1. Recuerde que la ecuación diferencial de la carga instantánea, q t , en el capacitor de un circuito en serie LRC , se describe con
L
d 2q 2
dt
R
dq dt
1 C
q E t
Use la transformada de Laplace para calcular q t cuando
L 1h, R 20, C 0.005 f , E t 150 V , t
0, q 0 0
e i 0 0 . ¿Cuál
es la corriente i t ? L 1h, R 20, C
0.005 f , E t 150 V ,t 0, q 0 0 e
Se tiene la ecuación : d 2q
d 2q
R
dq
20
dq
1
q E (t ) dt 2 dt c Reemp Reemplaza lazando ndo los valore valoress : L
1
dt 2 dt 0.005 Donde Donde los valor valor es :
q 150
q(0) 0 i (0) 0
q '(0) 0
Aplica Aplicando ndo Laplac Laplace e ‹
d 2q dq 1 dt 2 20 dt 0.005 q
‹
150
s 2Q(s) s*q(0) q'( 0) 20(sQ(s) q( 0)) 200Q(s 0Q(s)
( s 2 20s 200)Q(s ) Q(s)
150 s
150 s( s 2 20s 200)
1 Q(s) 150* 50 * s(s 2 20s 200) Por Fraccio Fracciones nes Parciale Parciale s :
1 s( s 2 20 s 200)
A s
Bs C s 2 20s 200
1 A(s 2 20s 200) Bs2 A B 0
Cs 20 A C 0
200 A 1 A
1 200
B
1 200
C
1 20
150 s
0
i 0
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Reem Reempla plazan zando do : Q(s) Q(s) Q(s)
1 1 s 1 1 10 * 2 200 200 s s 20 s 200 1 1 1 s 20 * * 200 s 200 (s 10)2 100
1 1 10 s 10 * 200 s 200 (s 10)2 100 (s 10)2 100 1
Aplic Aplicand ando o Laplac Laplace e Invers Inversa a: ‹
1
Q(s) 2100 *
q(t ) 2.
1 200
1
e 200
‹
10t
1
1 1 s 200
‹
1
s 10 10 ( s 10)2 100 (s 10)2 100
cos(10t ) e10t sen(10t )
Una batería de voltaje constante E 0 carga al capacitor de la figura. Divida la
ecuación 20 entre L y defina a 2
R L
y a
1 . Utilice la trasformada de Laplace LC 2
demostrar que la solución q t de q '' 2 q ' 2 q
E 0 L
E 0
L
R
para
C
, sujeta a q 0, i 0 0 , es
E0C 1 e t cos h 2 2 t sen h 2 2 t , 2 2 t q t E0C 1 e 1 t , t 2 2 2 2 E0C 1 e cos t 2 2 sen t ,
Solución:
q 0, i 0 0 Teniendo la ecuación diferencial con las condiciones iniciales entonces aplicamos la transformada de Laplace y la propiedad de linealidad a ambos miembros de la ecuación:
E 0 q L L q'' 2 q ' 2
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E 0 1 s Q sq ( 0 ) q'( 0 ) 2 ( sQ q ( 0 )) Q L s 2
2
Reemplazamos las condiciones iniciales entonces la ecuación resulta:
s 2Q
E 1 2 ( sQ ) 2 Q 0 L s
Factorizando el término ‘Q’, tenemos:
Q ( s2
2 s 2 )
E 0 1 L s
Despejamos ‘Q’, obteniendo:
E 0 Q 2 L s( s
2 2 s ) ……..(*) 1
Descomponemos la división que tenemos en dos sumandos, usando fracciones parciales; para ello hallamos ‘A’,’B’ y ‘C’.
