Problems and Discussion of the Inverse Laplace Transform - 1

Find \( h(t) \) from \( H(s) = \frac{s^2}{s^3 + 4s^2 + 4s} \)

Discussion:

We need to perform the inverse Laplace transform. Here are the steps to follow to obtain \( h(t) \) from the transfer function \( H(s) \):

Step 1: Factorize the denominator of \( H(s) \)

\[ H(s) = \frac{s^2}{s^3 + 4s^2 + 4s} = \frac{s^2}{s(s^2 + 4s + 4)} = \frac{s^2}{s(s + 2)^2} \]

Step 2: Convert the fraction into a simpler partial fraction form for easier inversion

\[ H(s) = \frac{s^2}{s(s + 2)^2} = \frac{A}{s} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2} \]

\[ s^2 = A(s + 2)^2 + Bs(s + 2) + Cs \]

\[ s^2 = A s^2 + 4A s + 4A + B s^2 + 2B s + C s \]

\[ s^2 = (A + B) s^2 + (4A + 2B + C) s + 4A \]

Step 3: Determine the Coefficients

\[ s^2 = (A + B) s^2 + (4A + 2B + C) s + 4A \]

By comparing coefficients, we get:

• \(1 = A + B\)
• \(0 = 4A + 2B + C\)
• \(0 = 4A \Rightarrow A = 0\)

From \(1 = A + B\), we get \(1 = 0 + B \Rightarrow B = 1\)

From \(0 = 4A + 2B + C\), we get \(0 = 0 + 2 \cdot 1 + C \Rightarrow 0 = 2 + C \Rightarrow C = -2\)

Step 4: Partial Fractions

Substitute \(A = 0\), \(B = 1\), and \(C = -2\) into \( H(s) \):

\[ H(s) = \frac{s^2}{s(s + 2)^2} = \frac{0}{s} + \frac{1}{s + 2} + \frac{-2}{(s + 2)^2} \]

\[ H(s) = \frac{1}{s + 2} - \frac{2}{(s + 2)^2} \]

Step 5: Inverse Laplace Transform

\[ H(s) = \frac{1}{s + 2} - \frac{2}{(s + 2)^2} \]

\[ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)} \right\} = e^{-2t} \]

\[ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)^2} \right\} = te^{-2t} \]

Thus:

\[ h(t) = e^{-2t} - 2t e^{-2t} \]

\( Graph: h(t) = e^{-2t} - 2t e^{-2t} \)

 [05220240602]