Holding your arms out straight in front of you, let your hands represent two signals, a and b. The vertical position of each hand represents the voltage present on each signal. Now flap your arms up and down in an opposite, or antipodal, fashion. This motion depicts the behavior of a differential-signal pair.
The distance between your hands represents the differential component of the signal pair [a, b], and the average position of your hands represents the common-mode component.
To make obvious the meaning of the common-mode component, first find a stick about three feet long. Mark the midpoint of the stick. Now, holding opposite ends of the stick in each hand, repeat the arm-flapping exercise. The midpoint of the stick, which is at all times the mechanical average of the positions of your two hands, represents the common-mode component. The inclination of the stick represents the differential component.
Moving your arms exactly opposite each other such that the midpoint of the stick remains fixed in space creates a perfectly differential-signal pair with no common-mode component. It may help to have someone else hold the midpoint fixed so you can get the feeling of a perfectly antipodal motion. You will find it difficult to create a purely differential signal. Any asymmetry in the amplitude or timing of your arms' motions creates a common-mode component.
In an electrical circuit, you care about the common-mode component of signal pair [a, b] because that component creates far more radiation and crosstalk than the differential component.
The relation between signal skew and the common-mode signal amplitude is fairly easy to derive. Suppose signal b is the same as negative a, only delayed by a small skew, Δt. You can then express the common-mode signal c≡(a+b)/2 as
Those of you familiar with calculus will recognize that the numerator of the right-hand term, for sufficiently small Δt, approximately equals Δt multiplied by the time derivative da/dt. Caution-this approximation works only when Δt remains smaller than the signal rise time.
The peak value of c coincides in time with the peak value of the derivative of a. Assuming a 10 to 90% rise time, tr, and a signal swing, Δv, on each wire, you can approximate the peak value of the derivative of a(t) as Δv/tr. After suitable rearrangement of terms, this calculation leads to the following expression for the peak common-mode amplitude:
This derivation assumes that signal a swings from zero to Δv, and signal b swings from Δv to zero. The differential signal (a-b) therefore swings from +Δv to -Δv. Therefore, the peak-to-peak amplitude of the differential signal is 2Δv, the peak amplitude of the differential signal is Δv, and the peak amplitude of the odd-mode signal (defined as half the differential-mode signal) is Δv/2.
Look at the peak amplitude of the common-mode signal as a percentage of the peak amplitude of the differential-mode signal, Δv. Also, imagine the skew, Δt, to be a percentage of the rise time, tr. With these two concepts in mind, you can reduce this entire discussion to the following rule: A 20% skew creates a 10% common-mode component.