Mathematical Models Of Physical Systems Practice Questions

Given : $\frac{\mathrm{dy}(\mathrm{t})}{\mathrm{dt}}+3 \mathrm{y}(\mathrm{t})=x(\mathrm{t})$ Taking Laplace transform,

We have impulse response $=\mathrm{L}^{-1} \text { (Transfer function) }$ So, taking inverse Laplace transform, $y(t)=2 e^{-3 t} u(t)$

Signal flow graph: Forward paths,

Loops: $L_1=\frac{1}{S}$ Using Masson's graph formula,

Transfer function of transportation lag system, $\mathrm{T}(\mathrm{s})=\mathrm{Ke}^{-\mathrm{ST} T_{\mathrm{d}}}$ From magnitude plot,

From angle plot, at $\omega=1 \mathrm{rad} / \mathrm{sec} ., \theta=-60^{\circ}$ We have,

Simplifying given signal flow graph

Transfer function, \frac{Y(s)}{U_1(s)}=\frac{\frac{k_1}{LJs^2}[1]}{1-\left[ -\frac{R}{Ls} -\frac{k_1k_2}{LJs^2}\right]\}$$\frac{Y(s)}{U_1(s)}=\frac{k_1}{JLs^{2+}JRs+k_1k_2}$

$\frac{Y(s)}{X(s)}=\frac{G_1+G_2}{1-(-G_1H-G_2H)}$ $\;\;\;=\frac{G_1+G_2}{1+G_1H+G_2H)}$

$\frac{dx_1}{dt}=2x_1+x_2+u\;\;\;...(i)$ $\frac{dx_2}{dt}=-2x_1+4\;\;\;\;...(ii)$ From equation (i), $sx_1=2x_1+x_2+u$ $sx_2=-2x_1+u$ $x_2=\left ( -\frac{2x_1+u}{s} \right )$ $sx_1=2x_1+\left ( -\frac{2x_1+u}{s} \right )+4$ $sx_1-2x_1+\frac{2x_1}{s}=\frac{4}{5}+4$ $x_1\left ( s-2+\frac{2}{s} \right )=u\left (1+\frac{1}{s} \right )$ $\frac{x_1}{4}=\frac{\left ( 1+\frac{1}{s} \right )}{\left ( s-2+\frac{2}{s} \right )}$ $\Rightarrow \;\;\frac{y}{u}=\frac{3x-1}{u}=\frac{3\left ( 1+\frac{1}{s} \right )}{\left ( s-2+\frac{2}{s} \right )}$ $\Rightarrow \;\;\frac{y}{u}=\frac{3(s+1)}{(s^{2-}2s+2)}$

$\frac{Y(s)}{X_2(s)}|_{X_1(s)=0}=\frac{G_2}{1-(-G_1G_2-G_1)} =\frac{G_2}{1+G_1(1+G_2)}$

Using Mason's gain fomlula, Transfer function, $G(s)=\frac{C(s)}{U(s)}=\frac{\Sigma P_k\Delta _k}{\Delta }$ Forward paths are: $P_1=\frac{h_0}{s^2} ,\; P_2=\frac{h_1}{s}$ Individual loops are: $L_1=\frac{-a_1}{s} , \; L_2=\frac{-a_0}{s^2}$ Non-touching loops=zero $\therefore \;\;\Delta _1=1,$ $\;\;\;\; \Delta _2=1-\left ( \frac{-a_1}{s} \right ) =\left ( 1+\frac{a_1}{s} \right ) =\left ( \frac{s+a_1}{s} \right )$ $\;\;\;\;\Delta =1-\left ( \frac{-a_1}{s}-\frac{a_0}{s^2} \right )=\frac{s^{2+}a_1s+a_0}{s^2}$ $\therefore \;\; G(s)=\frac{C(s)}{U(s)}=\frac{P_1\Delta _1+P_2\Delta _2}{\Delta }$ $\;\;\;\;=\frac{\left ( \frac{h_0}{s^2} \right )\times 1+\left ( \frac{h_1}{s} \right )\times \left ( \frac{s+a_1}{s} \right )}{\frac{s^{2+}a_1s+a_0}{s^2}}$ $\;\;\;=\left[ \frac{h_0+h_1(s+a_1)}{s^{2+}a_1s+a_0} \right]$ Given: $h_1=b_1$ and $h_0=b_0-b_1a_1$ Thus, $G(s)=\frac{(b_0-b_1a_1)+b_1(s+a_1)}{(s^{2+}a_1s+a_0)}$ $G(s)=\left ( \frac{b_1s+b_0}{s^{2+}a_1s+a_0} \right )$

