Question

Find the Laplace transform for the following time domain.

y(t) = -2te^-t + 4e^-t - 4e^-2t

1. \(\frac {2}{(s + 1)^2} + \frac {4}{(s + 1)} + 4(s + 2)\)
2. \(\frac {2s}{(s - 1)^2(s + 2)}\)
3. \(\frac {2}{(s + 1)^2} - \frac {4}{(s + 1)} + 4(s + 2)\)
4. \(\frac {2s}{(s + 1)^2(s + 2)}\)

Answer 1

Correct Answer - Option 4 : \(\frac {2s}{(s + 1)^2(s + 2)}\)

Formula Used:

f(t)	F(s)
δ (t)	1
u (t)	\(\frac{1}{s}\)
e-at	\(\frac{1}{s+a}\)
t	\(\frac{1}{s^2}\)
tn	\(\frac{n!}{s^{n+1}}\)
te-at	\(\frac{1}{(s+a)^2}\)
tne-at	\(\frac{n!}{(s+a)^{n+1}}\)

Application:

Given, y(t) = -2te-t + 4e-t - 4e-2t

From above formula,

Y(s) = \(\frac{-2}{(s+1)^2}+\frac{4}{(s+1)}-\frac{4}{(s+2)}\)

or, Y(s) = \(\frac{-2(s+2)+4(s+1(s+2)-4(s+1)^2}{(s+1)^2(s+2)}=\frac{N}{D}\) .... (1)

Consider numerator (N) for easier calculation,

N = -2(s + 2) + 4(s + 1)(s + 2) - 4(s + 1)(s + 1)

or, N = -2(s + 2) + 4(s + 1)[(s + 2) - (s + 1)]

or, N = -2(s + 2) + 4(s + 1)[1]

or, N = 2s

From equation (1),

Y(s) = \(\frac{N}{D}=\frac{2s}{(s+1)^2(s+2)}\)