What is the Laplace Transform of #tcosat+sinat#?
1 Answer
# ℒ \ {tcosat+sinat} = (s^2-a^2)/(s^2+a^2)^2 + a/(s^2+a^2) #
Which can also be written as:
# ℒ \ {tcosat+sinat} = (s^2-a^2 + as^2+a^3)/(s^2+a^2)^2 #
Explanation:
We seek:
# ℒ \ {tcosat+sinat} #
Firstly we note that the Laplace transformation has a linearity property, so that
# ℒ \ {tcosat+sinat} = ℒ \ {tcosat} + ℒ \ {sinat} #
We could just use a Laplace Transform lookup table and we would find:
# ℒ \ {tcosat} = (s^2-a^2)/(s^2+a^2)^2#
# ℒ \ {sinat} = a/(s^2+a^2)#
Thus:
# ℒ \ {tcosat+sinat} = (s^2-a^2)/(s^2+a^2)^2 + a/(s^2+a^2) #
# " " = (s^2-a^2)/(s^2+a^2)^2 + (a(s^2+a^2))/(s^2+a^2)^2 #
# " " = (s^2-a^2 + as^2+a^3)/(s^2+a^2)^2 #
I assume that the derivation of the above is required which we can easily gain using the rule for the Laplace Transformation of the derivative:
# ℒ \ (f''(t))=s^2 F(s)−s f(0)−f'(0) #
If we let:
# f(t) = sinat => f(0) = 0#
Differentiating we get:
# f'(t) = acosat => f'(0)=a#
# f''(t) = -a^2sinat #
# f''(t) = -a^2f(t) #
Taking Laplace transformations we get:
# ℒ \ {f''(t)} = -a^2ℒ \ {f(t)} #
# :. s^2 F(s)−s f(0)−f'(0) = -a^2F(s) #
# :. s^2 F(s)−0−a = -a^2F(s) #
# :. s^2 F(s)+a^2F(s) = a #
# :. (s^2 +a^2)F(s) = a #
# :. F(s) = a/(s^2 +a^2) #
Thus we have shown;
# ℒ \ {sinat} = a/(s^2+a^2)# QED
Similarly, Now suppose that we let
# f(t) = tcosat => f(0) = 0#
Then by the product rule:
# f'(t) = (t)(-asinat) + (1)(cosat) #
# " " = -atsinat+cosat => f'(0) = 1 #
Differentiating again we get:
# f''(t) = (-at)(acosat) + (-a)(sinat) -asinat#
# " " = -a^2tcosat -2asinat #
# " " = -a^2f(t) -2asinat => f''(0) = 0#
Taking Laplace transformations we get:
# ℒ \ {f''(t)} = -a^2ℒ \ {f(t)} - 2aℒ \ {sin at}#
# :. s^2 F(s)−s f(0)−f'(0) = -a^2F(s) - 2a a/(s^2+a^2)#
# :. s^2 F(s)−0−1 = -a^2F(s) - (2a^2)/(s^2+a^2)#
# :. s^2 F(s)+a^2F(s) = 1 - (2a^2)/(s^2+a^2)#
# :. (s^2 +a^2)F(s) = (s^2+a^2- 2a^2)/(s^2+a^2)#
# :. (s^2 +a^2)F(s) = (s^2-a^2)/(s^2+a^2)#
# :. F(s) = (s^2-a^2)/(s^2+a^2)^2#
Thus we have shown;
# ℒ \ {tcosat} = (s^2-a^2)/(s^2+a^2)^2# QED
