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8/17/2019 inversa 1
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7. F ( s)= 1
s2+2 s+5 (Migdalis González)
L−1
{
1
s2
+2 s
+5
} L
−1{ 1( s2+2 s+1 )+5−1 }
L−1{ 1( s+1 )2+4 }
¿e−t sin 2 t
2
L−1{ 1s2+2 s+5 }=e−t
sin 2 t
2
8/17/2019 inversa 1
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8. F ( S )= 6
S2−4 S−5
L−1{ 6S2−4 S−5}
Utilizando fracciones parciales
6
( S−5 ) (S+1 )=
A
( S−5 )+
B
( S+1 )
valores : A=1 y B=−1
6
( S−5 ) (S+1 )=
1
( S−5 )−
1
(S+1 )
L−1
{
6
S
2
−4 S−5 } ¿ L
−1{ 1( S−5 ) }− L−1 { 1(S+1 ) } ¿e5t −e−t
L−1{ 6S2−4 S−5 }=e5 t −e−t
8/17/2019 inversa 1
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9. F ( s)=
2
(s2+s+
1
2
)¿
2
( s2+s+ 14 )+12−14
¿ 2
(s+ 12 )2
+( 12 )2
¿2e−t 2
sinh( t 2 )1
2
¿4 e−t 2sinh( t 2 )
Lisseth Marín
8/17/2019 inversa 1
4/28
10. F (s)= 1
s2−6 s+10 ngel !e Gracia
L−1 1
(s2−6 s+9 )+1
L−1 1
(s−3)2+1
L−1{ 1(s−3)2 }+ L−1 {1 }
e3 t ∗t
8/17/2019 inversa 1
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18. L−1{ 1( s2+1 ) ( s2+4 ) }JesusChavez
A=0,B=1
3 , C =0, D=
−13
1
( s2+1 ) (s2+4) =
As+B
s2+1
+Cs+ D
s2+4
¿1
3 L
−1
{ 1(s2+1) }−13 L−1
{ 1(s2+4 ) }
¿1
3 sent −
1
3
sen2t
2
L−1{ 1( s2+1 ) ( s2+4 ) }=
1
3 sent −
2
3sen 2t
8/17/2019 inversa 1
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19. F ( s)=
1
( s+2 )3 Patricia Pimentel
Por el primer teoremade traslación
e−2
t t 2
2!
8/17/2019 inversa 1
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"0.
s+1¿¿¿3s2 ¿
F (t )=2 s−1¿
s
+1
¿¿s2 ¿
2 s−1¿
L−1¿
8/17/2019 inversa 1
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s+1¿¿¿3¿
s+1
¿¿¿2¿
s+1¿¿
s2¿1
¿
A=5, B=−1,C =−5, D=−4, E=−3
s+1¿¿
s+1¿¿¿1
¿¿−3 L−1¿
1
¿
5 L−1
{1
s
}− L
−1
{1
s2
}−5 L
−1
{ 1
s+1 }−4 L
−1
¿
s+1¿¿
s2¿=5−t −5e−t −4 t e−t −3
t 2
2 e
−t
2s−1¿
L−1¿
8/17/2019 inversa 1
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"#.
F ( S )=
s
(s2−1)2=
s
(s−1)2(s+1)2
F ( S )= s
(s2−1)2=
s
(s−1)2(s+1)2
s
(s−1)2(s+1)2=
A
(s−1)+
B
(s−1)2+
C
(s+1)+
D
(s+1)2 A=0 C =0
B=1
4 D=
−14
L−1 { F (S )}=1
4 L
−1
{ 1(s−1)2 }−14 L−1
{ 1(s+1)2 }¿ 1
4(t et )−
1
4(t e−t )
¿ t
2(
et −e−t
2 )
L−1 { F (S )}= t sinh(t )
2 ⋰⋰
$nthon% $da&es 9'7'91"
8/17/2019 inversa 1
10/28
8/17/2019 inversa 1
11/28
8/17/2019 inversa 1
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"9.
