Determine the representation in the frequency domain function f (t) below,
through the table of Laplace transforms:
f (t) = 6e ^ -(0,2+3t )
f (t) = e^-t sen(3t)
To determine the representation in the frequency domain for the given functions, we can use the table of Laplace transforms. The Laplace transform is defined as follows:
L{f(t)} = F(s) = ∫[0 to ∞] f(t) e^(-st) dt
Let's find the Laplace transform for each function.
1. f(t) = 6e^(-(0.2 + 3t))
To find the Laplace transform of this function, we'll use the property: L{e^(at) * f(t)} = F(s-a), where "a" is a constant.
In this case, a = 0.2 + 3t, so we can rewrite f(t) as: f(t) = 6 * e^(-0.2) * e^(-3t)
Now, we can find the Laplace transform using the exponential function property:
L{e^(-0.2)} = 1 / (s + 0.2) [using L{e^(at)} = 1 / (s - a)]
L{e^(-3t)} = 1 / (s + 3) [using L{e^(at)} = 1 / (s - a)]
L{f(t)} = 6 * [L{e^(-0.2)}] * [L{e^(-3t)}] = 6 / [(s + 0.2)(s + 3)]
Therefore, the representation in the frequency domain for f(t) = 6e^(-(0.2 + 3t)) is F(s) = 6 / [(s + 0.2)(s + 3)].
2. f(t) = e^(-t) * sin(3t)
To find the Laplace transform of this function, we'll use the property: L{e^(at) * f(t)} = F(s-a), where "a" is a constant.
In this case, a = 1, so we can rewrite f(t) as: f(t) = e^(-t + 3t) = e^(2t)
Now, we can find the Laplace transform using the exponential function property:
L{e^(2t)} = 1 / (s - 2) [using L{e^(at)} = 1 / (s - a)]
L{f(t)} = [L{e^(2t)}] = 1 / (s - 2)
Therefore, the representation in the frequency domain for f(t) = e^(-t) * sin(3t) is F(s) = 1 / (s - 2).
In summary:
1. f(t) = 6e^(-(0.2 + 3t)) -> F(s) = 6 / [(s + 0.2)(s + 3)]
2. f(t) = e^(-t) * sin(3t) -> F(s) = 1 / (s - 2)
Please note that these are the Laplace transform representations in the frequency domain for the given functions.