Consider the transfer function
$$G(s) = \dfrac{48000}{s(s+100)}$$
Now observe that the gain at frequency w = 1 Hz is 53.6244 dB. Completely reasonable, because this is exactly $$20\log10(48000/100) = 53.6244 dB$$
Now I change the system to:
$$G(s) = \dfrac{48000}{s(s+0.1)(s+100)}$$
MATLAB is telling me that the gain at frequency w = 1 is 53.5812 dB. This is roughly unchanged. Wouldn't the pole at 0.1 = 10^-1 contribute to greater drop in gain than the previous system? I should expect the Bode plot to show full 20 dB lower at w = 1 compared to the the first system, so according to this logic 53.6 - 20 = 33.6 dB, instead of 53.5812 dB.
Can anyone explain this? Why isn't the pole at 0.1 contributing to anything on the magnitude plot?
Code:
G = zpk([],[0,-0.1,-100],48000) bode(G) grid on
More craziness: $$G(s) = \dfrac{48000}{s(s+0.001)(s+0.1)(s+100)}$$
We have two poles before 10^0 = 1 Hz and contributing absolutely nothing to the magnitude plot!