2−tan−1
( s
2
) (Migdalis González)
L−1{ 2−tan−1( s2 )} tan−1 "= 11+ "2
8/17/2019 inversa 1
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−1t
L−1
{ dds ( tan−1( 2s ))}−1
t L
−1
{ 1
1+(2s )
2
#−2
s2
}−1
t L
−1
{ −2
s2(1+ 4s2 )}
2
t L
−1
{ 1
s2
(s2
+4
s2
)}2
t L
−1 { 1s2+4 }2
t ( sin 2t 2 )
¿ sin2 t t
L−1{! 2−tan−1( s2 )}=sin2 t t
2−tan−1 "=tan−1(
1 " )
*tan
−1
( 1
s
2)
8/17/2019 inversa 1
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+0. F ( S )=1S ln (1+
1
S2 )Cristo$al %onzalez
L
−1{ ln(1+ 1
S2 )
S }
Utilizando s,pti&a propiedad de traslaci-n
∫0
t
L−1{
ln(1+ 1
S2 )
S }¿ L−1{ln(1+ 1S2 )} Utilizando la carta propiedad de traslaci-n
¿−1
t L
−1
{ d
ds
(ln
(1+
1
S
2
))}¿−
1
t L
−1
{ 1
1+( 1S2 ) (−2S3 )}
¿−1
t L
−1
{
1
S2+1
S2
(−2S3 )
}¿−1t
L−1
{ S2
S2+1 (
−2
S3 )}
8/17/2019 inversa 1
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¿−1
t L
−1
{ −2(S2+1 ) S }
Utilizando fracciones parciales−2
( S2
+1 ) S=
A s+B
( S2
+1 )
+C
S valores: A=2 y B=-2
−2
( S2+1 ) S=
2S
( S2+1 )−
2
S
¿−1
t L
−1
{ −2(S2+1 ) S }
¿−
1
t L
−1
{ 2S
(S2+1 ) }−1
t L
−1
{−2
S
}¿−
2
t L
−1
{ S( S2+1 ) }+2t L−1
{1S }
¿−2
t cos t +
2
t
¿2
t (1−cost )
L
−1{ ln(1+ 1
S2 )
S }=∫
0
t 2
& (1−cos ( t −& ) ) d&
+1. F ( s)= 1
s3(s2+1)
'aul 'eina
A
s+ B
s2+C
s3+ D
s+ E
s2+1
A=−1B=0C =1 D=1
8/17/2019 inversa 1
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−1 L−1{1s }+1 L−1{ 1s3 }+ L−1{ s
s2+1 }
−1+ t
2
2 !
+cos t
−1+t 2
2 +cos t
8/17/2019 inversa 1
17/28
++. L
−1
{ 1
( s2+3 s−4 ) } /ess ha2es "+( 32 )
2
=−4, "=−4−9
4 , (s+32 )
2
−25
4
8/17/2019 inversa 1
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L−1
{ 1
( s+ 32 )2
−25
4 }= L
−1
{ 1
(s+ 32 )2
−( 52 )2 }
e−32
t senh 52 t
5
2
=2
5e−32
t senh
5
2t
+.
Patricia Pimentel F ( s)=
( s+1 ) e−!s
s2+s+1
( s+1 )e−s
s2+s+1
= As+Bs2+s+1
A=1 B=1( s+1 )e−!s
s2+s+1
= ss2+s+1
+ 1s2+s+1
L−1
{ s
s2+s+1 }+
L−1
{ 1
s2+s+1 }
L−1
{ s
( s2+s+1 )2+1−14 }+ L−1
{ 1
(s2+s+1)2+1−14 }
Patricia Pimentel
8/17/2019 inversa 1
19/28
L−1{ s+
1
2
(s2+12 )2
+3
4 }−12 L−1{ 1( s2+ 12 )2+ 34 }+ L−1{ 1( s2+ 12 )2+ 34 }
8/17/2019 inversa 1
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+.
s
(¿¿ 2+9)2
F (S )=s¿
s(¿¿ 2+9)2
s¿¿
L−1 ¿
*1
2t L
−1
{ 1s2+9 }
*1
2 t ( se n(3t )
3 )s
(¿¿ 2+9)2
s¿
L−1 ¿
*1
6 t se n (+t)
3ess% 4a2ante
8/17/2019 inversa 1
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1
s4+1
= AS+B
(S2−2 S+2)+
CS+ D
(S2+2S+2) A=
−18
C =1
8
B=1
4 D=
1
4
L−1 { F (S )}=−1
8 L
−1 { S( S2−2S+2 ) }+ 1
4 L
−1 { 1( S2−2S+2 ) }+1
8 L
−1{ S( S2+2 S+2) }+1
4 L
−1 { 1( S2+2S−1¿
S−1¿
S+1¿
S+1¿
¿−1
8 L
−1{ S(¿¿ 2+1 ) }+ 14 L−1 { 1(¿¿ 2+1 ) }+ 18 L−1{ S(¿¿ 2+1 ) }+ 14 L−1 { 1( ¿¿ 2+1 ) }
S−1¿
S+1
¿
¿−1
8 L
−1{( S−1 )+1(¿¿ 2+1 ) }+ et
4 sin t +
1
8 L
−1 {(S+1 )−1(¿¿ 2+1 ) }+ e−t
