For the given transfer function:$G\left( s \right) = \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{1}{{{s^2} + 3s + 2}}$the response y(t) for a step input r(t) = 5u(t) will be

Question

For the given transfer function:$G\left( s \right) = \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{1}{{{s^2} + 3s + 2}}$the response y(t) for a step input r(t) = 5u(t) will be

TuteeHUB · Accepted Answer

$\left[ {\frac{5}{2} - 5{e^{ - t}}} \right]u\left( t \right)$

TuteeHUB · Answer

$\left[ {\frac{5}{2} - 5{e^{ - t}} + \frac{5}{2}{e^{ - 2t}}} \right]u\left( t \right)$

TuteeHUB · Answer

$\left[ {\frac{5}{2} + \frac{5}{2}{e^{ - 2t}}} \right]u\left( t \right)$

TuteeHUB · Answer

$\left[ { - 5{e^{ - t}} + \frac{5}{2}{e^{ - 2t}}} \right]u\left( t \right)$

1.	For the given transfer function:\(G\left( s \right) = \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{1}{{{s^2} + 3s + 2}}\)the response y(t) for a step input r(t) = 5u(t) will be
A.	\(\left[ {\frac{5}{2} - 5{e^{ - t}} + \frac{5}{2}{e^{ - 2t}}} \right]u\left( t \right)\)
B.	\(\left[ {\frac{5}{2} - 5{e^{ - t}}} \right]u\left( t \right)\)
C.	\(\left[ {\frac{5}{2} + \frac{5}{2}{e^{ - 2t}}} \right]u\left( t \right)\)
D.	\(\left[ { - 5{e^{ - t}} + \frac{5}{2}{e^{ - 2t}}} \right]u\left( t \right)\)
Answer» B. \(\left[ {\frac{5}{2} - 5{e^{ - t}}} \right]u\left( t \right)\)

For the given transfer function:\(G\left( s \right) = \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{1}{{{s^2} + 3s + 2}}\)the response y(t) for a step input r(t) = 5u(t) will be

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