Question

Partial Fraction Decomposition problem: Exponentials in the numerator.

Answer 1

So here is the equation \[\frac{se^{-s}+e^{-2s}}{s^2+2s+1}\]

Answer 2

the bottom is easy to factor, but I feel like this is probably more complicated than I am thinking because of the exponentials in the numerator. I know to use PFD the degree of P(s) must be less than the degree of Q(s). ( for P(s)/Q(s)) but how do expoentials factor into the degree of the numerator?

Answer 3

looks like a laplace inverse

Answer 4

factor out the e^(-s) may be a start, or convert it to a taylor polynomial

Answer 5

oh what, its just decomp im thinking way to far ahead for this

Answer 6

It is actually an inverse laplace, we'ere supposed to solve using method of residues for partial fraction decomp. but the e's are tripping me up

Answer 7

\[\frac{se^{-s}+e^{-2s}}{s^2+2s+1}\]
\[\frac{se^{-s}+e^{-2s}}{(s+1)^2}=\frac{A}{(s+1)}+\frac{Bs}{(s+1)^2}\]
\[(s+1)^2\left[\frac{se^{-s}+e^{-2s}}{(s+1)^2}=\frac{A}{(s+1)}+\frac{Bs}{(s+1)^2}\right]\]
\[se^{-s}+e^{-2s}=A(s+1)+Bs\]
when s=0, solve for A
when s=-1, solve for B

Answer 8

Okay, so ultimately for doing a decomp, having e^s in the numerator won't effect the degree of the numerator?

Answer 9

nope, they are just a part that we will eventually define A and B with. e is just some number after all

Answer 10

would 3^s bother you?

Answer 11

right. right. lol. starting to over think things a bit. Might be a sign I should take a break? Thanks!

Answer 12

:) youre doing fine

Answer 13

so A=1 ... and B

\[-e^{1}+e^{2}=-B\]