Stability of Digital Control Systems Outline Asymptotic Stability

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Stability of Digital Control Systems Outline Asymptotic Stability Outline • Asymptotic stability. Stability of Digital • Input-output stability. Control Systems • Stability conditions. • Routh-Hurwitz criterion. M. Sami Fadali •Jury test. Professor of Electrical Engineering UNR 1 2 Bounded-Input-Bounded-Output Asymptotic Stability (BIBO) Stability Response due to any initial conditions Response due to any bounded input decays to zero asymptotically in the steady remains bounded. state, i.e. the response due to the initial conditions satisfies u()k < bu ,k = 012 ,,,... Limy(k) = 0 0 <<∞bu k→∞ Marginal Stability: response due to any y(k )< by ,k = 012 , , ,... initial conditions remains bounded but does not decay to zero. 0 <<∞by 3 4 1 Stable Z-Domain Pole Time Sequence Locations n n z f (k) = A pk , k = 0,1,2,... ←⎯→Z F(z) = A ∑ i i ∑ i z − p Sampled exponential and its z-transform i=1 i=1 i with p real or complex ¾ Bounded sequence for poles in the closed unit z disc (i.e. on or inside the unit circle). pk , k = 0,1,2,... ←⎯→Z⎯ z − p ¾ Sequence decays exponentially for poles in the open unit disc (i.e. inside the unit circle). ⎧0, p <1 ¾ Unbounded sequence for repeated poles on the k ⎪ unit circle. Limp → ⎨1, p =1 k→∞ ¾ For real time sequences poles and partial ⎪ ⎩∞, p >1 fraction coefficients are either real or complex conjugate pairs. 5 6 Proof y(k + n) + a y(k + n −1) + ...+ a y(k +1) + a y(k) Theorem: Asymptotic Stability n−1 1 0 = bmu(k + m) + bmu(k + m −1) +...+ b1u(k +1) + b0u(k) k = 0,1,2,... In the absence of pole-zero cancellation, a Response of the system due to ICs y(0), y(1), …, y(n) LTI digital system is asymptotically stable N ()z Yz()= if its transfer function poles are in the open zazn ++++n −1 ... aza unit disc and marginally stable if the poles n −1 10 are in the closed unit disc with no repeated 1. Denominator of the output z-transform = denominator of the z-transfer function for no pole-zero cancellation. poles on the unit circle. 2. Poles of Y(z) = poles of transfer function. 3. Y(z) due to initial conditions is bounded for system poles in the closed unit disc unit (no repeated poles on the unit circle) & decays exponentially for system poles 7 in the open unit disc (inside the unit circle). 8 2 Example Solution Determine the asymptotic stability of the • Use Theorem 1 can (a) and (b) without pole- following systems: zero cancellation. • Ignore zeros, (do not affect response due ICs) 4()z − 2 4()z − 0.2 a) H (z) = b) H (z) = a) Pole outside the unit circle ⇒ unstable ()()z − 2 z − 0.1 ()()z − 0.2 z − 0.1 b) All poles inside the unit circle ⇒ asymptotically stable. 5()z − 0.3 8()z − 0.2 c) H (z) = d) H (z) = ()()z − 0.2 z − 0.1 ()()z − 0.1 z −1 c) All poles inside the unit circle ⇒ asymptotically stable. d) One pole on the unit circle ⇒ marginally stable. 9 10 BIBO Stability Theorem 4.2:BIBO Stability k yk()=−∑ hk ( iui )(), k =012 ,,,... i = 0 A discrete-time linear system is BIBO Is system BIBO stable if its impulse stable if and only if its impulse response response h(k) is bounded? NO sequence is absolutely summable i.e. Bounded & strictly positive impulse response ∞ 0 < b < h(k) < b < ∞, k = 0,1,2,... h1 h2 hi() <∞ k k ∑ y(k) = ∑ h(k − i)u(i) > bh1∑u(i), k = 0,1,2,... i = 0 i=0 i=0 u(i) = 1,i = 0,1,L ⇒ unbounded output 11 12 3 Proof of Necessity (Only if) Proof of Sufficiency (If) • Assume the system is BIBO stable but the • Assume an absolutely summable impulse impulse response is not absolutely summable. response and show that the system is BIBO stable. • Unbounded output with the bounded input (contradiction) • Use the input bound bu in the convolution summation ⎧10,()hi ≥ k uk()−= i ⎨ ⎩−<10,()hi y(k) ≤ ∑ h(i) u(k − i) i=0 k k yk()=→∞→∞ hi (), yk () as k ∑ < bu ∑ h(i) < ∞, ∀k ≥ 0 i=0 i=0 13 14 Theorem 4.3: BIBO Stability Proof of Necessity n n z A discrete-time linear system is BIBO h(k) = A pk , k = 0,1,2,... ←⎯→Z⎯ H (z) = A ∑ i i ∑ i z − p stable if and only if the poles of its transfer i=1 i=1 i function lie inside the unit circle. • Impulse response is bounded if the poles of the transfer function are in the closed unit disc and decays exponentially if the poles are in the open unit disc. • Systems with a bounded impulse response that does not decay exponentially are not BIBO stable. 