# 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:
(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.