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Instituto Universitario Politécnico“Santiago Mariño”Extensión Maturín
Esc. Ingeniería Eléctrica y ElectrónicaEjercicios Transformada de Laplace
Profesora: Realizado Por:Mariangela Pollonais Rubén González C.I: 25.453.370 Teléfono: 0426-6895258
Enero 2017
EJERCICIOS LAPLACE
En los siguientes ejercicios determine la Transformada de Laplace de las siguientes funciones:1.) F(t) = 2sen t + 3 cos 2 t ∫ {f(t)} = 2 ∫ {sen t} + 3 ∫ {cos 2t}F (s) = 2 1 + 3 s
s2 + 1 s2 + 4.
2.) g (t) = t 2 e 4t ∫ {g(t)} = ∫ {t2 e4t}g(s) = 1
(s – 4)3
3.) h(t) = e -2t sen s t ∫ {h(t)} = ∫{e-2t sen s t}h(s) = 5
(s+2)2 + 25.
4.) p(t) = (t + a) 3 t3 + 3 t2 a + 3 t a2 + a3 ∫ {p(t)} = ∫{t3} + 3a ∫{t2} + 3a2 ∫ {t} + a3 ∫{1}p(s) = 3! + 3a 2! + 3a2
1! + a3 1 s4 s3 s2 sp(s) = 6 + 6a + 3a2 + a3
s4 s5 s2 s
5.) q(t) = sen 2 a t q(t) = ½ ( 1 – sen 2 a t)∫ {q (t)} = ½ ∫{1} - ½ ∫ {sen 2 a t 1}q(s) = ½ x ⅕ - ½ 2a s2 + 4 a2
6.) f(t) = 3 e -t + sen 6t ∫ {f(t)} = 3 ∫{e-3t} + ∫ {sen 6 t}f(s) = 3 1 + 6 s + 3 s2 + 36
7.) g(t) = t 3 – 3 t + cos 4 t ∫{ g(t) } = ∫ {t3} – ∫ {t} + ∫ {cos 4 t}g(s) = 3! - 3 1 + S s4 s2 s2 + 16
8.) h(t) = - 3 cos 2 t + s Sen 4 t ∫ { h(t) } = - 3 ∫ { cos 2 t} + s ∫ {sen 4t}h(s) = - 3 S + S 4
s2 + 4 s2 + 16
9.) q(t) = 4 cos 2 3 t. q(t) = 4 . ½ ( 1 + cos 6 t)
q(t) = 2 + 2 cos 6 t∫ { q(t) } = 2 ∫{ 1} + 2 { cos 6 t}q(s) = 2 1 + 2 S S S2 + 36q(s) = 2 + 2 S
S2 + 36
10.) r(t) = cos 3 t = cos 2 t cos t r(t) = ( 1 – Sen2 t) cos tr(t) = cos t – sen2 cos t
11.) f(t) = e -3t (t – 2) t e-3t – 2 e-3t ∫{ f(t) } = ∫{t e-3t } – 2 ∫ { e-3t}f(s) = 1 - 2 1 (s + 3)2 s + 3
12.) g(t) = e 4t [ 1 – cos t ] e 4t - e 4t cos t∫ { g(t) } = ∫ { e 4t} – ∫ { e 4t cos t }g(s) = 1 - S – 4 S – 4 ( S – 4)2 + 1
13.)h(t) = e -5t [ t 4 + 2t + 1] h(t) = t4 e-5t + 2 t e-5t + e-5t
∫ { h(t) } = ∫ { t4 e-5t } + 2 ∫ { e-5t }h(s) = 1 + 2 1 + 1 (s + 5)5 (s + 5)5 s + 5
14.)q(t) = t3 sen 2 t∫ { q(t) } = ∫{ t3 Sen 2 t }q(s) = 2 . 2 S = 4 S (S2 + 4 )2 (S2 + 4)4
En los siguientes ejercicios calcule la transformada inversa de Laplace de la función s dada
1.) f(s) = 2s + 3 S2 + 4s + 20S2 + 4S + 20(S2 - 4S + 22) + 20 – (2)2(S -2 )2 + 16
f(s) = 2S + 3 + 4 - 4 (S – 2)2 + 16f(s) = 25 – 4 + 7 (4/4) (S – 2)2 + 16 (S – 2)2 + 16
f(s) = 2 (S – 2) + 7/4 4 (S – 2)2 + 16 (S – 2)2 + 16
∫ -1 {S} = 2 ∫-1 { (S – 2) } + 7/4 ∫ -1 { 4 } (S – 2)2 + 16 (S – 2)2 + 16f(s) = 2 e2t cos 4t + 7/4 e2t Sen 4t
2.) G(s) = 1 S(s + 1) ∫ -1 {G(s)} = ∫ -1 {1/5} . ∫ -1 { 1/ S + 1}G(s) = 1 . e-t = e-t
3.) H(s) = 25 (S2 + 1)2 ∫ -1 H(s) = 2 ∫ -1 { s } (S2 + 1)2 H(t) = 2 t cos t
4.) P(s) = 1
S(S2 + 16) ∫ {p(s)} = ∫ -1 { ⅕} ∫ -1 { 1 } (S + 2)2 p(t) = 1 . t e-2t p(t) = t e-2t
5.) Q(s) = 1 , n€z (s – a)
∫ -1 { Q ( S !) } = ∫ -1 { 1 } (S – a)n
Q(t) = tn-1 eat (n – 1)!
