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?


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

  • \$\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


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).

  • \$\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\$ 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
  • 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
  • \$\begingroup\$ A first order pole cannot increase gain. \$\endgroup\$
    – Chu
    Dec 3, 2016 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.