partial fraction

Let 1/[s^2 (s - 3)] = A/s + B/s^2 + C/(s - 3) where A, B and C are constants, then:

1 = As(s - 3) + B(s - 3) + Cs^2

Sub s = 0, we have B = -1/2

Sub s = 3, we have C = 1/9

The comparing the coefficient of s^2, we have A + C = 0, giving A = -1/9

Therefore:

1/[s^2 (s - 3)] = -1/(9s) - 1/(2s^2) + 1/[9(s - 3)]

2010-11-11 15:56:06 補充：
B should be -1/3, thus:

1/[s^2 (s - 3)] = -1/(9s) - 1/(3s^2) + 1/[9(s - 3)]

2010-11-11 16:18:12 補充：
Usually when the denominator has a quadratic factor which is not able to be factorized into linear factors with real coefficients, we will use the mathod for taking natural log which is a faster method in this type of partial fractions.

1/s^2(s-3)= (As+B)/s^2 + C/(s-3)

=[(As+B)(s-3) +Cs^2]/(s^2)(s-3)

=[(A+C)s^2+(B-3A)s -3B ]/(s^2)(s-3)

then we have

A+C=0 -----(1)
B-3A=0----(2)
3B=-1-----(3)

therefore

B=-1/3, A=-1/9 C=1/9

1/s^2(s-3)= (-1/9s-1/3)/s^2 + 1/9/(s-3)
=-1/9s-1/3s^2+1/9(s-3)


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