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Consider the transfer function

$$G(s) = \dfrac{48000}{s(s+100)}$$

The Bode plot is given as enter image description here

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:

$$G(s) = \dfrac{48000}{s(s+0.1)(s+100)}$$

Consider the Bode Plot enter image description here

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! enter image description here

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  • \$\begingroup\$ At \$\omega=1 rad/sec\$ the integrator contributes 0dB to the gain, and the pole at s=-0.1 also contributes about 0dB. The pole at s=-0.001 will contribute even less... \$\endgroup\$
    – Chu
    Dec 3, 2016 at 9:18

2 Answers 2

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In the first calculation you correctly calculated the gain before the second pole to be 48000/100

On the second example the gain before the second pole is 48000/(100*0.1). So it is 20dB higher. This compensates with the higher attenuation of 20dB for the additional pole, so at w=1 the gain is unchanged.

To avoid these issues it is better to use a normalized notation where each pole is in the form (s/n + 1).

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  • \$\begingroup\$ Thanks, but when you say DC gain you meant approximation to the DC gain right? Because clearly the DC gain is infinity \$\endgroup\$
    – Fraïssé
    Dec 3, 2016 at 8:43
  • \$\begingroup\$ Sorry, you are right. DC gain is infinity, it is the gain before the pole as you calculated for the first example. I am changing the answer so it won't stay wrong. \$\endgroup\$
    – Claudio Avi Chami
    Dec 3, 2016 at 8:45
  • \$\begingroup\$ Also, it seems that the second example, at w = 0.1 the gain is 90.4 dB instead of 48000/(100*0.1) => 73.6 dB. So it is not just 20 dB higher and it is not just 40 dB higher (otherwise it would be 93.6 dB), but some seemingly random number. Can you explain this? \$\endgroup\$
    – Fraïssé
    Dec 5, 2016 at 12:33
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  • There is no pole at w = 0.1 firstly.
  • Moreover presence of a pole will increase the gain and won't contribute to drop as you have stated.
  • The magnitude is changed. Only thing is that the change is not huge
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  • \$\begingroup\$ A first order pole cannot increase gain. \$\endgroup\$
    – Chu
    Dec 3, 2016 at 10:01

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