Docstoc

1 Internal Model Principle

Document Sample
1 Internal Model Principle Powered By Docstoc
					1.

Internal Model Principle

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z),

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z), all polynomials in z −1

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z), all polynomials in z −1 • Let there be no common factors between α(z) and B(z)

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z), all polynomials in z −1 • Let there be no common factors between α(z) and B(z) • Can find a controller Gc(z) for servo/tracking (following Rc )

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z), all polynomials in z −1 • Let there be no common factors between α(z) and B(z) • Can find a controller Gc(z) for servo/tracking (following Rc) and regulation (rejection of disturbance V )

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z), all polynomials in z −1 • Let there be no common factors between α(z) and B(z) • Can find a controller Gc(z) for servo/tracking (following Rc) and regulation (rejection of disturbance V ) if Rc contains α,

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z), all polynomials in z −1 • Let there be no common factors between α(z) and B(z) • Can find a controller Gc(z) for servo/tracking (following Rc) and regulation (rejection of disturbance V ) if Rc contains α, say, Rc = αR1:

1.

Internal Model Principle
v r − e Gc (z) = Sc Rc u G(z) = B A y

• α(z) = least common multiple of the unstable poles of Rc(z) and of V (z), all polynomials in z −1 • Let there be no common factors between α(z) and B(z) • Can find a controller Gc(z) for servo/tracking (following Rc) and regulation (rejection of disturbance V ) if Rc contains α, say, Rc = αR1: v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

Digital Control

1

Kannan M. Moudgalya, Autumn 2007

2.

Internal Model Principle - Regulation

2.

Internal Model Principle - Regulation
v(n) r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

2.

Internal Model Principle - Regulation
v(n) r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

Sc 1 B Y (z) = R1 α A R Sc 1 B 1+ R1 α A

2.

Internal Model Principle - Regulation
v(n) r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

1 R1 α A Y (z) = R+ V Sc 1 B Sc 1 B 1+ 1+ R1 α A R1 α A

Sc 1 B

2.

Internal Model Principle - Regulation
v(n) r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

1 R1 α A Y (z) = R+ V Sc 1 B Sc 1 B 1+ 1+ R1 α A R1 α A Sc B = R R1Aα + ScB

Sc 1 B

2.

Internal Model Principle - Regulation
v(n) r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

1 R1 α A Y (z) = R+ V Sc 1 B Sc 1 B 1+ 1+ R1 α A R1 α A Sc B R1Aα bV = R+ R1Aα + ScB R1Aα + ScB αaV

Sc 1 B

2.

Internal Model Principle - Regulation
v(n) r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

1 R1 α A Y (z) = R+ V Sc 1 B Sc 1 B 1+ 1+ R1 α A R1 α A Sc B R1Aα bV = R+ R1Aα + ScB R1Aα + ScB αaV • Unstable pole present in α gets cancelled

Sc 1 B

2.

Internal Model Principle - Regulation
v(n) r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

1 R1 α A Y (z) = R+ V Sc 1 B Sc 1 B 1+ 1+ R1 α A R1 α A Sc B R1Aα bV = R+ R1Aα + ScB R1Aα + ScB αaV • Unstable pole present in α gets cancelled • Regulation problem verified
Digital Control

Sc 1 B

2

Kannan M. Moudgalya, Autumn 2007

3.

Internal Model Principle - Servo

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

Y (z) =

Sc B R1Aα + ScB

R+

R1Aα R1Aα + ScB

V

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) =

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z)

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B = 1− R1Aα + ScB R

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B = 1− R1Aα + ScB R= R1Aα bR

Sc B

R1Aα

R1Aα + ScB αaR

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B R1Aα bR R= = 1− R1Aα + ScB R1Aα + ScB αaR • Unstable poles of Rc are cancelled by zeros of α.

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B R1Aα bR R= = 1− R1Aα + ScB R1Aα + ScB αaR • Unstable poles of Rc are cancelled by zeros of α. • Can choose Rc and Sc such that R1Aα + ScB has roots within the unit circle

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B R1Aα bR R= = 1− R1Aα + ScB R1Aα + ScB αaR • Unstable poles of Rc are cancelled by zeros of α. • Can choose Rc and Sc such that R1Aα + ScB has roots within the unit circle (pole placement)

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B R1Aα bR R= = 1− R1Aα + ScB R1Aα + ScB αaR • Unstable poles of Rc are cancelled by zeros of α. • Can choose Rc and Sc such that R1Aα + ScB has roots within the unit circle (pole placement) • IM Principle:

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B R1Aα bR R= = 1− R1Aα + ScB R1Aα + ScB αaR • Unstable poles of Rc are cancelled by zeros of α. • Can choose Rc and Sc such that R1Aα + ScB has roots within the unit circle (pole placement) • IM Principle: unstable poles of V , Rc appear in loop

Sc B

R1Aα

3.

Internal Model Principle - Servo v(n)
r(n) − e(n) Gc = Sc αR1 u(n) G= B A y(n)

R+ V R1Aα + ScB R1Aα + ScB Servo problem: assume V = 0: Y (z) = E(z) = R(z) − Y (z) Sc B R1Aα bR R= = 1− R1Aα + ScB R1Aα + ScB αaR • Unstable poles of Rc are cancelled by zeros of α. • Can choose Rc and Sc such that R1Aα + ScB has roots within the unit circle (pole placement) • IM Principle: unstable poles of V , Rc appear in loop thro’ α
3
Kannan M. Moudgalya, Autumn 2007

Sc B

R1Aα

Digital Control

4.

Internal Stability

4.

Internal Stability
r(n) + e(n) gc(n) u(n) g(n) y(n)

−

4.

Internal Stability
r(n) + e(n) gc(n) u(n) g(n) y(n)

−

• Notion in UG classes: output has to be stable

4.

Internal Stability
r(n) + e(n) gc(n) u(n) g(n) y(n)

−

• Notion in UG classes: output has to be stable • Output being stable is not sufficient

4.

Internal Stability
r(n) + e(n) gc(n) u(n) g(n) y(n)

−

• Notion in UG classes: output has to be stable • Output being stable is not sufficient • Every signal in the loop should be bounded

4.

Internal Stability
r(n) + e(n) gc(n) u(n) g(n) y(n)

−

• Notion in UG classes: output has to be stable • Output being stable is not sufficient • Every signal in the loop should be bounded • If any signal is unbounded, will result in saturation / overflow / explosion

4.

Internal Stability
r(n) + e(n) gc(n) u(n) g(n) y(n)

−

• Notion in UG classes: output has to be stable • Output being stable is not sufficient • Every signal in the loop should be bounded • If any signal is unbounded, will result in saturation / overflow / explosion • When every signal is bounded, called internal stability

4.

Internal Stability
r(n) + e(n) gc(n) u(n) g(n) y(n)

−

• Notion in UG classes: output has to be stable • Output being stable is not sufficient • Every signal in the loop should be bounded • If any signal is unbounded, will result in saturation / overflow / explosion • When every signal is bounded, called internal stability • If output is stable and if there is no unstable pole-zero cancellation, internal stability
Digital Control

4

Kannan M. Moudgalya, Autumn 2007

5.

Unstb. Pole-Zero Cancel. = No Int. Stability

5.

Unstb. Pole-Zero Cancel. = No Int. Stability
r2 r1 + − e1 n2 Gc = d2 + + e2 G= n1 d1 y

5.

Unstb. Pole-Zero Cancel. = No Int. Stability
r2 r1 + − e1 n2 Gc = d2 + + e2 G= n1 d1 y



 1 + GG E1 = G c  c E2 1 + Gc G

1

−

1 + GGc  R1   R2 1 1 + Gc G

G



5.

Unstb. Pole-Zero Cancel. = No Int. Stability
r2 r1 + − e1 n2 Gc = d2 + + e2 G= n1 d1 y



 1 + GG E1 1 + GGc  R1   c = G  R2 1 c E2  1 + Gc G 1 + Gc G  n1d2 d 1 d2  n n + d d − n n + d d  R1 2 1 2 1 2 = 1 n d 1 2   R2 d 1 d2 2 1 n1n2 + d1d2 n1n2 + d1d2

1

−

G



5.

Unstb. Pole-Zero Cancel. = No Int. Stability
r2 r1 + − e1 n2 Gc = d2 + + e2 G= n1 d1 y



 1 + GG E1 1 + GGc  R1   c = G  R2 1 c E2  1 + Gc G 1 + Gc G  n1d2 d 1 d2  n n + d d − n n + d d  R1 2 1 2 1 2 = 1 n d 1 2   R2 d 1 d2 2 1 n1n2 + d1d2 n1n2 + d1d2 Suppose d1, n2 have a common factor:

1

−

G



5.

Unstb. Pole-Zero Cancel. = No Int. Stability
r2 r1 + − e1 n2 Gc = d2 + + e2 G= n1 d1 y



 1 + GG E1 1 + GGc  R1   c = G  R2 1 c E2  1 + Gc G 1 + Gc G  n1d2 d 1 d2  n n + d d − n n + d d  R1 2 1 2 1 2 = 1 n d 1 2   R2 d 1 d2 2 1 n1n2 + d1d2 n1n2 + d1d2 Suppose d1, n2 have a common factor: d1 = (z + a)d1 n2 = (z + a)n2
Digital Control

1

−

G



5

Kannan M. Moudgalya, Autumn 2007

6.

Unstable Pole-Zero Cancellation = No Internal Stability

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1.

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE = 1 1 + GGc

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE = 1 1 + GGc = d1 d2 d1d2 + n1n2

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE = 1 1 + GGc = d1 d2 d1d2 + n1n2

T.F. between R2 and Y can be shown to be unstable.

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE = 1 1 + GGc = d1 d2 d1d2 + n1n2

T.F. between R2 and Y can be shown to be unstable. Let R1 = 0.

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE = 1 1 + GGc = d1 d2 d1d2 + n1n2

T.F. between R2 and Y can be shown to be unstable. Let R1 = 0. Y R2 = G 1 + GGc = n1d2 (d1d2 + n1n2)(z + a)

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE = 1 1 + GGc = d1 d2 d1d2 + n1n2

T.F. between R2 and Y can be shown to be unstable. Let R1 = 0. Y R2 = G 1 + GGc = n1d2 (d1d2 + n1n2)(z + a)

It is unstable

Unstable Pole-Zero Cancellation = No Internal Stability Assume the cancellation of z + a, n1(z) (z + a)n2(z) G(z) = , Gc(z) = (z + a)d1(z) d2(z)
6.

with |a| > 1. Assume stability of TE = 1 1 + GGc = d1 d2 d1d2 + n1n2

T.F. between R2 and Y can be shown to be unstable. Let R1 = 0. Y R2
Digital Control

=

G 1 + GGc

=

n1d2 (d1d2 + n1n2)(z + a)
6
Kannan M. Moudgalya, Autumn 2007

It is unstable and a bounded signal injected at R2 will produce an unbounded signal at Y

7.

Forbid Unstable Pole-Zero Cancellation: Loop Variable

7.

Forbid Unstable Pole-Zero Cancellation: Loop Variable

• Internal stability = all variables in loop are bounded for bounded external inputs at all locations

7.

Forbid Unstable Pole-Zero Cancellation: Loop Variable

• Internal stability = all variables in loop are bounded for bounded external inputs at all locations • Can be checked by the following closed loop diagram

7.

Forbid Unstable Pole-Zero Cancellation: Loop Variable

• Internal stability = all variables in loop are bounded for bounded external inputs at all locations • Can be checked by the following closed loop diagram
r2 r1 + − e1 + n2 Gc = d2 + e2 n1 d1 y

G=

7.

