Transform Theory MCQs Question and Answer in Engineering Maths

A solution for the differential equation \[{\rm{\dot x}}\left( {\rm{t}} \right) + 2{\rm{x}}\left( {\rm{t}} \right) = \delta \left( {\rm{t}} \right)\]    with initial condition x(0^-) = 0 is

Consider the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}\left( {\text{t}} \right)}}{{{\text{dt}}}} + {\text{y}}\left( {\text{t}} \right) = \delta \left( {\text{t}} \right)$$      with $${\left. {{\text{y}}\left( {\text{t}} \right)} \right|_{{\text{t}} = 0}} = - 2$$   and $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 0.$$
The numerical value of $${\left. {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}}$$   is

A delayed unit step function is defined as \[{\text{u}}\left( {{\text{t}} - {\text{a}}} \right) = \left\{ {\begin{array}{*{20}{c}} {0,}&{{\text{for t}} < {\text{a}}} \\ {1,}&{{\text{for t}} \geqslant {\text{a}}} \end{array}} \right..\]      Its Laplace transform is

The Laplace Transform of f(t) = e^2t sin(5t) u(t) is

Let $${\text{X}}\left( {\text{s}} \right) = \frac{{3{\text{s}} + 5}}{{{{\text{s}}^2} + 10{\text{s}} + 21}}$$    be the Laplace Transform of a signal x(t). Then, x(0⁺) is

If f(t) is a function defined for all t ≥ 0, its Laplace transform F(s) is defined as

Laplace transform for the function f(x) = cosh(ax) is

If F(s) is the Laplace transform of function f(t), then Laplace transform of $$\int\limits_0^{\text{t}} {{\text{f}}\left( \tau \right){\text{d}}\tau } $$   is

The inverse Laplace transform of $$\frac{1}{{\left( {{{\text{s}}^2} + {\text{s}}} \right)}}$$  is

The Fourier series of the function,
\[\begin{array}{*{20}{c}} {{\text{f}}\left( {\text{x}} \right) = 0,}&{ - \pi < {\text{x}} \leqslant 0} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \pi - {\text{x,}}}&{0 < {\text{x}} < \pi } \end{array}\]      in the interval $$\left[ { - \pi ,\,\pi } \right]$$  is $${\text{f}}\left( {\text{x}} \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{\cos {\text{x}}}}{{{1^2}}} + \frac{{\cos {\text{3x}}}}{{{3^3}}} + \,...} \right] + \left[ {\frac{{\sin {\text{x}}}}{1} + \frac{{\sin {\text{2x}}}}{2} + \frac{{\sin {\text{3x}}}}{3} + \,...} \right]$$
The convergence of the above Fourier series at x = 0 gives