My question is why engineers developed theoretically something called common-mode voltage And why they chose it particularly as (v1+v2)/2. What is the benefit of all these?

If you have two voltages, you could specify them any number of ways. The most obvious way would be simply:

$$ \begin{align}
V_1 &= \text{something} \\
V_2 &= \text{something else}
\end{align} $$

However, this isn't the most *convienent* method for all applications. Consider this circuit, which is a rather typical application of a differential amplifier:

^{simulate this circuit – Schematic created using CircuitLab}

A real circuit doesn't *actually* have capacitors C1 or C2, or inductors L2 or L3, but these are the unintentional capacitive and inductive couplings to other stuff (adjacent cables, that nearby computer monitor, distant RF radiators, ...) that your circuit must necessarily have by virtue of existing in a real environment.

Now, given the two voltages \$V_1\$ and \$V_2\$, we have the problem of figuring out what \$V_{signal}\$ was.

Well, since this is a *differential* amplifier, that's easy. It's the difference between the voltages, or the *differential mode* voltage:

$$ V_{dm} = V_2 - V_1 $$

But that's not enough information to know what the two voltages actually are. We need something else. That something else is the *common mode* voltage:

$$ V_{cm} = \frac{V1 + V2}{2} $$

You might wonder why this, when something simpler (such as any one of simply \$V_1\$ or \$V_2\$ would also do). The reason is that this is the "average" or "middle" or "center" voltage of \$V_1\$ and \$V_2\$, or in other words, the difference from \$V_1\$ or \$V_2\$ to \$V_{cm}\$ is the same:

$$ |V_1 - V_{cm}| = |V_2 - V_{cm}| $$

This has the convenient property that if \$V_1\$ and \$V_2\$ are switched, \$V_{cm}\$ remains the same.

## Best Answer

For common mode signals you get two wires: one with a signal and one with the inverse of that signal. If you add them you'll get zero, if you'd take the difference you'll get the signal.

What's the cool thing about this. If the two wires (the pair of wires) picks up some noise. Than the noise will be added on both wires. If you add the two signals you only keep the noise.. But if you take the difference you'll get the signal without the noise.

Signal is S (without noise)

Cable a => S

Cable b => -S

The Difference is S - (-S) = 2*S

Signal is S (with noise)

Cable a => S + noise

Cable b => -S + noise

The Difference is (S+noise) - (-S+noise) = S+S+noise-noise = 2S

So the noise will be cancelled.

example image:

When Common mode signals are used they are mostly transferred with twisted pair cables like this: