Question 1 |
For the closed-loop system shown, the transfer function \dfrac{E\left ( s \right )}{R\left ( s \right )} is
\frac{G}{1+GH} | |
\frac{GH}{1+GH} | |
\frac{1}{1+GH} | |
\frac{1}{1+G} |
Question 1 Explanation:
\begin{aligned} \frac{E(s)}{R(s)} &=\frac{R(s)-H \times C(s)}{R(s)}=1-H \times \frac{C(s)}{R(s)} \\ &=1-\frac{H G}{1+G H}=\frac{1+G H-G H}{1+G H} \\ &=\frac{1}{1+G H} \end{aligned}
Question 2 |
Which of the options is an equivalent representation of the signal flow graph shown here?
A | |
B | |
C | |
D |
Question 2 Explanation:
Simplifying given signal flow graph
Question 3 |
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 __________.
0.5 | |
1 | |
1.5 | |
2.5 |
Question 3 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 4 |
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
\frac{k_1}{JLs^2+JRs+k_1k_2} | |
\frac{k_1}{JLs^2-JRs-k_1k_2} | |
\frac{k_1-U_2(R+sL)}{JLs^2+(JR-U_2L)s+k_1k_2-U_2R} | |
\frac{k_1-U_2(sL-R)}{JLs^2-(JR+U_2L)s-k_1k_2+U_2R} |
Question 4 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}
\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 5 |
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)
\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)
2e^{-2t}u(t)-7e^{-5t}u(t) | |
-2e^{-2t}u(t)+7e^{-5t}u(t) | |
7e^{-2t}u(t)-2e^{-5t}u(t) | |
-7e^{-2t}u(t)+2e^{-5t}u(t) |
Question 5 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 6 |
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
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
3(s+1)/(s^{2}-2s+2) | |
3(2s+1)/(s^{2}-2s+1) | |
(s+1)/(s^{2}-2s+2) | |
3(2s+1)/(s^{2}-2s+2) |
Question 6 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)}
\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 7 |
Find the transfer function \frac{Y(s)}{X(s)} of the system given below.
\frac{G_{1}}{1-HG_{1}}+\frac{G_{2}}{1-HG_{2}} | |
\frac{G_{1}}{1+HG_{1}}+\frac{G_{2}}{1+HG_{2}} | |
\frac{G_{1}+G_{2}}{1+H(G_{1}+G_{2})} | |
\frac{G_{1}+G_{2}}{1-H(G_{1}+G_{2})} |
Question 7 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 8 |
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}?
\frac{G_{1}}{1+G_{2}(1+G_{1})} | |
\frac{G_{2}}{1+G_{1}(1+G_{2})} | |
\frac{G_{1}}{1+G_{1}G_{2}} | |
\frac{G_{2}}{1+G_{1}G_{2}} |
Question 8 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 9 |
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
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
G(s)=\frac{b_{0}s+b_{1}}{s^{2}+a_{0}s+a_{1}} | |
G(s)=\frac{a_{1}s+a_{0}}{s^{2}+b_{1}s+b_{0}} | |
G(s)=\frac{b_{1}s+b_{0}}{s^{2}+a_{1}s+a_{0}} | |
G(s)=\frac{a_{0}s+a_{1}}{s^{2}+b_{0}s+b_{1}} |
Question 9 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 )
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 10 |
The signal flow graph for a system is given below. The transfer function \frac{Y(s)}{U(s)} for this system is
\frac{s+1}{5s^{2}+6s+2} | |
\frac{s+1}{s^{2}+6s+2} | |
\frac{s+1}{s^{2}+4s+2} | |
\frac{1}{5s^{2}+6s+2} |
Question 10 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}
\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}
There are 10 questions to complete.