A Bs C 2 s s 2 s 2
1 s( s 2
2 s 2 )
Efectuando el mínimo común múltiplo y cancelando denominadores:
A( s 2
2 s 2 ) s( Bs C ) 1
Agrupando términos semejantes:
s 2 ( A B ) s( 2 A C ) 2 A 1 Igualando coeficientes:
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A B 0 2 A C 0 2 A 1 Resolviendo las ecuaciones y hallando los valores de las incógnitas:
A
1
2
B C
1
2 2 2
Reemplazando en (*):
1 2 s 2 2 E 0 1 Q 2 2 2 L s s 2 s 1 2
Factorizamos en el segundo miembro
1 s 2 Q L2 s s 2 2 s 2 …….(**) E 0
2 Pero
1
LC , entonces reemplazando en (**) tenemos:
1 s 2 Q E0 C 2 2 s 2 s s
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1 s 2 Q E0 C 2 2 2 s s ( ) …..(***) Comenzamos a analizar los valores de
Si , entonces (***) tenemos:
2
2 :
2 2 es negativo; por lo que factorizamos el signo en
1 s 2 Q E0 C 2 2 2 s ( s ) ( )
Separamos el numerador de la ecuación y tenemos:
1 s Q E0 C s ( s ) 2 ( 2 2 )
2 2 2 ( s ) 2 ( 2 2 ) 2
Aplicamos transformada inversa de Laplace a ambos miembros y obtenemos:
2 2 s 1 L1 Q E0 CL 1 2 2 2 s ( s ) ( ) 2 2 ( s ) 2 ( 2 2 )
t q ( t ) E0 C 1 e cosh 2 2 t Si entonces tenemos en la ecuación (***)
2
2
senh
2
2
t
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1 s 2 Q E0 C 2 s ( s ) Desmembramos lo que está entre paréntesis y tenemos:
1 1 Q E0 C 2 s ( s ) ( s ) Aplicamos transformada inversa a ambos miembros y tenemos
L Q E0 CL 2 s s ( ) s ( ) 1
q( t )
1 1
1
E0 C 1 e t 1 t
2 2 Si , entonces es positivo entonces la ecuación (***) se mantiene
1 s 2 Q E0 C 2 2 2 s ( s )
Reescribiendo los numeradores y acomodando se tiene:
1 s Q E0 C s ( s ) 2 ( 2 2 ) Aplicamos transformada inversa y tenemos:
2 2 2 ( s ) 2 ( 2 2 ) 2
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s 1 L1 Q E0 CL 1 2 2 2 s ( s ) ( )
t q ( t ) E0 C 1 e cos 2 2 t
2 2 2 2 2 2 ( s ) ( ) 2
2
2
sen
2
t
3. Use la transformada de Laplace para determinar la carga q t en un circuito
0 y E t E0ekt , k 0 . Examine dos casos:
RC en serie, cuando q 0 k
1 1 y k RC RC
4. Usar la transformada de Laplace para determinar la carga q t en el capacitor en un circuito en serie RC , sujeto a las condiciones iníciales dadas: a) q 0 0 , R 2.5,
C 0.08 f , E t
E t
b) q 0 q0 , R 10
C 0.1 f , E t
E t
5
30et 30
3
t
1.5
t
2
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R
dq dt
2.5
dq
1 q E (t ) C
12.5q 5(t 3)
dt
Aplicamos transformada de Laplace
dq 12.5L q 5L (t 3) dt
2.5 L
e3 s 2.5Q( s) s q(0) 12.5Q( s) 5 s Como q(0) 0 despejamos:
Q(s)
2e3 s s(s 5)
Aplicamos la transformada inversa de Laplace 1
L
Q(s) L
1
2e3 s ( 5) s s
1 1 1 1 e3 s 5 s 5 s 5
Q(t ) 2L1
2 2 e 5(t 3) (t 3 ) 5 5
q(t )
30et ; 0 t 1.5 q 0 q0 , R 10 C 0.1 f , E t 0; 1.5 t b)
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10
dq dt
10q 30et (t 1.5) 30et
Aplicamos la transformada inversa de Laplace
dq 10L q 30L et (t 1.5) 30L e t dt
10 L
Q( s)s q(0) Q(s )
3 s 1
3e 1.5 s L et e1.5
1.5 s 1.5 1.5 3e1.5 Q(s) s 1 q0 s 1 s 1
3
5. (a) Con la transformada de Laplace determine la corriente i t en un circuito en serie LR con un solo bucle, cuando i 0 0, L 1h, R 10 y E t es la que muestra la figura.