6.) R(s) = 3s2
(s + 1)2 ∫ -1 {R(s)} = ∫ -1 { 3s2 } (s + 1)2
R(t) = 3 ∫ -1 { S2 } (s + 1)2
R(t) = 3 . t et
7.) R(s) = 2 [ 1 + 3 + 4 ] s s2 s6
R(s) = 2 + 6 + 8 S5 56 510
∫ -1 { R(s) } = 2 ∫ -1 { 1/s5 } + 6 ∫ -1 {1 /s6 } + 8 ∫ -1 {1 /s10 }R(t) = 2 t4/4! + 6 t5/s! + 8 t9/9!
8.) Q(s) ={ s – 4 } + { s } (S2 + 5)2 (S2 + 2
∫ -1 {Q(s)} = ∫ -1 { { s – 4 } + ∫ -1 { s } (S2 + 5)2 (S2 + 2)
Q(t) = ∫ -1 { s } - 4 ∫ - { 1 } + ∫ -1 { s } (S2 + 5)2 (S2 + 5)2 (S2 + 2) Q(t) = t cos √5 t - 4 t e-5t + cos √2 t 2√5
9.) P(s) = 1 S2 – 4s + 5(s2 – 4S + 22) + 5 - 22
(s – 2)2 + 1∫ -1 { P(s) } = ∫ -1 { 1 } (5 – 2)2 + 1P(t) = e 2t sen t
10.)H(s) = S – 3 S2 + 10s + 9 S – 3 (s + 9) (s + 1)∫ -1 { H(s) } = ∫ -1 { s – 3 } (s + 9) (s + 1) s – 3 = A + B
(s + 9) (s + 1) s + 9 s + 1
S – 3 = A(s + 1) + B(s + 9)(s – 9)( s + 1) (s – 9)( s + 1) S – 3 = A(s + 1) + B(s + 9)Si s – 1 - 1 – 3 = A (-1 +1) + B(- 1 + 9)- 4 = 8BB = - ¼Si s = - 9- 9 – 3 = A( -9 + 1) + B(-9 + 9) – 12 = - 8AA = 12/8 = 3/2∫ -1 { H(s) } = 3 / 2 ∫ -1 { 1 } - ¼ ∫ -1 { 1 } S + 9 s + 1H(s) = 3/2 e-9t - ¼ e-6
11.) G(s) = s S2 - 14s + 1(S2 – 14s + (7)2 ) + 1 - (7)2 (s – 7)2 – 48∫ -1 { G(s) } = ∫ -1 { s } ( 7 – 7) (s – 7)2 – 48
∫ -1 { s – 7 } - 7 ∫ -1 { 1 } √48 (s – 7)2 – 48 ( s – 7)2 – 48 √48∫ -1 { s – 7 } - 7/48 ∫ -1 { √48 } (s – 7)2 – 48 ( s – 7)2 - 48 G(s) = e7t cos √48 t – 7/√48 e7t sen √48 t
12.)F(s) = e-4s S ( s2 + 16)∫ -1 {f(s)} = ∫ -1 { e+4s } S( s2 + 16)F(t) = 1 ( s + 4)
13.) F(s) = { e-s } (s – s)3
∫ -1 F(s) = ∫ -1 { e-s } (s – s)3
F(t) = u (t2 + s)
14.) G(s) = s e-cos (s2 + 4)2 ∫ -1 {G(s)} = ∫ -1 s e-10s (s2 + 4)2
G(t) = t2 cos2 + 2 t ( s + 10)
15.)H(s) = s2 -2s + 3 S(s2 – 3s + 2) S2 – 2s + 3 = A + B + CS( s – 2) (s – 1) S s-2 S-1S2 – 2s + 3 = A( s-2)(s-1) + Bs(S-1) + cs(s-2)S = 24 – 4 + 3 = 2B B = 3/2S = 1 1 – 2 + 3 = - C C=2S = 0 3 = 2A A = 3/2
∫ -1 { F(s)}= 3/2 ∫ -1 {1/5} + 3/2 ∫ -1 { 1 } + 2 ∫ -1 { 1 } S – 2 s – 1 F(t) = 3/2 + 3/2 e2t + 2et
