Lecture7 - 1 Stability ODE y(n a1 y(n1 an1 y an y = b0 u(m...

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1 Stability ODE: y ( n ) + a 1 y ( n - 1) + · · · + a n - 1 ˙ y + a n y = b 0 u ( m ) + b 1 u ( m - 1) + · · · + b m - 1 ˙ u + b m u Characteristic equation: s n + a 1 s n - 1 + · · · + a n - 1 s + a n = 0 . Transfer function: G ( s ) = Y ( s ) U ( s ) = b 0 s m + b 1 s m - 1 + · · · + b m - 1 s + b m s n + a 1 s n - 1 + · · · + a n - 1 s + a n = N ( s ) D ( s ) Poles: D ( s ) = 0 . Zeros: N ( s ) = 0 For stability analysis: Poles: s n + a 1 s n - 1 + · · · + a n - 1 s + a n = 0 . (The system is stable if all poles are on LHP.) Zeros: b 0 s m + b 1 s m - 1 + · · · + b m - 1 s + b m = 0 . A system is stable if all initial conditions decay to zero. A system is unstable if any initial condition diverges. Example 1.1 Zero/pole cancellation. System 1 System 0 System 2 ODE ¨ y + 3 ˙ y + 2 y = ˙ u + u ˙ z + 2 z = v ¨ y + ˙ y - 2 y = ˙ u - u Characteristic Equ. s 2 + 3 s + 2 = 0 s + 2 = 0 s 2 + s - 2 = 0 Transfer function G 1 ( s ) = s + 1 ( s + 1)( s + 2) G 0 ( s ) = 1 s + 2 G 2 ( s ) = s - 1 ( s - 1)( s + 2) Poles - 1 , - 2 - 2 1 , - 2 Zeros - 1 N/A 1 Stability Stable Stable Unstable Initial condition ˙ y (0) = ε , y (0) = 0 z (0) = ε ˙ y (0) = ε , y (0) = 0 y ( t ) = ε ( e - t - e - 2 t ) 1 ( t ) z ( t ) = ε e - 2 t 1 ( t ) y ( t ) = 1 3 ε ( e t - e - 2 t ) 1 ( t ) decays decays diverges Impulse response y ( t ) = e - 2 t 1 ( t ) Response to input y ( t ) = L - 1 [ 1 s + 2 U ( s )] While considering the input-output relationship of the systems (with zero initial condition), G 1 ( s ) = G 0 ( s ) , G 2 ( s ) = G 0 ( s ) . However, while considering the stability of the systems, G 1 ( s ) = G 0 ( s ) , G 2 ( s ) = G 0 ( s ) . Transfer function: s + 1 ( s + 1)( s + 2) = 1 s + 2 s - 1 ( s - 1)( s + 2) = 1 s + 2 (for stability analysis) Unstable zero/pole cancellation is not allowed.
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