Forbid Unstable Pole-Zero Cancellation: Loop Variable

• Internal stability = all variables in loop are bounded for bounded external inputs at all locations • Can be checked by the following closed loop diagram
r2 r1 + − e1 + n2 Gc = d2 + e2 n1 d1 y

G=

• Can show that Output is stable + no pole-zero cancellation = internal stability
Digital Control

7

Kannan M. Moudgalya, Autumn 2007

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay:

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z B(z −1) −k A(z −1) =z b0 + b1z −1 + b2z −2 + · · · −k 1 + a1z −1 + a2z −2 + · · ·

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z B(z −1) −k A(z −1) =z b0 + b1z −1 + b2z −2 + · · · −k 1 + a1z −1 + a2z −2 + · · ·

• Controller not realizable ⇒

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z 1 + a1z −1 + a2z −2 + · · · • Controller not realizable ⇒ there is a common factor z −1 between plant and controller B(z −1) −k A(z −1) =z b0 + b1z −1 + b2z −2 + · · · −k

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z 1 + a1z −1 + a2z −2 + · · · • Controller not realizable ⇒ there is a common factor z −1 between plant and controller • z −1 = 0 ⇒ B(z −1) −k A(z −1) =z b0 + b1z −1 + b2z −2 + · · · −k

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z 1 + a1z −1 + a2z −2 + · · · • Controller not realizable ⇒ there is a common factor z −1 between plant and controller • z −1 = 0 ⇒ z = ∞, an unstable pole B(z −1) −k A(z −1) =z b0 + b1z −1 + b2z −2 + · · · −k

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z 1 + a1z −1 + a2z −2 + · · · • Controller not realizable ⇒ there is a common factor z −1 between plant and controller • z −1 = 0 ⇒ z = ∞, an unstable pole • If unstable pole-zero cancellation is forbidden while designing controllers, B(z −1) −k A(z −1) =z b0 + b1z −1 + b2z −2 + · · · −k

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z 1 + a1z −1 + a2z −2 + · · · • Controller not realizable ⇒ there is a common factor z −1 between plant and controller • z −1 = 0 ⇒ z = ∞, an unstable pole • If unstable pole-zero cancellation is forbidden while designing controllers, z −1 cannot appear in the denominator of the controller B(z −1) −k A(z −1) =z b0 + b1z −1 + b2z −2 + · · · −k

8.

Forbid Unstable Pole-Zero Cancellation ⇒ Get Causality

Not possible to realize this controller: 1 + z −1 Gc = z −1 All sampled systems have at least one delay: G(z −1) = z 1 + a1z −1 + a2z −2 + · · · • Controller not realizable ⇒ there is a common factor z −1 between plant and controller • z −1 = 0 ⇒ z = ∞, an unstable pole • If unstable pole-zero cancellation is forbidden while designing controllers, z −1 cannot appear in the denominator of the controller - i.e., controller is realizable
Digital Control

B(z −1) −k A(z −1)

=z

b0 + b1z −1 + b2z −2 + · · · −k

8

Kannan M. Moudgalya, Autumn 2007

9.

Delay Specification for Realizability

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay:

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z b0 = 0.
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

b0 = 0. Suppose that we use a feedback controller of the form Gc(z) = z −d Sc(z) Rc(z)

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

b0 = 0. Suppose that we use a feedback controller of the form Gc(z) = z
−d Sc (z)

Rc(z)

=z

−d s0

+ s1z −1 + · · ·

1 + r1z −1 + · · ·

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

b0 = 0. Suppose that we use a feedback controller of the form Gc(z) = z with s0 = 0
−d Sc (z)

Rc(z)

=z

−d s0

+ s1z −1 + · · ·

1 + r1z −1 + · · ·

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

b0 = 0. Suppose that we use a feedback controller of the form Gc(z) = z
−d Sc (z)

Rc(z)

=z

−d s0

+ s1z −1 + · · ·

1 + r1z −1 + · · ·

with s0 = 0 and d ≥ 0.

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

b0 = 0. Suppose that we use a feedback controller of the form Gc(z) = z
−d Sc (z)

Rc(z)

=z

−d s0

+ s1z −1 + · · ·

1 + r1z −1 + · · ·

with s0 = 0 and d ≥ 0. Closed loop transfer function:

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

b0 = 0. Suppose that we use a feedback controller of the form Gc(z) = z GGc 1 + GGc
−d Sc (z)

Rc(z)

=z

−d s0

+ s1z −1 + · · ·

1 + r1z −1 + · · ·

with s0 = 0 and d ≥ 0. Closed loop transfer function: T =

9.

