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Eigenvalue Problem: The Transfer Function

Module by: Steven Cox

Summary: (Blank Abstract)

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

The Transfer Function

One means by which to come to grips with Rs R s is to treat it as the matrix analog of the scalar function

1sb 1 sb (1)
This function is a scaled version of the even simpler function 11z 1 1z . This latter function satisfies the identity (just multiply across by 1z1z to check it)
11z=1+z+z2+...+zn1+zn1z 1 1z 1 z z2 ... z n1 zn 1z (2)
for each positive integer nn. Furthermore, if |z|<1 z 1 then zn0 z n 0 as n n and so Equation 2 becomes, in the limit, 11z=n=0zn 1 1z n 0 zn the familiar geometric series. Returning to Equation 1 we write 1sb=1s1bs=1s+bs2+...+bn1sn+bnsn1sb 1 sb 1 s 1 bs 1s b s 2 ... b n1 sn b n s n 1 sb and hence, so long as |s|>|b| s b we find, 1sb=1sn=0bsn 1 sb 1 s n 0 b s n This same line of reasoning may be applied in the matrix case. That is,
sIB-1=s-1IBs-11s+Bs2+...+Bn1sn+BnsnsIB-1 s I B s I Bs 1 s B s2 ... B n1 sn Bn sn sI B (3)
and hence, so long as |s|>B s B where B B is the magnitude of the largest element of BB, we find
sIB-1=s-1n=0Bsn sI B s n 0 B s n (4)
Although Equation 4 is indeed a formula for the transfer function you may, regarding computation, not find it any more attractive than the Gauss-Jordan method. We view Equation 4 however as an analytical rather than computational tool. More precisely, it facilitates the computation of integrals of Rs R s . For example, if Cρ Cρ is the circle of radius ρρ centered at the origin and ρ>B ρ B then
Cρ sIB-1ds=n=0Bn Cρ s-1nds=2πI s Cρ sI B n 0 B n s Cρ s -1n 2 I (5)
This result is essential to our study of the eigenvalue problem. As are the two resolvent identities. Regarding the first we deduce from the simple observation s2 IB-1 s1 IB-1= s2 IB-1 s1 IB s2 I+B s1 IB-1 s2 I B s1 I B s2 I B s1 I B s2 I B s1 I B that
R s2 R s1 = s1 s2 R s2 R s1 R s2 R s1 s1 s2 R s2 R s1 (6)
The second identity is simply a rewriting of sIBsIB-1=sIB-1sIB=I sI B sI B sI B sI B I namely,
BRs=RsB=sRsI B R s R s B s Rs I (7)

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