Group Delay

"EngNotes" are my engineering notes.  This is my way of creating a digital version of an engineering notebook.  This particular entry is on Group Delay.

I had one of those brain fart moments at work recently where you start to question everything and suddenly feel very small.  Why is the phase slope of a delay line negative!?  After a five minute mental battle I decided to give up and ask someone else - it was late and my brain had turned to mush.  After some really simple equations on a white board my confidant was stumped too.  All the usual sources confirmed what I had memorized, that it is negative, but I couldn't rationalize it to myself.

To my future self who begins to doubt again, don't think of phase slope induced by group delay this way:

If you increase frequency, the amount of phase traversed over a particular amount of time increases.  Therefore phase slope of a delay line is positive.

If you are stuck thinking the above, you can even write out an equation that is so simple and appears to contradict god and his uncle.  The statement above has a problem with its frame of reference.  Below is the right way to think about the problem:

Phase is always a relative measurement.  When characterizing the phase slope of a delay line, you are characterizing the phase of the output relative to the input at a particular point in time.  Think transfer function.  The output phase of a delay line will be retarded relative to the input.  Another way to put it is the output phase of a delay line will be behind (negative) relative to phase at the input of the delay line.

Equations to help set my brain straight for next time:

$ \phi(t)_{output} = 2 \Pi f(t-t_{0}) $

$ \phi(t)_{input} = 2 \Pi ft $

$ \tau = -\frac{\Delta\phi}{2\Pi \Delta f} $