1. ## 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

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

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

3. 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 $s^2+1$ if complex factors are used, and
the tables of LT still work with complex terms.

So:

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

RonL

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

Note $Cs+D$ 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

5. 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:

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

RonL

6. 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

7. $
\frac{1-s}{s^2(s^2+1)}=\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+ D}{s^2+1}
$

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

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

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

Thus
A + C = 0
B + D = 0
A = -1
B = 1

So
C = 1
D = -1

Thus:
$\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:
$\frac{1-s}{s^2(s^2+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+i} + \frac{D}{s-i}$

$\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+i} + \frac{D}{s-i} = \frac{As(s^2+1) + B(s^2+1) + Cs^2(s-i) + Ds^2(s+i)}{s^2(s^2+1)}$

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

So:
$\frac{1-s}{s^2(s^2+1)} = \frac{(A + C + D)s^3+ (B - iC + iD)s^2 + As + B}{s^2(s^2+1)}$

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

-Dan