Q21GATE 2004MCQ

For a tachometer, if $\theta \left ( t \right )$ is the rotor displacement in radians, e(t) is the output voltage and $K_{t}$ is the tachometer constant in V/rad/sec, then the transfer function, $\frac{E(s)}{\Theta(s)}$ will be

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📖 Explanation

$\theta \left ( t \right )=$ rotor displacement in radians $\omega \left ( t \right )=\frac{d\theta }{dt}=$ angular speed in rad/sec Output voltage; e(t)= $K_{t}\omega \left ( t \right )=\, K_{t}\frac{d\theta }{dt}$ Taking laplace transform on both sides $\; \; \; \; E\left ( s \right )\, =\, K_{t}s\theta \left ( s \right )$ $\Rightarrow \; \; \; \frac{E\left ( s \right )}{\theta \left ( s \right )}\, =\, K_{t}s$

Q23GATE 2003MCQ

The block diagram of a control system is shown in figure. The transfer function G(s) = Y(s)/U(s) of the system is

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📖 Explanation

Integrator are represented as 1/s in s-domain As per the block diagram, the corresponding signal flow graph is drawn One forward path $P_{1}\, =\, 2/s^{2}$ The individual loops are, $\; \; \; L_{1}\, =\, -\frac{3}{s},\: L_{2}\, =\, -\frac{12}{s} and L_{3}\, =\, -\frac{18}{s^{2}}$ $L_{1}$ and $L_{2}$ are non-touching loops $\; \; \; L_{1}L_{2}\, =\, \frac{36}{s^{2}}$ The loops touches the forward path $\Delta _{1}=1$ The graph determinant is $\; \; \; \Delta \, =\, 1-\left ( L_{1}+L_{2}+L_{3} \right )+L_{1}L_{2}$ $\; \; \; \; =1+\frac{3}{s}+\frac{12}{s}+\frac{18}{s^{2}}+\frac{36}{s^{2}}$ Applying mason's gain formula, $\; \; \; G\left ( s \right )=\frac{Y\left ( s \right )}{U\left ( s \right )}=\frac{P_{1}\Delta _{1}}{\Delta }$ $\; \; \; \; =\frac{2/s^{2}}{1+\frac{3}{s}+\frac{12}{s}+\frac{18}{s^{2}}+\frac{36}{s^{2}}}$ $\; \; \; \; =\frac{2}{s^{2}+15s+54}=\frac{2}{\left ( s+9 \right )\left ( s+6 \right )}$ $\; \; \; \; =\frac{1}{27\left ( 1+\frac{s}{9} \right )\left ( 1+\frac{s}{6} \right )}$

Q24GATE 2003MCQ

A control system is defined by the following mathematical relationship $\frac{d^{2}x}{dt^{2}}+6\frac{dx}{dt}+5x=12(1-e^{-2t})$ The response of the system as $t\rightarrow \infty$ is

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📖 Explanation

Taking (LT) on both sides $(s^{2+}6s+5)X(s)=12\left ( \frac{1}{s}-\frac{1}{s+2} \right ) =\frac{24}{s(s+2)}$ $X(s)=\frac{24}{s(s+2)(s+1)(s+5)}$ Responce at $t\rightarrow \infty$ Using final value theorem $\lim_{t\rightarrow \infty }x(t)=\lim_{s \rightarrow 0}sX(s)$ $=\lim_{s \rightarrow 0}\frac{s \times 24}{s(s+2)(s+1)(s+5)}=2.4$

Q25GATE 2003MCQ

A control system with certain excitation is governed by the following mathematical equation $\frac{d^{2}x}{dt^{2}}+\frac{1}{2}\frac{dx}{dt}+\frac{1}{18}x=10+5e^{-4t}+2e^{-5t}$ The natural time constant of the response of the system are

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📖 Explanation

Natural time constant of the response depends only on poles of the system. $T(s)=\frac{C(s)}{R(s)}=\frac{1}{s^{2+}s/2+1/18}$ $\;\;=\frac{18}{18s^{2+}9s+1}$ $\;\;=\frac{1}{(6s+1)(3s+1)}$ This is in the form $\frac{1}{(1+sT_1)(1+sT_2)}$ $\therefore \;\; T_1,T_2=6sec, 3 sec$

Q26GATE 2002MCQ

The transfer function of the system described by $\frac{d^{2}y}{dt^{2}}+\frac{dy}{dt}=\frac{du}{dt}+2u$ with u as input and y as output is

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📖 Explanation

$\frac{d^2y}{dt^2}+\frac{dy}{dt}=\frac{du}{dt}+2u$ $\Rightarrow s^2Y(s)+sY(s)=sU(s)+2U(s)$ $\therefore \;\; \frac{Y(s)}{U(s)}=\frac{s+2}{(s^{2+}s)}$