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

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

$$s^2=A(s+2{)}^2+Bs(s+2)+Cs$$

$$s^2=A{s}^2+4As+4A+B{s}^2+2Bs+Cs$$

$$s^2=(A+B)s^2+(4A+2B+C)s+4A$$

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

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

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

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

Thus:

$$h(t)=e^{-2t}-2t{e}^{-2t}$$

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

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