The open-loop transfer function of a $$dc$$ motor is given as $${{\omega \left( s \right)} \over {{V_a}\left( s \right)}} = {{10} \over {1 + 10s}}.$$ When connected in feedback as shown below, the approximate value of $${K_a}$$ that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

The Bode plot of a transfer function $$G(s)$$ is shown in the figure below.

The gain is $$\left( {20\log \left| {G\left( s \right)} \right|} \right)$$ is $$32$$ $$dB$$ and $$–8$$ $$dB$$ at $$1$$ $$rad/s$$ and $$10$$ $$rad/s$$ respectively. The phase is negative for all $$\omega .$$ Then $$G(s)$$ is

The state variable formulation of a system is given as
$$\left[ {\matrix{ {\mathop {{x_1}}\limits^ \bullet } \cr {\mathop {{x_2}}\limits^ \bullet } \cr } } \right] = \left[ {\matrix{ { - 2} & 0 \cr 0 & { - 1} \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr } } \right] + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u,\,\,{x_1}\left( 0 \right) = 0,$$
$${x_2}\left( 0 \right) = 0$$ and $$y = \left[ {\matrix{ 1 & 0 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr } } \right]$$

The response $$y(t)$$ to a unit step input is

The state variable formulation of a system is given as
$$\left[ {\matrix{ {\mathop {{x_1}}\limits^ \bullet } \cr {\mathop {{x_2}}\limits^ \bullet } \cr } } \right] = \left[ {\matrix{ { - 2} & 0 \cr 0 & { - 1} \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr } } \right] + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u,\,\,{x_1}\left( 0 \right) = 0,$$
$${x_2}\left( 0 \right) = 0$$ and $$y = \left[ {\matrix{ 1 & 0 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr } } \right]$$

The system is