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Lesson 10: I/O representation (Deterministic and Stochastic)

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Systems and Control Theory Lecture Notes

Laura Giarr´ e

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Lesson 10: I/O representation (Deterministic and Stochastic)

 I/O representation

 CT case

 DT case

 AR(n)

 MA(m)

 ARMA(n,m)

 From I/O to state representation

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I/O representation (SISO p = m = 1)

 Let us define an operator: D  dt d (TC) and D  Forward shift operator (TD).

 Then, D k y(t) = d

k

dt y(t)

k

and D 0 y(t) = y(t) in the CT case, or Dy(t) = y(t + 1), D k y(t) = y(t + k) in the DT case.

 Then, the transfer function H(s) (CT) or H(z) (DT) can be rewritten substituting s or z with the operator D or (z −1 with D −1 ):

H(D) = B(D)

A(D) = b 0 D m + b 1 D m−1 + . . . + b m−1 D + b m

D n + a 1 D n−1 + . . . + a n−1 D + a n

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I/O representation

 Correspondingly, the relation between input and output (zero i.c. or steady state) is:

A(D)y(t) = B(D)u(t)

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I/O representation

 (CT) A differential equation:

(D n + a 1 D n−1 + . . . + a n−1 D + a n )y(t) (b 0 D m + b 1 D m−1 + . . . + b m−1 D + b m )u(t)

 Such that:

d n y(t)

dt n + a 1 d n−1 y(t)

dt n−1 + . . . + a n y(t) = b 0 d m u(t)

dt m + b 1 d m−1 u(t)

dt m−1 + . . . + b m u(t)

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I/O representation (DT)

 (DT) A differential equation:

y(t + n) + a 1 y(t + n − 1) + . . . + a n y(t) = b 0 u(t + m) + b 1 u(t + m − 1) + . . . + b m u(t)

 If we consider the Backward operator, D −1 then:

y(t) + a 1 y(t − 1) + . . . + a n y(t − n) =

b 0 u(t) + b 1 u(t − 1) + . . . + b m u(t − m)

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From I/O to I/S/0: Realization 

case 1: deg B = n − 1 < deg A = n ( Strictly proper)

 A(D) = D n + a 1 D n−1 + . . . + a n−1 D + a n , B(D) = b 1 D m−1 + . . . + b n−1 D + b n

 where A(D)y(t) = B(D)u(t),

 We define a intermediate state as ξ(t) = A(D) u(t) and then y(t) = B(D)ξ(t)

 This leads to a realization in state space called the Reachability Canonical Form:

A =

⎢ ⎢

0 1 0 . . . 0

0 0 1 . . . 0

0 0 . . . . . . 0

−a n −a n−1 −a n−2 . . . −a 1

⎥ ⎥

⎦ B =

⎢ ⎢

⎢ ⎢

⎣ 0

.. . .. . 1

⎥ ⎥

⎥ ⎥

C = [b n b n−1 . . . b 1 ], D = 0

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From I/O to I/S/0: Realization

 case 2: deg B = deg A = n ( proper)

 Step 1: rewrite the transfer function as H(D) = A(D) ¯B(D) + β 0

 Then the realization is the same for the matrices A and B, but

 C = [β n β n−1 . . . β 1 ], D = β 0

 Being β i the coefficients of polynomial ¯ B(D)

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AR(n) processes

 Let e(k) be a stochastic process

 Assume that H(D) = A(D 1

−1

)

 where A(D −1 ) = 1 +  n

i=1 a i D −i that is to say A(z −1 ) = 1 +  n

i=1 a i z −i is a polynomial with zeros such that

|z i | < 1 and H(z) = A(z 1

−1

)

 The process y(k) = H(z)e(k) satisfies the recursive equation Y (k) + a 1 Y (k − 1) + . . . + a n Y (k − n) = e(k)

 Y (k) is called an Auto Regressive process of order n, AR(n).

 The covariance function R y (l) satisfies the recursion:

R y (l) + a 1 R y (l − 1) + ... + a n R y (l − n) = 0, forl > n

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MA(m) processes

 Assume that H(z) = C (z −1 )

 with C (z −1 ) = 1 +  n

i=1 c i z −i

 The process y(k) = H(z)e(k) satisfy the recursive equation Y (k) = e(k) + c 1 e(k − 1) + . . . + c m e(k − m)

 Y (k) is called a Moving Average process of order m, MA(m).

 The covariance function R y (l) is given by:

R y (l) =

c l + c 1 c l+1 + . . . c m−l c m , for l ≤ m

0 for l > m

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ARMA(n, m) processes

 We prefer AR processes because 1)more parsimonious, 2) easier to evaluate the covariance function.

 For a complete description of processes with rational spectrum we need to introduce the general description:

H(z) = C (z −1 ) A(z −1 )

 Y (k) satisfies the recursive equation:

Y (k) + a 1 Y (k − 1) + . . . + a n Y (k − n) = e(k) + c 1 e(k − 1) + . . . + c m e(k − m)

 Y (k) is called an Auto Regressive Moving Average process of

order n, m, ARMA(n, m)

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Thanks

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