If every Bounded Input produces Bounded Output, system is externally stable equivalently, system is BIBO stable

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Clyde Burke
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1 1. External (BIBO) Stability of LTI Systems If every Bounded Input produces Bounded Output, system is externally stable equivalently, system is BIBO stable g(n) < BIBO Stability Don t care about what unbounded input does... CL 692 Digital Control, IIT Bombay 1 c Kannan M. Moudgalya, Autumn Condition for BIBO Stability g(n) < BIBO Stability y(n) k g(k)u(n k) Suppose g(n) <. Bounded u with u(n) < M Evaluate one term: y(n) M k n g(k) u(n k) k g(k) As M is bounded, so is y(n). This is true for all n. Is it possible for system to be BIBO stable and g(i)? Produce the following signal: { g( k) g( k) g( k) 0, u(k) 0 g( k) 0. {u(n)} {..., 1, 1, 1, 1, 1,...} y(0) k k k u(k)g(n k) n0 u(k)g( k) [g( k)] 2 g( k) So, not BIBO stable k g( k) CL 692 Digital Control, IIT Bombay 2 c Kannan M. Moudgalya, Autumn 2006

2 3. Z-transform - Motivation y(n) u(k)g(n k) k {u} {u(0), u(1), u(2)} {g} {g(0), g(1), g(2)} Let these be ero at all other times. y(0) u(0)g(0) y(1) u(0)g(1) + u(1)g(0) y(2) u(0)g(2) + u(1)g(1) + u(2)g(0) y(3) u(1)g(2) + u(2)g(1) Also carryout multiplication: (u(0) + u(1) 1 + u(2) 2 ) (g(0) + g(1) 1 + g(2) 2 ) u(0)g(0)+ (u(0)g(1) + u(1)g(0)) 1 + (u(0)g(2) + u(1)g(1) + u(2)g(0)) 2 + (u(1)g(2) + u(2)g(1)) 3 + Think of as a position marker - coefficient of i occurs at i th instant u(0) + u(1) 1 + u(2) 2 - a way of representing a sequence with three terms: {u(0), u(1), u(2)} The polynomial can often be represented by a compact expression even if the original sequence has infinitely many nonero terms CL 692 Digital Control, IIT Bombay 3 c Kannan M. Moudgalya, Autumn Two Sequences to Show Importance of Convergence u 1 (n) a n 1(n) u 2 (n) a n 1( n 1) n 1(n) n 1( n 1) Time (n) 1 a 0. 9 ; 2 n 10: 20; 3 y eros ( s i e ( n ) ) ; 4 f o r i 1 : length ( n ) 5 i f n ( i )>0, 6 y ( i ) aˆn ( i ) ; 7 end 8 end 9 A axes ( F o n t S i e, 1 4 ) ; 10 set ( get (A, X l a b e l ), F o n t S i e, 1 4 ) ; 11 o stem ( n, y ) ; 12 set ( o ( 1 ), Marker,. ) ; 13 x l a b e l ( Time ( n ) ) ; 14 y l a b e l ( 0.9ˆ n1 ( n ) ) ; Time (n) 1 a 0. 9 ; 2 n 10: 20; 3 y eros ( s i e ( n ) ) ; 4 f o r i 1 : length ( n ) 5 i f n ( i )< 1, 6 y ( i ) (aˆn ( i ) ) ; 7 end 8 end 9 A axes ( F o n t S i e, 1 4 ) ; 10 set ( get (A, X l a b e l ), F o n t S i e, 1 4 ) ; 11 o stem ( n, y ) ; 12 set ( o ( 1 ), Marker,. ) ; 13 x l a b e l ( Time ( n ) ) ; 14 y l a b e l ( 0.9ˆn1( n 1) ) ; CL 692 Digital Control, IIT Bombay 4 c Kannan M. Moudgalya, Autumn 2006

3 5. Z-transform - Convergence Condition for Uniqueness u 1 (n) a n 1(n) Take Z-transform: U 1 () a n n n0 (a 1 ) n n0 If the sum exists, 1 1 a 1 a u 2 (n) a n 1( n 1) U 2 () a n 1( n 1) n 1 n 1 a n n (a 1 ) m m0 If the sum exists, 1 a m m m1 1 1 a 1 a Any difference? Clue: If the sum exists. Convergence Condition: a 1 < 1 a 1 < 1 a < < a CL 692 Digital Control, IIT Bombay 5 c Kannan M. Moudgalya, Autumn Z-transform - Convergence Condition for Uniqueness u 1 (n) a n 1(n) U 1 () a a 1 < 1 May want a complex, would want complex So make the convergence absolute u 2 (n) a n 1( n 1) U 2 () a a 1 < 1 a 1 < 1 a < a 1 < 1 < a CL 692 Digital Control, IIT Bombay 6 c Kannan M. Moudgalya, Autumn 2006

