Download - asignacion 1 rengel
-
8/16/2019 asignacion 1 rengel
1/15
-
8/16/2019 asignacion 1 rengel
2/15
1. DISCRETIZACION DE LA ECUACION DE PRIMER ORDEN
dy
dt +b . y (t )=a . r (t )
SY ( s)+b . Y (s )=a . R (s )
Sa/in%& , $a r$aci2n %$ %&3ini& S a Z "!
S= 2
T (1− z−1
1+ z−1 )S,"tit,4n%&
2T (
1− z−1
1+ z−1 )Y (Z )+a .Y ( Z )=b . R( z)
(1− z−1 )Y (Z )+b T
2(1+ z−1 )Y (Z )=a
T
2(1+ z−1 ) R( z)
Y ( Z )− z−1
Y ( Z )+b T
2(Y (Z )+Y (Z ) z
−1)=a T
2( R( z )+ R( z) z
−1)
Y ( Z )−Y ( Z ) z−1+b
T
2Y ( Z )+b
T
2Y ( Z ) z
−1=a T
2( R( z )+ R( z) z
−1)
(1+b T 2 )Y ( Z )+(−1+b T 2 )Y ( Z ) z−1=a T 2 ( R( z )+ R( z) z−1)
(1+b T 2 )Y k +(−1+b T 2 )Y k −1=a T 2 ( R k + R k −1)
D"p5an%& Y k
Y k =(1−b T
2 )(1+b T 2 )
Y k −1+a T 2( Rk + Rk −1)
(1+b T 2 )Dn&3ina3&"!
-
8/16/2019 asignacion 1 rengel
3/15
α =[1−b T 2 ]
[1+(b T 2 )] 6 β=
aT
2
[1+b T 2 ]7,%an%& $a "i8,int c,aci2n % rc,rrncia!
yk 9 α yk −1 : β ( rk +rk −1 )
Matri- % rc,rrncia
A ; <
[ y
0 r
1+r
0
y1
⋮
yn−1
r2+r
1
⋮
rn+rn−1][α β ]=[
y1
y2
⋮
yn]
A; 9 <
= 9 ( AT A )−1
AT <
P$antan%& $ Si"t3a % Ec,aci&n" para Dtr3inar $&" c&>icint" a. /' Tn3&"!
[1−b T 2 ]−[1+b T 2 ]α =0 Ec,aci2n 1
aT
2−[1+(b T 2 )] β=0 Ec,aci2n 0
-
8/16/2019 asignacion 1 rengel
4/15
1'1 Ra$i-aci2n %$ a$8&rit3& para $a i%nti>icaci2n % "i"t3a" % pri3r &r%n
-
8/16/2019 asignacion 1 rengel
5/15
En $a i3a8n antri&r rpr"nta >,nci2n para >ct,ar $ c?$c,$& % a$>a 4 /ta
A c&ntin,aci2n. n $a "i8,int >i8,ra " &/"r@a $ r",$ta%& % $a c&3paraci2n % $&""i"t3a" % pri3r &r%n "i"t3a p$anta%& @" "i"t3a i%nti>ica%&B'
-
8/16/2019 asignacion 1 rengel
6/15
-
8/16/2019 asignacion 1 rengel
7/15
4 (1−2 z−1+ z−2 ) Y (Z ) : 2bT (1− z−1 ) (1+ z−1 )Y (Z ) : c . (1+2 z
