Classic control theory - Negative gain required for a finite velocity error?

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 ?

• I can't speak for the number you mentioned of $K_v=20$, since you don't mention how much velocity error is desired; you only specify that it be finite, which virtually any $K_v$ value would satisfy. – Zulu Apr 16 '15 at 3:08

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 $|G(s)P(s)|=0\rm{db}$).