Asked by SILVA
                By the method of the fractions partial determine the function f (t) of the following functions F (s).
F(s)= s^2+7s+12/(s+2)(s+4)(s+6)
F(s)= (s+3)(s+6)/s(s^2+10s+24)
F(s)= 5(s+2)^2/s(s+1)^3
F(s)= s+6/s(s+1)^2
F(s)= 27/(s+1)^3(s+4)
            
            
        F(s)= s^2+7s+12/(s+2)(s+4)(s+6)
F(s)= (s+3)(s+6)/s(s^2+10s+24)
F(s)= 5(s+2)^2/s(s+1)^3
F(s)= s+6/s(s+1)^2
F(s)= 27/(s+1)^3(s+4)
Answers
                    Answered by
            Steve
            
    I'll do one, and you use the same method to do the others. Come on back if you get stuck, and show where things go south.
F(s)= 5(s+2)^2/s(s+1)^3
F(s) = A/s + B/(s+1) + Cx+D/(s+1)^2 + Es^2+Fs+G/(s+1)^3
place all over a common denominator s(s+1)^3 and expand the numerator, then equate coefficients of like powers of s.
You will find that most of the coefficients are zero, and you end up with
A = 20
B = -20
D = -15
G = -5
F(s) = 20/s - 20/(s+1) - 15/(s+1)^2 - 5/(s+1)^3
    
F(s)= 5(s+2)^2/s(s+1)^3
F(s) = A/s + B/(s+1) + Cx+D/(s+1)^2 + Es^2+Fs+G/(s+1)^3
place all over a common denominator s(s+1)^3 and expand the numerator, then equate coefficients of like powers of s.
You will find that most of the coefficients are zero, and you end up with
A = 20
B = -20
D = -15
G = -5
F(s) = 20/s - 20/(s+1) - 15/(s+1)^2 - 5/(s+1)^3
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