−1+ z−2 )T 2Y (Z ) 9
a (1+2 z−1+ z−2 )T 2 R( z )
(4−8 z−1+4 z−2 )Y (Z ) : 2bT (1− z−2 ) Y (Z ) : c . (1+2 z
−1+ z−2 ) T 2 Y (Z ) 9 a
(1+2 z−1+ z−2 )T 2 R( z )
(4−8 z−1+4 z−2 )Y (Z ) : (2bT −2bT z−2) Y (Z ) : (c T
2+2c T 2 z−1+c T 2 z−2 ) Y (Z ) 9
a (1+2 z−1+ z−2 )T 2 R( z )
(4−2bt +c T 2 ) Y (Z ) : (−8+2c T 2 ) z−1 Y (Z ) : (4−2bT +c T
2 ) z−2 Y (Z ) 9 a
T 2 ( R( z )+2 R( z ) z−
1+ R( z) z−2)
(4
−2
bt +c T 2
) y k : (−8
+2
c T
2
) y k −1 : (4
−2
bT +c T 2
) y k −2 9 aT
2 ( yk +2 y k −1+ yk −2 )
D"p5an%& Y k
yk 9(−8+2c T 2)
(4−2bt +c T 2) yk −1
(4−2bT +c T 2 )(4−2bt +c T 2 )
yk −2 aT
2 (rk +2 rk −1+rk −2 )(4−2bt +c T 2 )
Dn&3ina3&"
α 9(−8+2cT 2)
(4
−2
bt +cT
2
)
6 β 9(4−2bT +c T 2 )
(4
−2
bt +c T
2
)
6 γ 9aT
2
(4−2bt +cT 2)
7,%an%& $a "i8,int c,aci2n % rc,rrncia! yk 9 α yk −1 β yk −2 γ ( rk +2 rk −1+rk −2 )
Matri- % Rc,rrnciaA ; <
[
y1
y0 (r2+2 r1+r0 )
y2
y1 (r3+2 r2+r1 )
⋮ ⋮ ⋮
yn−1 yn−2 (rn+2 rn−1+rn−2 )
][
α
β
γ ] 9
[
y2
y3
⋮
yn
]A; 9
-
8/16/2019 asignacion 1 rengel
8/15
α 9(−8+2cT 2)
(4−2bt +cT 2)
(−8+2cT 2 ) (4−2bt +c T
2 ) α 9 + Ec,aci2n 1
β 9 (4
−2
bT +c T 2
)(4−2bt +c T 2 )
(4−2bT +c T 2 ) (4−2bt +c T 2 ) β 9 + Ec,aci2n 0
γ 9a T
2
(4−2bt +c T 2)
a T 2
(4−2bt +cT 2 ) γ 9 + Ec,aci2n
0'1 Ra$i-aci2n %$ a$8&rit3& para $a i%nti>icaci2n % "i"t3a" % "8,n%& &r%n'
-
8/16/2019 asignacion 1 rengel
9/15
-
8/16/2019 asignacion 1 rengel
10/15
La $(na n8ra 4 $&" c(rc,$&" rpr"ntan $ 3&%$& % ,n "i"t3a p$anta%& 4 ,n "i"t3a8nra%& p&r i%nti>icaci2n % "8,n%& &r%n r"pcti@a3nt'
3. DISCRETIZACION DE LA ECUACION DE TERCER ORDEN
d3
y
dt 3 +b
d2
y
dt 2 : c
dy
dt : % y(t ) 9 a' r(t )
S3
Y (s)+bS2
Y (s) : c S Y (s ) : % Y (s) 9 a' R(s)
Sa/in%& , $a r$aci2n %$ %&3ini& S a Z "!
