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## Q. 2.7.2

One Negative Root and Two Zero Roots

The inverse Laplace transform of

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

## Verified Solution

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

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

Using the coefficient formulas (2.7.4), (2.7.8), and (2.7.9) with p = 2 and $r_1$ = 0, we obtain

$C_i=\underset{s \rightarrow -r_i}{\text{lim}}[X(s)(s+r_i)] \quad (2.7.4)$

$C_1=\underset{s \rightarrow -r_1}{\text{lim}}[X(s)(s+r_1)^{p}] \quad (2.7.8)$

$C_2=\underset{s \rightarrow -r_1}{\text{lim}} \{\frac{d}{ds}[X(s)(s+r_1)^p] \} \quad (2.7.9)$

$C_1= \underset{s \rightarrow 0}{\text{lim}}[s^2 \frac{5}{3s^2(s+4)}]= \underset{s \rightarrow 0}{\text{lim}}[\frac{5}{3(s+4)}]=\frac{5}{12}$

$C_2= \underset{s \rightarrow 0}{\text{lim}} \frac{d}{ds}[s^2 \frac{5}{3s^2(s+4)}]= \underset{s \rightarrow 0}{\text{lim}} \frac{d}{ds} [\frac{5}{3(s+4)}]= \underset{s \rightarrow 0}{\text{lim}} [-\frac{5}{3} \frac{1}{(s+4)^2}]=-\frac{5}{48}$

$C_3 = \underset{s \rightarrow -4}{\text{lim}} [(s+4)\frac{5}{3s^2(s+4)}]= \underset{s \rightarrow -4}{\text{lim}} (\frac{5}{3s^2})=\frac{5}{48}$

The inverse transform is

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