# Control Systems

 Question 1
The asymptotic magnitude Bode plot of a minimum phase system is shown in the figure. The transfer function of the system is $(s)=\frac{k(s+z)^{a}}{s^{b}(s+p)^{c}}$, where $k, z, p, a, b$ and $c$ are positive constants. The value of $(a+b+c)$ is ___
(rounded off to the nearest integer).

 A 3 B 4 C 8 D 10
GATE EC 2023      Frequency Response Analysis
Question 1 Explanation:
From the Bode magnitude plot, it is clear that there is one pole at origin,
$\therefore$ $b=1$
and at frequency $\omega_{1}$, system has a zero
$\therefore$ $a=1$

and at frequency $\omega_{2}$, system has a two poles
\begin{aligned} \therefore \quad c&=2 \\ \therefore \quad a+b+c&=1+1+2 \\ a+b+c&=4 \end{aligned}
 Question 2
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      Basics of Control Systems, Block Diagram and SFGs
Question 2 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 3
A closed loop system is shown in the figure where $k \gt 0$ and $\alpha \gt 0$. The steady state error due to a ramp input $\left(R(s)=\alpha / s^{2}\right)$ is given by

 A $\frac{2 \alpha}{k}$ B $\frac{\alpha}{k}$ C $\frac{\alpha}{2 k}$ D $\frac{\alpha}{4 k}$
GATE EC 2023      Time Response Analysis
Question 3 Explanation:
Given, $\quad$ input is $r(t)=\alpha t u(t)$
$R(s)=\frac{\alpha}{s^{2}}$

From the figure,
$G(s) H(s)=\frac{K}{s(s+2)}$
Now steady state error for Ramp input is
$e_{s s} =\frac{\alpha}{K_{v}}, \text { where } \alpha \text { is the magnitude of Ramp input }$
\begin{aligned} K_{v} & =\lim _{s \rightarrow 0}[s G(s) H(s)] \\ K_{v} & =\lim _{s \rightarrow 0}\left[\frac{s \times K}{s(s+2)}\right]=\frac{K}{2}\\ \therefore \quad e_{s s} & =\frac{\alpha \times 2}{K} \\ e_{s s} & =\frac{2 \alpha}{K} \end{aligned}
 Question 4
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      Basics of Control Systems, Block Diagram and SFGs
Question 4 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 5
The block diagram of a closed-loop control system is shown in the figure. R(s),Y(s) and D(s) are the Laplace transforms of the time-domain signals $r(t),y(t),$ and $d(t)$, respectively. Let the error signal be defined as $e(t)=r(t)-y(t)$. Assuming the reference input $r(t)=0$ for all $t$, the steady-state error $e(\infty )$, due to a unit step disturbance $d(t)$, is _________ (rounded off to two decimal places).

 A -0.2 B -0.1 C 0 D -0.3
GATE EC 2022      Time Response Analysis
Question 5 Explanation:
$Y(s)=\frac{R(s)\cdot \frac{10}{s(s+10)}}{1+\frac{10}{s(s+10)}} +\frac{D(s)\cdot \frac{10}{s(s+10)}}{1+\frac{10}{s(s+10)}}$
When, $r(t)=0\; and \; d(t)=U(t)\underleftrightarrow{L.T.}\frac{1}{s}$
$e(\infty )=-\lim_{s \to 0}\frac{s\cdot \frac{1}{s} \times 1}{s(s+10)+10} =-1/10=-0.1\;\;\;(\because e(t)=r(t)-y(t))$

There are 5 questions to complete.