Delay Specification for Realizability

Closed loop delay has to be ≥ open loop delay: G(z) = z
−k B(z)

A(z)

=z

b0 + b1z −1 + · · · −k 1 + a1z −1 + · · ·

b0 = 0. Suppose that we use a feedback controller of the form Gc(z) = z GGc 1 + GGc b0s0 + (b0s1 + b1s0)z −1 + · · ·
9
Kannan M. Moudgalya, Autumn 2007

−d Sc (z)

Rc(z)

=z

−d s0

+ s1z −1 + · · ·

1 + r1z −1 + · · ·

with s0 = 0 and d ≥ 0. Closed loop transfer function: T =

= z −k−d
Digital Control

1 + (s1 + r1)z −1 + · · · + z −k−d(b0s0 + · · · )

10.

Delay Specification for Realizability

10.

Delay Specification for Realizability

T = z −k−d

b0s0 + (b0s1 + b1s0)z −1 + · · · 1 + (s1 + r1)z −1 + · · · + z −k−d(b0s0 + · · · )

• Closed loop delay = k + d

10.

Delay Specification for Realizability

T = z −k−d

b0s0 + (b0s1 + b1s0)z −1 + · · · 1 + (s1 + r1)z −1 + · · · + z −k−d(b0s0 + · · · )

• Closed loop delay = k + d ≥ k

10.

Delay Specification for Realizability

T = z −k−d

b0s0 + (b0s1 + b1s0)z −1 + · · · 1 + (s1 + r1)z −1 + · · · + z −k−d(b0s0 + · · · )

• Closed loop delay = k + d ≥ k = open loop delay.

10.

Delay Specification for Realizability

T = z −k−d

b0s0 + (b0s1 + b1s0)z −1 + · · · 1 + (s1 + r1)z −1 + · · · + z −k−d(b0s0 + · · · )

• Closed loop delay = k + d ≥ k = open loop delay. • Can make it less only by d < 0,

10.

Delay Specification for Realizability

T = z −k−d

b0s0 + (b0s1 + b1s0)z −1 + · · · 1 + (s1 + r1)z −1 + · · · + z −k−d(b0s0 + · · · )

• Closed loop delay = k + d ≥ k = open loop delay. • Can make it less only by d < 0, but controller is unrealizable.

Digital Control

10

Kannan M. Moudgalya, Autumn 2007

11.

Example: What if Wrong Delay is Specified?

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay:

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function:

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function: T = GGc/(1+ GGc).

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function: T = GGc/(1+ GGc). Solving for Gc,

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function: T = GGc/(1+ GGc). Solving for Gc, and substituting for T , G

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function: T = GGc/(1+ GGc). Solving for Gc, and substituting for T , G Gc = 1 T G1 − T

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function: T = GGc/(1+ GGc). Solving for Gc, and substituting for T , G Gc = 1 T G1 − T = 1 1 − 0.5z −1 z −1 1 − (a + 1)z −1

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function: T = GGc/(1+ GGc). Solving for Gc, and substituting for T , G Gc = 1 T G1 − T = 1 1 − 0.5z −1 z −1 1 − (a + 1)z −1

This controller is unrealizable,

11.

Example: What if Wrong Delay is Specified?

Design a controller for the plant 1 −2 G=z 1 − 0.5z −1 so that the overall system has smaller delay: T =z
−1

1 1 − az −1

Recall the standard closed loop transfer function: T = GGc/(1+ GGc). Solving for Gc, and substituting for T , G Gc = 1 T G1 − T = 1 1 − 0.5z −1 z −1 1 − (a + 1)z −1
11
Kannan M. Moudgalya, Autumn 2007

This controller is unrealizable, no matter what a is.
Digital Control

12.

Specifications - Step Response

12.

Specifications - Step Response

Give unit step input in r. Output y should have

12.

Specifications - Step Response

Give unit step input in r. Output y should have 1. small rise time

12.

