TALLER MECANISMOS Y AUTOMATIZACION

BRAYAN ALEXIS MONTAÑEZ SIERRA CODIGO 1056799009

Docente:Leonel Alberto Gomez Perez

UNIVERSIDAD MANUELA BELTRANINGENIERIA INDUSTRIAL

MECANISMOS Y AUTOMATIZACIONBOGOTA D.C.

2014

TALLER 1 CALCULAR LA CONSTANTE DEL TIEMPO A UN SISTEMA DE PRIMER ORDEN ANTE UNA ENTRADA IMPULSO.

V (s )=1 G (s )= kτs+1

V ( t )=U 0(t)

G (s )=Y (s)V (s)

Y (s )=G(s)V (s )

Y (s )=[ kτs+1 ]∗1

Y (s )= kτs+1

Y (s )= k

s+1τ

L−1 [Y ( s ) ]= k

s+1τ

Y (t )=k e−1τ

∗t

k=1ta=1t= 0:10y=k*exp(-t/ta)figure (1)plot(y)

TALLER 2

F t=Fex−KX

ma=f ( t )−ky

mdvdt

= f ( t )−k∫ x

m Ldvdt

=L f ( t )−k L∫ x

msV (s )−v ( 0 )=F ( s )− ksV (s)

msV (s )+ ksV (s )=F (s )

V (s )(ms+ ks )=F ( s )

V ( s )F ( s)

= 1

ms+ks

G (s )= s

m s2+k

M K G(S)10 130 s

10 s2+13020 130 s

20 s2+13030 130 s

30 s2+13040 130 s

40 s2+13050 130 s

50 s2+130

TALLER 3

RCL

R C L G(s)2.2 3.1 2.7 1

8.37 s2+6.82 s+12.2 3.3 2.9 1

9.57 s2+7.26 s+12.2 3.6 3.2 1

11.52s2+7.92 s+12.5 4.5 3.5 1

15.75 s2+11.25 s+12.7 4.5 3.8 1

17.1 s2+12.15 s+12.9 4.5 4.1 1

18.45 s2+13.05 s+13.3 5.1 4.4 1

22.44 s2+16.83 s+13.7 5.7 4.4 1

25.08 s2+21.09 s+13.9 5.9 4.4 1

25.96 s2+23.01 s+1

TALLER 4

I 1R1+R2 ( I 1−I2 )+C2 ( I 1 )=V 1(s)

I 2 (C1 )+R2 ( I2−I 1 )=0

I 2 (C1+R2 )=I 1∗R2

I 2=I1∗R2

C1+R2

I 1 (R1+R2+C2 )−R2 I 2=V 1(s )

I 1 (R1+R2+C2 )−R2( I 1∗R2

C1+R2)=V 1(s)

I 1(R1+R2+C2−(R2)2

C1+R2)=V 1(s)

I 1=V 1(s)

(R1+R2+C2−(R2 )2

C1+R2)I 1=

V 1(s)1

( R1 (C1+R2 )+R2 (C1+R2 )+C2 (C1+R2)−(R2 )2

C1+R2)

I 1=V 1(s) (C1+R2 )

(R1 (C1+R2 )+R2 (C1+R2 )+C2 (C1+R2)−(R2 )2 )

V 2 (s )=I 1∗C2

V 2 (s )=( V 1(s) (C1+R2)(R1 (C1+R2 )+R2 (C1+R2 )+C2 (C1+R2)−(R2 )2 ) )∗C2

G (s )=V 2(s )V 1(s )

=ZC 2 [ZC 1+ZR2 ]

[ZC 1+R2 ] (ZR1+ZR2+ZC 2 )−Z (R2 )2

G (s )=V 2(s )V 1(s )

=ZC 2 [ZC 1+ZR2 ]

[ZC 1+R2 ] (ZR1+ZR2+ZC 2 )−Z (R2 )2

ZC ( s)= 1CS

ZR=R ZL=SL

G (s )=

1SC2 [ 1

SC1

+R2][ 1SC1

+R2](R1+R2+1

SC2)−(R2 )2

G (s )=

1S2C1C2

+R2

SC2

[ 1+R2SC1

SC1]( R1SC2+R2SC2+1

SC2)−(R2 )2

G (s )=

SC2+R2S2C1C2

S3C1 (C2 )2

[1+R2SC1 ] [R1SC2+R2SC2+1 ]S2C1C2

−(R2 )2

G (s )=

SC2+R2S2C1C2

S3C1 (C2 )2

[1+R2SC1 ] [R1SC2+R2SC2+1 ]−[ (R2 )2S2C1C2 ]S2C1C2

G (s )=(S2C1C2) (S C2+R2S

2C1C2 )

(S3C1 (C2 )2 )[ [1+R2SC1 ] [R1SC2+R2SC2+1 ]−[ (R2)2S2C1C2 ]

S2C1C2]

G (s )=(SC2+R2S

2C1C2 )

(SC2 ) [ [1+R2SC1 ] [R1S C2+R2SC2+1 ]−[ (R2 )2S2C1C2 ]S2C1C2

]TALLER 5

G (S )=H (S )Qi (S )

V=Ah ;Q= hR

Vd=Qi−Q0

Adhdt

=Qi−Q0

Adhdt

=Qi−hR

A Ldhdt

+ 1RLh=LQi

A (Sh (S )−h(0))+ 1Rh(s)=Qi(S)

ASh (S )+ 1Rh ( s )=Qi (S )

h (S )(AS+ 1R )=Qi (S )

h (S )( ARS+1R )=Qi (S )

G (S )=H (S )Qi (S )

= RARS+1

R=Cambio en la altura(h)[m ]

Cambio enel caudal(qi)[m3/ s]

G (S )=Q0 (S )Qi (S )

C=dhdt

=Cantidad de liquido acumulado ,dondeQ 0 (t ) esel caudal de salida .

RCdhdt

+h=RQi

CR=Ld Q0 (t )+LQ0 (t )=Qi ( t )

CRSQ0 (S )+Q 0 (S )=Qi (S )

1.G (S )=Q 0 (S )Q i (S )

= 1CRS+1

2.G (S )=H (S )Q i (S )

= RARS+1

R C m3 G(s)10% 1 1

0.10S+120% 1 1

0.20S+130% 1 1

0.30S+140% 1 1

0.40S+150% 1 1

0.50S+1

R Am G(S)10% 1 0.10

0.10S+120% 1 0.20

0.20S+130% 1 0.30

0.30S+140% 1 0.40

0.40S+150% 1 0.50

0.50S+1

TALLER 6

1

2

3

4

G1+GH

=

8 s+ks3+2 s2

1+ 8 s+ks3+2 s2

¿(s3+2 s2 )+(8 s+4k )

s3+2 s2

¿

8 s+ks3+2 s2

( s3+2 s2 )+(8 s+4 k )s3+2 s2

¿ 8 s+k(s3+2 s2 )+(8 s+4k )

s3+2 s2+8 s+4 k=0

a0+a1+a2+a3=0

S3 18

S2 2 4k

S1 16−4 k2

S0 4 k

16−4k2

>0

16−4 k>0

−4k>−16

−k>−164

k<4

0<k<4

%Estabilidadnum=[8 4]

den=[1 2 8 4]printsys(num,den)num1=roots(num)den1=roots(den)%mapa zeros y polospzmap(num,den)%polosc=pzmap(num,den)

G1+GH

=

ks+ks3+2 s2−8 s

1+ ks+ks3+2 s2−8 s

¿(s3+2 s2−8 s )+(ks+k )

s3+2 s2−8 s

¿

ks+ks3+2 s2−8 s

( s3+2 s2−8 s )+(ks+k)s3+2 s2−8 s

¿ ks+k(s3+2 s2−8 s )+(ks+k )

s3+2 s2−8 sk+k=0

a0+a1+a2+a3=0

S3 1−8

S2 2k

S1 −16−k2

S0 k

−16−k2

>0

−16−k>0

−k>−16

−k>−161

k<16

0<k<16

num=[1 1]den=[1 2 -8 1]printsys(num,den)num1=roots(num)den1=roots(den)%mapa zeros y polospzmap(num,den)%polosc=pzmap(num,den)

kW n2

s2+2ξWn+W n2 =θ0

θi

=G(s)

G(s)=

ks(s+2)

1+[(1+k ' s) ks (s+2) ]

G(s)=

ks(s+2)

s ( s+2 )+k+k ' kss(s+2)

G(s)= kW n2

s2+2 s+(k+k ' s)

16

s2+5.6 s+16= k

s2+s ( 2+k ' k )+k

k=16

5.6=2+16k '

5.6−2=16k '

k '=(5.6−2)

16

k '=0.225

TALLER 7

G (s )= 400

S (S2+30 S+1)

G (s )= 400

(S3+30 S2+S)

1.1

400 Kp

(S3+30 S2+S)

1.2

G1+GH

= 400Kp

(S3+30 S2+S+Kp)

2.

S3+30 S2+S+Kp=0

2.1

S3+30 S2+S+Kp=0

a0+a1+a2+a3=0

S3 11

S2 30kp

S1 30−kp6

S0 kp

kp>030−kp

6>0

30−kp6

>0

30−kp>0

−kp>−30

k p<30

0<kp<30

KCR.

K=0 K=5 K=10 K=15 K=20 K=25 K=30

%Diseño controladorn=[1]d=[1 30 1 0]figure(1)step(n,d)c=pzmap(n,d) %calculo pcr1kcr=0n1=[kcr]d1=[1 30 1 kcr]figure(2)step(n1,d1)grid %calculo pcr2kcr=5n1=[kcr]d1=[1 30 1 kcr]figure(3)step(n1,d1)grid %calculo pcr3kcr=15n1=[kcr]

d1=[1 30 1 kcr]figure(4)step(n1,d1)grid %calculo pcr4kcr=20n1=[kcr]d1=[1 30 1 kcr]figure(5)step(n1,d1)grid %calculo pcr5kcr=25n1=[kcr]d1=[1 30 1 kcr]figure(7)step(n1,d1)grid

PCR = 6.279

3.

KCR= 30 PCR=6.279

3.1

S=JW

J3W 3+30J 2W 2+JW+30=0

(30J 2W 2+30)(J ¿¿3W 3+JW )=0¿

30J 2W 2+30=0

30J 2W 2=−30

30 (√−1 )2W 2=−30

W 2=3030

W 2=1

W=±√1

W=1

PCR=2πW

PCR=2π1

PCR=6.283

4.

PCR=6.28 KCR=30

4.1

P= 0,6KCR

P=18

I=0,5PCR

I=3.14

D=0,125PCR

D=0.785

P I D18 18/3.14 18*0.785