Time Response Analysis Practice Questions

In system -1

Settling time $-t_{s 1}=\frac{4}{\xi \omega_{n}}=\frac{4}{0.5 \times 3}=\frac{8}{3}=2.67 \mathrm{sec}$ In system -2

Settling time $-t_{s 2}=\frac{4}{\xi \omega_{n}}=\frac{4}{\cos 70^{\circ} \times 1}=11.69 \mathrm{sec}$ So, $t_{s 2} > t_{s 1}$

Reduced the block diagram: Transfer function, $\frac{C(s)}{R(s)}=\frac{100/(s(s+10))}{1+(100/s(s+10))}=\frac{100}{s^{2+}10s+100}$ Standard form, $T.F.=\frac{\omega _n^2}{s^{2+}2\xi \omega _ns+\omega _n^2}$ On comparison : $\omega =\sqrt{100}=10rad/sec$ and $2\xi \omega _n=10$ $\Rightarrow \xi=\frac{10}{2 \times 10}=0.5$

Given system is fourth order system and unstable. stablity status: since it has one missing term of 's' thus undoubtedly given transfer function is unstable.

CE is $1+G(s)H(s)=0$ $\Rightarrow \, 1+\frac{s^{2}+s+1}{s^{3}+2s^{2}+2s+k}=0$ $\Rightarrow \, s^{3}+3s^{2}+3s+(1+K)=0$ R.H. criteria: $9 - (1 + K) = 0$ $\Rightarrow \, \, K=8$

Steady state error for type-0 and step input, $e_{ss}=\frac{1}{1+k+p}$ $k_p=\lim_{s \to 0}\frac{1}{(s+1)(s+2)}=\frac{1}{2}$ $e_{ss}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$ $\;\;=0.66$ uinits

$P=\frac{15}{s^{2+}5s+15}$ $\omega _n=\sqrt{15}=3.872$ rad/sec $2 \xi \times 3.872=5$ $\xi=\frac{5}{2 \times 3.872}=0.64 \;\;\;\;(Underdamped)$ $Q=\frac{25}{s^{2+}10s+25}$ $\omega _n=\sqrt{25}=5$ rad/sec $2 \xi \times 5=10$ $\xi=1 \;\;\;\;\;(Critically \; damped)$ Observing all the options, option (C) is correct.

Closed loop transfer function, $\frac{C(s)}{R(s)}=\frac{\frac{K}{(s+1)^2(s+2)}}{1+\frac{K}{(s+1)^2(s+2)}}$ $\frac{C(s)}{R(s)}=\frac{K}{(s+1)^2(s+2)+K}$ Given $R(s)=\frac{1}{s}$ $C(s)=\frac{K}{s((s+1)^2(s+2)+K)}$ $\lim_{s \to 0}sC(s)=0.8$ $\frac{K}{2+K}=0.8$ $\Rightarrow \; K=8$

For maximum peak over shoot $M_P\propto \frac{1}{\xi}$ $\xi=0.25$ for option (C) which is least among all options. Therefore correct option is C.

Closed loop transfer function $=\frac{Ks+b}{s^{2+}as+b}$ Open loop transfer function $= G(s)=\frac{Ks+b}{s^{2+}as+b-Ks-b}$ $G(s)=\frac{Ks+b}{s^{2+}as-Ks} =\frac{Ks+b}{s(s+a-K)}$ Steady state error for ramp input given to type-1 system $=1/K_V$ where, velocity error coefficient, $K_V=\lim_{s \to 0}s\cdot \frac{Ks+b}{s(s+a-K)} =\frac{b}{a-K}$ Steady state error, $e_{ss}=\frac{a-K}{b}$

Damping ratio $\xi=0.5$ Undamped natural frequency $\omega _n=10$ rad/sec Steady state output toa unit step input $C_{ss}=1.02$ Hence steady state error $e_{ss}=1.02-1.00=0.02$ $\because$ Characteristic equation is, $s^{2+}2 \xi \omega _ns+\omega _n^2=0$ $s^{2+}2 \times 0.5\times 10 s+100=0$ $s^{2+}10s+100=0$ From options, if we take option (B) then $C_{ss}=\lim_{s \to 0}s\cdot C(s)$ $\;\; \; \; \;=\lim_{s \to 0 }s \times \frac{1}{s} \times \frac{102}{s^{2+}10s+100}$ $C_{ss}=1.02$ Hence, Option (B) is correct answer.

