# Does the integrator in a PID controller removes the disturbance?

I am reading "Small Unmanned Aircraft: Theory and Practice" (page 7). We have the following Transfer Function (TF):

$\phi(s)=\frac{a_{\phi 2}}{s(s+a_{\phi 1})}(\delta_a(s)+\frac{1}{a_{\phi 2}}d_{\phi 2})$

where $\delta_a$ is the output signal to the actuator and $d_{\phi 2}(s)$ is an unwanted disturbance.

Initially, the disturbance is neglected, and only a PD controller is employed, and the resulting 2nd order TF (with $k_p$ and $k_d$)is compared with the general second-order TF:

$H(s)=\frac{\omega_n^2}{s^2+2\zeta \omega_ns+\omega_n^2}$

From which the value of $k_d$ is obtained (and $k_p=\frac{\delta_{a,max}}{e_{max}}$, where $e_{max}$ is the maximum error).

Then, the disturbance is considered, and an integrator is added to get a full PID. The resulting TF (with $k_p$,$k_d$ and $k_i$) will be different, actually it will become 3rd order. However, the book does not update the values of the gains after the integrator has been added, and somehow claims that the integrator will basically nullify the disturbance, so no change is needed for the gains.

Can anyone explain this to me?

• I know that higher order systems (e.g. 3rd order) will have a behavior that is basically determined by the 2nd order poles. Is this the reason why the 3rd order system is ignored and only the 2nd order one is considered when obtaining the gains? – student1 May 8 '14 at 2:03
• Actually the third pole is very close the the first and second poles so neglecting it is not justifiable (see page 11 in the link above) – student1 May 8 '14 at 2:37