16.) P(s) = 4s – 5 S3 – S2 – 5s – 3∫ -1 { P(s) } = ∫ -1 { 4s – 5 } S3 – S2 – 5s – 3
S3 – S2 – 5s – 3
1 -1 -5 -3 -1 -1 2 3 1 -2 -3 0 -1 -1 3 1 -3 0 3 3
1 0
4s – 5 4s – 5 = A + B + C (s + 1) (s + 1)(s – 2) = (s + 1)2 (s – 2) s + 1 (s + 1) s – 2
4s – 5 = A(s+1) (s – 2 + (s - 2) + C (s + 1)2 4s – 5 = A (S2 – s + s – 2) + B5 – 2B + c (S2 + 2s + 1)4s – 5 = A s2 – 2s +2A + B5 – 2B – 2B + CS2 + 2cs + c4s – 5 = (A+ C) s2 + (-A + B +2C)S + (-2A – 2B +C)
A + C = 0 - A + B + 2c + 4 - 2A + 2B + 4C = 8- 2A – 2B + C = - 5 - 2A – 2B + C = -5 - 4A + 5 C = 5
A + C = 0 4A + 4C = 0-4A +5C + 2s - 4A + 5C = 5 9C = 5 C = 5/9
A = - C = – 5/9B = A – 2C + 4B = - 5/9 – 10/9 + 4 = 7/3
∫ -1 {P(s) } = 5/9 ∫ -1 { 1 } + 7/3 ∫ -1 { 1 } + 5/9 ∫ -1 { 1 } S + 1 (s + 1)2 S – 2P(t) = - 5/9 e-t 7/3 t e-t 5/9 e2t
17.) Q(s) = -S (s – 4)2 (s – 5)
- S = A + B + C (s – 4)2 (s – 5) s – 4 (s-4)2 s – 5
-S = A(S – 4(S – 5) + B(s – 5) + C(S – 4)2
-S = A(S2 - 5S -4S +20) + B5 – 5B – C(S2 – 8S + 16)-S = A S2 – 9 AS + 20A + B5 – 5B + Cs2 – 8cs + 16 C-S = (A +C) S2 + ( -9A +B – 8C) + S (20A – 5B +16C)
A +C = 0 -9A + B – 8C = -1 - 45A + 5B – 40C = -520A – 5B+ 16C = 0 20A – 5B + 16 C = 0 -25 A -24C = -5 25 A + C = 0 C = - 5A = - C = - ( - 5) = 5B = - 1 + 8C + 9A = - 1 + 8 (-5) + 9 (5) + 9(5) = 4∫ -1 {Q(s) } = S ∫ -1 { 1 } +4 ∫ -1 { 1 } -5 ∫ -1 { 1 } S - 4 (S – 4)2 s – 5Q(t) = 5 e4t + 4 t e4t -5 e5t
18.) R(S) = S2 + 4s + 1 (S – 2)2 ( 5+ 3) S2 + 4s +1 = A + B + C(s – 2)2 (s + 3) s – 2 (s – 2)2 s + 3
S2 + 4s + 1 = A( s – 2) (s + 3) + B( s + 3) + (s + 2)2
S2 + 4s + 1 = A S2 + As – 6 A + B5 + 3B +CS2 – 4Cs +4CS2 + 4s + 1 = (A +C) S2 + (A + B – 4C)S + (- 6A + 3B + 4C)
A+C = 1-3 A + B – 4C = 4 - 3A – 3B +12C = -12
-6 A + 3B +4C = 1 - 6A + 3B + 4C = 1 - 9A + 16 C = - 11 9 A + C = 1 -9A + 16 C = -11 9A + 9C = 9 25 C = 2 C = 2/25