$T.F.=\frac{1-2s}{1+s}$ $C(s)=\frac{1-2s}{1+s}\cdot \frac{1}{s}$ $\;\;=\frac{A}{1+s}+\frac{B}{s}$ For A, $A=\lim_{s \to -1}(s+1)\cdot \frac{1-2s}{s(1+s)}$ $\;\;=\frac{(1-2(-1))}{(-1)}=-3$ For B, $B=\lim_{s \to 0}s\cdot \frac{1-2s}{s(1+s)}=1$ $C(s)=\frac{1}{s}-\frac{3}{1+s}$ $C(t)=(1-3e^{-t})u(t)$

$\frac{5}{(s+2)^{2+}5^2}+\frac{(s+2)}{(s+2)^{2+}5^2}$ Putting $s=0,$ $\;\;=\frac{5}{2^{2+}5^2}+\frac{2}{2^{2+}5^2}$ $\;\;=\frac{5}{29}+\frac{2}{29}=\frac{7}{29}=0.241$

Let the system is represnted as $\frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s)H(s)}$ $H(s)=1 (unity \; feedback)$ $Error=E(s)=\frac{X(s)}{1+G(s)}$ Steady state error for $X(s)=1/s, \; e_{ss}=0.1$ $e_{ss}=\lim_{s \to 0}sE(s)$ $\;\; =\lim_{s \to 0}\frac{sX(s)}{1+G(s)}$ $\Rightarrow \; 0.1=\lim_{s \to 0}\frac{s\cdot \frac{1}{s}}{1+G(s)}$ $\Rightarrow \; 0.1=\lim_{s \to 0}\frac{1}{1+G(s)}$ When input, $X(t)=10[u(t)-u(t-1)]$ $X(s)=10\left ( \frac{1}{s}-\frac{e^{-s}}{s} \right )$ $\;\;=\frac{10}{s}(1-e^{-s})$ $e_{ss}=\lim_{s \to 0}sE(s)$ $\;\;=\lim_{s \to 0}\frac{sX(s)}{1+G(s)}$ \;\;=\lim_{s \to 0}\frac{s \times \frac{10}{s}\[1-e^{-s}]$}{1+G(s)}$$e_{ss}=\lim_{s \to 0}\frac{10(1-e^{-s})}{1+G(s)}=0$

$G(s)=\frac{1}{s(s+1+k)}$ and $H(s)=1$ Characteristic equation $1+G(s)H(s)=0$ $\Rightarrow \;\; 1+\frac{1}{s(s+1+k)}=0$ $\Rightarrow \;\; s(s+1+k)+1=0$ $\Rightarrow \;\; s^{2+}(R+1)s+1=0$ Comapring with, $s^{2+}2 \xi \omega _ns+\omega _n^2$ Natural frequency $\omega _n=1$ remains constant and does not depend on k. $2 \xi \omega _n=k+1$ $\xi=\frac{k+1}{2}$ Damping ratio depends on k Peak overshoot $=m_p=e^{-\xi \pi/\sqrt{1-\xi^2}}$ Since, $m_p$ depends on $\xi$ which depends on k. Hence, peak overshoot is influenced by gain (k) of techo-generator.

$\frac{C(s)}{R(s)}=\frac{2}{s+1}$ $R(s)=\frac{1}{s}(step-input)$ $C(s)=R(s)\left ( \frac{2}{s+1} \right )$ $\;\;=\frac{2}{s(s+1)} =2\left ( \frac{1}{s} -\frac{1}{s-1}\right )$ $C(t)=L^{-1}[C(s)]=2[-1-e^{-t}]$ Final value of $C(t)=C_{ss}=2$ 98% of $C_{ss}=0.98 \times 2=1.96$ Let, $t=T$ , the response reaches 98% of its final values. $1.96=2[1-e^{-T}]$ $T\approx 4 sec$

Steady state value of response = 0.75 Input is unit step So steady state error $e_{ss}=1-0.75=0.25$ $Error=E(s)=\frac{R(s)}{1+G(s)H(s)}$ $where, \;\; G(s)=\frac{k}{(s+1)(s+2)} \;\; H(s)=1$ $R(s)=\frac{1}{s}$ Steady state erro using find value theorem, $e_{ss}=\lim_{s \to 0}sE(s)$ $\;\;\lim_{s \to 0}\frac{sR(s)}{1+G(s)H(s)}$ $\;\;\lim_{s \to 0}\frac{s\cdot \frac{1}{s}}{1+\frac{k}{(s+1)(s+2)}}$ $\;\;=\frac{1}{1+k/2}$ $\Rightarrow\;\; 0.25=\frac{1}{1+k/2}$ $\Rightarrow \;\; k=6$