There are 3 main types of receivers used to detect "differential signals":
DC coupled differential signals
RS-485, RS-422, CANbus, LVDS, USB, SATA, PCI Express, etc. directly connect differential signals to the receiver chip -- "DC-coupled".
They require a ground connection to keep the signal at the receiver's end of the bus within the common-mode range of the receiver chip.
Often such systems stop working when the voltage offset is more than a few volts, and can be permanently damaged if the voltage offset ever reaches a few dozen volts.
(That is, the voltage offset between the system "ground" at one end of the cable and the system "ground" at the other end of the cable).
Often 2 boxes with a cable between them carrying such a protocol (or a single-ended protocol such as SPI or RS232) seem to work fine in the lab sitting next to each other,
but have intermittent communication or stop communicating entirely when placed in the field with long distances between them.
When that happens people often end up buying 2 "isolators" that internally use one of the following approaches, and putting the long cable between those isolators.
opto-isolator coupled differential signals
Systems like MIDI connect more-or-less differential signals to the LED of an opto-isolator at the receiver.
With proper design,
similar systems can and sometimes do work just fine with kilovolts of offset between the system "ground" at one end of the cable and the system "ground" at the other end of the cable.
transformer-coupled and capacitor-coupled differential signals
Analog audio, LonWorks(a), etc. connect differential signals to DC-blocking capacitors.
Ethernet, etc. connect differential signals to DC-blocking transformers.
Broadband-over-powerline receivers typically have both DC-blocking capacitors and DC-blocking transformers.
With proper design,
they can and sometimes do work just fine with kilovolts of offset between the system "ground" at one end of the cable and the system "ground" at the other end of the cable.
These systems block the DC offset with a transformer or capacitors or both to carry the signal across the isolation boundary.
(To reduce EMI and protect against cable discharge events, many systems also connect each cable wire with resistors or capacitors or both -- a Bob Smith AC termination -- to the chassis ground (b) (c) (d) (Intel AP-434); often with additional capacitors to support power over ethernet (e). )
Such offset voltages are the main reason behind
" 2kV capacitor on ethernet? ".
Differential over a cable
How is this normally accomplished in practice?
When sending Ethernet, LonWorks, opto-isolated data, etc. over a cable, a ground wire is not required. All the wires in the cable can be used for data transmission.
(PoE systems often end up pulling the two system grounds close together anyway; non-PoE systems allow the two system grounds to float apart).
When sending RS-485, CANbus, etc. over a cable, typically at least 1 wire in the cable is reserved for the ground wire, which pulls the system ground at one end of the cable and the system ground at the other end of the cable closer -- hopefully close enough to allow communication or at least to prevent permanent damage.
Many people use exactly the same (unshielded) CAT5 cable with standard RJ45 plugs at both ends for both kinds of systems.
When using a shielded cable, some people are very careful to design the system with the socket where that cable plugs in to have a separate "chassis ground"/"frame ground" and connect it to the shielding in the cable, and separate from the "data ground"/"signal ground" on, for example, pin 9 of a DB9 connector carrying RS232 data.
Other people simply connect all the grounds together.
I'm not going to say more here about that raging controversy.
But where is the common ground for the differential input case?
It can be anywhere! Assume that the average voltage of the two input voltages is at common ground, so 0 Volt. Now what is the differential input voltage ? It is:
\$V_{in,diff} = (V_{inp} - 0) - (V_{inn} - 0) = V_{inp}-V_{inn}\$
Now I use a different common ground which is at + 12.3 V, what do we get now:
\$V_{in,diff} = (V_{inp} - 12.3) - (V_{inn} - 12.3) = V_{inp}-V_{inn}\$
See, no difference!
Since it is the difference between \$V_{inp}\$ and \$V_{inp}\$ what counts, whatever the common ground is, it is added and subtracted so the net result is zero.
The common ground is irrelevant. Also they do not need to be tied although in practice they often are. Ethernet (network) cables for example use differential mode signals. The ground does not need to be connected. On both sides of the Ethernet cables small isolation transformers are used to re-define the ground level so that it suits the local circuit.
Also in practice the voltage between inputs and the ground of the circuit will be limited for example by the input voltage range of the amplifier.
Best Answer
"Ground" as a concept needs some clarification. If you have just one signal line and one ground, then yes, it's hard to tell the difference. But if there's anything else going on, it matters.
All AC signalling involves current flow, even if it's being measured as a voltage on the receiver. At a minimum, you have to charge/discharge the parasitic capacitance of the receiver, and the signal wire will have a capacitance to ground as well. Note that the common mode behavior only holds if the two wires have the same length and are a constant distance from each other.
So if you have a signal wire and a ground wire, then the current in the signal wire must be matched by a corresponding current in the other direction in the ground wire. If you have lots of signals, the ground wire will contain a mashed-together copy of all the signals. Therefore it's advantageous for each signal to have its own ground. If you look at VGA, you'll notice that each signal gets its own ground because of this. If you look at 80 pin IDE, each pair of signals has a ground wire between it in the ribbon cable. Those are to prevent the signal wires inducing currents in each other ("crosstalk").
Once you've accepted that each signal must have its own matched ground, it's more natural to embrace the two as a matched pair, disconnect one from ground and connect the two of them together via a termination resistor network, and drive / read them as a differential signal.