A = 1 – CA = 1 – 2/25 = 23/25B = 4 + 4C – AB = 4 + 4 ( 2/25) -25/25 =17/5
∫ -1 {R(s)} = 25/25 ∫ -1 { 1 } + 17/5 ∫ -1 { 1 } 2/25 ∫ -1 { 1 } S – 2 (S – 2)2 S + 3 R(t) = 23/25 e2t + 17/5 e2t + 2/25 e-3t
19.) R(s) = S (S2 + a2 (S2 – b2) S = A5+ B + C5 + D (S2 + A2)(S2 – b2) (S2 + A2) (S2 – b2)
S = (As + B) (S2 – B2) + (CS + D)(S2 + A2)S = AS3 – B2 A S + B S2 – b2 B + CS3 + C A2 S + Ds2 + A2 DS = (A + C) S2 + (B +D) S2 + (- b2 A + C A2) S ( - b2 B + A2 D )
A + C = 0B + D = 0-b2 A + C a2 = 1 -b2 B + A2 D = 0
B2 A + C = b2 A + b2 C = 0 -b2 A + CA2 = 1 -b2 A + C A2 = 1 C = ( B2 + a) = 1 A = - C = - 1 C = 1 b2 + a2 b2 + a2
B2 B + D = 0 b2 B + b2 D = 0 -B2 + A2 D = 0 -B2 B + a2 D = 0 D(B2 + a2 ) = 0D = 0B = -D = 0∫ -1 {R(s)} = - 1 ∫ -1 { 1 } ∫ -1 { a/b + 1 } ∫ -1 { 1 } B2 a2 s2 + a2 b2 + a2 S2 – B2
R(t) = - { 1 } ∫ -1 { a } + { 1 } ∫ -1 { S } a (b2 + a2) S2 + a2 b2 + a2 S2 – b2R(t) = - 1 sen a t + 1 cos h (bt) A(b2 + a2) b2 + a2
20.) Q(s) = 1
(s + 2) (S2 – 9) 1 = A + B + C
(s + 2) ( s – 3) ( s + 3) s + 2 s – 3 s + 3
1 = A(S – 3) (s + 3) + B (S + 2) (S + 3) + C( S + 2) (S – 3)S = 3 1 = 30B B = 1/30S = - 31 = 6C C = 1/6S = - 21 = - 5A A = -1/5
∫ -1 {R(s)} = -1/5 ∫ -1 { 1 } + 1/30 ∫ -1 { 1 } 1/6 ∫ -1 { 1 } S + 2 S – 3 S – 3 Q(t) = -1/5 e-2t + 1/30 e3t + 1/6 e-3t
21.) P(s) = 2 S3 (S2 + 5)
2 = A + B + C + Ds + ES3 (S2 + 5) S S2 S3 S2 + 52 = AS2 (S2 + 5) + B5 (S2 + 5) + C(S2 + 5) + (Ds + E) S3
2 = AS4 + 5 As2 + bs3 + 5Bs + CS2 + 5C + DS4 + ES3 2 2 = (A +D) S4 + (B + E) S3 + ( SA + C) S2 (5B) S +(5C)
A + D = D = - A D = 2/25 B + E = 0 E = - B = 0 5 A + C = A = - C/5 = -2/5/5 = 2/25 5B = 0 B = 0 5C = 2 E = 2/5 ∫ -1 {R(s)} = - 2/25 ∫ -1 { 1/5 } + 2/5 ∫ -1 { 2/2 } + ∫ -1 { 4 25 S } S 2 + 5P(t) = - 2/25 ∫ -1 { 1/5 } + 1/5 ∫ -1 { 2/s3 } + 2/25 ∫ -1 { S } S 2 + 5
P(t) = -2/25 + 1/5 t2 + 2/25 cos √5 t
En los Siguientes problemas resuelva las siguientes ecuaciones diferenciales.
