Question

Consider a standard negative-feedback loop with $G(s)=\dfrac{1}{(s-2)(s-3)}$ and a PID controller $C(s)=K_P+\dfrac{K_I}{s}+K_Ds$. The root-locus of $G(s)C(s)$ (shown) has a double zero at $s=-1$. The gain $|C(j\omega)|=2$ at $\omega=1$ rad/s. The value of $K_D$ is______ (round off to one decimal place).

Hint:

A PID can be written $C(s)=\dfrac{K_D(s+z_1)(s+z_2)}{s}$. If the root-locus shows a double zero at $-1$, then $z_1=z_2=1\Rightarrow K_P=2K_D,\ K_I=K_D$, making $|C(j1)|=2K_D$ immediately.

Answer 1

Correct Answer: 0.9

From the given root-locus, $C(s)$ has two coincident zeros at $s=-1$. Thus the controller numerator is proportional to $(s+1)^2$, i.e., \[ C(s) = K_P + \frac{K_I}{s} + K_Ds = \frac{K_Ds^2 + K_Ps + K_I}{s} = \frac{K_D(s+1)^2}{s}, \] so the coefficient ratios are $K_P:K_I:K_D = 2:1:1 \Rightarrow K_P=2K_D,\ K_I=K_D$. At $\omega=1$, \[ |C(j\omega)|=\left|K_P + \frac{K_I}{j\omega} + K_D j\omega\right| = \sqrt{K_P^2 + \left(K_D\omega-\frac{K_I}{\omega}\right)^2} = \sqrt{(2K_D)^2 + (K_D-K_D)^2}=2K_D. \] Given $|C(j1)|=2 \Rightarrow 2K_D=2 \Rightarrow K_D=1.0$.