# Basics of Control Systems, Block Diagram and SFGs

 Question 1
In the following block diagram, $R(s)$ and $D(s)$ are two inputs. The output $Y(s)$ is expressed as $Y(s)=G_{1}(s) R(s)+G_{2}(s) D(s)$.
$G_{1}(s)$ and $G_{2}(s)$ are given by

 A $G_{1}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)}$ and $G_{2}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)}$ B $G_{1}(s)=\frac{G(s)}{1+G(s)+H(s)}$ and $G_{2}(s)=\frac{G(s)}{1+G(s)+H(s)}$ C $G_{1}(s)=\frac{G(s)}{1+G(s)+H(s)}$ and $G_{2}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)}$ D $G_{1}(s)=\frac{G(s)}{1+G(s)+G(s) H(s)}$ and $G_{2}(s)=\frac{G(s)}{1+G(s)+H(s)}$
GATE EC 2023   Control Systems
Question 1 Explanation:

$Y(s)=\underbrace{G_{1}(s) R(s)}_{Y_{1}(s)}+\underbrace{G_{2}(s) D(s)}_{Y_{2}(s)}$ Considering first $R(s)$ only, then $Y(s)$ is $Y_{1}(s)$

\begin{aligned} \frac{Y_{1}(s)}{R(s)}&=\frac{\frac{G(s)}{1+G(s) H(s)}}{1+\frac{G(s)}{1+G(s) H(s)}} \\ \frac{Y_{1}(s)}{R(s)}&=\frac{G(s)}{1+G(s) H(s)+G(s)} \\ Y_{1}(s)&=\left[\frac{G(s)}{1+G(s)+G(s) H(s)}\right] R(s) \\ G_{1}(s)&=\frac{G(s)}{1+G(s)+G(s) H(s)} \end{aligned}
Now considering $D(s)$ only, then $Y(s)$ is $Y_{2}(s)$

\begin{aligned} \frac{Y_{2}(s)}{D(s)} & =\frac{G(s)}{1+G(s)[1+H(s)]} \\ Y_{2}(s) & =\left[\frac{G(s)}{1+G(s)+G(s) H(s)}\right] D(s) \\ G_{2}(s) & =\frac{G(s)}{1+G(s)+G(s) H(s)} \end{aligned}
Hence, $G_{1}(s)$ and $G_{2}(s)$ both are equal.
 Question 2
The open loop transfer function of a unity negative feedback system is $G(s)=\frac{k}{s\left(1+s T_{1}\right)\left(1+s T_{2}\right)}$, where $k, T_{1}$ and $T_{2}$ are positive constants. The phase crossover frequency, in rad/s, is
 A $\frac{1}{\sqrt{T_{1} T_{2}}}$ B $\frac{1}{T_{1} T_{2}}$ C $\frac{1}{T_{1} \sqrt{T_{2}}}$ D $\frac{1}{T_{2} \sqrt{T_{1}}}$
GATE EC 2023   Control Systems
Question 2 Explanation:
We know phase crossover frequency is that frequency at which phase of the open loop transfer function is $-180^{\circ}$.

\begin{aligned} \therefore \quad G(s) & =\frac{K}{s\left(1+s T_{1}\right)\left(1+s T_{2}\right)} \\ G(j \omega) & =\frac{K}{(j \omega)\left(1+j \omega T_{1}\right)\left(1+j \omega T_{2}\right)} \\ \text{Phase of }G(j \omega) & =\phi =-90-\tan ^{-1}\left(\omega T_{1}\right) - \tan ^{-1}\left(\omega T_{2}\right) \\ \therefore \quad \text { At } \omega &=\omega_{p c}, \phi =-180 \\ -180 & =-90-\tan ^{-1}\left(\omega_{p c} T_{1}\right)-\tan ^{-1}\left(\omega_{p c} T_{2}\right) \\ 9 \quad 90 & =\tan ^{-1}\left(\omega_{p c} T_{1}\right)+\tan ^{-1}\left(\omega_{p c} T_{2}\right) \\ \tan ^{-1}\left(\frac{\omega_{p c} T_{1}+\omega_{p c} T_{2}}{1-\omega_{p c}^{2} T_{1} T_{2}}\right) & =90 \\ 1-\omega_{p c}^{2} T_{1} T_{2} & =0 \\ \omega_{p c} & =\frac{1}{\sqrt{T_{1} T_{2}}} \end{aligned}

 Question 3
The block diagram of a feedback control system is shown in the figure

The transfer function $\dfrac{Y{\left ( s \right )}}{X {\left ( s \right )}}$ of the system is
 A $\frac{G_{1}+G_{2}+G_{1}G_{2}H}{1+G_{1}H}$ B $\frac{G_{1}+G_{2}}{1+G_{1}H+G_{2}H}$ C $\frac{G_{1}+G_{2}}{1+G_{1}H}$ D $\frac{G_{1}+G_{2}+G_{1}G_{2}H}{1+G_{1}H+G_{2}H}$
GATE EC 2021   Control Systems
Question 3 Explanation:

$\frac{Y}{R}=\frac{G_{1}(1-0)+G_{2}(1-0)}{1-\left[-G_{1} H\right]}=\frac{G_{1}+G_{2}}{1+G_{1} H}$
 Question 4
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 4 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 5
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 5 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}

There are 5 questions to complete.