Specifications - Step Response

Give unit step input in r. Output y should have 1. small rise time 2. small overshoot

12.

Specifications - Step Response

Give unit step input in r. Output y should have 1. small rise time 2. small overshoot 3. small settling time

12.

Specifications - Step Response

Give unit step input in r. Output y should have 1. small rise time 2. small overshoot 3. small settling time 4. small steady state error

12.

Specifications - Step Response
y(n)

Give unit step input in r. Output y should have 1. small rise time 2. small overshoot 3. small settling time 4. small steady state error

n e(n)

n

12.

Specifications - Step Response
y(n)

Give unit step input in r. Output y should have 1. small rise time 2. small overshoot 3. small settling time 4. small steady state error

n e(n)

n

Error e(n) of the following form satisfies the requirements:

12.

Specifications - Step Response
y(n)

Give unit step input in r. Output y should have 1. small rise time 2. small overshoot 3. small settling time 4. small steady state error

n e(n)

n

Error e(n) of the following form satisfies the requirements: e(n) = ρn cos ωn, 0<ρ<1

12.

Specifications - Step Response
y(n)

Give unit step input in r. Output y should have 1. small rise time 2. small overshoot 3. small settling time 4. small steady state error

n e(n)

n

Error e(n) of the following form satisfies the requirements: e(n) = ρ cos ωn,
n

Im(z) × ω Re(z) ×

ρ

0<ρ<1

Digital Control

12

Kannan M. Moudgalya, Autumn 2007

13.

Specifications - Step Response

13.
y(n)

Specifications - Step Response

n e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one

n e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero

e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure:

e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure: User will specify the following:

e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure: User will specify the following: 1. a maximum allowable fall time < Nr

e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure: User will specify the following: 1. a maximum allowable fall time < Nr 2. a maximum allowable undershoot < ε

e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure: User will specify the following: 1. a maximum allowable fall time < Nr 2. a maximum allowable undershoot < ε 3. a minimum required decay ratio < δ

e(n)

n

Im(z) × ω Re(z) ×

ρ

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure: User will specify the following: 1. a maximum allowable fall time < Nr 2. a maximum allowable undershoot < ε 3. a minimum required decay ratio < δ

e(n)

n

Im(z) × ω Re(z) ×

ρ

• We will develop a method to determine ρ a ω satisfying the above requirements

e(n) = ρn cos ωn 0<ρ<1

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure: User will specify the following: 1. a maximum allowable fall time < Nr 2. a maximum allowable undershoot < ε 3. a minimum required decay ratio < δ

e(n)

n

Im(z) × ω Re(z) ×

ρ

• We will develop a method to determine ρ a ω satisfying the above requirements

e(n) = ρn cos ωn 0<ρ<1

• Calculate trans. fn. between e(n) - r(n

13.
y(n)

Specifications - Step Response 1. initial error is one 2. decaying oscillations about zero
n

3. steady state error is zero Procedure: User will specify the following: 1. a maximum allowable fall time < Nr 2. a maximum allowable undershoot < ε 3. a minimum required decay ratio < δ

e(n)

n

Im(z) × ω Re(z) ×

ρ

• We will develop a method to determine ρ a ω satisfying the above requirements • Back calculateKannan controllerAutumn 2007 the M. Moudgalya, Gc(z) 13

e(n) = ρn cos ωn 0<ρ<1
Digital Control

• Calculate trans. fn. between e(n) - r(n

14.

Small Fall Time in Error

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e.,

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e., e(n) = 0.

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e., e(n) = 0. for the first time when

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e., e(n) = 0. for the first time when π ωn = 2

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e., e(n) = 0. for the first time when π ωn = 2 π ⇒n= 2ω

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e., e(n) = 0. for the first time when π ωn = 2 π ⇒n= 2ω As want n < Nr , some given value, we get

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e., e(n) = 0. for the first time when π ωn = 2 π ⇒n= 2ω As want n < Nr , some given value, we get π < Nr 2ω

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1

Error becomes zero, i.e., e(n) = 0. for the first time when π ωn = 2 π ⇒n= 2ω As want n < Nr , some given value, we get π π < Nr ⇒ ω > 2ω 2Nr

14.

