# Mathematical Models of Physical Systems

 Question 1
Which of the options is an equivalent representation of the signal flow graph shown here?
 A A B B C C D D
GATE EE 2020   Control Systems
Question 1 Explanation:
Simplifying given signal flow graph

 Question 2
For a system having transfer function $G(s)=\frac{-s+1}{s+1}$, a unit step input is applied at time t=0. The value of the response of the system at t=1.5 sec is __________.
 A 0.5 B 1 C 1.5 D 2.5
GATE EE 2017-SET-1   Control Systems
Question 2 Explanation:
\begin{aligned} G(s)&=\frac{-s+1}{s+1} \\ &\text{System output,} \\ Y(s)&=G(s)\cdot \frac{1}{s} \\ &=\frac{-s+1}{s+1}\cdot \frac{1}{s} \\ &=\frac{1}{s}-\frac{2}{s+1} \\ y(t)&=u(t)-2e^{-t}u(t)\\ y(1.5)&=1-2e^{-1.5} \\ &=1-0.466=0.554 \end{aligned}
 Question 3
In the system whose signal flow graph is shown in the figure, $U_1(s)$ and $U_2(s)$ are inputs. The transfer function $\frac{Y(s)}{U_{1}(s)}$ is
 A $\frac{k_1}{JLs^2+JRs+k_1k_2}$ B $\frac{k_1}{JLs^2-JRs-k_1k_2}$ C $\frac{k_1-U_2(R+sL)}{JLs^2+(JR-U_2L)s+k_1k_2-U_2R}$ D $\frac{k_1-U_2(sL-R)}{JLs^2-(JR+U_2L)s-k_1k_2+U_2R}$
GATE EE 2017-SET-1   Control Systems
Question 3 Explanation:
Transfer function,
$\frac{Y(s)}{U_1(s)}=\frac{\frac{k_1}{LJs^2}[1]}{1-\left [ -\frac{R}{Ls} -\frac{k_1k_2}{LJs^2}\right ]}$
$\frac{Y(s)}{U_1(s)}=\frac{k_1}{JLs^2+JRs+k_1k_2}$
 Question 4
Let a causal LTI system be characterized by the following differential equation, with initial rest condition
$\frac{d^{2}y}{dt^{2}}+7\frac{dy}{dt}+10y(t)=4x(t)+5\frac{dx(t)}{dt}$
Where x(t) and y(t) are the input and output respectively. The impulse response of the system is (u(t) is the unit step function)
 A $2e^{-2t}u(t)-7e^{-5t}u(t)$ B $-2e^{-2t}u(t)+7e^{-5t}u(t)$ C $7e^{-2t}u(t)-2e^{-5t}u(t)$ D $-7e^{-2t}u(t)+2e^{-5t}u(t)$
GATE EE 2017-SET-1   Control Systems
Question 4 Explanation:
\begin{aligned} \frac{d^2y}{dt^2}+7\frac{dy}{dt}+10y&=4x+5\frac{dx}{dt} \\ (s^2+7s+10)Y(s)&=(4+5s)X(s) \\ \frac{Y(s)}{X(s)}&=\frac{5s+4}{s^2+7s+10} \\ \text{Impulse response}&=L^{-1}(\text{Transfer function}) \\ &=L^{-1}\frac{5s+4}{(s+2)(s+5)} \\ &=L^{-1}\left [ -\frac{2}{s+2} +\frac{7}{s+5}\right ] \\ &=-2e^{-2t}u(t)+7e^{-5t}u(t) \end{aligned}
 Question 5
For the system governed by the set of equations:
$dx_{1}/dt=2x_{1}+x_{2}+u$
$dx_{2}/dt=-2x_{1}+u$
$y=3x_{1}$
the transfer function Y(s)/U(s) is given by
 A $3(s+1)/(s^{2}-2s+2)$ B $3(2s+1)/(s^{2}-2s+1)$ C $(s+1)/(s^{2}-2s+2)$ D $3(2s+1)/(s^{2}-2s+2)$
GATE EE 2015-SET-2   Control Systems
Question 5 Explanation:
$\frac{dx_1}{dt}=2x_1+x_2+u\;\;\;...(i)$
$\frac{dx_2}{dt}=-2x_1+4\;\;\;\;...(ii)$
From equation (i),
$sx_1=2x_1+x_2+u$
$sx_2=-2x_1+u$
$x_2=\left ( -\frac{2x_1+u}{s} \right )$
$sx_1=2x_1+\left ( -\frac{2x_1+u}{s} \right )+4$
$sx_1-2x_1+\frac{2x_1}{s}=\frac{4}{5}+4$
$x_1\left ( s-2+\frac{2}{s} \right )=u\left (1+\frac{1}{s} \right )$
$\frac{x_1}{4}=\frac{\left ( 1+\frac{1}{s} \right )}{\left ( s-2+\frac{2}{s} \right )}$
$\Rightarrow \;\;\frac{y}{u}=\frac{3x-1}{u}=\frac{3\left ( 1+\frac{1}{s} \right )}{\left ( s-2+\frac{2}{s} \right )}$
$\Rightarrow \;\;\frac{y}{u}=\frac{3(s+1)}{(s^2-2s+2)}$
 Question 6
Find the transfer function $\frac{Y(s)}{X(s)}$ of the system given below.
 A $\frac{G_{1}}{1-HG_{1}}+\frac{G_{2}}{1-HG_{2}}$ B $\frac{G_{1}}{1+HG_{1}}+\frac{G_{2}}{1+HG_{2}}$ C $\frac{G_{1}+G_{2}}{1+H(G_{1}+G_{2})}$ D $\frac{G_{1}+G_{2}}{1-H(G_{1}+G_{2})}$
GATE EE 2015-SET-1   Control Systems
Question 6 Explanation:

$\frac{Y(s)}{X(s)}=\frac{G_1+G_2}{1-(-G_1H-G_2H)}$
$\;\;\;=\frac{G_1+G_2}{1+G_1H+G_2H)}$
 Question 7
For the signal-flow graph shown in the figure, which one of the following expressions is equal to the transfer function $\frac{Y(s)}{X_{2}(s)}|_{X_{1}(s)=0}$?

 A $\frac{G_{1}}{1+G_{2}(1+G_{1})}$ B $\frac{G_{2}}{1+G_{1}(1+G_{2})}$ C $\frac{G_{1}}{1+G_{1}G_{2}}$ D $\frac{G_{2}}{1+G_{1}G_{2}}$
GATE EE 2015-SET-1   Control Systems
Question 7 Explanation:

$\frac{Y(s)}{X_2(s)}|_{X_1(s)=0}=\frac{G_2}{1-(-G_1G_2-G_1)} =\frac{G_2}{1+G_1(1+G_2)}$
 Question 8
The signal flow graph of a system is shown below. U(s) is the input and C(s) is the output

Assuming $h_{1}=b_{1} \; and \; h_{0}=b_{0}-b_{1}a_{1}$, the input-output transfer function $G(s)=\frac{C(s)}{U(s)}$ of the system is given by
 A $G(s)=\frac{b_{0}s+b_{1}}{s^{2}+a_{0}s+a_{1}}$ B $G(s)=\frac{a_{1}s+a_{0}}{s^{2}+b_{1}s+b_{0}}$ C $G(s)=\frac{b_{1}s+b_{0}}{s^{2}+a_{1}s+a_{0}}$ D $G(s)=\frac{a_{0}s+a_{1}}{s^{2}+b_{0}s+b_{1}}$
GATE EE 2014-SET-3   Control Systems
Question 8 Explanation:
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, $G(s)=\frac{(b_0-b_1a_1)+b_1(s+a_1)}{(s^2+a_1s+a_0)}$
$G(s)=\left ( \frac{b_1s+b_0}{s^2+a_1s+a_0} \right )$
 Question 9
The signal flow graph for a system is given below. The transfer function $\frac{Y(s)}{U(s)}$ for this system is
 A $\frac{s+1}{5s^{2}+6s+2}$ B $\frac{s+1}{s^{2}+6s+2}$ C $\frac{s+1}{s^{2}+4s+2}$ D $\frac{1}{5s^{2}+6s+2}$
GATE EE 2013   Control Systems
Question 9 Explanation:
Using mason's gain formula,
$\Delta =1-[-2s^{-1}-2s^{-2}-4-4s^{-1}]$
$\Delta =1+\frac{2}{s}+\frac{2}{s^2}+4+\frac{4}{s}$
$\Delta =\frac{5s^2+6s+2}{s^2}$
$P_1=s^{-2}=\frac{1}{s^2}$
$P_2=s^{-1}=\frac{1}{s}$
$\Delta _1=1$
$\Delta _2=2$
$\frac{Y(s)}{U(s)}=\frac{\Sigma P_k\Delta _k}{\Delta }$
$\;\;\;\;=\frac{\frac{1}{s^2} \times 1+\frac{1}{s} \times 1}{\frac{5s^2+6s+2}{s^2}}$
$\frac{Y(s)}{U(s)}=\frac{s+1}{5s^2+6s+2}$
 Question 10
The transfer function $\frac{V_{2}(s)}{V_{1}(s)}$ of the circuit shown below is
 A $\frac{0.5s+1}{s+1}$ B $\frac{3s+6}{s+2}$ C $\frac{s+2}{s+1}$ D $\frac{s+1}{s+2}$
GATE EE 2013   Control Systems
Question 10 Explanation:

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