# partial fraction question

• Nov 22nd 2006, 03:16 PM
feage7
partial 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 PM
ThePerfectHacker
Quote:

Originally Posted by feage7
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

You have,
$\displaystyle \frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+ D}{s^2+1}$

Note $\displaystyle Cs+D$ because it belongs to an irreducible quadradic.
• Nov 22nd 2006, 08:52 PM
CaptainBlack
Quote:

Originally Posted by feage7
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

What ImPerfectHacker says.

But you can factorise $\displaystyle s^2+1$ if complex factors are used, and
the tables of LT still work with complex terms.

So:

$\displaystyle s^2+1=(s+\bold{i})(s-\bold{i})$

RonL
• Nov 23rd 2006, 04:27 AM
feage7
Quote:

Originally Posted by ThePerfectHacker
You have,
$\displaystyle \frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+ D}{s^2+1}$

Note $\displaystyle Cs+D$ because it belongs to an irreducible quadradic.

Ive tried that and spent ages trying to find away to then equate the terms on top but i still end up finding the need to use i and every time ive involved i i seem to get it wrong when i put in values at the end to check my answer
• Nov 23rd 2006, 04:37 AM
CaptainBlack
Quote:

Originally Posted by feage7
Ive tried that and spent ages trying to find away to then equate the terms on top but i still end up finding the need to use i and every time ive involved i i seem to get it wrong when i put in values at the end to check my answer

When I do it I get:

$\displaystyle \frac{1-s}{s^2(s^2+1)}=-\frac{1}{s}+\frac{1}{s^2}+\frac{s-1}{s^2+1}$

RonL
• Nov 23rd 2006, 05:18 AM
feage7
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 AM
topsquark
$\displaystyle \frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+ D}{s^2+1}$

$\displaystyle \frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+1} = \frac{As(s^2+1) + B(s^2+1) + (Cs + D)s^2}{s^2(s^2+1)}$

= $\displaystyle \frac{(A+C)s^3 + (B+D)s^2 + As + B}{s^2(s^2+1)}$

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:
$\displaystyle \frac{1-s}{s^2(s^2+1)} = -\frac{1}{s}+\frac{1}{s^2}+\frac{s-1}{s^2+1}$

================================================== ==

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