The cut in voltage of both Zener diode $${D_z}$$ and diode $$D$$ shown in Figure is $$0.7$$ $$V,$$ while break down voltage of the zener is $$3.3$$ $$V$$ and reverse breakdown voltage of $$D$$ is $$50$$ $$V.$$ the other parameters can be assumed to be the same as those of an ideal diode. The values of the peak output voltage ($${V_0}$$) are

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

Obtain a state variable representation of the system governed by the differential equation: $${{{d^2}y} \over {d{t^2}}} + {{dy} \over {dt}} - 2y = u\left( t \right){e^{ - t}},\,\,\,$$ with the choice of state variables as $${x_1} = y,$$ $${x_2} = \left( {{{dy} \over {dt}} - y} \right){e^t}.$$ Aso find $${x_2}\left( t \right),$$ given that $$u(t)$$ is a unit step function and $${x_2}\left( 0 \right) = 0.$$

For the system $$\mathop X\limits^ \bullet = \left[ {\matrix{ 2 & 0 \cr 0 & 4 \cr } } \right]X + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u;\,\,\,y = \left[ {\matrix{ 4 & 0 \cr } } \right]X,\,$$ with u as unit impulse and with zero initial state, the output, $$y$$, becomes