Unstable Pole-Zero Cancelled System 1. e ( n ) 1 y ( n ) E ( z ) z - - PowerPoint PPT Presentation

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Unstable Pole-Zero Cancelled System 1. e ( n ) 1 y ( n ) E ( z ) z - - PowerPoint PPT Presentation

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Unstable Pole-Zero Cancelled System 1. e ( n ) 1 y ( n ) E ( z ) z + 2 U ( z ) 1 Y ( z ) z + 0 . 5 z + 0 . 5 z + 2 Solution of cancelled system: x s ( k + 1) = 0 . 5 x s ( k ) + e ( k ) y s ( k ) = x s ( k ) Iteratively solving, k 1

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SLIDE 1

1.

Unstable Pole-Zero Cancelled System

U(z) Y (z) E(z) z + 2 z + 0.5 1 z + 2

y(n) e(n) 1 z + 0.5

Solution of cancelled system: xs(k + 1) = −0.5xs(k) + e(k) ys(k) = xs(k) Iteratively solving, ys(k) = (−0.5)kxs(0) +

k−1

  • m=0

(−0.5)me(k − m − 1) Stable.

Digital Control

1

Kannan M. Moudgalya, Autumn 2007

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SLIDE 2

2.

System without Pole-Zero Cancellation

U(z) Y (z) E(z) z + 2 z + 0.5 1 z + 2

Can verify for first block: First block: x1(k + 1) = −0.5x1(k) + 1.5e(k) u(k) = x1(k) + e(k) Second block: x2(k + 1) = −2x2(k) + u(k) y(k) = x2(k) Substituting, x2(k + 1) = −2x2(k) + x1(k) + e(k) x1(k + 1) x2(k + 1)

  • =

−0.5 1 −2 x1(k) x2(k)

  • +

1.5 1

  • e(k)

y(k) =

  • 0 1

x1(k) x2(k)

  • Digital Control

2

Kannan M. Moudgalya, Autumn 2007

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SLIDE 3

3.

System without Pole-Zero Cancellation

U(z) Y (z) E(z) z + 2 z + 0.5 1 z + 2

x1(k + 1) x2(k + 1)

  • =

−0.5 1 −2 x1(k) x2(k)

  • +

1.5 1

  • e(k)

y(k) =

  • 0 1

x1(k) x2(k)

  • = CAkx(0) +

k−1

  • m=0

CAmbe(k − m − 1) Can show that this is equal to

3 2

  • (−0.5)k (−2)k

x1(0) −x1(0) + 1.5x2(0)

  • +

k−1

  • m=0

(−0.5)me(k − m − 1)

x(0) = 0, results in y(k) being unbounded! Whence −2?

Digital Control

3

Kannan M. Moudgalya, Autumn 2007

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SLIDE 4

4.

Diagonalization A: square matrix λj: jth eigenvalue xj: jth eigenvector Ax1 = λ1x1 . . . Axn = λnxn Stacking these side by side, A   | | x1 · · · xn | |   =   | | x1 · · · xn | |     λ1 ... λn   AS = SΛ Assume the eigenvectors to be independent ⇒ S−1 exists A = SΛS−1

Digital Control

4

Kannan M. Moudgalya, Autumn 2007

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SLIDE 5

5.

Diagonalization A = SΛS−1 A2 = SΛS−1SΛS−1 = SΛ2S−1 A3 = SΛ2S−1SΛS−1 = SΛ3S−1 . . . Ak = SΛkS−1 Easy to evaluate RHS: Λk =   λ1 ... λn  

k

=   λk

1

... λk

n

  This approach is used to arrive at the solution.

Digital Control

5

Kannan M. Moudgalya, Autumn 2007

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SLIDE 6

6.

Condition for Cancellation of Poles and Zeros

U(z) Y (z) E(z) z + 2 z + 0.5 1 z + 2

Solution of system after cancellation: ys(k) = (−0.5)mxs(0) +

k−1

  • m=0

(−0.5)me(k − m − 1) Solution of system if there is no cancellation:

3 2

  • (−0.5)k (−2)k

x1(0) −x1(0) + 1.5x2(0)

  • +

k−1

  • m=0

(−0.5)me(k − m − 1)

Two solutions are identical only if

  • x1(0) = 1.5x2(0) or
  • x1(0) = x2(0) = 0.

Digital Control

6

Kannan M. Moudgalya, Autumn 2007

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SLIDE 7

7.

Aryabhatta’s Identity We need to solve the polynomial equation X(z)D(z) + Y (z)N(z) = C(z) for X and Y , with D, N and C specified.

  • Has a solution iff the GCD of D(z) and N(z) divides C(z)
  • There are infinitely many solutions to Aryabhatta’s identity
  • Unique solution under special conditions. Suppose D and N

are coprime with degrees dD > 0 and dN > 0. If 0 ≤ dC < dD + dN there is a unique least degree solution given by dX(z) < dN(z) dY (z) < dD(z)

Digital Control

7

Kannan M. Moudgalya, Autumn 2007

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SLIDE 8

8.

