Classic control theory – Negative gain required for a finite velocity error

control system

I have an assignment to design a control system with 2 degrees of freedom (Prefilter + Controller) to control a plant with the following transfer function:

$$P(s)= \frac{2}{s(s-3)}$$
I am required to have \$K_v = 20\$

I do not need to add any integrator since the plant already has one (since i need a finite non zero error for a ramp input).

The gain required is: $$K_v=\lim_{s\rightarrow 0}sP(s)$$
I get a gain of \$K = -30\$, this is where I got confused since I'm used to get a positive gain.

Is this correct ? What is the meaning if a negative velocity error ?

Best Answer

The transfer function \$P(s)\$ has a RHP pole at \$s=3\rm{rad/s}\$. As with any RHP pole, this means the plant is naturally unstable.

You will need to use a PD controller (transfer function \$G(s)=K_P+sK_D\$). You can select the proportional gain \$K_P\$ to achieve the desired velocity error, and select \$K_D\$ such that the zero occurs before crossover for stability (i.e., before \$|G(s)P(s)|=0\rm{db}\$).

Related Topic