E t
1
sen t ,0 t 2
3 2
3 2
t
1
(b) Use un programa de computo para gráficas, para trazar la de i t en el intervalo 0 t 6 . Con la gráfica estime imax e imin , los valores máximo y mínimo de la corriente. DATOS
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di Ri E (t ) dt di 10i sin t sin t (t 3 ) dt 2
L
di L10i L sin t L sin t (t 3 ) 2 dt
L
sI ( s ) i( s ) 10 I (s ) sI ( s ) i( s ) 10 I (s )
sI ( s ) i( s ) 10 I ( s )
sI( s ) i( s) sI( s ) i( s)
10I ( s) 10I ( s)
I ( s ) (s 10) I ( s )
1 s 2 1
2
s 2 1 1 s 2 1 1 s 2 1
s
2
1 (s 10)
e
e
s 2 1
1 s 2 1
e
1 (s 10)
e
1
1
1
s
1
e
As B
s
2
1
3 s 2
3 s 2
3 s 2
3 s 2
3 s 2
e e
3 s 2
L sin(t 3 ) 2
L sin t cos 3
L
cos t
L
cos t s
s2 1
s s2 1
3 s 2
s
s
2
1 (s 10)
C
s 10
As2 10 As Bs 10B Cs2 C 1 s 2 ( A C ) 0
A C......(I )
sin 3 cos t 2 2
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B A
10 101 1 101 1
C
101
s 10 1 2 101 s 1 s 10 1
e
3
s
2
s 2 1 (s 10) s
s
2
1 (s 10)
e
3 s
Ds E
s
2
1
2
Ds E F 2 s 1 s 10 F
s 10
Ds2 10Ds Es 10E Fs2 F s 2 ( D F ) 0 D F ......(I ) s (10D E ) 1. 1........(II )
10 E F
0.........(III )
de ( II ) tenemos que E 10 F
1........( IV )
( IV ) 10( III )
1 101 10 D 101 10
E
s
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3 s
3 2
I ( s )
Ds E e 2 F 2 s 1 s 10 101
1
s
2
1 (s 10)
e
3 s 2
10s 1 10 2 s 1 s 10 s
s
2
1 (s 10)
3 s 1 s 10 1 e 2 10s 1 10 I ( s ) 101 s 2 1 s 10 101 s 2 1 s 10
aplicandoL1
1
L
I ( s )
3 s 2 1101 L1 2 s L1 210 L1 21 e101 s 1 s 1 s 1
10s 1 10 1 1 1 L L 2 L 2 s 1 s 1 s 1
3 10 (t ) 3 3 10t 2 10 sin t e (t 3 ) 10 cos(t ) sin(t ) 10e i( s ) cos t 10 101 2 2 2
1
0 t
3 2
3 10 ( t ) 3 3 10t 2 i( s ) cos t 10 sin t e sin t 10 cos(t ) sin(t ) 10e 101 2 2
1
t
3 2
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(4) y EI 4 W ( x) y (4) y L EI 4 Wo WoH ( x ) y 2 Aplicamos Laplace EI s Y (s) s Y (0) s Y (0) sY (0) Y (0) 4
3
2 '
EI s Y (s) C1s C2 4
Wo 1
''
Wo s
'''
Wo s
L
e
s Wo 2
L
e
s Wo 2
s
L
1 e 2 C1s C 2 EI s s L Wo 1 2 s 1 C 1 C 2 Y ( s) 5 e 5 3 4 EI s s s s
s Y (s) 4
s
AplicamosTransformada AplicamosTrans formada Inversade Inversa de Laplace Wo x 4
L Y ( x) x EI 4! 2
4
L 1 C1 2 C 2 3 x x 2 4! 2! 3!