Small Fall Time in Error e(n) = ρn cos ωn, 0<ρ<1
y(n)

Error becomes zero, i.e., e(n) = 0. for the first time when π ωn = 2 π ⇒n= 2ω As want n < Nr , some given value, we get π π < Nr ⇒ ω > 2ω 2Nr
Digital Control
n e(n)

n

Desired region:
Im(z) Re(z)

14

Kannan M. Moudgalya, Autumn 2007

15.

Small Undershoot

15.

Small Undershoot e(n) = ρn cos ωn

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.?

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx.

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π e(n)|ωn=π = ρn cos ωn|ωn=π

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π e(n)|ωn=π = ρn cos ωn|ωn=π = −ρn|ωn=π

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π e(n)|ωn=π = ρn cos ωn|ωn=π = −ρn|ωn=π = −ρπ/ω

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π e(n)|ωn=π = ρn cos ωn|ωn=π = −ρn|ωn=π = −ρπ/ω User specified maximum deviation = ε:

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π e(n)|ωn=π = ρn cos ωn|ωn=π = −ρn|ωn=π = −ρπ/ω User specified maximum deviation = ε: ρπ/ω < ε

15.

Small Undershoot e(n) = ρn cos ωn

When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π e(n)|ωn=π = ρn cos ωn|ωn=π = −ρn|ωn=π = −ρπ/ω User specified maximum deviation = ε: ρπ/ω < ε, ρ < εω/π

15.

Small Undershoot
y(n)

e(n) = ρn cos ωn When does it reach first min.? de/dn = 0 • Not applicable, because n is an integer • Look for a simpler expression • Reaches min. approx. when ωn = π e(n)|ωn=π = ρ cos ωn|ωn=π = −ρn|ωn=π = −ρπ/ω User specified maximum deviation = ε: ρπ/ω < ε,
Digital Control
n e(n) n

n

Desired region:
Im(z) Re(z)

ρ < εω/π
15
Kannan M. Moudgalya, Autumn 2007

16.

Small Decay Ratio

16.

Small Decay Ratio e(n) = ρn cos ωn

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn • First overshoot occurs at ωn 2π π

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ:

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π <δ e(n)|ωn=π

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay.

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay. δ = 0.25 1/8 decay.

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay. δ = 0.25 1/8 decay. ρ2π/ω <δ π/ω ρ

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay. δ = 0.25 1/8 decay. ρ2π/ω < δ ⇒ ρπ/ω < δ ρπ/ω

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay. δ = 0.25 1/8 decay. ρ2π/ω < δ ⇒ ρπ/ω < δ ⇒ ρ < δ ω/π ρπ/ω

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay. δ = 0.25 1/8 decay. ρ2π/ω < δ ⇒ ρπ/ω < δ ⇒ ρ < δ ω/π ρπ/ω • Small undershoot: ρ < εω/π .

16.

Small Decay Ratio

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small • First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay. δ = 0.25 1/8 decay. ρ2π/ω < δ ⇒ ρπ/ω < δ ⇒ ρ < δ ω/π ρπ/ω • Small undershoot: ρ < εω/π . Usually ε < δ

16.

Small Decay Ratio
y(n)

e(n) = ρn cos ωn Ratio of two successive peak/trough to be small

• First undershoot in e(n) occurs at ωn π • First overshoot occurs at ωn 2π Want this ratio to be less than user specified δ: e(n)|ωn=2π ρn|ωn=2π <δ⇒ n <δ e(n)|ωn=π ρ |ωn=π δ = 0.5 1/4 decay. δ = 0.25 1/8 decay. Desired region: Im(z) ρ2π/ω < δ ⇒ ρπ/ω < δ ⇒ ρ < δ ω/π Re(z) ρπ/ω
e(n)

n

n

• Small undershoot: ρ < εω/π . Usually ε < δ • Small undershoot satisfies fast decay
Digital Control

16

Kannan M. Moudgalya, Autumn 2007

17.

Overall Requirements

17.

