Question

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

Answer 1

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