25.) y” + y = e-2t sen t y(0) = 0 y´ (0) = 0∫ { y”} * ∫ -1 { y } = ∫ { e-2t sen t}s y s – s (0) – y´(0) + y s = 1 (s + 2)2 + 1
ys (s + 1) = 1 (s + 2)2 + 1ys = 1 = a + bs + c (s + 1) [(s + 1)2 +1] S + 1 ( S + 2)2 + 1
1 = a [ ( s + 2)2 + 1] + (bs + c) (S +1)1 = a(s + 2)2 + a + bs2 + BS + cs + + c1 = a( S2 + 4s + 4) + a + BS2 + bs + cs + c1 = aS2 + 4as + 4a + a bs2 + bs + cs + c1 = (a + b) s2 + (4a + b + c) S + (5a +c)
A + B = 0-1 4A + B + C = 0 -4A – B – C = 0 5A + C = 1 5A + C = 1 A - B = 1 A + B = 0 2A = 1 A = ½ B = - A = -1/2 C = 1-5A = 1 – S (1/2) = 1 – s/2 = -3/2
∫ -1 {F(s)} = ½ ∫ -1 {1 / s + 1} + ∫ -1 -1/5 – 3/2 (s + 2) + 2} F(s) = ½ ∫ -1 { 1 } -1/2 ∫ -1 { S – 3 } s + 1 s + 2 )2 + 1 F(t) = ½ ∫ -1 { 1 } -1/2 ∫ -1 { s + 2 – 2 } + 3/2 ∫ -1 { 1 } S + 1 ( s + 2)2 + 1 (s + 2)2 + 1F(t) = ½ ∫ -1 { 1 } -1/2 ∫ -1 { S + 2 } - ½ ∫ -1 { 1 } ( S + 2)2 + 1 (s + 2)2 + 1F(t) = ½ e-t -1/2 e-t cos t – ½ e-t Sen t
26.) y” + 4 y´ + 5 y + 2y = 10 cos t s (0) = y´ (0) = s” (0) = 3∫ { y”} + 4 ∫ { y”} + 5 ∫ { y´ } + ∫ { y } = 10 ∫cos E
s3 ys – S2 y(0) – S y´(0) – y“(0) + 4 { S2 y s - S y (0) – y´(0) } + 5 [ 5 y s – y(0)] + 2 y s = 10 * s / s2 + 1s3 y s – 3 + 4 S2 y s + 5 s y s + 2 y s + 2 y s = 10 S / s2 + 1s (S3 + 4 S2 + 5s + 2) = 10S / s2 +1 + 3
ss = 10S / S2 +1 +3 = 10s + 3 (s2 + 1) / s2 + 1 S3 + 4S2 + 5 S + 2 s3 + 4s2 5s + 2ss = 10S + 3S2 + 3(s2 + 1) (s3 + 4s2 + 5s + 2)
1 4 5 3 -2 -2 -4 -2 1 2 1 0 -1 -1 -1 1 1 0 -1 -1 0
10S + 3S2 +3 = AS + B + C + D + t(S2 + 1) ( s+ 2) ( S + 1)2
S2 + 1 S + 2 (S + 1) (S + 1)2
10S + 3S2 +3 = (AS + B) ( s+ 2) ( S + 1)2 + C (S2 + 1) ( S + 1)2 + D (S2 + 1) ( s+ 2) ( S + 1) E (S2 + 1) ( s+ 2)
(S2 + 1) ( s+ 2) ( S + 1)2 (S2 + 1) ( s+ 2) ( S + 1)2
10S + 3S2 +3 =(AS + B) ( s+ 2) (S2 + 2s + 1) + C (s2 + m1 (S2 + 2s + 1) + D (S3 + 2s2 + S + 2) + E (S2 + 2s2 + S + 2)
10S + 3S2 +3 = (AS2 + 2AS + BS + 2B) (S2 + 2s + 1) + C (S4 + 2S3 + s2+ S6 + 2S + 1) + D (S4 + S3 + 2S3 + 2s2
+ S2 + S + 25 + 2) + E (S2 + 2S2 + S + 2).
10S + 3S2 +3 = AS4 + 2AS3 +AS2 +2AS3 + 2AS + BS3 + 2BS2 + 2B+ 2BS2 + 2BS + 2B + CS4 + 2C S3 + CS2 + CS2 + 2CS + C + DS4 + DS3 +2DS3 +2DS2 +DS2 + DS + 2DS+ 2D + ES2 + 2 ES2 + ES + 2E.
