Question

Hi everyone! Can someone tell me why they set up the partial fraction this way? I posted a pic. Thanks! :o)

Answer 1

@Kainui

Answer 2

Hi Kainui!

Answer 3

Well the reason they set it up like that is cause a polynomial that's of s^2 might not be able to fully divide out a polynomial of a degree less than it, so we have to include all lower terms, so we have to have an s term.

For instance if the bottom one was As^3+Bs^2+Cs+D we'd have to put a Es^2+Fs+G term on top of it.

When you solve it though, it might turn out that the coefficient is zero though, it just depends. =)

Answer 4

"For instance if the bottom one was As^3+Bs^2+Cs+D we'd have to put a Es^2+Fs+G term on top of it. "
are you talking about As^3+Bs^2+Cs+D being in the denominator left of the = or right?

Answer 5

@Kainui could you write out your example for me? I don't know where you meant to put your example exactly? :o)

Answer 6

Yeah sure one sec

Answer 7

Just a slightly altered example

\[\frac{3s-2}{s^2(As^3+Bs^2+Cs+D)}=\frac{H}{s}+\frac{K}{s^2}+\frac{Es^2+Fs+G}{As^3+Bs^2+Cs+D}\]

Answer 8

wait I realized that this might not work out since I just made it up

Answer 9

um, if we process the linear complex root, itll add up into what you have

Answer 10

can you maybe give me a quick short cut on how to remember this...my test is very soon :o)

Answer 11

if you have an irreducable quantity in the demoniator, just do like Ds+E above it where like the "s" is one degree less than the bottom?

Answer 12

s^2 + 4 = (s-2i )(s+2i)

\[\frac{A}{s-2i}+\frac{B}{s+2i}\]
\[\frac{A(s+2i)}{s^2+4}+\frac{B(s-2i)}{s^2+4}\]
\[\frac{(A+B)s+2i(A-B)}{s^2+4}\]
i cant work withthe lag in this ...

Answer 13

i have never done imaginary partial fraction before

Answer 14

A/(s+2i) + B/(s-2i)

A(s-2i)/(s^2+4) + B(s+2i)/(s^2+4)

(As-2Ai + Bs +2Bi)/(s^2+4)


As-2Ai + Bs +2Bi = Ds + E

(A+B)s -2i(A-B) = Ds + E

A+B = D
A-B = Ei/2

2A = D + Ei/2

A = D/2 + Ei/4
B = D/2 - Ei/4

A and B are complex conjugates, complex coeffs to complex linear roots

Answer 15

this website is going crazy

Answer 16

\[\frac{2D+Ei}{4(s+2i)}+\frac{2D-Ei}{4(s-2i)}=\frac{Ds+E}{s^2+4}\]

Answer 17

yeah, its breaking for sure

Answer 18

complex conjugate roots, have complex conjugate coeffs :)

Answer 19

hey amistre...

Answer 20

D/2(s+2i) + Ei/4(s+2i) + D/2(s-2i) - Ei/4(s-2i)

D(2s/2(s^2+4) + Ei(1/4(s+2i)-1/4(s-2i)

D(2s/2(s^2+4) + Ei((s-2i-(s+2i))/4(s^2+4))

Ds/(s^2+4) + Ei(-4i/4(s^2+4))

Ds/(s^2+4) + Ei(-i/(s^2+4))

Ds/(s^2+4) + E/(s^2+4)

(Ds + E)/(s^2+4)