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