# Given f (t) = t^2e^-2x cos (3t). The value of L {f(t)} is __________________

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Given f (t) = t^2e^-2x cos (3t). The value of L {f(t)}  is __________________

(a) $\frac{2(s+2)(s^2+4s-23)}{(s^2+4s+13)^3}$

(b) $\frac{2(s-2)(s^2-4s-23)}{(s^2+4s+13)^3}$

(c) $\frac{2(s+2)(s^2+4s+23)}{(s^2+4s+13)^3}$

(d) $\frac{2(s-2)(s^2+4s-23)}{(s^2+4s-13)^3}$

I had been asked this question by my school principal while I was bunking the class.

Origin of the question is Common Laplace Transforms topic in division Laplace Transform and System Design of Signals and Systems

by (42.1k points)
Right answer is (a) $\frac{2(s+2)(s^2+4s-23)}{(s^2+4s+13)^3}$

Explanation: Let g (t) = cos (3t); h (t) = e^-2x cos (3t) = e^-2x g (t)

Then, f (t) = t^2h (t)

Let G(s) = L {g (t)}, H(s) = L {h (t)}, F(s) = L {f (t)}

So, G(s) = $\frac{s}{s^2+9}$

And H(s) = $\frac{s+2}{(s+2)^2+9}$

∴ F(s) = $-\frac{d}{ds}[-\frac{d}{ds} H(s)] = \frac{2(s+2)(s^2+4s-23)}{(s^2+4s+13)^3}$.

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