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Chapter 2

Q. 2.7.2

One Negative Root and Two Zero Roots

The inverse Laplace transform of

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

Step-by-Step

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}