Feedback Characteristics Of Control Systems Practice Questions

Transfer function $\rightarrow \frac{40\left(1+\frac{s}{40}\right)}{20\left(1+\frac{s}{40}\right)(1+s)}$ Answer $=\frac{2}{s+1}$

Given plant: So, $\quad$ OLTF $=\frac{(3 s+1)}{s(s-1)}$ Closed loop transfer function,

Final value of plant, $Y(\infty )=\operatorname{Lims}_{s \rightarrow 0} Y(s)=1$ From plant,

$\therefore$ Final value of controller, $x(\infty )=\operatorname{LimsX}_{s \rightarrow 0} \mathrm{~s}(\mathrm{~s})=-1$

Closed loop transfer function, $T(s)=\frac{4}{(s^{2+}0.4s+4)}$ or, $\frac{G(s)}{1+G(s)}$ $=\left ( \frac{4}{s^{2+}0.4s+4} \right )\;\;\;\;[H(s)=1]$ or, $(s^{2+}0.4s+4)=4+4G(s)$ or, $G(s)=\frac{4}{s(s+0.4)}$ =Open llop transfer function ( for unity feedback system) Given, input =u(t)=r(t) $\therefore \;\;\; R(s)=1/s$ Steady state error $e_{ss}=\lim_{s \to 0}\left[ \frac{sR(s)}{1+G(s)H(s)} \right]$ $\;\;=\lim_{s \to 0}\left[ \frac{sR(s)}{1+G(s)} \right]\;\;\; [For H(s)=1]$ $\;\;=\lim_{s \to 0}\left[ \frac{s \times \frac{1}{s}}{1+\frac{4}{s(s+0.4)}} \right]$ \;\;=\lim_{s \to 0}\left[ \frac{s(s+0.4)}{s^{2+}0.4s+4} \right]\=0$$\therefore$Steady state error for step input,$e_{ss}=0$

For all given values of G(s) in options, G(s) does not satisfy the desired condition. But option (B) is official answer given by IIT.

$\frac{\omega (s)}{V_a(s)}=\frac{10}{1+10s}=\frac{10}{1+\tau s}$ $\tau =10$ Closed loop systems, $\frac{\omega (s)}{R(s)}=\frac{\frac{10K_a}{1+10s}}{1+\frac{10K_a}{1+10s}}$ $=\frac{10K_a}{1+10K_a+10s}$ $=\frac{\frac{10K_a}{1+10K_a}}{1+\frac{10}{1+10K_a}s}$ According to quetion, $\frac{10}{1+10K_a}=\frac{1}{100}\times 10$ $1+10K_a=100$ $K_a=9.9$

$G=100\pm 10$ % $\frac{\Delta G}{G}=10$ % or 0.1 $H=\frac{9}{100}$ Overall gain, $T=\frac{G}{1+GH}$ $T=\frac{100}{1+100 \times \frac{9}{100}}=10$ Sensitivity w.r.t. $'G'=\frac{1}{1+GH} \times 10$ % $\;\;\;=\frac{1}{1+10 \times 1}\times 10$ % $\;\;\;=\frac{10}{11}\cong 1$ %