# The relationship between the I control and the stability margin

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'?

• Please quote the exact words, as what you have put is not very clear. Sep 4, 2018 at 6:02
• Can you add a reference? Sep 4, 2018 at 13:10
• The reference is "From PID to Active Disturbance Rejection Control" which is written by Jingqing Han. Sep 5, 2018 at 4:25
• 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. Sep 5, 2018 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()): 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.