Decibel adjustment on Bode diagram

Question

Say we have the system $G(s) = 1/(s+1)^3$ with break frequency $\omega_b = 1$. Can someone explain to me why we should expect $|G(\omega_b)|$ to be $3$ dB below the low frequency asymptote, rather than 9 dB? I thought we were correcting once, with -3dB each, for every factor.

Answer 1

You are absolutely correct. If you do the calculation $20log_{10}(|G(j1)|)=−9.03$ i.e. -3dB for each pole. See the Bode below.