solución tarea n°1 - udec · ©udec - die - 2019 0 2 4 6 8 0 2 4 eigenvalsa(...

© UdeC - DIE - 2019Solución Tarea N°1

if

Lt

Rt

ie = d·if

ef = d·e +

-

+

- es

tc

v

Problema Estudiar el control de un péndulo

Parámetros

es 50:= Lt 500 103−

⋅:= Rt 5:=

kt 2:= tt 0.05:= τt 0.5:=

Parte A F de T

tc s( )

vRt s( )

tc s( )

Rt if s( )⋅= kt

1

τt s⋅ 1+⋅ e

tt− s⋅⋅= d es⋅ Lt

dif

dt⋅ Rt if⋅+=

tc s( ) kt1

τt s⋅ 1+⋅ e

tt− s⋅⋅ Rt⋅ if s( )⋅= d es⋅ Lt s⋅ if⋅ Rt if⋅+=

if s( )

d s( )

es

Lt s⋅ Rt+= if s( )

es

Lt s⋅ Rt+d s( )⋅=

tc s( ) kt1

τt s⋅ 1+⋅ e

tt− s⋅⋅ Rt⋅

es

Lt s⋅ Rt+⋅ d s( )⋅= kt es⋅

1

τt s⋅ 1+⋅

1

Lt

Rt

s⋅ 1+

⋅ ett− s⋅

⋅ d s( )⋅=

Modelo Promedio.

τt

dtc

dt⋅ tc+ kt Rt⋅ if⋅= d es⋅ Lt

dif

dt⋅ Rt if⋅+=

dif

dt

Rt−

Lt

if⋅es

Lt

d⋅+=dtc

dt

kt Rt⋅

τt

if⋅1−

τt

tc⋅+= y tc= u d=

Al

Rt−

Lt

kt Rt⋅

τt

0

1−

τt

:= bl

es

Lt

0

:= cl 0 1( ):= dl 0:=

Parte B Punto de operación d0 0.0:= if0 0:= tcd0 60:=

Given 0Rt−

Lt

if0⋅es

Lt

d0⋅+= 0kt Rt⋅

τt

if0⋅1−

τt

tcd0⋅+= Sol Find d0 if0, ( )0.6

6

=:=

Tarea N°1 - Solución 1 de 15 Sistemas de Control - 543 244

© UdeC - DIE - 2019d0 Sol

1:= d0 0.6= if0 Sol

2:= if0 6=

Ganancia DC: kt es⋅ 100= cl Al−( ) 1−⋅ bl⋅ 100= Φ t( ) if t 0≥ 1, 0, ( ):=

Parte C Simulación.

tf 8:= nf 3000:= m 0 nf..:= tr1 2:= tr2 15:= d t( ) d0 Φ t tr1−( )⋅ d0 0.5⋅ Φ t tr2−( )⋅−:=

t 0 tf nf1−

⋅, tf..:=

D t x, ( ) Al

x1

x2

⋅ bl d t( )⋅+:=Zc rkfixed

0

0

0, tf, nf, D,

:=

0 2 4 6 80

0.2

0.4

0.6

0.8

d

0 2 4 6 80

2

4

6

if

0 2 4 6 80

20

40

60

tc

eigenvals Al( )2−

10−

=

Parte D Simulación Sistema en L.C. con hc(s) = kc. kc 0.2:= kst1

10:= ka 1:=

eigenvals Al bl ka⋅ kc⋅ kst⋅ cl⋅−( )6− 4.899i+

6− 4.899i−

=CI

0

0

:=

tcd t( ) tcd0 kst⋅ Φ t tr1−( )⋅ 0.5 tcd0⋅ kst⋅ Φ t tr2−( )⋅−:=


© UdeC - DIE - 2019

D t x, ( ) Al bl ka⋅ kc⋅ kst⋅ cl⋅−( )x

1

x2

⋅ bl kc⋅ ka⋅ tcd t( )⋅+:= Za3 rkfixed CI 0, tf, nf, D, ( ):=

w m( ) kc tcd mtf

nf

⋅

kst Za3m 3,

⋅−

⋅:= d m( ) ka w m( )⋅:= va0 ka kc⋅ θ1 θ0 kst⋅−( )⋅=

0 2 4 60

0.5

1

1.5

d

0 2 4 6 80

2

4

6

8

10

if

0 2 4 6 80

10

20

30

40

50

tc

0 2 4 6 80

2

4

6

tcd

0 2 4 6 80

0.5

1

1.5

w

Parte E Simulación Sistema en L.C. con hc(s) = kc/s. kc 0.2:=

Ac augment stack Al kc− kst⋅ cl⋅, ( ) stack bl ka⋅ 0, ( ), ( ):= bc stack bl 0⋅ kc, ( ):=