Using mason's gain formula, $\Delta =1-[-2s^{-1}-2s^{-2}-4-4s^{-1}]$ $\Delta =1+\frac{2}{s}+\frac{2}{s^2}+4+\frac{4}{s}$ $\Delta =\frac{5s^{2+}6s+2}{s^2}$ $P_1=s^{-2}=\frac{1}{s^2}$ $P_2=s^{-1}=\frac{1}{s}$ $\Delta _1=1$ $\Delta _2=2$ $\frac{Y(s)}{U(s)}=\frac{\Sigma P_k\Delta _k}{\Delta }$ $\;\;\;\;=\frac{\frac{1}{s^2} \times 1+\frac{1}{s} \times 1}{\frac{5s^{2+}6s+2}{s^2}}$ $\frac{Y(s)}{U(s)}=\frac{s+1}{5s^{2+}6s+2}$

$\frac{V_2(s)}{V_1(s)}=\frac{R+\frac{1}{Cs}}{\frac{1}{Cs}+R+\frac{1}{Cs}}$ $=\frac{1+RCs}{2+RCs}$ $=\frac{1+10 \times 10^3 \times 100 \times 10^{-6}s}{2+10 \times 10^3 \times 100 \times 10^{-6}s}$ $=\frac{s+1}{s+2}$

Transafer function of system is impulse response of the system with zero initial conditions. Transfer function= $H(s)=L(e^{-t}+e^{-2t})$ $\;\;=\frac{1}{s+1}+\frac{1}{s+2}$ $H(s)=\frac{C(s)}{R(s)}=\left ( \frac{1}{s+1}+\frac{1}{s+2} \right )$ $R(s)=\frac{1}{s}$ (step input) $C(s)=R(s)\cdot H(s)=\frac{1}{s}\left ( \frac{1}{s+1}+\frac{1}{s+2} \right )$ $C(s)=\frac{1}{s(s+1)}+\frac{1}{s(s+2)}$ $C(s)=\left ( \frac{1}{s}-\frac{1}{s+1} \right )+\frac{1}{2}\left ( \frac{1}{s}-\frac{1}{s+2} \right )$ $C(s)=\frac{1.5}{s}-\frac{1}{s+1}-\frac{0.5}{s+2}$ $Response=C(t)=L^{-1}[C(s)]$ $C(t)=L^{-1}\left[ \frac{1.5}{s}-\frac{1}{s+1}-\frac{0.5}{s+2} \right]$ $C(t)=(1.5-e^{-t}-0.5e^{-2t})u(t)$

Taking (L.T.) on both sides $Y(s)(s+1)=1$ $\therefore \;\; Y(s)=\frac{1}{s+1}$ Taking inverse laplace transform $y(t)=e^{-t}u(t)$

