Basics of Control Systems, Block Diagram and SFGs

Question 1
The block diagram of a system is illustrated in the figure shown, where X(s) is the input and Y(s) is the output. The transfer function H(s)=\frac{Y(s)}{X(s)} is
A
H(s)=\frac{s^2+1}{s^3+s^2+s+1}
B
H(s)=\frac{s^2+1}{s^3+2s^2+s+1}
C
H(s)=\frac{s+1}{s^2+s+1}
D
H(s)=\frac{s^2+1}{2s^2+s+1}
GATE EC 2019   Control Systems
Question 1 Explanation: 
Using block diagram reduction, we get,


\frac{Y(s)}{X(s)}=H(s)=\frac{s^{2}+1}{s^{3}+2 s^{2}+s+1}
Question 2
For the system shown in the figure, Y(s)/X(s)= __________.
A
1
B
2
C
3
D
4
GATE EC 2017-SET-2   Control Systems
Question 2 Explanation: 


\begin{aligned} Y(s) &=[X(s)-Y(s)] G(s)+X(s) \\ [1+G(s)] Y(s) &=[G(s)+1] X(s) \\ \frac{Y(s)}{X(s)} &=1 \end{aligned}
Question 3
The block diagram of a feedback control system is shown in the figure. The overall closed-loop gain G of the system is
A
G=\frac{G_{1}G_{2}}{1+G_{1}H_{1}}
B
G=\frac{G_{1}G_{2}}{1+G_{1}G_{2}+G_{1}H_{1}}
C
G=\frac{G_{1}G_{2}}{1+G_{1}G_{2}H_{1}}
D
G=\frac{G_{1}G_{2}}{1+G_{1}G_{2}+G_{1}G_{2}H_{1}}
GATE EC 2016-SET-3   Control Systems
Question 3 Explanation: 


\begin{aligned} \frac{Y(s)}{X(s)} &=\frac{\frac{G_{2} G_{1}}{1+G_{1} H_{1}}}{1+\frac{G_{2} G_{1}}{1+G_{1} H_{1}}} \\ &=\frac{G_{2} G_{1}}{1+G_{1} G_{2}+G_{1} H_{1}} \end{aligned}
Question 4
The position control of a DC servo-motor is given in the figure. The values of the parameters are K_{T}=1 N-m/A, R_{a}=1 \Omega , L_{a}=0.1H,J=5kg-m^{2} , B=1 N-m(rad/sec) and K_b=1V/(rad/sec) . The steady-state position response (in radians) due to unit impulse disturbance torque T_{d} is_______.
A
0
B
-0.5
C
-1
D
-1.5
GATE EC 2015-SET-3   Control Systems
Question 4 Explanation: 
The transfer function due to the disturbance torque T_{d}(s) \text { is }
\begin{aligned} \frac{\theta(s)}{T_{d}(s)}&=\frac{-\frac{1}{(J s+B)} \times \frac{1}{s}}{1+\left(\frac{1}{J s+B}\right)\left(\frac{K_{b} K_{T}}{R_{a}+L_{a} s}\right)} \\ &=\frac{-\left(R_{a}+L_{a} s\right) \cdot \frac{1}{s}}{\left(R_{a}+L_{a} s\right)(J s+B)+K_{b} K_{T}} \end{aligned}
The steady value of response for unit impulse input
\begin{aligned} &=\frac{-s\left(R_{a}+L_{a} s\right) \cdot \frac{1}{s}}{\left(R_{a}+L_{a} s\right)(J s+B)+K_{b} K_{T}}.T_{d}\\ &=-\frac{R_{a}}{R_{a} B+K_{b} K_{T}} \cdot 1\\ \text{Given:}\\ K_{T}&=1 \mathrm{N}-\mathrm{m} / \mathrm{A} R_{a}=1 \Omega \\ B&=1 \mathrm{N}-\mathrm{m} / \mathrm{rad} / \mathrm{sec}\\ \text{and }\quad K_{b}&=1 \mathrm{V} / \mathrm{rad} / \mathrm{sec} \end{aligned}
Substituting the given values into above equation, we get
\theta(0)=-\frac{1}{2}=-0.5 \mathrm{rad}
Question 5
For the signal flow graph shown in the figure, the value of \frac{C(s)}{R(s)} is