10S + 3S2 +3 = ( A + C +D) S4 + ( 4A + B + 2C + 3D) S3 + (3A + 4B + 2C +3D + 3E) S2 + (2A + 2B+ 2C+ 3D+ E) S + (4B + C+ 2D +2E)
A + C + D = 0 4A + B+ 2C+ 3A= 0 3A+ 4B+ 2C + 3D + 3E =3 2A + 2B + 2C + 3C + 3ª E = 10 4B + C +2D + 2E = 3
-4 4A + B + 2C+ 3D + = 0 3A + 4B + 2C + 3D + 3E = 3 -12 A – 4B – 8C – 12D = 0
3A + 4B + 2C + 3D + 3E = 3 -13A – 6C – 9D + 3E = 3
-2 2A + 2B +2C +3D + E =10 4B + C + 2D +2E = 3 -4A – 4B – 4C – 6 D – 2E = - 20 - 4B + C + 2D + 2E = 3 -4A – 3C – 4D = -17 -1 3A + 4B +2C + 3D + 3E = 3 4B + C + 2D + 2E = 3 -3A + 4B + 2C +3D – 3E = -3 4B + C + 2D + 2E = 33 -3A – C – D – E = 0 -13A – 6C – 9D + 3E =3 - 9A – 3 C 3D -3E = 0 - 13A – 6C – 9d + 3E = 3 - 22A – 9C – 12D = 3-3 -4A – 3C – 4D = -17 - 22A – 9C -12D = 3 12A + 9C + 12D = 51 10A = 54 A = - 27/5
-12 A + C + D = 0 - 12A -12C -12D = 0 12A + 9C + 12D = 51 12ª + 9C + 12D = 51 -3C = 51 C = - 17 D = - A –C = 27/5 + 17 = 112/5B = 4A -2C – 3D B = -4 (-27/5) – 2(-17) – 3 (112/5) = -58/5E = 3 – 4B – C – 2D = 3 – 4 ( -58/5) – (-17) – 2(112/5)
2 2E = 54/5
∫ -1 -27/s s – 58/5 -17 ∫ -1 {1 / S+2} + 112/5 ∫ -1 {1 / S+1} + 54/5 ∫ -1 {1 / S+1}2
s2 + 1
-27/5 ∫ -1 {S / S2+2} - 58/5 ∫ -1 {1 / S2+2} -17 ∫ -1 {1 / S+2}
y (t) = -27/5 Cos t – 58/5 Sen t – 17 e-t + 112/5 et + 54/5 t et27) y ‘’ + 4 y’+8y= sen t y (0) =1 y (0) =0∫ {y’’} + 4∫ {y ‘} + 8∫ {y} = ∫ {sen t}
s2y - sy (0) – y’(0) + 4 [sy - y (0)] + 8ys = 1 s2 + 1s2∫s - s∫(0) - ∫1(0) + 4s∫ - 4∫(0) + 8ys = 1 s2 + 1∫s (s2 – s + 4 + 8) = 1 s2 + 1∫s= 1 (s2 + 1) (s2 – s + 12) 1 = as + b + cs + d (s2 + 1) (s2 + 12) s2 + 1 s2 –s + 121 = (as + b) (s2 – s + 12) + (cs + d) (s2 + 1)1 = as3 – as2 + 12 as + bs2 – bs + 12 b + cs3 + cs + ds2 + d1 = (a + c) s3 + (- a + b + d) s2 + (12a – b + c) s + (12b + d)
{ a+c=0−a+b+d=0 } (-1) {–a+b+d=012b+d=1 }
12a−b+c=0 A−b−d=012b+d=1 12b+d=1
11{ a+11b=112a−b+c=0} a+11b=1
132a – 11b + 11c =0 133a + 11c =1
-11{ a+c=0133a+11c=1} -11a – 11c =0
133a + 11c =0 122a=1 a= 1 122
c= -a= -1 122
-12{−a+b+d=012b+d=1 }12a−12b−12d=012b+d=112a−11 d=1d=12a-1 = 12(1/122) – 1= - 5 11 11 61
b=12a+c=12( 1122 )− 1122
= 11122
∫-1 { 1122 s+ 11122s2+1 } + ∫-1{−1122 s− s61
s2−s+12 } 1/122 ∫-1 { S
s2+1} + 11/122 ∫-1 { 1s2+1} - 1/22 ∫-1 { S−12 +
12
(S−12 )2
+ 474 }
s/61 ∫-1 { 1
(S−12 )2