Algorithm for Aryabhatta’s Identity X(z)D(z) + Y (z)N(z) = C(z) is the same as

  • X(z) Y (z)

D(z) N(z)

  • = C(z)

Solved by comparing the coefficients of powers of z−1. Same as V (z)F (z) = C(z) Can be written as [V0 + V1z−1 + · · · + Vvz−v][F0 + F1z−1 + · · · + FdFz−dF] = C0 + C1z−1 + · · · + CdCz−dC v is an unknown. All Vi are also unknowns.

Digital Control

8

Kannan M. Moudgalya, Autumn 2007

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SLIDE 9

9.

Algorithm for Aryabhatta’s Identity [V0 + V1z−1 + · · · + Vvz−v][F0 + F1z−1 + · · · + FdFz−dF] = C0 + C1z−1 + · · · + CdCz−dC is the same as [V0 V1 · · · Vv]    F0 F1 · · · FdF · · · F0 F1 · · · FdF · · · . . . 0 · · · F0 F1 · · · FdF    = [C0 C1 · · · CdC] with maximal v. Ensures all combination of products are used. Multiplying, first two equations are obtained as V0F0 = C0, V0F1 + V1F0 = C1 Matrix equation is written as V F = C

Digital Control

9

Kannan M. Moudgalya, Autumn 2007

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SLIDE 10

10.

Algorithm for Aryabhatta’s Identity Recall from the previous slide: [V0 + V1z−1 + · · · + Vvz−v][F0 + F1z−1 + · · · + FdFz−dF] = C0 + C1z−1 + · · · + CdCz−dC Can be written as, V F = C IF F is right invertible, the solution is, V = C F −1

  • For this, rows of F have to be linearly independent.
  • Choose v as the largest integer, satisfying this requirement.
  • From V , V (z) and then X(z), Y (z) can be determined,

because, V =

  • X Y
  • Digital Control

10

Kannan M. Moudgalya, Autumn 2007

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SLIDE 11

11.

Summary: Algorithm for Aryabhatta’s Identity X(z)D(z) + Y (z)N(z) = C(z)

  • X(z) Y (z)

D(z) N(z)

  • = C(z)

V (z)F (z) = C(z) Can be written as

  • V0 V1 · · · Vv

   F0 F1 · · · FdF · · · F0 F1 · · · FdF · · · . . . 0 · · · F0 F1 · · · FdF     =

  • C0 C1 · · · CdC
  • Write this as V F = C , solve for V as V = C F −1

Digital Control

11

Kannan M. Moudgalya, Autumn 2007

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SLIDE 12

12.

Example: Algorithm for Aryabhatta’s Identity Solve for X(z) and Y (z) satisfying X(z)D(z) + Y (z)N(z) = C(z) with D(z) = 1 − 5z−1 + 4z−2 N(z) = z−1 + z−2 C(z) = 1 − z−1 + 0.5z−2 Construct F : F (z) = 1 − 5z−1 + 4z−2 z−1 + z−2

  • =

1

  • +

−5 1

  • z−1 +

4 1

  • z−2

which is of the form F0 + F1z−1 + F2z−2.

Digital Control

12

Kannan M. Moudgalya, Autumn 2007

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SLIDE 13

13.

Example: Algorithm for Aryabhatta’s Identity Recall F (z) = 1 − 5z−1 + 4z−2 z−1 + z−2

  • =

1

  • +

−5 1

  • z−1 +

4 1

  • z−2

which is of the form F0 + F1z−1 + F2z−2. Explore v = 1: F =

  • F0 F1 F2
  • =

1 −5 4 1 1

  • Although independent, v is not the largest. Explore v = 2:

F = F0 F1 F2 0 0 F0 F1 F2

  • =

    1 −5 4 1 1 1 −5 4 1 1     These rows are independent. Hence v is still not the maximum.

Digital Control

13

Kannan M. Moudgalya, Autumn 2007

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SLIDE 14

14.

Example: Algorithm for Aryabhatta’s Identity Explore the possibility of v = 3: F =   F0 F1 F2 0 0 F0 F1 F2 0 0 F0 F1 F2   =         1 −5 4 1 1 1 −5 4 1 1 1 −5 4 1 1         No longer independent. Remove one row and solve: V =

  • 1 −1 0.5 0 0

     1 −5 4 1 1 1 −5 4 1 1 1 −5 4      

−1

Digital Control

14

Kannan M. Moudgalya, Autumn 2007

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SLIDE 15

15.

Example: Algorithm for Aryabhatta’s Identity

  • V is a 1 × 5 vector.
  • To account for the row removed from F, we add a zero to

V . We obtain V =

  • 1 3.25 | 0.75 −3 | 0 0
  • where we have separated coefficients of powers of z−1 with

vertical lines.

  • We have V0 =
  • 1 3.25
  • and V1 =
  • 0.75 −3
  • .
  • We obtain

X(z) = 1 + 0.75z−1 Y (z) = 3.25 − 3z−1 These form the solution to Aryabhatta’s identity

Digital Control

15

Kannan M. Moudgalya, Autumn 2007