using the method of partial fractions obtain the expansion of
1-s/ s^2(s^2+1)
Im not sure how to go about it as you cant factorise the s^2+1
Im just going wrong everywehre
When i put i in i get
1-i = (Ci+D)(i)(-1)
1-i = (-c+Di)(-1)
1-i = C-Di
so i get C=1 and D = 1 which isnt right
aslo when i do A and B i end up with
1-s= As^4 + As^2 = Bs^3 +Bs + Cs^4 + Ds^3
equating terms in s gives me B=-1 and equating either s^4 or s^2 gives me A as 0
$\displaystyle
\frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+ D}{s^2+1}
$
$\displaystyle \frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+1} = \frac{As(s^2+1) + B(s^2+1) + (Cs + D)s^2}{s^2(s^2+1)}$
= $\displaystyle \frac{(A+C)s^3 + (B+D)s^2 + As + B}{s^2(s^2+1)}$
So:
$\displaystyle \frac{1-s}{s^2(s^2+1)} = \frac{(A+C)s^3 + (B+D)s^2 + As + B}{s^2(s^2+1)}$
Thus
A + C = 0
B + D = 0
A = -1
B = 1
So
C = 1
D = -1
Thus:
$\displaystyle \frac{1-s}{s^2(s^2+1)} = -\frac{1}{s}+\frac{1}{s^2}+\frac{s-1}{s^2+1}
$
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In terms of the complex denominators:
$\displaystyle \frac{1-s}{s^2(s^2+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+i} + \frac{D}{s-i}$
$\displaystyle \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+i} + \frac{D}{s-i} = \frac{As(s^2+1) + B(s^2+1) + Cs^2(s-i) + Ds^2(s+i)}{s^2(s^2+1)}$
= $\displaystyle \frac{(A+ C + D)s^3+ (B - iC + iD)s^2 + As + B}{s^2(s^2+1)}$
So:
$\displaystyle \frac{1-s}{s^2(s^2+1)} = \frac{(A + C + D)s^3+ (B - iC + iD)s^2 + As + B}{s^2(s^2+1)}$
Thus:
A + C + D = 0
B - iC + iD = 0
A = -1
B = 1
So
C + D = 1
-iC + iD = -1
So
C = (1/2)(1-i)
D = (1/2)(1+i)
Thus:
$\displaystyle \frac{1-s}{s^2(s^2+1)} = -\frac{1}{s} + \frac{1}{s^2} + \frac{\frac{1}{2}(1-i)}{s+i} + \frac{\frac{1}{2}(1+i)}{s-i}$
-Dan