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I'm very new to this page and anything electrical in general. I considered putting this on the music SE page but I thought this site may prove more helpful.


I have a small personal "studio" (put very lightly) for recording my own music as a personal hobby. I have a mic hooked up through to a vocoder synth through 2 XLR cables, as my mic uses a phantom power supply. The total distance from the mic to the synth is about 6 feet, but the smallest XLR cables I have are 25 feet and 10 feet, respectively.

My question(s) are thus:

If I coil up the excess slack in my XLR cable, will I experience unwanted interference (either from the loops or the unneccesary length of the cable)? Does this apply to any audio-carrying cable?

Would buying 2 shorter XLR cables give me a cleaner signal than keeping my current ones?

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    \$\begingroup\$ If you do have problems picking up interference on the longest run, look into "star quad" cable (on the same XLR conns) for better interference rejection. \$\endgroup\$ – Brian Drummond Aug 19 '17 at 10:02
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An XLR cable is a "balanced" cable type, which means that it is inherently noise-canceling. For the most-part anyways (nothing is perfect.)

It achieves this by not just sending one signal down the cable, but sending two identical signals - except that one is electrically reversed or opposite from the other. This is called a "differential" mode - when one wire goes positive, the other goes negative. This is beneficial because any extraneous noise that is picked up by the cable is picked up evenly by both wires. This "common-mode" noise is easily rejected.

So long story short, no, a hundred feet of XLR cable probably will make little difference in noise. The signal strength might be reduced slightly due to singal losses, but it should not be noisier because it is long.

That said, XLR cables are great at eliminating the E part of EMI - Electro-Magnetic Interference - or things like radio waves, the "usual" sources of interference. But they are not as good at rejecting the M part - magnetic interference. Luckily, few things produce a magnetic field large enough to be concerned about, so unless you have this cord wrapped around a vacuum cleaner, it shouldn't be a problem.

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    \$\begingroup\$ I once ran over 1000 ft. of XLR down a river bank for a fireworks show. The only downside was the signal was a few dB weaker at the last stack. \$\endgroup\$ – Matt Young Aug 19 '17 at 5:50
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Using shorter XLR cables would give you a cleaner floor, but shouldn't have any effect on the signal.

In a sound system I work with, we normally have 25 ft or 25 ft + 12 ft cables on stage, then a 75 ft snake to the audio console. In a tv studio I worked at the mic cables from the studio floor boxes to the console must have been over 100 ft in some cases, and they may have used 50 - 100 ft of cable in the studio.

I don't know what the recommended maximum length for XLR mic cable is, but think it must be a few hundred feet.

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Lets compute some Magnetic Field interference into "balanced" audio cable.

Assume the extra cabling is thrown on top of a power-brick, a switching power supply brick, with 1 amp being switched in 100 nanosecond. Thus di/dT is 10^+7 amps per second. We'll need this number in just a few paragraphs.

How much of this flux escapes the brick's sheet metal shield? assume 10% (the shield has gaps at the edges).

Assume the imbalance in the "balanced" audio cable is 10%; it may be an old cable, often crushed under foot, or just poorly manufactured.

Assume the cables, with reference to the RTN path --- the shield, have 3.3mm spacing. This is the "loop height" which we'll need.

Assume the cable has length of 30 meters, looped uniformly around the brick. An assumption, I know, but you'll see the uniformity does not matter.

Assume "distance" from brick to cable is 0.1 meter.

We need Distance, Loop Area, and dI/dT.

Formula:

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

This becomes 2e-7 * Area/Distance * dI/dT [ we'll include the derating factors after we compute the worstcase Vinduce

Vinduce = 2e-7 * [3.3mm * 30meter]/0.1meter * 10^+7

Vinduce = 2e-7 * [3.3e-3 *3e1] * 1e-1 * 10^+7 = 2 * 9.9 *10^[-7 -3 +1 -1 +7]

Vinduce = 20e-3 = 20 milliVolts = 0.02 volts, worstcase

However we assume the metal shield inside switcher give 10:1 reduction. We also assume the "twin-lead balanced audio cable" gives 10:1 reduction. Total 100;1

Thus the 20 milliVolts becomes 200 microVolts.

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  • \$\begingroup\$ Kudos for the rigorous mathematics. The signal levels on XLR cables can vary by orders of magnitude - a "strong", line-level, pro-audio, +4dBm signal is 1.228vRMS, while "normal", line-level is −10 dBV / 0.316vRMS, while a "pro", lo-Z microphone may output 20dB less; 200uV may be perceptible in that case. It certainly could be if the cable were resting on the power brick and not 10cm away. Hmm, maybe I'll get out the scope and investigate. \$\endgroup\$ – rdtsc Aug 28 '17 at 1:47
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The type of signal that is transmitted in a "XLR" is a balanced signal. The benefit of balanced signals is that they are quite resilient against interference.

The theory is well explained here: http://www.aviom.com/blog/balanced-vs-unbalanced/

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There is an assumption that all the noise is from external signals. That is the way the balanced lines cancel.

However, noise also comes from resistance. It is emitted fairly randomly. The noise emitted from two lengths of wire due to random unequal impulses will not cancel. What is the resistance? Enough to cause the above-mentioned 'few dB' of loss. At the same time that the losses move the signal towards the receiver noise floor, thermal noise is added. This is commonly known in RF and microwaves. It matters at audio frequencies if the series resistance is near that of the source/receiver. The 'few dB' of loss is a sign of the larger problem. A low output dynamic mic would be that much closer to receiver noise and that much more susceptible to the thermal noise of 100+100 feet of wire (the full circuit).

There is a way to deal with this, however: if a preamp is put the cable run, with a lower noise level, it will push the signal further above the noise. Amplifying afterwards is too late: you amplify noise and signal. Even at low digital rates, "line drivers" are used to add to the acceptable cable length.

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