segunda práctica calificadacon2

8/20/2019 Segunda práctica calificadacon2

1/11

Segunda práctica califcada

Ingeniería de control 2

G (s ) H (s )= (s +1 )2

s3 s= jw

G ( jw ) H ( jw )= ( jw+1 )2

( jw)3 = ( jw+1)

2

( jw)3 =( (1+ jw ) (1+ jw ) j w . jw . jw )

|G ( jw) H ( jw)|=|(1 + jw ) (1 + jw ) j w . jw . jw |= ¿

90

√ w2 +1 angle (arctan (w )√ w 2+1 .angle (arctan (w ) ))√ w2 angle (¿√ w2 angle (90 ) )√ w

2angle (90 )

¿

|(1+w2)w3 |angle (2arctan (w)− 270 )

En el cruce por 0dB modulo=1:

|(1+w2)

w3 |= 1→w 1= 1.4656 rad /seg MF = 180 +(2 arctan (w )− 270 )


2/11

MF = 180 +(2 arctan (1.4656 )− 270 )

MF = 21.38

El margen de ganancia cuando MF=0:180 +(2 arctan (w )− 270 )= 0

arctan (w)= 45 → w = 1

Entonces:

|(1+w2)

w3

|= (1+1

2)

13 = 2

En decibelios:

MG= 20log (2)=− 6.0203 dB


3/11

AB=( 0 4 0−3 0 20 0 − 2)∗(002)=( 04− 4) y= (1 0 0 )(

x 1

x 2 x 3)→ y = x

1

Si es controlable entonces:

PC = (B AB)=(0 00 42 − 4)→ esde rango = 2 implicaque pPC ≠ pA

No es controlable

Po =( ACA)=(1 0 00 4 0 )→es derango = 2Es observable


4/11


5/11

!rimera apro#$

|G 1( p)|∗|G 1(c )|= 1 en&onces|G 1 ( p )| ' 1

| 200

s (s+10 )|= 1

| 200 jw ( jw+10 )|= 20 0w∗angle (90 )∗√ x2+100∗angle (arctan ( w10

))

200

w∗√ w2

+100angle (− 90 − arctan ( w

10))

200

w∗√ w2+100= 1 →w= 12,4962

MF = 180 +(− arctan ( w10)− 90 )

MF = 180 +(− arctan (12.496210 )− 90 )

MF = 38.668

!or condici%n del problema MF( 75 en&onces )al&a paracumplir lacondicion

75 − 38.668 = 36.332

El sobrante debe proporcionarlo el G 1 (c )

Entonces su MF = 36.332 +5 = *


6/11

a = 1− sen*1+sen* = 0.2045

|G 1 (c )|= 1√ a

= 2.2113

|G 1 ( p)|∗|G 1 (c )|=|G 1 ( p)|∗2.2113 = 1

|G 1 ( p)|= 0.4522 =| 200s (s +10 )|→ 200w∗√ w2 +100 = 0.4522 → w= 15.8621

√ a" = 15.862 →" = 0.1394

+ = 1"

= 7.174 y p= 1a"

= 35.08 y = a∗ c → c = 978

!or lo tanto el controlador toma la &orme de :

G (c )= c (s + !)(s+ p) =

978 (s+7.174 )(s+35.08 )

G ( p )= 1s (s+1)(0.5 s+1)

y G (c )= c (s + ! )(s+ p )

Donde p=1"

! = 1a"

para 0


7/11

G (c )= c (a"s +1 )

a ("s +1 ) donde = c

a

G (s )= G ( p )G (c )= 1

s (s+1)(0.5 s+1)

(a "s +1)

("s +1)

Ordenamos tal ue:

s (s +1)(0.5 s +1)

∗(a"s +1)

("s +1) = G 1( p )∗G 1 (c )

G 1 ( p)= s (s +1 )(0.5 s +1 )

# donde $= lims → 0

sG (s)

$ = lims→ 0

sG (s)= lims → 0

s 1

s (s +1 )(0.5 s +1 ) (a"s +1 )

("s +1 ) = = 5 %!→ = 5

G 1 ( p)= 5s (s +1)(0.5 s +1)

!ara el c"lculo de MF:

|G (s )|=|G ( p )G (c)|=|G 1( p )|∗|G 1(c)|= 1

!rimera apro#$

|G 1 ( p)|∗|G 1 (c )|= 1 en&onces|G 1 ( p )| ' 1

| 5s (s+1 )(0.5 s+1 )|= 1

| 5 jw ( jw+1 )(0.5 jw +1 )|= 5w∗angle (90 )∗√ w 2 +1∗angle (arctan w )∗√ 0.25 w 2+1. angle (arctan ( w2 ))


8/11

5

w∗√ w2 +1∗√ 0.25 w 2 +1.angle (− 90 − arctan (w )− arctan ( w

2 ))

5

w∗√ w2

+1∗√ 0.25 w2

+1.= 1 →w = 1.8022

MF = 180 +(− arctan (w )− 90 − arctan ( w2 ))

MF = 180 +(− arctan (1.8022 )− 90 − arctan ( 1.80222

))

MF =− 13

!or condici%n del problema MF( 40 en&onces )al&a paracumplir la condicion

40 − 13 = 27

El sobrante debe proporcionarlo el G 1 (c )

Entonces su MF = 27 +5= *

a =1 − sen*1 +sen* =

0 .3073

|G 1(c )|= 1

√ a= 1.8039

|G 1 ( p)|∗|G 1 (c )|=|G 1 ( p)|∗1.8039 = 1

|G 1 ( p)|= 0.5544 =| 5s (s +1 ) (0.5 s+1 )|→ 5w∗√ w 2 +1∗√ 0.25 w2 +1.


9/11

→ w= 2.3245

1

√ a " = 2.3245 →" = 0.776

p= 1"

= 1.289 y ! = 1a"

= 4.195 y = ca

→ c = 0.5365

!or lo tanto el controlador toma la &orme de :

G (c )= c (s + !)(s+ p) =

0.5365 (s+4.195 )(s+1.289 )

x 1= y→ x 1, = y , = ox 1+ x2+0 x 3+u

x 2= y , → x 2 , = y ,, = ox 1+0 x2+ x 3+u

x 3= y ,, → x 3, = y , ,, =− 2 x− 3 x2− 5 x 3+u


10/11

x , =( 0 1 00 0 1− 2 − 3 −5) x+(001)u y= (

1 0 0

) x

Conx =( x 1 x 2 x 3

)

- =( - 1 - 2 - 3 ) u=− -x

x , = Ax +Bu x , = Ax− B-x

!ero :

A− B- =( 0 1 00 0 1− 2 − 3 − 5)−(0

0

1)(- 1 - 2 - 3 )

A =( 0 1 00 0 1−2− - 1 − 3− - 1 −5− - 3) /e& (01 − A )= 0

01 − A =(s 0 00 s 00 0 s)−( 0 1 00 0 1− 2− - 1 − 3− - 1 − 5− - 3) A (s )= /e& (01 − A )= s 3 +(5 +- 3 ) s2 +(3 +- 2 )s +(2 +- 1 )= 0 ''''''$(1)

!or la condici%n del problema:


11/11

"s = 1

2(n)=

1

0.8 (6 )= 1 →

s3 +(14.4 )s 2 +(82.15 )s +(172.8 )= 0 ''''''''''$(*)

+gualando (1), (*) tenemos:

-./ =1 $ / =2$

./*=3*$1 /*=42$1

*./1=14*$3 /1=140$3

- =(

- 1 - 2 - 3) = (

170.8 79.1 9.4

)

segunda práctica calificadacon2

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