S 92
T ( 1− z−1
1+ z−1 )S,"tit,4n%&
[ 2T ( 1− z−1
1+ z−1 )]3
Y (Z )+b [ 2T ( 1− z−1
1+ z−1 )]2
Y (Z ) :2c
T (1− z−1
1+ z−1 )Y (Z ) : d ¿Y (Z ) 9 a' R( z)
8(1− z−1
)3
Y (Z )+4bT (1− z−1
)2
(1+ z−1
)Y (Z )+2cT 2
(1− z−1
)(1+ z−1
)2
Y (Z )+dT 3
(1+ z−1
)3
Y (Z )=aT 3
(1+ z−1
)3
( Z )=¿a T 3 (1+ z−1 )3
R (Z )
8(1− z−1)3Y (Z )+4bT (1− z−2 ) (1− z−1 ) Y (Z )+2c T
2 (1− z−2 ) (1+ z−1 )Y (Z )+d T 3 (1+ z−1 )
3
Y ¿
8 (1−3 z−1+3 z−2− z−3 )Y (Z )+4 bT (1− z−1− z−2+ z−3 )Y (Z )+2cT
2 (1+ z−1− z−2− z−3 )Y (Z )+dT 3 (1+3 z−1+3
-
8/16/2019 asignacion 1 rengel
11/15
(8−24 z−1+24 z−2−8 z−3 ) Y (Z )+(4bT −4 bT z−1−4 bTz−2+4bTz−3 )Y (Z )+(2c T
2+2c T 2 z−1−2c T 2 z−2−
(Z )=¿aT 3 (1+3 z−1+3 z−2+ z−3 ) R(Z )(Z )+¿ (−8+4bT −2cT 2+dT 3 ) z−3Y ¿
(8+4bT +2cT 2
+dT 3
)Y (Z )+(−24−4bT +2cT 2
+3d T 3
) z−1
Y (Z )+(24−4 bT −2cT 2
+3d T 3
) z−2
Y ¿
(8+4bT +2c T 2+d T 3 )Y k −(+24+4 bT −2c T 2−3d T 3 )Y k −1+(24−4bT −2c T
2+3d T 3 ) Y k −2−(8−4bT
Y k =(24+4bT −2c T 2−3d T 3)Y k −1
(8+4bT +2c T 2+d T 3 ) −
(24−4 bT −2c T 2−3d T 3 ) Y k −2(8+4bT +2c T 2+d T 3 )
+(8−4 bT +2c T 2−d T 3 )Y k −3
(8+4bT +2c T 2+d T 3 ) +
Dn&3ina3&"
α 9(24+4bT −2c T 2−3 d T 3)
(8+4 bT +2c T 2+d T 3 )
β 9−(24−4bT −2cT 2−3d T 3 )
(8+4bT +2cT 2+dT 3 )
γ 9
(8−4bT +2c T 2−d T 3 )(8+4bT +2c T 2+d T 3 )
δ =aT
3 (rk +3 rk −1+3 rk −2+rk −3 )(8+4 bT +2c T 2+d T 3 )
7,%an%& $a "i8,int c,aci2n % rc,rrncia!
yk 9 α yk −1 β yk −2 γy k −3+δ (rk +3 rk −1+3 rk −2+rk −3 )
Matri- % rc,rrnciaA ;
-
8/16/2019 asignacion 1 rengel
12/15
[ y
3 y
2 y
1
y4
y3
y2
⋮ ⋮ ⋮
yn−1 yn−2 yn−3
r4+3 r
3+3 r
2+r
1
r5+3 r
4+3 r
3+r
2
⋮
rn+3 rn−1+3 r n−2+r n−3] [
α
β
γ δ
] 9 [ y
4
y5
⋮
yn]
A; 9 <
= 9 ( AT A )−1 AT <
P$antan%& $ Si"t3a % Ec,aci&n" para Dtr3inar $&" c&>icint" a. /. c 4 %' Tn3&"!
α 9(24+4bT −2c T 2−3 d T 3)
(8+4 bT +2c T 2+d T 3 )
(24
+4
bT −2
c T
2
−3
d T
3
)−α (8
+4
bT +2
c T
2
+d T 3
)=0
Ec,aci&n 1
β 9−(24−4bT −2c T 2−3d T 3 )
(8+4bT +2c T 2+d T 3 )
(24−4 bT −2cT 2−3dT 3 )+ β (8+4bT +2c T 2+d T 3)=0 Ec,aci2n 0
γ 9(8−4bT +2c T 2−d T 3 )(8+4bT +2c T 2+d T 3 )
(8−4 bT +2cT 2−dT 3 )−γ (8+4bT +2c T 2+dT 3 )=0 Ec,aci2n
δ = a T
3
(8+4bT +2c T 2+d T 3 )
a T 3−δ (8+4bT +2c T 2+d T 3 )=0 Ec,aci2n
'1 Ra$i-aci2n %$ a$8&rit3& para $a i%nti>icaci2n % "i"t3a" % trcr &r%n'
-
8/16/2019 asignacion 1 rengel
13/15
-
8/16/2019 asignacion 1 rengel
14/15
-
8/16/2019 asignacion 1 rengel
15/15