Topic: Control Systems
For a feedback system shown below, If $K_t=0$ and $K_a=5$, then steady state error for unit ramp input is 0.2. What will be the new value of $K_t$ and $K_a$ if damping ratio is increased to 0.5 without affecting steady state error:
- A. $K_t=1.5, K_a=12.5$
- B. $K_t=15, K_a=1.25$
- C. $K_t=1.5, K_a=1.25$
- D. $K_t=15, K_a=12.5$
Answer: A. $K_t=1.5, K_a=12.5$
Explanation: Simplifying the block diagram, the forward transfer function is $G(s) = \frac{2K_a}{s(s+2+2K_t)}$. The steady state error for a unit ramp input is $e_{ss} = \frac{1}{K_v} = \frac{2+2K_t}{2K_a}$. Setting $e_{ss} = 0.2$ means $\frac{2+2K_t}{2K_a} = 0.2$. The closed-loop characteristic equation is $s^2 + (2+2K_t)s + 2K_a = 0$. Here, $\omega_n^2 = 2K_a$ and $2\zeta\omega_n = 2+2K_t$. Substituting these into the error equation gives $\frac{2\zeta\omega_n}{\omega_n^2} = \frac{2\zeta}{\omega_n} = 0.2$. With a new required damping ratio $\zeta = 0.5$, we get $\frac{2(0.5)}{\omega_n} = 0.2 \implies \omega_n = 5$. Thus, $2K_a = 5^2 = 25 \implies K_a = 12.5$. Finally, $2+2K_t = 2\zeta\omega_n = 2(0.5)(5) = 5 \implies 2K_t = 3 \implies K_t = 1.5$.