Find K and Kt for ζ=0.6, ts=0.1 s

Step 1: Reduce the minor (tachometer) loop

The minor loop has forward block $\dfrac{100}{1 + 0.2s}$ and feedback $K_t$ taken from its own output:

$$G_{in}(s) = \frac{\frac{100}{1 + 0.2s}}{1 + \frac{100K_t}{1 + 0.2s}} = \frac{100}{0.2s + 1 + 100K_t}$$

Step 2: Open-loop transfer function

$$G(s) = K\cdot\frac{100}{0.2s + 1 + 100K_t}\cdot\frac{1}{20s} = \frac{5K}{s(0.2s + 1 + 100K_t)} = \frac{25K}{s\left[s + 5(1 + 100K_t)\right]}$$

Step 3: Closed-loop characteristic equation (unity feedback)

$$s^2 + 5(1 + 100K_t)s + 25K = 0$$

Comparing with $s^2 + 2\zeta\omega_ns + \omega_n^2 = 0$:

$$\omega_n^2 = 25K, \qquad 2\zeta\omega_n = 5(1 + 100K_t)$$

Step 4: Use the specifications

Settling time (2% criterion):

$$t_s = \frac{4}{\zeta\omega_n} = 0.1 \;\Rightarrow\; \zeta\omega_n = 40$$ $$\omega_n = \frac{40}{0.6} = 66.67\ \text{rad/s}$$

Step 5: Solve for $K$ and $K_t$

$$\begin{aligned} 25K &= \omega_n^2 = (66.67)^2 = 4444.4 \;\Rightarrow\; K = 177.8 \\ 5(1 + 100K_t) &= 2\zeta\omega_n = 80 \;\Rightarrow\; 1 + 100K_t = 16 \;\Rightarrow\; K_t = 0.15 \end{aligned}$$

Answer: $K \approx 177.8$ and $K_t = 0.15$ (using the 2% settling criterion).

If the 5% criterion $t_s = 3/(\zeta\omega_n)$ is used instead, $\zeta\omega_n = 30$, $\omega_n = 50$ rad/s, giving $K = 100$ and $K_t = 0.11$.

Note how the tachometer feedback $K_t$ sets the damping, while $K$ sets the natural frequency, so both specifications can be met independently.