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

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- Nov 22nd 2006, 03:16 PMfeage7partial fraction question
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 - Nov 22nd 2006, 03:58 PMThePerfectHacker
- Nov 22nd 2006, 08:52 PMCaptainBlack
- Nov 23rd 2006, 04:27 AMfeage7
- Nov 23rd 2006, 04:37 AMCaptainBlack
- Nov 23rd 2006, 05:18 AMfeage7
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 - Nov 23rd 2006, 06:10 AMtopsquark
$\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