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).
(rounded off to the nearest integer).
3 | |
4 | |
8 | |
10 |
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}
\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
G_{1}(s) and G_{2}(s) are given by
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)} | |
G_{1}(s)=\frac{G(s)}{1+G(s)+H(s)} and G_{2}(s)=\frac{G(s)}{1+G(s)+H(s)} | |
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)} | |
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)} |
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
\frac{2 \alpha}{k} | |
\frac{\alpha}{k} | |
\frac{\alpha}{2 k} | |
\frac{\alpha}{4 k} |
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}
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
\frac{1}{\sqrt{T_{1} T_{2}}} | |
\frac{1}{T_{1} T_{2}} | |
\frac{1}{T_{1} \sqrt{T_{2}}} | |
\frac{1}{T_{2} \sqrt{T_{1}}} |
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}
\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).
-0.2 | |
-0.1 | |
0 | |
-0.3 |
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))
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.