Question

Consider the following function in the s-domain. $F(s) = \frac{(8 \cdot s^3 + 24) \cdot e^{-4s}}{s^4} + \frac{3 \cdot s - 6}{s^2 - 4 \cdot s + 5}$ Calculate the inverse transform of F(s) by following these steps. If you need to enter the Heaviside step function for any part of this question you can type it as H(t). a. Calculate the following inverse Laplace transform: $\mathcal{L}^{-1} \left\{ \frac{8 \cdot s^3 + 24}{s^4} \right\} =$ Now use your previous answer and the second shift theorem to find the inverse Laplace transform of the first term. $\mathcal{L}^{-1} \left\{ \frac{(8 \cdot s^3 + 24) \cdot e^{-4s}}{s^4} \right\} =$ b. Find the inverse Laplace Transform of $\frac{3 \cdot s - 6}{s^2 - 4 \cdot s + 5}$. Hint: Write the denominator in the form $(s - a)^2 + b$. You may need to factorise the numerator. $\mathcal{L}^{-1} \left\{ \frac{3 \cdot s - 6}{s^2 - 4 \cdot s + 5} \right\} =$ c. Finally, find f(t). $f(t) = \mathcal{L}^{-1} \{ F(s) \}=$

          Consider the following function in the s-domain.
\(F(s) = \frac{(8 \cdot s^3 + 24) \cdot e^{-4s}}{s^4} + \frac{3 \cdot s - 6}{s^2 - 4 \cdot s + 5}\)
Calculate the inverse transform of F(s) by following these steps. If you need to enter the Heaviside step function for any
part of this question you can type it as H(t).
a. Calculate the following inverse Laplace transform: \(\mathcal{L}^{-1} \left\{ \frac{8 \cdot s^3 + 24}{s^4} \right\} =\)
Now use your previous answer and the second shift theorem to find the inverse Laplace transform of the first term.
\(\mathcal{L}^{-1} \left\{ \frac{(8 \cdot s^3 + 24) \cdot e^{-4s}}{s^4} \right\} =\)
b. Find the inverse Laplace Transform of \(\frac{3 \cdot s - 6}{s^2 - 4 \cdot s + 5}\). Hint: Write the denominator in the form \((s - a)^2 + b\). You
may need to factorise the numerator.
\(\mathcal{L}^{-1} \left\{ \frac{3 \cdot s - 6}{s^2 - 4 \cdot s + 5} \right\} =\)
c. Finally, find f(t).
\(f(t) = \mathcal{L}^{-1} \{ F(s) \}=\)

$Consider the following function in the s-domain. F(s) = ((8 · s^3 + 24) · e^-4s)/(s^4) + (3 · s - 6)/(s^2 - 4 · s + 5) Calculate the inverse transform of F(s) by following these steps. If you need to enter the Heaviside step function for any part of this question you can type it as H(t). a. Calculate the following inverse Laplace transform: ℒ^-1{(8 · s^3 + 24)/(s^4)} = Now use your previous answer and the second shift theorem to find the inverse Laplace transform of the first term. ℒ^-1{((8 · s^3 + 24) · e^-4s)/(s^4)} = b. Find the inverse Laplace Transform of (3 · s - 6)/(s^2 - 4 · s + 5). Hint: Write the denominator in the form (s - a)^2 + b. You may need to factorise the numerator. ℒ^-1{(3 · s - 6)/(s^2 - 4 · s + 5)} = c. Finally, find f(t). f(t) = ℒ^-1{ F(s) }=$

Added by Michael B.

Question

Please give Ace some feedback