+ 474 }
1/122 ∫-1 { Ss2+1}+ 1/122 ∫-1 { S
s2+1} – 1/22 ∫-1 { s−1
(s−12 )2
+ 474 }
71/122 ∫-1 { 1
(s−12 )2
+ 474 }
1/122 cos t + 1/122 sen t -1/22 t cos √ 472 t + 71/122 t sen √ 472 t28) y '−2 y=1−t y (0 )=1
∫ { y ' } - 2 ∫{ y }= ∫ {1 } - ∫{t }
s y s – y(0) – 2ys = 15− 152
y s (s – 2) = S−1S2
+ 1 ys (s – 2) = S−1+S2S2
ys = S2+S−1S2 (S−2 )
s2+s−1s2 (s−2 )
=as+ bs2
+ cs−2
s2+s−1=as ( s−2 )+b ( s−2 )+cs2 s=0-1 = -2b b=1/2s=2
4+2-1= 4c c=5/4s2+s−1=as2−2as+bs−2b+cs2
s2+s−1=(a+c ) s2+(−2a+b ) s−2 ca+c=1 = a=1-c-2a+b=1 a=1- S4=−14-2b= -1
−14∫−1{1s }+ 12 ∫−1 { 1s2 }+ 54 ∫−1{ 1
s−2 }−14
+ 12t+ s4e2 t
29) y ' '−4 y '+4 y=1 y (0 )=1 y ' (0 )=4
∫ {y ' ' }−4 ∫ { y ' }+4 ∫ { y }= ∫ {1 }
s2 ys−sy (0 )− y ' (0 )−4 ( sys− y (0))+4 ys=15
s2 ys−s−4−4 sys+4+4 ys=15
ys ( s2−4 s+4 )=15+s
ys= 1+s2
s ( s2−4 s+4 )
1+s2
s ( s−2 )(s−2)= 1+s2
s (s−2 )2=a5+ bs−2
+ c(s−2 )2
1+s2=a (s−2 )2+bs (S−2 )+cs1+s2=a ( s2−4 s+4 )+bs2−2bs+cs1+s2=as2−4as+4 a+bs2−2bs+cs1+s2=(a+b ) s2+(−4 a−2b+c ) s+4a
a+b=1b=1−a=1− 14=34
−4a−2b+c=04 a=1 c=4a+2b
a=14
c=4 ( 14 )+2( 34 )c=1+ 6
4=104
14 ∫
−1{15 }+ 34 ∫−1{ 1s−2 }+ 104 ∫−1{ 1(s−2 )2 }
14+ 34e2 t+ 10
4t e2 t
30) y ' '+9 y=t y (0 )= y ' (0 )=0
∫ {y ' ' }+9 ∫ { y }= ∫ {t }
s2 ys−sy (0 )− y ' (0 )+9 ys= 1S2
ys ( s2+9 )= 1s2
ys= 1s2 ( s2+9 )
=as+ bs2
+ cs+ds2+9
1=as ( s2+9 )+b (s2+9 )+(cs+d )s2
1=as3+9aS+bs2+9b+cs3+ds2
1= (a+c )S3+ (b+d ) s2+9as+9b
a+c=0c=0
b+d=0d=−19
9a=0a=09b=1 b=19
0 ∫−1{1s }+ 19 ∫−1{ 1s2 }−19 ∫−1 { 1s2+9 }3/3
19∫−1 { 1s2 }− 1
2 t∫−1 { 3
s2+9 }19t− 12 tSen3 t
31) y ' '−10 y '+26 y=4 y (0 )=3 y ' (0 )=15
∫ {y ' ' }−10 ∫ { y ' }+26 ∫ { y }=4 ∫ {1 }
s2 ys−sy (0 )− y2 (0 )−10 {sys− y (0)}+26 ys=4 15s2 ys−35−15−10Sys+30+26 ys=4
5
s2 ys−10sys+26 ys=45+35+15
ys ( s2−10S+26 )=4+352+15 ss
ys= 4+352+15 ss ( s2−10s+26 )
=as+ bs+cs2−10 s+26
4+352+15 s=a ( s2−10 s+26 )+ (bs+c ) s4+352+15 s=as2−10 as+26a+bs2+cs4+352+15 s=(a+b ) s2+(−10a+c ) s+26aa+b=3b=3−a
−10a+c=15b=3− 213
=3713
26a=4
a= 213
c=15+10a=15+10 ( 213 )=21513
213 ∫
−1{15 }+ ∫−1{ 3713 s+ 21513(s−5 )2+1 }213 ∫
−1{15 }+ 3713 ∫−1{ s−5+5(s−5 )2+1 }+ 21513 ∫−1 { 1(s−5 )2+1 }
213 ∫
−1{15 }+37 ∫−1{ s−5(s−5 )2+1 }+ 28013 ∫−1{ 1
(s−5 )2+1 }213
+ 3713t cos t+ 280
13t Sen t
32) y ' '−6 y '+8 y=e t y (0 )=3 y ' (0 )=9
