Example 3.2-No Cance lation H()=a+a1q+a2=-16059+06065 Example 3. 3-Continuous time We have b+=1 andb-=B Bn(0)=B+b=4+一 G(s ose Dio phantine equation (q2+a19+a2)(q+71)+(bq+b1)(s0q+s1) Ao(s=s+a, Put q =-b1/bo solve for T1 B Xob+x1b061+ Dio phantine eq s(s+a(s+r1)+b(sos+s1) X1 Then solve fo Example 3. 3- Cont Identification of coefficients of powers of aT1+ bso =w+2cwao Inte rpre tation of poly nomial Ao What happens if we solve the pro blem wit h fb≠ state space met ho ds? What is the interpret ation of pole-zero cancel- T1=2Cw+ao-a C K.J. Astrom andB.WittenmarkExample 3.2 - No Cancellation H(q) = b0q + b1 q2 + a1q + a2 = 0:1065q + 0:0902 q2 ￾ 1:6065q + 0:6065 We have B+ = 1 and B￾ = B = b0q + b1 Hm(q) =  b0q + b1 q2 + am1q + am2 = bm0q + bm1 q2 + am1q + am2 Diophantine equation (q 2 + a1q + a2)(q + r1)+(b0q + b1)(s0q + s1) = (q 2 + am1q + am2)(q + ao) Put q = ￾b1=b0 solve for r1 r1 = X0b 2 0 + X1b0b1 + X2b 2 1 b 2 1 ￾ a1b0b1 + a2b 2 0 where X0 = aoam2 X1 = a2 ￾ am2 ￾ aoam1 X2 = ao + am1 ￾ a1 (5) Then solve for s0 and s1. Example 3.3 - Continuous time G(s) = b s(s + a) Choose Ao (s) = s + ao and Bm(s) Am(s) = !2 s2 + 2!s + !2 Diophantine equation s(s + a)(s + r1) + b(s0s + s1) = (s 2 + 2!s + !2 )(s + ao ) Example 3.3 - Cont. Identication of coecients of powers of s a + r1 = 2! + ao ar1 + bs0 = !2 + 2!ao bs1 = !2ao (6) If b 6= 0 r1 = 2! + ao ￾ a s0 = ao2! + !2 ￾ ar1 b s1 = !2ao b (7) Interpretation of Polynomial A0 What happens if we solve the problem with state space methods? What is the interpretation of pole-zero cancel￾lations? c K. J. Åström and B. Wittenmark 3