# Average of a set of transfer functions: how should I treat the phase?

I have a set of transfer functions obtained by impacting an instrumentation hammer against a mass, measuring the acceleration on a set of accelerometers; the result is shown below. I want to get a curve to generalize the behavior of the system, in magnitude it is as simple as averaging the magnitude of the signals, however I wonder how can I treat the phase since by just calculating the angle of the mean per frequency the resulting curve is meaningless (which is explained in Phonon's answer).

With this in mind, is there a way to go? Additionally, I am working straight with the frequency domain signals of the measurement (calculated by the device); would this be solved by switching instead to the original time series --> averaging --> transforming manually to frequency?

• You’d need to average the complex arrays resulting from the DFTs, then compute the phase of that average.
– Jdip
Commented Jul 18, 2023 at 9:23
• note that both "averaging" and "DFT" are linear operations, DFT->element-wise average is the same as element-wise average->DFT Commented Jul 18, 2023 at 9:32
• That was my first approach, written in Matlab reads as angle( mean( H_hammer_test(:,2:end), 2 ) ), however it plots plenty of blank segments and if I unwrap it, frequency range is reduced to less than 1 Hz. Commented Jul 18, 2023 at 9:38
• @MarcusMüller Correct. Julio, how are the individual transfer functions computed?
– Jdip
Commented Jul 18, 2023 at 9:40
• @Julio then I'd suggest you use the original time series, and do your calculations on that, because there you at least know what happens. Commented Jul 18, 2023 at 10:14

Consider computing the average of two phase angles, $$\alpha = 7\pi/8$$ radians and $$\beta =$$ -$$7\pi/8$$ radians. Due to the directional nature of phase angles, we know the average of $$\alpha$$ and $$\beta$$ is an angle exactly halfway between $$7\pi/8$$ radians and -$$7\pi/8$$ radians, or $$\pm\pi$$ radians ($$\pm180^\circ$$). However, standard numerical averaging of the two scalar radian values $$7\pi/8$$ and -$$7\pi/8$$ results in zero radians ($$0^\circ$$), which is obviously incorrect.