1
GATE ECE 2013
+2
-0.6
A system is described by the differential equation $${{{d^2}y} \over {d{t^2}}} + 5{{dy} \over {dt}} + 6y\left( t \right) = x\left( t \right)$$$Let x(t) be a rectangular pulse given by $$x\left( t \right) = \left\{ {\matrix{ {1\,\,\,\,\,\,\,\,\,0 \le \,t\, \le 2} \cr {0\,\,\,\,\,otherwise} \cr } } \right.$$$

Assuming that y(0) = 0 $${{dy} \over {dt}} = 0$$ at t = 0, the Laplace transform of y(t) is

A
$${{{e^{ - 2s}}} \over {s\left( {s + 2} \right)\left( {s + 3} \right)}}$$
B
$${{1 - {e^{ - 2s}}} \over {s\left( {s + 2} \right)\left( {s + 3} \right)}}$$
C
$${{{e^{ - 2s}}} \over {\left( {s + 2} \right)\left( {s + 3} \right)}}$$
D
$${{1 - {e^{ - 2s}}} \over {\left( {s + 2} \right)\left( {s + 3} \right)}}$$
2
GATE ECE 2011
+2
-0.6
If $$F\left( s \right) = L\left[ {f\left( t \right)} \right] = {{2\left( {s + 1} \right)} \over {{s^2} + 4s + 7}}$$ then the initial and final values of f(t) are respectively
A
0, 2
B
2, 0
C
0, 2/7
D
2/7, 0
3
GATE ECE 2010
+2
-0.6
Given f(t) = $${L^{ - 1}}\left[ {{{3s + 1} \over {{s^3} + 4{s^2} + \left( {K - 3} \right)s}}} \right].$$ If $$\matrix{ {Lim\,f\,\left( t \right) = 1,} \cr {t \to \infty } \cr } \,\,$$ then the value of K is
A
1
B
2
C
3
D
4
4
GATE ECE 2009
+2
-0.6
Given that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of $$\int\limits_0^t {f\left( \tau \right)\,d\tau }$$ is
A
$$sF\left( s \right) - f\left( 0 \right)$$
B
$$\,{1 \over s}F\left( s \right)\,$$
C
$$\int\limits_0^s {F\left( \tau \right)d\tau }$$
D
$${1 \over s}\left[ {F\left( s \right) - f\left( 0 \right)} \right]$$
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