Fal!2001 16.3122-9 Derivation of the srl The closed-loop poles are given by the eigenvalues of A Ruu B H C/ Recz so solve det(sI-h)=0 a B . If a is invertible: det det(a)det(D-Ca B det(sI-H)=det(sI-A)det[(sI+A)-Ci RuCz(sI- A)BuR BUT det(sI-A)det(sI+A')det [I-Ci RuCi(sI-A)-BuRul B (sI+A)-1I Note that det(I +abc)=det(I + CAB), and if a(s)=det(sI A),then a-s)=det(-sI -A)=(1)n det(sI +A det(sI-H)=(1)a(s)a(-s)det [I+ Ru BG-SI-A)-CZ R2Cz(sI-A)-Bu IfGzu(s)=C2(sI-A)-'Bu, then GZ(s)=BI(-SI-AT-CI so for SISo systems det(sI-H)=(1) a(s)a(-s)det [I+Ru Gau(s)Ru Gau(s) R (-1)a(sa(-s)I+2G(-s)G(s R R 1a(s)a(-s)+言b(s)b( RFall 2001 16.31 22—9 Derivation of the SRL • The closed-loop poles are given by the eigenvalues of " # A −BuR−1 uu Bu T −Cz TRzzCz −AT H , so solve det(sI − H)=0 " # = det(A) det(D − CA−1B) A B C D • If A is invertible: det £ (sI + AT ) − Cz TRzzCz(sI − A) −1 BuR−1 u ¤ uu BT = det(sI − A) det(sI + AT ) det ⇒ det(sI − H) = det(sI − A) det £ I − Cz TRzzCz(sI − A) −1 BuR−1 u (sI + AT ) −1 ¤ uu BT • Note that det(I + ABC)=det(I + CAB), and if a(s)=det(sI − A), then a(−s)=det(−sI − AT )=(−1)n det(sI + AT ) det(sI−H) = (−1)na(s)a(−s) det £ I + R−1 u (−sI − AT ) −1 Cz TRzzCz(sI − A) −1 Bu ¤ uu BT • If Gzu(s) = Cz(sI−A) −1 Bu, then GT zu(−s) = Bu T (−sI−AT ) −1Cz T , so for SISO systems £ I + R−1 zu(−s)RzzGzu(s) ¤ uu GT = (−1)na(s)a(−s) I + Rzz Gzu(−s)Gzu(s) ¸ Ruu det(sI − H)=(−1)na(s)a(−s) det ∙ ∙ Rzz = (−1) a(s)a(−s) + n Ruu b(s)b(−s) ¸ = 0