Question 1 |
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 1 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))
Question 2 |
Two linear time-invariant systems with transfer functions
G_1(s)=\frac{10}{s^2+s+1}, G_2(s)=\frac{10}{s^2+s\sqrt{10}+10}
have unit step responses y_1(t) and y_2(t) , respectively. Which of the following statements is/are true?
G_1(s)=\frac{10}{s^2+s+1}, G_2(s)=\frac{10}{s^2+s\sqrt{10}+10}
have unit step responses y_1(t) and y_2(t) , respectively. Which of the following statements is/are true?
y_1(t) and y_2(t) have the same percentage peak overshoot. | |
y_1(t) and y_2(t) have the same steady-state value. | |
y_1(t) and y_2(t) have the same damped frequency of oscillation. | |
y_1(t) and y_2(t) have the same 2% settling time. |
Question 2 Explanation:
\begin{aligned}
G_1(s)&=\frac{10}{s^2+s+1}\\
\omega _{n1}^2&=1\\
2\xi _1 \times 1&=1\\
\xi _1&=1/2\\
G_2(s)&=\frac{10}{s^2+s\sqrt{10}+10}\\
\omega _{n2}^2&=10\\
2\xi _1 \times \sqrt{10}&=\sqrt{10}\\
\xi _2&=1/2\\
\end{aligned}
\because M_P depends on \xi only
y_1(t) \; and \; y_2(t) have same percentage peak overshoot.
\omega _{d_1}=\omega _{n_1}\sqrt{1-\xi ^2}
\Rightarrow \omega _{d_1}\neq \omega _{d_2} \;\; \because \omega _{n_1}\neq \omega _{n_2}
damped frequency of oscillation is not same.
\begin{aligned} T_s &=\frac{4}{\xi \omega _n}(2%\; setting \; time) \\ T_{s_1} &\neq T_{s_2} \;\because \omega _{n_1}\neq \omega _{n_2}\\ C_1(s) &= \frac{10 \times 1/s}{s^2+s+1}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_1(s) \\ &= 10\\ C_2(s) &= \frac{10 \times 1/s}{s^2+\sqrt{10}s+10}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_2(s) \\ &= \frac{10}{10}1=1 \end{aligned}
Steady - state value is not same.
\because M_P depends on \xi only
y_1(t) \; and \; y_2(t) have same percentage peak overshoot.
\omega _{d_1}=\omega _{n_1}\sqrt{1-\xi ^2}
\Rightarrow \omega _{d_1}\neq \omega _{d_2} \;\; \because \omega _{n_1}\neq \omega _{n_2}
damped frequency of oscillation is not same.
\begin{aligned} T_s &=\frac{4}{\xi \omega _n}(2%\; setting \; time) \\ T_{s_1} &\neq T_{s_2} \;\because \omega _{n_1}\neq \omega _{n_2}\\ C_1(s) &= \frac{10 \times 1/s}{s^2+s+1}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_1(s) \\ &= 10\\ C_2(s) &= \frac{10 \times 1/s}{s^2+\sqrt{10}s+10}\\ C_{ss} &=\lim_{s \to 0}s\cdot C_2(s) \\ &= \frac{10}{10}1=1 \end{aligned}
Steady - state value is not same.
Question 3 |
Consider the following closed loop control system
where G(s)=\frac{1}{s(s+1)} and C(s)=K\frac{s+1}{s+3}. If the steady state error for a unit ramp input is 0.1, then the value of K is _______.
where G(s)=\frac{1}{s(s+1)} and C(s)=K\frac{s+1}{s+3}. If the steady state error for a unit ramp input is 0.1, then the value of K is _______.
10 | |
20 | |
30 | |
40 |
Question 3 Explanation:
Open loop transfer function for the system =C(s)\times G(s)=\frac{K(s+1)}{ (s+3)}\times \frac{1}{s(s+1)}
Since the system is type-1, so far a given unit ramp input steady state
e_{ss}=\frac{1}{K_{V}}
where, K_{V}=\lim_{s\rightarrow 0}S\times \frac{K}{S(S+3)}=\frac{K}{3}
so, \; e_{ss}=\frac{1}{K/3}=\frac{3}{K}
Given that, \; e_{ss}=0.1
So, 0.1=\frac{3}{K}\Rightarrow K=30
Since the system is type-1, so far a given unit ramp input steady state
e_{ss}=\frac{1}{K_{V}}
where, K_{V}=\lim_{s\rightarrow 0}S\times \frac{K}{S(S+3)}=\frac{K}{3}
so, \; e_{ss}=\frac{1}{K/3}=\frac{3}{K}
Given that, \; e_{ss}=0.1
So, 0.1=\frac{3}{K}\Rightarrow K=30
Question 4 |
A system with transfer function G(s)=\frac{1}{(s+1)(s+a)}, a \gt 0 is subjected an input 5\cos 3t.
The steady state output of the system is \frac{1}{\sqrt{10}} \cos (3t-1.892). The value of a is ____.
