ME5659_LecNotes_RouthHurwitz

# ME5659_LecNotes_RouthHurwitz - EXAMPLE : T (s) = 2 . Is T...

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Unformatted text preview: EXAMPLE : T (s) = 2 . Is T (s) stable? s 2 + 3s + 2 2 T (s) = , (s + 1)(s + 2) Roots at s = -1, s = -2. Stable! 10s + 24 EXAMPLE : T (s) = . Is T (s) stable? s 3 + 2s 2 - 11s - 12 10(s + 2.4) T (s) = , (s + 1)(s - 3)(s + 4) Roots at s = -1, s = +3, s = -4. Unstable! s EXAMPLE: T (s) = . Is T (s) stable? s2 + 1 s T (s) = , (s - j)(s + j) Roots at s = j. MARGINALLY stable. Use input = sin(t). Y (s) = s 1 s = 2 s 2 + 1 s 2 + 1 (s + 1)2 y(t) = t sin(t) . . . unbounded. MARGINALLY stable = unstable (bounded impulse response, but unbounded output for some inputs.) Routh Hurwitz Stability Criterion Factoring high-degree polynomials to find roots is tedious and numerically not well conditioned. Want other stability tests, and also MARGINS of stability. "Case 1" Once we have determined that ai 0 i, we need to run the Routh test. Very mechanical, not intuitive. Proof difficult. a(s) = s n + an-1s n-1 + + a1s + a0. Form "Routh Array". sn s n-1 s n-2 s n-3 . . . s1 s0 an an-2 an-4 an-1 an-3 an-5 b1 b2 c1 c2 j1 k1 b1 = -1 an an-2 an-1 an-1 an-3 -1 an an-4 an-1 an-1 an-5 -1 an-1 an-3 b1 b1 b2 -1 an-1 an-5 b1 b1 b3 b2 = c1 = c2 = TEST : Number of unstable roots = number of sign changes in left column. a(s) = s 3 + s 2 + 2s + 8. b1 = c1 = -1 1 2 1 1 8 = -(8 - 2) = -6 = 1 (0 + 68) = 8 6 EXAMPLE : s3 1 2 s2 1 8 s 1 -6 s0 8 -1 1 8 -6 -6 0 Two sign changes (1 -6, -6 8) Two roots in RHP. ECE4510, STABILITY ANALYSIS 57 EXAMPLE : a(s) = s 2 + a1s + a0. b1 = -1 1 a0 a1 a1 0 = -1 (0 - a0 a1) = a0 a1 s 2 1 a0 s 1 a1 s 0 a0 Stable iff a1 > 0, a0 > 0. EXAMPLE : a(s) = s 3 + a2s 2 + a1s + a0. 1 a2 a0 a1 a0 b1 = -1 1 a1 a2 a2 a0 = -1 (a0 - a1 a2) a2 a0 a2 s3 s2 s0 s 1 a1 - a0 a2 = a1 - c1 = -1 a2 a0 b1 b1 0 = -1 (-a0b1 ) = a0. b1 Stable iff a2 > 0, a0 > 0, a1 > a0 . a2 "Case 2" 0 Sometimes find an element in first column = 0 (!) 0 Replace with as 0. EXAMPLE : a(s) = s 5 + 2s 4 + 2s 3 + 4s 2 + 11s + 10. b1 = -1 1 2 2 2 4 = -1 (4 - 4) = 0 2 -1 (10 - 22) = 6 2 -1 -12 b2 = -1 1 11 2 2 10 -1 2 4 6 = s5 s4 s3 new s 3 s2 s1 s0 If If 1 2 0 -12 6 10 2 11 4 10 6 6 10 c1 = = (12 - 4 ) c2 = -1 2 10 0 6 10 = -1 (0 - 10 ) = 10 > 0, two sign changes. < 0, two sign changes. Two poles in RHP. d1 = 12 -12 = 12 (10 + 72 )6 -1 e1 = 6 -12 6 10 0 = -1 (0 - 60) = 10 6 "Case 3" Sometimes an entire row in Routh array = 0. This means that polynomial factors such that one factor has conjugate-MIRROR-roots. (s) (s) (s) (s) (s) (s) In any case, system is unstable. But how many RHP roots? Complete Routh array by making polynomial a1(s) from last non-zero row in array. Poly has even orders of s only!! a1(s) is a factor of a(s). It is missing some orders of s, so IS NOT STABLE (by "case 0".) We want to see if it has roots in the RHP or only on the j-axis. da1(s) Replace zero row with and continue. ds EXAMPLE : a(s) = s 4 + s 3 + 3s 2 + 2s + 2. 3 2 2 2 0 b1 = -1 1 3 1 1 2 -1 1 2 1 1 0 = 1 s4 s3 s2 s1 new s 1 s0 1 1 1 0 2 2 b2 = = 2 No sign changes in Routh array. So, no RHP roots. Roots of a1(s) at s = j 2. Marginally stable. -1 1 2 c1 = 1 1 2 -1 1 2 2 2 0 = 0 a1(s) = s 2 + 2 da1(s) = 2s ds d1 = = 2 EXAMPLE : a(s) = s 4 + 4 (Hard). 1 0 4 0 0 0 0 4 4 4 a1 (s) = s 4 + 4; -1 1 0 4 4 0 -1 1 4 4 4 0 -1 4 0 4 4 0 da1(s) = 4s 3. ds = 0 s4 s3 new s 3 s2 new s 2 s1 s0 b1 = -16 4 b2 = = 4 Two sign changes. Therefore 2 RHP poles. Other two poles are mirrors in LHP. c1 = = -16 d1 = 16 -16 = 4 Design Tool Consider the system: r (t) K s+1 s(s - 1)(s + 6) y(t) s+1 K s(s-1)(s+6) Y (s) T (s) = = s+1 R(s) 1 + K s(s-1)(s+6) = K (s + 1) s(s - 1)(s + 6) + K (s + 1) a(s) = s(s - 1)(s + 6) + K (s + 1) = s 3 + 5s 2 + (K - 6)s + K . s3 s2 s1 s0 1 K -6 5 K 4K - 30 5 K -1 1 K - 6 5 5 K -1 (K - 5(K - 6)) 5 b1 = = = (4K - 30)/5. -1 5 K b1 b1 0 -1 (-b1 K ) = K b1 c1 = = For stability of the closed-loop system, K > 0, and K > 30/4. Step response for different values of K . 2.5 2 Amplitude 1.5 1 0.5 0 -0.5 0 K =25 K =13 K =7.5 2 4 6 8 10 12 Time (sec.) EXAMPLE : r (t) K 1+ 1 TI s 1 (s + 1)(s + 2) y(t) Y (s) s(s+1)(s+2) T (s) = = K TI s+K R(s) 1 + s(s+1)(s+2) = K TI s + K . s(s + 1)(s + 2) + K TI s + K K TI s+K a(s) = s 3 + 3s 2 + (2 + K TI )s + K . s3 s2 s1 s 0 1 2 + K TI 3 K 6 + 3K TI - K 3 K b1 = -1 1 2 + K TI 3 3 K -1 3 K b1 b1 0 = -1 (K - 3(2 + K TI )) 3 -1 (-b1 K ) = K b1 c1 = = For stability of the closed-loop system, 0 < K < 1 K > 0 for TI > . 3 1.2 1 6 1 for TI < or 1 - 3TI 3 K = 3, TI = 0.5 K = 1, TI = 0.25 K = 1, TI = Amplitude 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 Time (sec.) ...
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