The block diagram can be redrawn as Signal flow graph of the block diagram, There are two forward paths: $P_1=1 \times c_0 \times \frac{1}{s}\times \frac{1}{s}\times P=\frac{c_0P}{s^2}$ $P_2=1 \times c_1 \times 1\times \frac{1}{s}\times P=\frac{c_1P}{s}$ These are four individual loops $L_1=-a_1 \times \frac{1}{s}=-\frac{a_1}{s}$ $L_2=-a_0 \times \frac{1}{s}\times \frac{1}{s}=-\frac{a_0}{s^2}$ $L_3=b_0 \times \frac{1}{s}\times \frac{1}{s} \times P=-\frac{b_0P}{s^2}$ $L_4=b_1 \times \frac{1}{s}\times P=\frac{b_1P}{s}$ All the loop touch forward paths $\Delta _1=\Delta _2=1$ $\Delta =1-(L_1+L_2+L_3+L_4)$ $\;\;=1+\frac{a_0}{s^2}+\frac{a_1}{s}-\frac{b_0P}{s^2}-\frac{b_1P}{s}$ Using Masson's gain formula $\frac{C(s)}{R(s)}=\frac{P_1\Delta _1+P_2\Delta _2}{s}$ $\frac{C(s)}{R(s)}=\frac{\frac{c_0P}{s^2}+\frac{c_1P}{s}}{1+\frac{a_0}{s^2}+\frac{a_1}{s}-\frac{b_0P}{s^2}-\frac{b_1P}{s}}$ $\frac{C(s)}{R(s)}=\frac{c_0P+c_1Ps}{s^{2+}(a_0-b_1)s+(a_0-b_0P)}$ $\frac{C(s)}{R(s)}=\frac{P(c_0+c_1s)/(s^2a_1s+a_0)}{1-(b_0+b_1s)P/(s^{2+}a_1s+a_0)} \;\;\;...(i)$ $\frac{C(s)}{R(s)}=\frac{xyP}{1-yzP} \;\;\;...(ii)$ Comparing equation (i) and (ii), we get $xy=\frac{c_0+c_1s}{s^{2+}a_1s+a_0}$ $yz=\frac{b_0+b_1s}{s^{2+}a_1s+a_0}$ Hence, Option (D)is correct.

Using signal flow graph, Forward path gains $P_{1}=\left ( -1 \right )\times \frac{2}{s+2}=\frac{-2}{s+2}$ and $P_2=\frac{3}{s}\frac{2}{s+2}=\frac{6}{s(s+2)}$ Individual loop, $L_1=-1 \times \frac{3}{s}\times \frac{2}{s+2}=\frac{-6}{s(s+2)}$ Loop touches forward paths, therefore, $\Delta _1=1 \text{ and } \Delta _2=1$ $D=1-L_1 =1+\frac{6}{s(s+2)}=\frac{s(s+2)+6}{s(s+2)}$ Using Mason's gain formula, $\frac{Y(s)}{R(s)}=\frac{P_1\Delta _1+P_2\Delta _2}{\Delta } =\frac{-\frac{2}{s+2}\times 1+\frac{6}{s(s+2)}\times 1}{\frac{s(s+2)+6}{s(s+2)}}$ $\frac{Y(s)}{R(s)}=\frac{6-2s}{s^{2+}2s+6}$ For unit step input, $R(s)=\frac{1}{s}$

Steady state value of error, using final value theoram,

Transfer function of a system is the unit impulse response of the system. Transfer function= $\frac{C\left ( s \right )}{R\left ( s \right )}=\left[ 12.5e^{-6t}\sin 8t \right]$ $\; \; \; \; =12.5\times \frac{8}{\left ( s+6 \right )^{2}+8^{2}}=\frac{100}{\left ( s+6 \right )^{2}+8^{2}}$ when input is unit step, R(s)= $\frac{1}{s}$ $\; \; C\left ( s \right )=\frac{1}{s}.\frac{100}{\left ( s+6 \right )^{2}+8^{2}}$ Steady-state value of response, using final value theorem $C_{steady-state}=\lim_{s\rightarrow \infty }C\left ( t \right )=\lim_{s\rightarrow 0}s\left ( s \right )$ \; \; \; \; =\lim_{s\rightarrow 0}s\left[ \frac{1}{s}.\frac{100}{\left ( s+6 \right )^{2}+8^{2}} \right]\=1$