1. **Problem Statement:** Design a compensator for the given control system with open-loop transfer function $$G(s) = \frac{16}{s(s+4)}$$ such that the static velocity error constant $$k_v = 20$$ sec$$^{-1}$$ without significantly changing the location of the complex-conjugate closed-loop poles. 2. **Recall the static velocity error constant formula:** $$k_v = \lim_{s \to 0} sG_c(s)G(s)$$ where $$G_c(s)$$ is the compensator transfer function. 3. **Calculate the current $$k_v$$ without compensator:** $$k_v = \lim_{s \to 0} s \times \frac{16}{s(s+4)} = \lim_{s \to 0} \frac{16}{s+4} = \frac{16}{4} = 4$$ 4. **Required $$k_v$$ is 20, so the compensator must increase $$k_v$$ by a factor of $$\frac{20}{4} = 5$$.** 5. **Choose a compensator that increases $$k_v$$ without moving poles significantly:** A lag compensator is suitable: $$G_c(s) = \frac{s + z}{s + p}$$ with $$p < z$$ and both close to zero. 6. **Set $$k_v$$ with compensator:** $$k_v = \lim_{s \to 0} s \times \frac{s+z}{s+p} \times \frac{16}{s(s+4)} = \lim_{s \to 0} \frac{16(s+z)}{(s+p)(s+4)} = \frac{16z}{4p} = 4 \times \frac{z}{p}$$ 7. **Since $$k_v$$ must be 20, then:** $$4 \times \frac{z}{p} = 20 \implies \frac{z}{p} = 5$$ 8. **Choose $$z = 0.05$$ and $$p = 0.01$$ (both small to avoid moving poles much).** 9. **Final compensator:** $$G_c(s) = \frac{s + 0.05}{s + 0.01}$$ 10. **Closed-loop transfer function:** $$T(s) = \frac{G_c(s)G(s)}{1 + G_c(s)G(s)} = \frac{\frac{s+0.05}{s+0.01} \times \frac{16}{s(s+4)}}{1 + \frac{s+0.05}{s+0.01} \times \frac{16}{s(s+4)}}$$ 11. **Plotting the unit-step response can be done in MATLAB using:** ```matlab s = tf('s'); Gc = (s + 0.05)/(s + 0.01); G = 16/(s*(s + 4)); T = feedback(Gc*G,1); step(T) title('Unit Step Response of Compensated System') ``` **Answer:** The compensator is $$G_c(s) = \frac{s + 0.05}{s + 0.01}$$ which achieves $$k_v = 20$$ without appreciably changing the closed-loop poles.