MATH319 Slides

136 Partial fractions

where here q𝐂 is a constant since f𝒮 is proper, and all the λj have λj<0. So we can take linear combinations of such f, and stay in 𝒮. Also f1(s)+f2(s)=(g1(s)h2(s)+h1(s)g2(s))/h1(s)h2(s)𝒮.

(C) Commutativity of multiplication follows from the corresponding property for polynomials;

(ID2) likewise;

(Diff) Also, we can differentiate

f(s)=j=1N-njaj(s-λj)nj+1,

and the poles are at λj in open left half plane.

(ii) Whereas 1/(s+1) belongs to 𝒮, the inverse s+1 is not proper, hence not in 𝒮.