# Effect on stability margins when an integrator is introduced in the loop of a closed loop system

I was asked a question in a control system test which is as follows:

*Introduction of integral action in the forward path of a unity feedback system results in a

• a) marginal stability
• b) system with no steady state error
• c) system with increase stability margin
• d) system with better speed of response*

Since we were told that only one of the option is correct I immediately hit (b) because an integrator is known to make steady state error 0. However, I wondered why the other options would have been incorrect. Now I could rule out marginal stability because a closed loop will stabilize the open loop pole at 0. I also took a random function (s+2)/(s+1) and investigated its phase margin which was higher than that of (s+2)/(s(s+1)). This is enough to rule out option (c) but I am unable to get a physical sense of why an integrator should reduce stability of a system. Also, what defines speed of response of a system, or which parameter I should be looking for to quantify how fast my system is?

• The integrator will add a low-frequency pole, and poles added before the unity gain frequency will tend to make the system unstable. Sep 21, 2022 at 7:34
• Cancelling lower frequency pole(s) is how you may get a better response speed, as the next higher frequency pole then determines the speed. Adding an integrator adds a low frequency pole, so this is antithetical to (d). That eliminates (d) as an answer.
– jonk
Sep 21, 2022 at 7:35

Since we were told that only one of the option is correct I immediately hit (b) because an integrator is known to make steady state error 0

A real-life integrator does not make the steady-state error zero; nothing has infinite gain and that includes an integrator. An integrator is only as good as the op-amp it is built around and, if the open-loop gain of that op-amp is 1 million then, the integrator is stuck with that maximum gain.

Apart from anything else, a real-life system has offset errors and these errors apply to integrators too so no, it will not produce a zero steady-state error.

I think option (a) is the likeliest to be more correct than the others because, adding an integrator could degrade the phase margin. Given that we know nothing about the "control system" that you have specified i.e. how close to instability it might be, it seems that (a) is the better choice.

Clearly (c) is opposite to (a) so it cannot be (c) and (d) cannot be possible because an integrator will tend to slow things down. But, all of this is totally dependent on what the "control system" is.