For the open loop system in Fig. 2.51 (a) find the percentage change in the steady-state value of C(s) for a unit step input and for a 10% change in K. For the closed loop system in Fig. 2.51 (b), find the percentage change in the steady-state value of C(s) for a unit step input for the same increase in the value of K.

a) For the open loop system:

Step 1: Determine the transfer function $C(s)/R(s)$. From Fig. 2.51 (a), the open-loop transfer function is $G(s) = \frac{K}{0.1s+1}$. For a unit step input, $R(s) = \frac{1}{s}$. The output $C(s)$ is given by: $C(s) = G(s)R(s) = \frac{K}{0.1s+1} \cdot \frac{1}{s}$

Step 2: Calculate the original steady-state value of $C(s)$ using the Final Value Theorem. $C_{ss1} = \lim_{s \to 0} sC(s) = \lim_{s \to 0} s \left( \frac{K}{0.1s+1} \cdot \frac{1}{s} \right)$ $C_{ss1} = \lim_{s \to 0} \frac{K}{0.1s+1} = \frac{K}{0.1(0)+1} = K$

Step 3: Calculate the new value of K and the new steady-state value. K changes by 10%, so $K_{new} = K + 0.1K = 1.1K$. The new steady-state value $C_{ss2}$ is: $C_{ss2} = \lim_{s \to 0} s \left( \frac{K_{new}}{0.1s+1} \cdot \frac{1}{s} \right) = \lim_{s \to 0} \frac{K_{new}}{0.1s+1} = K_{new} = 1.1K$

Step 4: Calculate the percentage change in the steady-state value. $\text{Percentage Change} = \frac{C_{ss2} - C_{ss1}}{C_{ss1}} \times 100\%$ $\text{Percentage Change} = \frac{1.1K - K}{K} \times 100\% = \frac{0.1K}{K} \times 100\% = 0.1 \times 100\%$ \text{Percentage Change} = \text{*10%*}

b) For the closed loop system:

Step 1: Determine the closed-loop transfer function $C(s)/R(s)$. From Fig. 2.51 (b), this is a unity feedback system with $G(s) = \frac{K}{0.1s+1}$ and $H(s)=1$. The closed-loop transfer function $T(s)$ is: