Question 2.7.5: Four Repeated Roots Choose the most convenient method for ob...
Four Repeated Roots
Choose the most convenient method for obtaining the inverse transform of
X(s)=\frac{s^2+2}{s^4(s+1)}
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_1and 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}