A
\frac{G_{1}G_{2}G_{3}G_{4}}{1-G_{1}G_{2}H_{1}-G_{3}G_{4}H_{2}-G_{2}G_{3}H_{3}+G_{1}G_{2}G_{3}G_{4}H_{1}H_{2}}
B
\frac{G_{1}G_{2}G_{3}G_{4}}{1+G_{1}G_{2}H_{1}+G_{3}G_{4}H_{2}+G_{2}G_{3}H_{3}+G_{1}G_{2}G_{3}G_{4}H_{1}H_{2}}
C
\frac{1}{1+G_{1}G_{2}H_{1}+G_{3}G_{4}H_{2}+G_{2}G_{3}H_{3}+G_{1}G_{2}G_{3}G_{4}H_{1}H_{2}}
D
\frac{1}{1-G_{1}G_{2}H_{1}-G_{3}G_{4}H_{2}-G_{2}G_{3}H_{3}+G_{1}G_{2}G_{3}G_{4}H_{1}H_{2}}
GATE EC 2015-SET-2   Control Systems
Question 5 Explanation: 
Transfer function
\begin{aligned} \frac{C(s)}{R(s)}&=\frac{\Sigma P_{k} \Delta_{k}}{\Delta} \\ =& \frac{G_{1} G_{2} G_{3} G_{4}}{1-\left(-G_{1} G_{2} H_{1}-G_{2} G_{4} H_{2}-G_{2} G_{3} H_{3}\right)+G_{1} G_{2} H_{1} \cdot G_{3} G_{4} H_{2}} \\ \frac{C(s)}{R(s)}&= \frac{G_{1}G_{2}G_{3}G_{4}}{1+G_{1}G_{2}H_{1}+G_{3}G_{4}H_{2}+G_{2}G_{3}H_{3}+G_{1}G_{2}G_{3}G_{4}H_{1}H_{2}} \end{aligned}
Question 6
By performing cascading and/or summing/differencing operations using transfer function blocks G_{1}(s) \; and \; G_{2}(s), one CANNOT realize a transfer function of the form
A
{G_{1}(s)}{G_{1}(s)}
B
\frac{G_{1}(s)}{G_{1}(s)}
C
G_{1}(s)(\frac{1}{G_{1}(s)}+G_{2}(s))
D
G_{1}(s)(\frac{1}{G_{1}(s)}-G_{2}(s))
GATE EC 2015-SET-2   Control Systems
Question 6 Explanation: 
Given blocks has transfer function G_{1}(s) and G_{2}(s)
In cascade connection G_{1}(s) \cdot G_{2}(s)
In parallel connection G_{1}(s)+G_{2}(s)
From the given options:
(For cascading/ summing/differncing)
(A) G_{1}(s) G_{2}(s) \Rightarrow Realization is possible.
(B) \frac{G_{1}(s)}{G_{2}(s)} \Rightarrow Realization is not possible because
the gain of blocks given as G_{1}(s) and G_{2}(s)
(C) G_{1}(s)\left(\frac{1}{G_{1}(s)}+G_{2}(s)\right)=1+G_{1}(s) G_{2}(s)
Realization is possible.
(D) G_{1}(s)\left(\frac{1}{G_{1}(s)}-G_{2}(s)\right)=1-G_{1}(s) G_{2}(s)
Realization is possible.
Question 7
Negative feedback in a closed-loop control system DOES NOT
A
reduce the overall gain
B
reduce bandwidth
C
improve disturbance rejection
D
reduce sensitivity to parameter variation
GATE EC 2015-SET-1   Control Systems
Question 8
Consider the following block diagram in the figure.
The transfer function \frac{C(s)}{R(s)} is
A
\frac{G_{1}G_{2}}{1+G_{1}G_{2}}
B
G_{1}G_{2}+G_{1}+1
C
G_{1}G_{2}+G_{2}+1
D
\frac{G_{1}}{1+G_{1}G_{2}}
GATE EC 2014-SET-3   Control Systems
Question 8 Explanation: 


Converting the block d iagram into signal flow


graph on
Forward paths,
P_{1}=G_{1} G_{2} ; P_{2}=G_{2} \cdot 1 ; P_{3}=1 \cdot 1=1
So, the transfer function is
\frac{C(s)}{R(s)}=G_{1} G_{2}+G_{2}+1
Question 9
For the following system,

when X_{1}(s) = 0, the transfer function \frac{Y(s)}{X_{2}(s)} is
A
\frac{s+1}{s^{2}}
B
\frac{1}{s+1}
C
\frac{s+2}{s(s+1)}
D
\frac{s+1}{s(s+2)}
GATE EC 2014-SET-2   Control Systems
Question 9 Explanation: 
Redrawing the block diagram with X_{1}(s)=0


The transfer function
T(s)=\frac{Y(s)}{X_{2}(s)}=\frac{G(s)}{1+G(s) H(s)} \quad\ldots(i)
Here, G(s)=\frac{1}{s} and H(s)=\frac{s}{s+1}
\frac{Y(s)}{X_{2}(s)}=\frac{1 / s}{1+\frac{1}{s} \times \frac{s}{s+1}}=\frac{(s+1)}{s(s+2)}
Question 10
The signal flow graph for a system is given below. The transfer function \frac{Y\left ( s \right )}{U\left ( s \right )} 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 EC 2013   Control Systems
Question 10 Explanation: 
Using Mason's gain formula,
\begin{aligned} \Delta &=1-\left[-2 s^{-1}-2 s^{-2}-4-4 s^{-1}\right] \\ \Delta &=1+\frac{2}{s}+\frac{2}{s^{2}}+4+\frac{4}{s} \\ \Delta &=\frac{5 s^{2}+6 s+2}{s^{2}} \\ P_{1} &=s^{2}=\frac{1}{s^{2}} \\ P_{2} &=s^{-1}=\frac{1}{s} \\ \Delta_{1} &=1 \\ \Delta_{2} &=1 \\ \frac{Y(s)}{U(s)} &=\frac{\sum P_{k} \Delta_{k}}{\Delta}=\frac{\frac{1}{s^{2}} \times 1+\frac{1}{s} \times 1}{\frac{5 s^{2}+6 s+2}{s^{2}}} \\ \frac{Y(s)}{U(s)} &=\frac{s+1}{5 s^{2}+6 s+2} \end{aligned}
There are 10 questions to complete.
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