Laplace Transform of f(s) | Calculation of 𝜓(s)

[darrenabrown]

[tex]

1) \hat f(s) = 7/(s+2)(s^2+8s=41) \exp3s

 

[Integral]

It would help everyone if you would spend a few minutes learning to usehttps://www.physicsforums.com/showthread.php?t=8997" for equations.

 

[ravenprp]

The Laplace Transform of a function f(s) is denoted by \hat f(s) and is defined as the integral of the function multiplied by the exponential function e^{-st}, where s is a complex number. In this case, the function f(s) is given as 7/(s+2)(s^2+8s+41) \exp3s. To calculate the Laplace Transform, we first need to simplify the function by factoring the denominator.

The denominator can be factored as (s+2)(s^2+8s+41) = (s+2)(s+5)(s+8). Therefore, the function can be rewritten as 7/(s+2)(s+5)(s+8) \exp3s. Now, we can use the property of the Laplace Transform that states that the transform of a product of two functions is equal to the product of their individual transforms. Applying this property, we get:

\hat f(s) = 7 \hat g(s) \hat h(s) \hat k(s) \exp3s, where g(s) = 1/(s+2), h(s) = 1/(s+5), and k(s) = 1/(s+8).

Using the Laplace Transform tables, we can find the transforms of g(s), h(s), and k(s) as \hat g(s) = e^{-2s}, \hat h(s) = e^{-5s}, and \hat k(s) = e^{-8s}. Substituting these values in the above equation, we get:

\hat f(s) = 7 e^{-2s} e^{-5s} e^{-8s} \exp3s.

Using the property of exponents, we can simplify this to:

\hat f(s) = 7 e^{-12s}.

Therefore, the Laplace Transform of the given function f(s) is \hat f(s) = 7 e^{-12s}.

In conclusion, the Laplace Transform of f(s) can be calculated by factoring the denominator, using the property of the Laplace Transform, and simplifying the resulting expression. This method can be applied to calculate the Laplace Transform of any function with a given expression.