Overall Requirements
Im(z)

Re(z)

17.

Overall Requirements
Im(z)
Im(z)

Re(z)

Re(z)

17.

Overall Requirements
Im(z)
Im(z)
Im(z)

Re(z)

Re(z)

Re(z)

Desired region by the current approach

17.

Overall Requirements
Im(z)
Im(z)
Im(z)

Re(z)

Re(z)

Re(z)

Desired region by the current approach
Im(z)

Re(z)

Obtained by discretization of continuous domain result (Astrom and Wittenmark)
Digital Control

17

Kannan M. Moudgalya, Autumn 2007

18.

Desired Transfer Function

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform,

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2

For step input R(z).

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2

For step input R(z). Transfer function between R(z)-E(z):

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2 E(z) R(z)

For step input R(z). Transfer function between R(z)-E(z): TE (z) =

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2 E(z) R(z) z(z − r cos ω) z 2 − 2zr cos ω + r 2 z−1 z

For step input R(z). Transfer function between R(z)-E(z): TE (z) = =

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2 E(z) z(z − r cos ω) z−1 z

For step input R(z). Transfer function between R(z)-E(z): TE (z) = = R(z) z 2 − 2zr cos ω + r 2 (z − 1)(z − r cos ω) z 2 − 2zr cos ω + r 2 =

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2 E(z) z(z − r cos ω) z−1 z

For step input R(z). Transfer function between R(z)-E(z): TE (z) = = R(z) z 2 − 2zr cos ω + r 2 (z − 1)(z − r cos ω) =

z 2 − 2zr cos ω + r 2 One approach: equate this to 1/(1 + GGc)

18.

Desired Transfer Function

Desired error to a step input is e: e(n) = r n cos ωn Taking Z-transform, E(z) = z(z − r cos ω) z 2 − 2zr cos ω + r 2 E(z) z(z − r cos ω) z−1 z

For step input R(z). Transfer function between R(z)-E(z): TE (z) = = R(z) z 2 − 2zr cos ω + r 2 (z − 1)(z − r cos ω) =

z 2 − 2zr cos ω + r 2 One approach: equate this to 1/(1 + GGc) and calculate Gc
Digital Control

18

Kannan M. Moudgalya, Autumn 2007

19.

Desired Transfer Function

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z):

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z)

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2 −1 ) −1 ψ(z =z φcl (z −1)

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2 −1 ) −1 ψ(z =z φcl (z −1) • One approach: equate this to GGc/(1 + GGc),

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2 −1 ) −1 ψ(z =z φcl (z −1) • One approach: equate this to GGc/(1 + GGc), find Gc

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2 −1 ) −1 ψ(z =z φcl (z −1) • One approach: equate this to GGc/(1 + GGc), find Gc • Can’t do it exactly - will use part of this approach

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2 −1 ) −1 ψ(z =z φcl (z −1) • One approach: equate this to GGc/(1 + GGc), find Gc • Can’t do it exactly - will use part of this approach • We will mainly make use of φcl and not ψ

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2 −1 ) −1 ψ(z =z φcl (z −1) • One approach: equate this to GGc/(1 + GGc), find Gc • Can’t do it exactly - will use part of this approach • We will mainly make use of φcl and not ψ • TY (1) = 1

19.

Desired Transfer Function

Transfer function between setpoint R(z) and actual output Y (z): TY (z) = 1 − TE (z) z 2 − zr cos ω − z + r cos ω =1− z 2 − 2zr cos ω + r 2 z(1 − r cos ω) + (r 2 − r cos ω) = z 2 − 2zr cos ω + r 2 −1 ) −1 ψ(z =z φcl (z −1) • One approach: equate this to GGc/(1 + GGc), find Gc • Can’t do it exactly - will use part of this approach • We will mainly make use of φcl and not ψ • TY (1) = 1 ⇒ No steady state offset between Y and R
Digital Control

19

Kannan M. Moudgalya, Autumn 2007


				
DOCUMENT INFO
Shared By:
Stats:
views:56
posted:1/19/2010
language:English
pages:194
Description: 1 Internal Model Principle