3 | |
4 | |
5 | |
6 |
Question 4 Explanation:
Given that,
\begin{aligned}G(j\omega )&=\frac{1}{(1+j\omega )(\alpha +j\omega )} \\ \left |G(j\omega ) \right |&=\frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}} \\ \text{According to }& \text{question,}\\ \left | G(j\omega ) \right |_{\omega =3}&=\frac{1}{5\sqrt{10}}\\ \Rightarrow \frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}}&=\frac{1}{5\sqrt{10}} \\ \Rightarrow \frac{1}{\sqrt{10(a^{2}+9)}}&=\frac{1}{5\sqrt{10}} \alpha ^{2}+9=25 \\ \alpha ^{2}&=16 \\ \alpha &=4\end{aligned}
\begin{aligned}G(j\omega )&=\frac{1}{(1+j\omega )(\alpha +j\omega )} \\ \left |G(j\omega ) \right |&=\frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}} \\ \text{According to }& \text{question,}\\ \left | G(j\omega ) \right |_{\omega =3}&=\frac{1}{5\sqrt{10}}\\ \Rightarrow \frac{1}{\sqrt{(\omega ^{2}+1)(\omega ^{2}+\alpha ^{2})}}&=\frac{1}{5\sqrt{10}} \\ \Rightarrow \frac{1}{\sqrt{10(a^{2}+9)}}&=\frac{1}{5\sqrt{10}} \alpha ^{2}+9=25 \\ \alpha ^{2}&=16 \\ \alpha &=4\end{aligned}
Question 5 |
The loop transfer function of a negative feedback system is
G(s)H(s)=\frac{K(s+11)}{s(s+2)(s+8)}
The value of K, for which the system is marginally stable, is ________.
G(s)H(s)=\frac{K(s+11)}{s(s+2)(s+8)}
The value of K, for which the system is marginally stable, is ________.
160 | |
120 | |
180 | |
200 |
Question 5 Explanation:
Characteristic equation q(s) for the given open loop system will be
q(s)=s^{3}+10s^{2}+16s+Ks+11K=0
Using R-H criteria,
For System to be Marginally Stable
\begin{aligned}\frac{10(16+K)-11K}{10}&=0 \\ 160+10K-11K&=0 \\ K&=160\end{aligned}
q(s)=s^{3}+10s^{2}+16s+Ks+11K=0
Using R-H criteria,
For System to be Marginally Stable
\begin{aligned}\frac{10(16+K)-11K}{10}&=0 \\ 160+10K-11K&=0 \\ K&=160\end{aligned}
Question 6 |
Consider a causal second-order system with the transfer function
G(s)=\frac{1}{1+2s+s^2}
with a unit-step R(s)=\frac{1}{s} as an input. Let C(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value \lim_{t \to \infty }c(t), rounded off to two decimal places, is
G(s)=\frac{1}{1+2s+s^2}
with a unit-step R(s)=\frac{1}{s} as an input. Let C(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value \lim_{t \to \infty }c(t), rounded off to two decimal places, is
5.25 | |
4.5 | |
3.89 | |
2.81 |
Question 6 Explanation:
\begin{aligned} G(s)&=\frac{1}{s^{2}+2 s+1}=\frac{1}{(s+1)^{2}} \\ R(s)&=\frac{1}{s} \\ \alpha (s)&=G(s) R(s)=\frac{1}{s(s+1)^{2}} \end{aligned}
Using partial fraction expansion, we get,
\begin{aligned} C(s) & =\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+1)^{2}} \\ A\left(s^{2}+2 s+1\right)&+B\left(s^{2}+s\right)+C s=1 \\ A&=1 \\ A+B & =0 \Rightarrow B=-1 \\ 2 A+B+C & =0 \Rightarrow C=-1 \\ \therefore \quad C(s) & =\frac{1}{s}-\frac{1}{(s+1)}-\frac{1}{(s+1)^{2}} \\ \text{and}\quad c(t) & =\left(1-e^{-t}-t e^{-t}\right) u(t) \\ \lim _{t \rightarrow \infty} c(t) & =1 \end{aligned}
In order to reach 94 % of its steady-state value,
\left(1-e^{-t}-t e^{-t}\right)=0.94
By trial and error, we get,
t \approx 4.50 \mathrm{sec}
Using partial fraction expansion, we get,
\begin{aligned} C(s) & =\frac{A}{s}+\frac{B}{(s+1)}+\frac{C}{(s+1)^{2}} \\ A\left(s^{2}+2 s+1\right)&+B\left(s^{2}+s\right)+C s=1 \\ A&=1 \\ A+B & =0 \Rightarrow B=-1 \\ 2 A+B+C & =0 \Rightarrow C=-1 \\ \therefore \quad C(s) & =\frac{1}{s}-\frac{1}{(s+1)}-\frac{1}{(s+1)^{2}} \\ \text{and}\quad c(t) & =\left(1-e^{-t}-t e^{-t}\right) u(t) \\ \lim _{t \rightarrow \infty} c(t) & =1 \end{aligned}
In order to reach 94 % of its steady-state value,
\left(1-e^{-t}-t e^{-t}\right)=0.94
By trial and error, we get,
t \approx 4.50 \mathrm{sec}
Question 7 |
The open loop transfer function G(s)=\frac{(s+1)}{s^{P}(s+2)(s+3)}
Where p is an integer, is connected in unity feedback configuration as shown in figure.
Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter p is _________.
Where p is an integer, is connected in unity feedback configuration as shown in figure.
Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter p is _________.
1 | |
2 | |
3 | |
4 |
Question 7 Explanation:
Steady state error of type 1 system, for step input
is zero, for ramp input is \frac{1}{K_{v}} and for parabolic input is infinity.
So, the given system must be of type- 1 and p=1
So, the given system must be of type- 1 and p=1
Question 8 |
For the unity feedback control system shown in the figure, the open-loop transfer function G(s) is given as
G(s)=\frac{2}{s(s+1)}
. The steady state error e_{ss} due to a unit step input is
G(s)=\frac{2}{s(s+1)}
. The steady state error e_{ss} due to a unit step input is
0 | |
0.5 | |
1 | |
\infty |
Question 8 Explanation:
\begin{aligned} E(s) &=\frac{R(s)}{1+G(s) H(s)} \\ R(s) &=\frac{1}{s} ; G(s)=\frac{2}{s(s+1)}; \\ H(s) &=1 \\ E(s) &=\frac{\frac{1}{s}}{1+\frac{s}{s(s+1)}}=\frac{s+1}{s^{2}+s+2} \\ e_{s s} &=\lim _{s \rightarrow 0} s E(s) \\ &=\lim _{s \rightarrow 0} \frac{s(s+1)}{s^{2}+s+2} \\ e_{s s} &=0 \end{aligned}
Question 9 |
In the feedback system shown below G(s)=\frac{1}{(s^{2}+2s)}. The step response of the closed-loop system should have minimum settling time and have no overshoot.
The required value of gain k to achieve this is ________
The required value of gain k to achieve this is ________
0.5 | |
1 | |
2 | |
3 |
Question 9 Explanation:
\begin{aligned} G(s)&=\frac{1}{s^{2}+2 s} \\ \frac{Y(s)}{R(s)}&=\frac{\frac{K}{s^{2}+2 s}}{1+\frac{K}{s^{2}+2 s}}=\frac{K}{s^{2}+2 s+K} \end{aligned}
Minimum settling time and no overshoot implies critical damping i.e.
\begin{aligned} \xi &=1 \\ \omega_{n} &=\sqrt{K} \\ \therefore \quad 2 \times \xi \cdot \omega_{n} &=2 \\ \omega_{n} &=1 \\ \Rightarrow \quad \sqrt{K} &=1 \text { or } K=1 \end{aligned}
Minimum settling time and no overshoot implies critical damping i.e.
\begin{aligned} \xi &=1 \\ \omega_{n} &=\sqrt{K} \\ \therefore \quad 2 \times \xi \cdot \omega_{n} &=2 \\ \omega_{n} &=1 \\ \Rightarrow \quad \sqrt{K} &=1 \text { or } K=1 \end{aligned}
Question 10 |
The response of the system G(s)=\frac{s-2}{(s+1)(s+3)} to the unit step input u(t) is y(t). The value of \frac{dy}{dt} at t=0^{+} is _________
0 | |
1 | |
2 | |
3 |
Question 10 Explanation:
\begin{aligned} L\left(\frac{d y}{d t}\right) &=(s Y(s)-y(0)) \\ Y(s) &=G(s) \times \frac{1}{s}=\frac{s-2}{s(s+1)(s+3)} \\ y(0)&=\lim _{s \rightarrow \infty} s Y(s) \\ \text{(Applying } & \text{initial value theorem) } \\ &=\lim _{s \rightarrow \infty} \frac{s-2}{(s+1)(s+3)} \\ &=\frac{\left(1-\frac{2}{s}\right)}{s\left(1+\frac{1}{s}\right)\left(1+\frac{3}{s}\right)} \\ y(0)&=0 \\ L\left(\frac{d y}{d t}\right) &=s Y(s)=\frac{s \times(s-2)}{s(s+1)(s+3)} \\ &=\frac{s-2}{(s+1)(s+3)} \\ \left.\frac{d y}{d t}\right|_{t=0} &=\lim _{s \rightarrow \infty} s L\left(\frac{d y}{d t}\right) \\ &=\lim _{s \rightarrow \infty} \frac{s \times(s-2)}{(s+1)(s+3)}\\ &=\frac{\left(1-\frac{2}{s}\right)}{\left(1+\frac{1}{s}\right)\left(1+\frac{3}{s}\right)}=1 \end{aligned}
There are 10 questions to complete.