4 sin t
8/17/2019 inversa 1
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S−1¿
S−1¿
S+1¿
S+1¿−18
L−1 { ( S−1 )(¿¿ 2+1 ) }−18 L−1 { 1(¿¿ 2+1 ) }+ e
t
4 sin t +
1
8 L
−1{ (S+1 )(¿¿ 2+1 ) }−18 L−1 { 1(¿¿ 2+1 ) }+ e−t
4 sin t
t +¿e−t
8 cos t −
e−t
8 sin t +
e−t
4 sin t
¿−e
t
8 cos t −
et
8 sin t +
et
4 sin¿
L−1 { F (S )}=−e
t
8 cos t +
et
8 sin t +
e−t
8 cos t +
e−t
8 sin t ⋰⋰
$nthon% $da&es 9'7'91"
0. F (S )= 1
s4+1
8/17/2019 inversa 1
23/28
". 5(s) *S
2
s4+64
An(el De %'acia
S2
s4+64
= As+B
s2−4 s+8
+ Cs+ D
s2+4 s+8 6 $* 18 6 3* 0 6 * '18 !*0
ntonces
s
s
8(¿¿ 2+4s+8)
8(¿¿ 2−4 s+8)−s
¿s
¿=¿
L−1{ s
2
s4+64 }= L−1¿
* 18 [ L−1 { ss2−4 s+8− ss2+4 s+8 }]
8/17/2019 inversa 1
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* 18
s
(¿¿ 2−4 s+4)+8−4s¿¿s
(¿¿ 2−4 s+4)+8−4s¿
L−1¿¿
* 18
s−2¿¿
s+2¿¿¿s¿
¿− L−1¿s¿
L−1 ¿¿
8/17/2019 inversa 1
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*18
s−2¿¿
s−2¿¿
s−2¿¿
s+2¿¿¿1
¿¿+2 L−1¿
s+2¿
¿− L−1¿
1¿
¿−2 L−1 ¿s−2¿
L−1¿¿
*18 [ e2t cos2t +e2 t sent −e−2 t cos2 t +e−2 t sen2t ]
*18 [cos2 t (e2 t
−e−2t
)+sen2 t (e2t
+e−2t
) ]
*18 [2 senh2 t (cos2 t )+2cosh2 tsen2t ]
8/17/2019 inversa 1
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+. F ( S )= s
2−6 s+4
s3−3 s2+2 s 3ess% 4a2ante
L−1{ s
2−6 s+4s3−3 s2+2 s }= L−1{ s
2−6 s+4s ( s−2 ) ( s−1 ) }
*
L−1{ AS }+ L−1{ Bs−2 }+ L−1{ C s−1 }
* L−1{2S }− L−1{ 2s−2 }+ L−1 { 1s−1 }
* 2−2 (e2 t )+et
L−1{ s
2−6 s+4s3−3 s2+2 s }=2−2e2 t +et
s2−6 s+4
s (s−2)(s−1)=
A
s +
B
s−2+
C
s−1
$*"3*'"
8/17/2019 inversa 1
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7. y) ) ) −3 y ) ) +3 y ) − y=t 2 e t y (0 )=1 y ) (0)=0 y ) ) (0 )=−2
L { y ) ) ) }−3 L { y ) ) }+3 L { y ) }− L { y }= L {t 2 e t }
s3
* ( s)−s2 y (0 )−s y ) (0 )− y ) ) (0 )−3 s2* ( s)−3 sy (0 )−3 y ) (0 )+3 s* ( s)+3 y (0 )−* ( s)= 2
( s−1 )3
s3
* ( s)−s2(1)−s (0 )−(−2 )−3 s2* ( s)−3 s (1 )−3 (0 )+3 s* ( s )+3 (1 )−* (s )= 2
(s−1 )3
s3
* ( s)−s2
+2−3 s2
* ( s)−3 s+3 s* (s )+3−* ( s)= 2
( s−1 )3
s3
* ( s)−3 s2* (s )+3 s* ( s)−* ( s)−s2−3 s+5= 2
( s−1 )3
( s3−3 s2+3 s−1 )* ( s)= 2( s−1 )3
+s2+3 s−5
* ( s )= 2
(s−1 )3 ( s3−3s2+3 s−1 )+ s
2
+3
s−5
(s3−3 s2+3 s−1 )
* (s )= 2
(s−1 )3 (s−1 )3+ s
2+3 s−5( s−1 )3
8/17/2019 inversa 1
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* (s )= 2(s−1 )6
+ s
2
( s−1 )3+
3 s
(s−1 )3−
5
(s−1 )3
* (s )= 2
(s−1 )6+
s2
( s−1 )3+3
[s−1+1
(s−1 )3
]−
5
(s−1 )3
* (s )= 2(s−1 )6
+ s
2
( s−1 )3+3
(s−1)
(s−1)3+
3
(s−1 )3−
5
(s−1)3
* (s )= 2
(s−1 )6+
s2
( s−1 )3+3
(s−1)
(s−1)3−
2
( s−1 )3
* (
s)=
2
(s−1 )6+
s2
( s−1 )3+
3
(s−1 )2−
2
(s−1 )3
aplicandolela place inversa o$tenemos
y (t )= t 5
et
60 +e t ( t
2
2 +2 t +1)+3 t et + 3 t
2e
t
2 −t 2 e t
Lisseth Marín
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