Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.