While I'm reading the paper, it said
"The I control can makes the phase lag, so it can makes stability margin reduced."
Words are not same as that, but the meaning is the same.

  1. How the I control makes the phase lag in the PID controller?
    If the plant's transfer function is $G_p (s)$, the open-loop transfer function $G(s)$ is
    $G(s) = \left(\frac{k_i}{s} + k_p + k_d s\right)G_p (s)$
    when we use the PID for the controller. After substitute $s=j\omega$,
    $G(j\omega) = \frac{\left(k_i-k_d\omega^2 \right)j+jk_p\omega}{\omega} G_P (s)$
    So, how can I find that the I controller makes phase lag?
  2. How the phase lag makes stability margin reduced?
    In the paper, it wrote "stability margin". I think that's phase margin. Anyway, how can I get the relation between 'stability margin' and 'phase lag'?

Thank you for advising!

  • $\begingroup$ Please quote the exact words, as what you have put is not very clear. $\endgroup$ – Solar Mike Sep 4 '18 at 6:02
  • $\begingroup$ Can you add a reference? $\endgroup$ – Karlo Sep 4 '18 at 13:10
  • $\begingroup$ The reference is "From PID to Active Disturbance Rejection Control" which is written by Jingqing Han. $\endgroup$ – Kim Jaewoo Sep 5 '18 at 4:25
  • $\begingroup$ The exact words are those: The integral term, while critical to rid of steady-state error, introduces other problems such as saturation and reduced stability margin due to phase lag. $\endgroup$ – Kim Jaewoo Sep 5 '18 at 4:30

In this answer stability is covered first, thereafter the effect of the PID-controller on the stability.

Consider the following simple control system

which has a closed loop transfer function of $$\Gamma(s)= \frac{C(s)G(s)}{1+C(s)G(s)}=\frac{L(s)}{1+L(s)}.$$ Here $L(s) = C(s)G(s)$ is called the loop.

The system is unstable if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|>1(=0$dB$)$.

Similarly, the system is stable if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|<1(=0$dB$)$.

Furthermore, neutral (or marginal) stability is at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|=1(=0$dB$)$.

The stability margin is ''the space'' you have until the system becomes unstable (i.e. robustness). Since stability is determined by using two measures: the gain and the phase, consequently there are also two stability margins (the gain margin and the phase margin).

The gain margin is ``the space'' you have at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until $|L(j\omega)| = 0$.

The phase margin is ``the space'' you have at frequency $\omega$ where $|L(j\omega)| = 0$ until $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until.

The gain, phase and corresponding margins can be computed by hand, however I would recommend using software (e.g. MATLAB), since they can be very extensive. To find effect your controller $C(s)$, you can draw Bodediagram and see the effect of the various parameters (MATLAB command bode()): Bodediagram PID-controlers The margins of the complete system can also be visualized using MATLAB (command margin()), which should give more insight in the effect of the chosen controller and your stability margins. Stability margins


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