Find the Laplace transform of 2/(s + 1) – 4/(s + 3).
2 e^(-t) – 4 e^(-3t)
e^(-2t) + e^(-3t)
e^(-2t) – e^(-3t)
[2 e^(-t)][1 – 2 e^(-3t)]
The value of (1 + j)^6 is equal to
j4
-j8
-j12
j6
The Laplace transform of i(t) is given by I(s) = 2 / [2(s+1)] . As t -> infinity, the value of i(t) tends to approach
0
1
2
infinity