Question

Anyone have any idea how to expand this with partial fractions? s/(s+1)^2(s^2+w^2)

Answer 1

\[\frac{ s }{ (s+1)^2(s^2+\omega^2) }\]

Answer 2

\[\frac{ s }{ (s+1)^2(s^2+w^2) }=s\times \frac{ 1 }{ (s+1)^2(s^2+w^2) }\]
\[=s\times \frac{ 1}{ (s+1)^2 }\times \frac{ 1 }{ s^2+w^2 }\]
\[=s\times (s+1)^{-2}\times (s^2+w^2)^{-1}\]

Answer 3

Is this what you need to do?

Answer 4

no.

Answer 5

hmm I got to \[\frac{ s }{ (s+1)^2(s^2+\omega^2) }= \frac{ A }{ (s+1)^2 }+\frac{ B }{ (s+1) }....\]

Answer 6

I don't know how to expand s^2 + w^2?

Answer 7

Ah okay...

Answer 8

\[= \frac{ A }{ (s+1) }+\frac{ B }{ (s+1)^2 }+\frac{ Cx+D }{ (s^2+w^2) }\]

Answer 9

so that I can solve for a b and c given that c will be the expansion for s^2+w^2

Answer 10

Yep Okay.

Answer 11

oh snap that's what I wanted. How'd you know that it was cx+d?

Answer 12

all polynomials are written in the form Cx+D.

Answer 13

so if I want to solve for c and d would I just multiply both sides by cx+d? and let s=-cx-d?

Answer 14

No.

Answer 15

You are solving for A, B, C, and D.

Answer 16

I know how to solve for b and a

Answer 17

hmm....I usually get rid of the denominators, group like terms and factor out...then I solve for each individual variable, or I make a system of equations.

Answer 18

You can try your method though see if it works

Answer 19

would you consider omega as a constant?

Answer 20

Oh I like your method better actually

Answer 21

the degree of numerator polynomial is always 1 less than the degree of denominator, in partial fractions.
thats why when we have linear expression in denominator, we write constants A,B,C... in numerator,
so, when we have quadratic expression in denominator, we'll write a linear expression in numerator, like Ax+B,Cx+D,...and so on...

Answer 22

omega?! where did Ω come from? Lol

Answer 23

to find C and D, comparing the co-efficients is the best method...

Answer 24

w <----omega

Answer 25

and yes, its constant

Answer 26

oOoooooH. LOL thought it was a w. Lol. yeah constant.

Answer 27

oh so multiply to get same denominator and get rid of it then group terms based on degrees to solve for variables. Yeah gotcha thanks for your help I shall try that

Answer 28

thanks for all your help

Answer 29

yeah.