One Negative Root and Two Zero Roots

The inverse Laplace transform of

$X(s)=\frac{5}{s^2(3s+12)}$

Question

One Negative Root and Two Zero Roots

The inverse Laplace transform of

[latex] X(s)=\frac{5}{s^2(3s+12)}[/latex]

Accepted Answer

The denominator roots are s= −12/3= −4, s=0, and s=0. Thus, the partial-fraction expansion has the form

[latex]X(s)=\frac{5}{s^2(3s+12)} = \frac{1}{3} \frac{5}{s^2(s+4)} = \frac{C_1}{s^2}+\frac{C_2}{s}+\frac{C_3}{s+4}[/latex]

Using the coefficient formulas (2.7.4), (2.7.8), and (2.7.9) with p = 2 and [latex]r_1[/latex] = 0, we obtain

[latex]C_i=\underset{s ightarrow -r_i}{ ext{lim}}[X(s)(s+r_i)] \quad (2.7.4)[/latex]

[latex]C_1=\underset{s ightarrow -r_1}{ ext{lim}}[X(s)(s+r_1)^{p}] \quad (2.7.8)[/latex]

[latex]C_2=\underset{s ightarrow -r_1}{ ext{lim}} \{\frac{d}{ds}[X(s)(s+r_1)^p] \} \quad (2.7.9)[/latex]

[latex]C_1= \underset{s ightarrow 0}{ ext{lim}}[s^2 \frac{5}{3s^2(s+4)}]= \underset{s ightarrow 0}{ ext{lim}}[\frac{5}{3(s+4)}]=\frac{5}{12}[/latex]

[latex]C_2= \underset{s ightarrow 0}{ ext{lim}} \frac{d}{ds}[s^2 \frac{5}{3s^2(s+4)}]= \underset{s ightarrow 0}{ ext{lim}} \frac{d}{ds} [\frac{5}{3(s+4)}]= \underset{s ightarrow 0}{ ext{lim}} [-\frac{5}{3} \frac{1}{(s+4)^2}]=-\frac{5}{48}[/latex]

[latex]C_3 = \underset{s ightarrow -4}{ ext{lim}} [(s+4)\frac{5}{3s^2(s+4)}]= \underset{s ightarrow -4}{ ext{lim}} (\frac{5}{3s^2})=\frac{5}{48}[/latex]

The inverse transform is

[latex]x(t)=C_1t+C_2+C_3e^{-4t}=\frac{5}{12}t-\frac{5}{48}+\frac{5}{48}e^{-4t}[/latex]

Question 2.7.2: One Negative Root and Two Zero Roots The inverse Laplace tra...

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