© UdeC - DIE - 2019

eigenvals Ac( )10.453−

0.774− 1.797i+

0.774− 1.797i−

=CI

0

0

0

:=

D t x, ( ) Ac

x1

x2

x3

⋅ bc tcd t( )⋅+:= Za3 rkfixed CI 0, tf, nf, D, ( ):=

w m( ) Za3m 4,

:= d m( ) ka w m( )⋅:=

0 2 4 60

0.2

0.4

0.6

0.8

1

d

0 2 4 6 80

2

4

6

8

10

if

0 2 4 6 80

20

40

60

80

tc

0 2 4 6 80

2

4

6

tcd

0 2 4 6 80

0.2

0.4

0.6

0.8

1

w

Given

Re eigenvals augment stack Al kc− kst⋅ cl⋅, ( ) stack bl ka⋅ 0, ( ), ( )( )2( ) 0=

( )Tarea N°1 - Solución 4 de 15 Sistemas de Control - 543 244

© UdeC - DIE - 2019kcc Find kc( ):= kcc 1.2=

Ac augment stack Al kcc− kst⋅ cl⋅, ( ) stack bl ka⋅ 0, ( ), ( ):= Tosc2 π⋅

Im eigenvals Ac( )2( ):= Tosc 1.405= Periodo de la

oscilación en s.

Frecuencia de la

oscilación en Hz.

Im eigenvals Ac( )2( )2 π⋅

0.712=

Simulación Sistema en L.C. con hc(s) = kc/s. kc kcc:=

Ac augment stack Al kc− kst⋅ cl⋅, ( ) stack bl ka⋅ 0, ( ), ( ):=

bc stack bl 0⋅ kc, ( ):=

eigenvals Ac( )12−

4.472i

4.472i−

=CI

0

0

0

:=

D t ∆x, ( ) Ac

∆x1

∆x2

∆x3

⋅ bc tcd t( )⋅+:= Za3 rkfixed CI 0, tf, nf, D, ( ):=

w m( ) Za3m 4,

:= d m( ) ka w m( )⋅:=

0 2 4 61−

0

1

2

3

d

0 2 4 6 810−

0

10

20

if

0 2 4 6 80

50

100

150

tc


© UdeC - DIE - 2019

0 2 4 6 80

2

4

6

tcd

0 2 4 6 81−

0

1

2

3

w

Parte F Ecuaciones de Estado Discretas Equivalentes. Tm 200 103−

⋅:= T eigenvecs Al( ) 1−:=

Φc t( ) T1−

exp eigenvals Al( )1 t⋅( )0

0

exp eigenvals Al( )2 t⋅( )

⋅ T⋅:= mf

tf

Tm

:= k 0 mf..:=

Ad Re Φc Tm( )( ):= Ad

0.135

1.337

0

0.67

= cd cl:=

bd Re0

Tm

τΦc Tm τ−( ) bl⋅( )1

⌠⌡

d

0

Tm

τΦc Tm τ−( ) bl⋅( )2

⌠⌡

d

:= bd

8.647

19.593

= xo

0

0

:=

Simulación de Ecuaciones de Estado Discretas Equivalentes y Comparación en L.A..

d t( ) d0 Φ t tr1−( )⋅ d0 0.5⋅ Φ t tr2−( )⋅−:=

w t( )d t( )

ka

:= wd j( ) w j Tm⋅( ):= kf

tf

Tm

:= k 0 kf..:= xo

0

0

:=


© UdeC - DIE - 2019

xd k( ) if k 0= xo, Adk

xo⋅

0

k 1−

j

Adk j− 1−

bd⋅ ka⋅ wd j( )⋅

∑

=

+,

:=

w t( ) Re wd trunct

Tm

:= d t( ) w t( ) ka⋅:=

D t ∆x, ( ) Al

∆x1

∆x2

⋅ bl ka⋅ w t( )⋅+:= CI0

0

:= Zp rkfixed CI 0, tf, nf, D, ( ):=

0 2 4 6 80

0.2

0.4

0.6

0.8

w

0 2 4 6 80

0.2

0.4

0.6

0.8

d

0 2 4 6 80

2

4

6

if

0 2 4 6 80

20

40

60

tc

eigenvals Ad( )0.67

0.135

=

Parte G Simulación de Ecuaciones de Estado Discretas Equivalentes y Comparación en L.C., hc(z) = kc.


© UdeC - DIE - 2019tcd t( ) tcd0 kst⋅ Φ t tr1−( )⋅ 0.5 tcd0⋅ kst⋅ Φ t tr2−( )⋅−:= tcdd j( ) tcd j Tm⋅( ):=

kf

tf

Tm

:= k 0 kf..:= kc 0.2:= xo

0

0

:=

xd k( ) if k 0= xo, Ad bd cd⋅ ka⋅ kc⋅ kst⋅−( )kxo⋅

0

k 1−

j

Ad bd cd⋅ ka⋅ kc⋅ kst⋅−( )k j− 1−bd⋅ kc⋅ ka⋅ tcdd j( )⋅

∑

=

+

...,

:=

wd k( ) kc tcdd k( ) kst cd⋅ xd k( )⋅−( )⋅:= w t( ) Re wd trunct

Tm

:= d t( ) ka w t( )⋅:=

D t ∆x, ( ) Al

∆x1

∆x2

⋅ bl d t( )⋅+:= CI0

0


0 2 4 6 80

0.5

1

1.5

w

0 2 4 6 80

0.5

1

1.5

d

0 2 4 6 80

5

10

15

if

0 2 4 6 80

10

20

30

40

50

tc

6

tcd


© UdeC - DIE - 2019

0 2 4 6 80

2

4

6

eigenvals Ad bd cd⋅ ka⋅ kc⋅ kst⋅−( )0.207 0.476i+

0.207 0.476i−

= eigenvals Ad bd cd⋅ ka⋅ kc⋅ kst⋅−( )2 0.519=

Given eigenvals Ad bd cd⋅ ka⋅ kc⋅ kst⋅−( )2 1= kcc Find kc( ):= kcc 1.02=

Parte H Simulación de Ecuaciones de Estado Discretas Equivalentes y Comparación en L.C., hc(z) = kc, incluyendo retardo. kc 0.2:=

Adr augment stack Ad cd 0⋅, ( ) stack bd 0, ( ), ( ):= bdr stack bd 0⋅ 1, ( ):= cdr augment cd 0, ( ):=

tcd t( ) tcd0 kst⋅ Φ t tr1−( )⋅ 0.5 tcd0⋅ kst⋅ Φ t tr2−( )⋅−:= tcdd j( ) tcd j Tm⋅( ):=

kf

tf

Tm

:= k 0 kf..:= xo

0

0

0

:=

xd k( ) if k 0= xo, Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )kxo⋅

0

k 1−

j

Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )k j− 1−bdr⋅ kc⋅ ka⋅ tcdd j( )⋅

∑

=

+

...,

:=

wd k( ) kc tcdd k 1−( ) kst cdr⋅ xd k 1−( )⋅−( )⋅:= w t( ) Re wd trunct

Tm

:= d t( ) ka w t( )⋅:=

D t x, ( ) Al

x1

x2

⋅ bl d t( )⋅+:= CI0

0



© UdeC - DIE - 2019

0 2 4 6 80.5−

0

0.5

1

1.5

w

0 2 4 6 80.5−

0

0.5

1

1.5

d

0 2 4 6 85−

0

5

10

15

if

0 2 4 6 80

20

40

60

80

tc

0 2 4 6 80

2

4

6

tcd

eigenvals Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )0.242−

0.524 0.68i+

0.524 0.68i−

= eigenvals Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )1 0.242=

Given eigenvals Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )2 1= kcc Find kc( ):= kcc 0.311=


© UdeC - DIE - 2019Simulación de Ecuaciones de Estado Discretas Equivalentes y Comparación en L.C., hc(z) = kcc, incluyendo retardo.

kc kcc:= Tosc

2 π⋅ Tm⋅

Im ln eigenvals Adr bdr cdr⋅ ka⋅ kcc⋅ kst⋅−( )2( )( ):= Tosc 1.258=

Tosc

Tm

6.291=

xd k( ) if k 0= xo, Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )kxo⋅

0

k 1−

j

Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )k j− 1−bdr⋅ kc⋅ ka⋅ tcdd j( )⋅

∑

=

+

...,

:=

wd k( ) kc tcdd k 1−( ) kst cdr⋅ xd k 1−( )⋅−( )⋅:= w t( ) Re wd trunct

Tm

:= d t( ) ka w t( )⋅:=

D t x, ( ) Al

x1

x2

⋅ bl d t( )⋅+:= CI0

0


0 2 4 6 82−

1−

0

1

2

3

w

0 2 4 6 82−

1−

0

1

2

3

d

0 2 4 6 810−

0

10

20

if

0 2 4 6 850−

0

50

100

tc

6

tcd


© UdeC - DIE - 2019

0 2 4 6 80

2

4

eigenvals Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )0.277−

0.541 0.841i+

0.541 0.841i−

= eigenvals Adr bdr cdr⋅ ka⋅ kc⋅ kst⋅−( )2 1=

Parte I Simulación de Ecuaciones de Estado Discretas Equivalentes y Comparación en L.C., hc(z) = kc/(z-1), incluyendo retardo.

Adr augment stack Ad cd 0⋅, ( ) stack bd 0, ( ), ( ):= bdr stack bd 0⋅ 1, ( ):= cdr augment cd 0, ( ):= Modelo con retardo kc 0.03:=

tcd t( ) tcd0 kst⋅ Φ t tr1−( )⋅ 0.5 tcd0⋅ kst⋅ Φ t tr2−( )⋅−:= tcdd j( ) tcd j Tm⋅( ):= kc 0.03=

AdI augment stack Adr cdr 0⋅, ( ) stack bdr 1, ( ), ( ):= bdI stack bdr 0⋅ 1, ( ):= cdI augment cdr 0, ( ):= Modelo con integrador

kf

tf

Tm

:= k 0 kf..:= xo

0

0

0

0

:=

xd k( ) if k 0= xo, AdI bdI cdI⋅ ka⋅ kc⋅ kst⋅−( )kxo⋅

0

k 1−

j

AdI bdI cdI⋅ ka⋅ kc⋅ kst⋅−( )k j− 1−bdI⋅ kc⋅ ka⋅ tcdd j( )⋅

∑

=

+

...,

:=

dd k( ) xd k( )3

:= wd k( )dd k( )

ka

:= d t( ) Re dd trunct

Tm

:= w t( )d t( )

ka

:=

D t x, ( ) Al

x1

x2

⋅ bl d t( )⋅+:= CI0

0



© UdeC - DIE - 2019

0 2 4 6 80

0.5

1

1.5

w

0 2 4 6 80

0.5

1

1.5

d

0 2 4 6 80

5

10

15

teta grados

0 2 4 6 80

20

40

60

80

100

omega

0 2 4 6 80

2

4

6

tcd

eigenvals AdI bdI cdI⋅ ka⋅ kc⋅ kst⋅−( )

0.915 0.272i+

0.915 0.272i−

0.012− 0.171i+

0.012− 0.171i−

= eigenvals AdI bdI cdI⋅ ka⋅ kc⋅ kst⋅−( )1 0.954=

Given eigenvals AdI bdI cdI⋅ ka⋅ kc⋅ kst⋅−( )1 1= kcc Find kc( ):= kcc 0.046=


© UdeC - DIE - 2019Parte J Simulación de Ecuaciones de Estado Discretas Equivalentes y Comparación en L.C., hc(z) = kcc/(z-1), incluyendo retardo.

eigenvals AdI bdI cdI⋅ ka⋅ kcc⋅ kst⋅−( )

0.941 0.338i+

0.941 0.338i−

0.038− 0.199i+

0.038− 0.199i−

= eigenvals AdI bdI cdI⋅ ka⋅ kcc⋅ kst⋅−( )1 1=

Tosc

2 π⋅ Tm⋅

Im ln eigenvals AdI bdI cdI⋅ ka⋅ kcc⋅ kst⋅−( )1( )( ):=

Tosc 3.648=Tosc

Tm

18.241=kc kcc:=

xd k( ) if k 0= xo, AdI bdI cdI⋅ ka⋅ kc⋅ kst⋅−( )kxo⋅

0

k 1−

j

AdI bdI cdI⋅ ka⋅ kc⋅ kst⋅−( )k j− 1−bdI⋅ kc⋅ ka⋅ tcdd j( )⋅

∑

=

+

...,

:=

dd k( ) xd k( )3

:= wd k( )dd k( )

ka

:= d t( ) Re dd trunct

Tm

:= w t( )d t( )

ka

:=

D t x, ( ) Al

x1

x2

⋅ bl d t( )⋅+:= CI0

0



© UdeC - DIE - 2019

0 2 4 6 80.5−

0

0.5

1

1.5

w

0 2 4 6 80.5−

0

0.5

1

1.5

d

0 2 4 6 85−

0

5

10

15

teta grados

0 2 4 6 850−

0

50

100

150

omega

0 2 4 6 80

2

4

6

tcd


solución tarea n°1 - udec · ©udec - die - 2019 0 2 4 6 8 0 2 4 eigenvalsa(...

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