be 100,000,000/377 or 200,000X worse.

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  • \$\begingroup\$ here's some attempt to give something concrete +1 for reminding. \$\endgroup\$ – user287001 Sep 25 '18 at 7:55

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