4 7. Z-transform Definition Z-transform of a sequence {u(n)}, denoted by U(), calculated using: U() u(n) n where is such that U() u(n) n < Stronger condition of absolute convergence allows u and to be complex. The main advantage in taking Z-transform is that cumbersome convolution calculations can be done easier. Suitable for studying stability, causality and oscillatory behaviour of even infinitely long signals. Will use this fact while designing controllers. There is an added burden of converting the polynomial to a closed form expression and inverting this procedure to arrive at the resultant polynomial. It is possible to develop techniques to simplify these operations. CL 692 Digital Control, IIT Bombay 7 c Kannan M. Moudgalya, Autumn Region of Convergence (ROC) - Property 1 ROC consists of a ring centred around the origin 0 0 U() u(n) n 0 ROC u(n) n 0 < > u(n) n 0 u(n) n 0 u(n) 0 n 0 u(n) n u(n) n Thus, arbitrary also belongs to ROC CL 692 Digital Control, IIT Bombay 8 c Kannan M. Moudgalya, Autumn 2006

5 9. ROC Property 2 - Poles are Outside ROC ROC does not have any pole At a pole, transfer function is infinite, does not converge CL 692 Digital Control, IIT Bombay 9 c Kannan M. Moudgalya, Autumn ROC - 3: Larger Circles are in ROC for Causal Signals For sequence {u(n)}, with u(n) 0 for n < 0, if 0 ROC, all such that > 0 will belong to ROC U() 0 0 ROC 0 u(n) n n0 u(n) n 0 < > > n0 n0 u(n) n 0 u(n) n 0 u(n) 0 n n0 n0 n0 u(n) 0 n u(n), > 0 n u(n) n n0 Thus, arbitrary also belongs to ROC CL 692 Digital Control, IIT Bombay 10 c Kannan M. Moudgalya, Autumn 2006

6 11. ROC - 4: Poles of Causal, Stable Systems in Unit Circle Causal and stable system Impulse response g(n) Z-transform of g(n), namely G(), will have poles inside unit circle As g(k) is causal and stable, g(k) < k0 G() > g(k) k k0 g(k) k0 k0 g(k) k 1 Thus, there is absolute convergence at 1. Equivalently, the unit circle belongs to the region of convergence. As it is a causal sequence, by Property 3, all points outside the unit circle also belong to the ROC From property 2, the poles cannot lie in ROC - they have to lie within the unit circle CL 692 Digital Control, IIT Bombay 11 c Kannan M. Moudgalya, Autumn ROC - 5: Z-transform of Causal Signal is Proper u(k) is a causal sequence U() N() D() with N() is a polynomial of degree n D(Z) is a polynomial of degree m n m. Since u(k) is causal, by property 3, is in ROC If n > m, U() will diverge at Thus n m. CL 692 Digital Control, IIT Bombay 12 c Kannan M. Moudgalya, Autumn 2006

7 13. Z-transform Theorems - Linearity Z [α{u 1 (n)} + β{u 2 (n)}] αu 1 () + βu 2 () where, u 1 (n) U 1 () u 2 (n) U 2 (). Here α and β are arbitrary scalars. LHS [αu 1 (n) + βu 2 (n)] n αu 1 (n) n + αu 1 () + βu 2 () RHS βu 2 (n) n CL 692 Digital Control, IIT Bombay 13 c Kannan M. Moudgalya, Autumn Example 1 - Linearity Find the Z-transform of u 1 (n) 2δ(n) 3δ(n 2) + 4δ(n 5) U 1 () 2 δ(n) n δ(n 2) n δ(n 5) n finite > 0. Find the Z-transform of {u 2 (n)} [2 + 4( 3) n ] {1(n)} U 2 () [2 + 4( 3) n ] n n , > 1 > ( 1)( + 3), > 3 CL 692 Digital Control, IIT Bombay 14 c Kannan M. Moudgalya, Autumn 2006

8 15. Example 2 - Linearity Find the Z-transform of cos wn 1(n) and sin wn 1(n). cos ωn + j sin ωn e jωn Z [cos ωn + j sin ωn] Z { e jωn} Z [ e jωn 1(n) ] e jωn n 1(n) e jωn n n0 e, > jω ejω 1 ( e jω ) ( cos ω + j sin ω) ( cos ω) + j sin ω Comparing real and imaginary parts, Z [cos ωn] Re [ Z { e jωn}] Z [sin ωn] Im [ Z { e jωn}]. ( cos ω) (cos ωn)1(n) sin ω (sin ωn)1(n) CL 692 Digital Control, IIT Bombay 15 c Kannan M. Moudgalya, Autumn Z-transform - Shifting Example: Z [u(k + d)] d U() Z [u(k + d)] u(k + d) k k d u(k + d) (k+d) k d U() If then {u(n)} U(), {u(n + 3)} 3 U() {u(n 2)} 2 U() CL 692 Digital Control, IIT Bombay 16 c Kannan M. Moudgalya, Autumn 2006

1. Linearity of a Function A function f(x) is defined linear if. f(αx 1 + βx 2 ) = αf(x 1 ) + βf(x 2 )

1. Linearity of a Function A function f(x) is defined linear if f(αx 1 + βx 2 ) αf(x 1 ) + βf(x 2 ) where α and β are scalars. Example of a linear function: f(x) 2x A nonlinear function: What about f(x)