∫ {y ' ' }−6 ∫ { y ' }+8 ∫ {y }=∫ {e t }
S2 ys−Sy (0 )− y ' (0 )−6 {Sys− y (0)}+8 ys= 1(s−1 )
S2 ys−3 s−9−6 sys+18+8 ys= 1( s−1 )
S2 ys−6Sys+8 ys−3 s+9= 1(s−1 )
ys ( s2−6 s−8 )= 1S−1
+35−9
ys ( s2−6 s+8 )=1+3 s ( s−1 )−9 ( s−1 )s−1
ys=1+3 s2−3 s−9 s+9
( s−1 ) ( s2−6 s+8 )
ys= 3 s2−12 s+10(s−1 ) (s−4 ) ( s−2 )
= as−1
+ bs−4
+ cs−2
3 s2−12 s+10−a (s−4 ) (s−2 )+b ( s−1 ) (s−2 )+c (s−1 ) (s−4 )
s=4
10=6b=b=106
s=2−2=−2c=c=1s=1
1=6 a=a=16
16∫−1 { 1
s−1 }+ 106 ∫−1 { 1s−4 }+ ∫−1{ 1
s−2 }16e t+ 10
6e4 t+e2 t
33) y ' '+4 y=e−t Sen t y (0 )=1 y ' (0 )=4
∫ {y ' ' }+4 ∫ {y }=∫ {e−t Sen t }
s2 ys−sy (0 )− y ' (0 )+4 ys= 1( s+1 )2+1
ys ( s2+4 )=1+s {(s+1 )2+1}+4 {(s+1 )2+1}( s+1 )2+1
ys=1+s( s2+25+1 )+s+4 (s2+2 s+1 )+4
{( s+1 )2+1} ( s2+4 )
ys=1+s3+2 s2+s+s+4 s2+8 s+4+4
{(s+1 )2+1 }( s2+4 )
Ys= s3+6 s2+10 s+9{(s+1 )2+1} (s2+4 )
s3+6 s2+10 s+9{( s+1 )2+1 } (s2+4 )
= as+b(s+1 )2+1
+ cs+ds2+4
s3+6 s2+10 s+9=(as+b ) (s2+4 )+ (cs+d ) {( s+1 )2+1}s3+6 s2+10 s+9=as3+4as+bs2+4b+ (cs+d ) ( s2+2 s+2 ) s36 s2+10 s+9=as3+4 as+bs2+4 c+cs3+2cs2+2cs+ds2+2ds+2ds36 s2+10 s+9=(a+c ) s3+ (b+2c+d ) s2+(4a+2c+2d ) s+ (4b+2d )
a+c=1b+2c+d=64 a+2c+2d=104 b+2d=9
−4 {b+2c+d=64b+2d=9 }−4b−8e-4 d=−244 b+2d=9−8 c−2d=−15
{ a+c=14 a+2c+2d=10}−4
−4a−4 c=−44 a+2c+2d=102c+2d=6−8 c−2d=−15−6 c=−9
c=96
a=1−c=1−96=−36
d=−8 c+152
=−8 ( 96 )+15
2=32
b=9−2d4
=9−2( 32 )4
=64
∫−1{ 36 s
+ 64
(s+1 )2+1 }+ ∫−1{ 96 s + 32s2+4 }−36 ∫−1{ s+1−1(s+1 )2+1 }+ 64 ∫−1 { 1
( s+1 )2+1 }+ 96 ∫−1 { ss2+4 }+ 32 ∫−1 { 1
s2+4 }22−36 ∫−1{ s+1
(s+1 )2+1 }+12 ∫−1 { 1( s+1 )2+1 }+ 96 ∫−1 { s
s2+4 }+ 34 ∫−1{ 2s2+4 }
−36e cos t+ 1
2e−t Sen t+ 9
6cos2 t+ 3
4Sen2 t
Dado el siguiente circuito determine la corriente I aplicando Transformada de Laplace
Malla I: VR+VL=E IR1+L
DILDT
=E L DILDT
+ IR=E (¿ L )
∫ {I 1 }+RL ∫−1 { I }= ∫ {EL } SI+ R
LI= EL. 1S
I [S+ RL ]=EL . 1 IS
I 1=
EL
S (S+RL )= AS
+ B
S+RL
EL
=A(S+ RL )+BSS=0
EL
=A RLA= E
R
S=−RL
EL
=B (−RL )B=−ER
I 1=ER∫−1{1S }− ER ∫−1 { 1
S+RL }
I 1=ER
−ERe
−RL t
Malla II: VC+VR=0
VC+ I 2R=0 IC=I R2
ICI 2+ I2
1 R=0
I 21R+ I
CI 2 = 0
I 21+ 1CR
I 2=0
∫−1 {I21 }+ 1CR∫ {I 2 }=∫ {0 }
S I2+1CR
I 2=0
I 2(S+ 1CR )=1
∫−1 {I 2 }= ∫−1 { 1
S+ 1CR }
I 2=e−1CR t
I=e−1CR t− E
R+ ERe
−RL t
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