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GATE 2015 EE - SET 2 - Question 24

Solution :

c(t)= e^{-2t}(sin5t+cos5t)
L\left [ c(t) \right ] = L\left [ e^{-2t}sin5t + e^{-2t} cos5t\right ]

G(s) = \frac{5}{(s+2)^{2}+5^{2}} + \frac{(s+2)}{(s+2)^{2}+5^{2}}

Formulae :

L\left [ e^{at} sinbt\right ] = \frac{b}{(s-a)^{2}+b^{2}}

L\left [ e^{at} cosbt\right ] = \frac{(s-a)}{(s-a)^{2}+b^{2}}

G(s) = \frac{5}{(s+2)^{2}+5^{2}} + \frac{(s+2)}{(s+2^{2})+5^{2}}

G(s) = \frac{5}{(s+2)^{2}+25} + \frac{(s+2)}{(s+2^{2})+25}

G(s)= \frac{s+7}{(s+2)^{2}+ 25}

DC gain  = \lim_{s\rightarrow 0}G(s)

\lim_{s\rightarrow 0} \frac{s+7}{(s+2)^{2}+ 25} = \frac{7}{4+25} = \frac{7}{29} = 0.241

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