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## Q. 2.7.5

Four Repeated Roots

Choose the most convenient method for obtaining the inverse transform of

$X(s)=\frac{s^2+2}{s^4(s+1)}$

## Verified Solution

There are four repeated roots (s = 0) and one distinct root, so the expansion is

$X(s)=\frac{C_1}{s^4}+\frac{C_2}{s^3}+\frac{C_3}{s^2}+\frac{C_4}{s}+\frac{C_5}{s+1}$

Because there are four repeated roots, use of (2.7.10) to find the coefficients would require taking the first, second, and third derivatives of the ratio (s² + 2)/(s + 1). Therefore, the LCD method is easier to use for this problem. Using the LCD method we obtain

$C_i=\underset{s \rightarrow -r_1}{\text{lim}} \{ \frac{1}{(i-1)!} \frac{d^{i-1}}{ds^{i-1}}[X(s)(s+r_1)^p] \} \quad (2.7.10)$

$X(s)=\frac{C_1(s+1)+C_2s(s+1)+C_3s^2(s+1)+C_4s^3(s+1)C_5s^4}{s^4(s+1)} \\ = \frac{(C_5+C_4)s^4+(C_4+C_3)s^3+(C_3+C_2)s^2+(C_2+C_1)s+C_1}{s^4(s+1)}$

Comparing numerators we see that

$s^2+2=(C_5+C_4)s^4+(C_4+C_3)s^3+(C_3+C_2)s^2+(C_2+C_1)s+C_1$

and thus $C_1 = 2, C_2 +C_1 = 0, C_3 +C_2 = 1, C_4 +C_3 = 0$, and $C_5 +C_4 = 0$. These give $C_1 = 2, C_2 = −2, C_3 = 3, C_4 = −3$, and $C_5 = 3$. So the expansion is

$X(s)=\frac{2}{s^4}-\frac{2}{s^3}+\frac{3}{s^2}-\frac{3}{s}+\frac{3}{s+1}$

The inverse transform is

$x(t)=\frac{1}{3}t^3-t^2+3t-3+3e^{-t}$