15 16 4 Proof of Sufficiency Example 4.2 Assume exponentially decaying impulse Investigate the BIBO stability of systems response (i.e. poles inside the unit circle). with the impulse response Ar is the coefficient of largest magnitude |ps | < 1 is the largest pole magnitude. ⎧ Kkm, 0 ≤≤< ∞ hk()= ⎨ The impulse response is bounded by ⎩ 0, elsewhere n n k k k hk( )=≤∑∑ Apii Aii p ≤ nArs p , k =012 , , ,... i ==11i where K is a finite constant. ∞ ∞ i 1 ∑∑hi() ≤= nArs p n Ar <∞ i = 00i = 1− ps 17 18 Solution Example 4.3 BIBO stable since the impulse response Investigate BIBO stability for Example 4.1 satisfies ∞ m hi()==+<∞ hi () ( m1) K 4(z − 2) 4()z − 0.2 ∑∑ a) H (z) = b) H (z) = ii= 00= ()()z − 2 z − 0.1 ()()z − 0.2 z − 0.1 • Let K= upper bound for any impulse 5(z − 0.3) 8()z − 0.2 response of finite duration. c) H (z) = d) H (z) = ()()z − 0.2 z − 0.1 ()()z − 0.1 z −1 • Any FIR system is BIBO stable. 19 20 5 Solution Z-plane Stable Pole Locations After pole-zero cancellation Im[z] a) BIBO stable, all poles inside unit circle. Unit Circle b) BIBO stable, all poles inside unit circle. c) BIBO stable, all poles inside unit circle. STABLE d) Not BIBO stable, a pole on unit circle. Re[z] 21 22 MATLAB Stability Determination MATLAB: ddamp Obtain roots of polynomial: Gives the pole locations, ζ and ωn » roots(den) % denominator coeffts. den Eigenvalue Magnitude Equiv. Damping Equiv. Freq. (rad/sec) 0.2306 + 0.7428I 0.7778 0.1941 12.9441 » zpk(g) %g = transfer function 0.2306 - 0.7428I 0.7778 0.1941 12.9441 -0.6612 0.6612 0.1306 31.6871 Stable for roots inside the unit circle. Closed-loop transfer function >> H = feedback(gforward, gfeedback, ±1) 23 24 6 Routh-Hurwitz Criterion Advantages/Disadvantages 1. Transform the inside of the unit circle to • Easy stability test for low-order polynomials. the LHP (bilinear transformation). • Difficult for high order z-polynomials. 2. Use the Routh-Hurwitz criterion for the • For high order polynomials, use symbolic investigation of discrete-time system manipulation. stability. n n−1 Fz( )=+ azn an−1 z ++ ... a0 1 + w z − 1 z w 1+w n n−1 = ⇔= z= ⎛1+ w⎞ ⎛1+ w⎞ 1 − w z + 1 ⎯→⎯1⎯−w a ⎜ ⎟ + a ⎜ ⎟ ++... a n ⎝1− w⎠ n−1 ⎝1− w⎠ 0 25 26 Example 4.4 Solution: 1st order Find stability conditions for • Solve for the root. a) The first order polynomial • Stability conditions a1 z + a0 , a1 > 0 a 0 < 1 b) The second order polynomial a1 2 az2 ++ az10 a, a 2 >0 27 28 7 nd Bilinear Transformation Solution: 2 order 2 ⎛1+ w⎞ ⎛1+ w⎞ a2 ⎜ ⎟ + a10⎜ ⎟ + a 2 ⎝1− w⎠ ⎝1− w⎠ −±aaaa11 −4 02 z12, = 2 a2 2 (aaaw210−+) +2( aawaaa20 −) +( 210 ++) ¾ Stability determination by solving for roots is difficult. Routh-Hurwitz criterion: poles of 2nd order w-polynomial ¾ Monic polynomial remain in the LHP iff its coefficients are all positive. aaa− +>0 ¾ constant term = product of poles 210 ¾ For pole magnitudes < 1 aa20−>0 ¾ Necessary stability condition ¾ Sufficient for complex conjugate poles ) aaa210++ >0 a Adding the first and third conditions gives the condition 0 < 1 ⇔ −a < a & a < a obtained earlier 0 2 0 2 aa+ > 0 a2 29 20 30 Stable Parameter Range for 2nd Comments Order z-polynomial • Recall: The condition is sufficient for a0 a0=1 complex conjugate roots and only 1 necessary for real roots. stable a0= a1−1 a0= −a1−1 • For real roots, if the three conditions are a1 −1 1 substituted in the expression (4.14) we a2 =1 obtain roots between −1 and +1. −1 1− a1 + a0 > 0 1− a0 > 0 1+ a1 + a0 > 0 31 32 8 Jury Test Jury Table Row z0 z1 z2 … zn − k … zn − 1 zn 1 a0 a1 a2 … an − k … an − 1 an The roots of the polynomial are inside the unit 2 an an − 1 an − 2 … ak … a1 a0 3 b b b … b … b circle iff 0 1 2 n − k n − 1 4 bn − 1 bn − 2 bn − 3 … bk … b0 Fz()=+ azn a zn −1 +++= az a00 , a > n n −1 L 10 n 5 c0 c1 c2 … … cn − 2 6 c c c … … c (1) F(1) > 0 n − 2 n − 3 n − 4 0 n . … … (2) (−1) F(−1) > 0 . … … . … … (3) | a0 | < an 2 n − 5 s s s s (4) | b0 | > | bn − 1 | 0 1 2 3 2 n − 4 s3 s2 s1 s0 (5) | c0 | > | cn − 2 | 2 n − 3 r r r M 0 1 2 (n+1) | r | > | r | 0 2 33 34 Table Entries Remarks 1.
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