H x
4 L 4 L C1 2 C 2 3 Y ( x) x x H x x x 24 EI 2 2 2 6 L Wo 4 C1 2 C 2 3 0x 24 EI x 2 x 6 x , 2 Y ( x) 4 L Wo x4 x L C1 x2 C 2 x 3 , x Wo
s
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Para los valores en la Frontera y '' (L ) 0 y ''' (L ) 0 4 L C1 2 C 2 3 x x x x Y ( x) 24 EI 6 2 2 3 Wo L ' 3 x x C1 x C 2 x 2 Y ( x) 6 EI 2 2 2 Wo L '' 2 x x C1 C2 x 0 Y ( x) 2 EI 2 L Wo ''' Y ( x) x x C2 0 EI 2 Wo L x x C 2 0 2 EI
Wo
C 2
4
WoL
2 EI
L L L 2 EI 2 Wo
2
2
WoL2 C 0 1 EI 2
3L2 WoL2 C 1 2 EI 0 2 EI 4 Wo
C 1
WoL2
8 EI
Y ( x)
Wo
24
x
4
4 L WoL2 x 2 16
x
2
2
WoL
12
x3
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4 EI y (x) 0 (k 0 0))H(x (x L L// 3) 3) (0 k) Y H(x 2 L/ L/ 3) 3)
Aplicando L 4 EI L{ y (x)} k L{H(x L/ 3) 3) H(x 2 L/ 3) 3)}
1 Ls3 1 23Ls EI s Y (s) s y(0) s y '(0) sy''(0) y'''(0) k e e s s como y (0) 0 , y '(0) 0, entonces : 4
3
2
1 Ls3 1 23Ls EI s Y (s) sc1 c2 k e e s s 4
DespejamosY DespejamosY (s) k s Y (s) EI 4
1 Ls3 1 2 3Ls e e sc1 c2 s s
ke Ls / 3 ke2 Ls / 3 c1 c2 Y (s) 3 4 EIs5 EI s 5 s s 1 Aplican Aplicand o L 1
L
{Y(s)}
k 1 Ls / 3 1 1 2 Ls / 3 1 1 1 1 1 L e L e c L c L 4 1 2 3 E I s 2 s 2 s s
4 2 L 2L L x H (x ) 4 H(x ) k L x2 x3 3 3 3 c c2 y (x) x 3 1 2 EI 4! 4! 6
Sabem
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Derivando Derivando y (x)para x 2 L / 3, ya queneces que necesitamos itamosevalua evaluarr y ''(L) 0, y'''(L) 0 :
y'(x) 3 3 2 k 4 x L 4 x 2 L c x c x ; 2L / 3 x 2 1 24 EI 3 3 2 y''(x) 2 2 k 12 x L 12 x 2L c c x ; 2L / 3 x 1 2 24 EI 3 3 y'''(x) k x L x 2 L c2 ; 2 L / 3 x EI 3 3 Evaluamos Evaluamos : y '''(L) (L) 0 k L 2L L L c2 0 EI 3 3 k 2 L L c2 0 EI 3 3 k L
kL
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8. Encuentre la flexión y x de una viga en voladizo, empotrada en su extremo izquierdo y libre en el derecho, cuando la carga se define como
2 w0 1 L x , w x 0 ,
0x L
2
L
2
x L
9. Una viga esta empotrada en su extremo izquierdo y simplemente apoyado en el derecho. Calcule la flecha y x , cuando la carga es como en el problema 6. problema 6. W(x)
w0 ; 0
<
x < L/2
0 ; L/2 ≤ x < L
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EId (4 ) W ( x ) escribiendoen formacompacta L transfor transforma mada da delapl de laplace ace funcionde dehea heavisid viside e funcion EIdy EIdy 4 w( x) escribie escribiendoen ndoen forma compac compacta ta w( x)
w( x) g (t ) h(t ) g (t )
(t a)
w( x) w0 w0 (t l / 2) EIy EIy (4 )
w0 w0(t l / 2)
aplicam aplicamos oslapl laplace acea a am ambos bos miemb miembros ros
L w
EIL EIL y (4 )
0
w0 (t l / 2)
l s 1 e 2 EI s 4Y ( s) s3 y (0) s 2 y '(0) sy ''(0) y '''(0) w0 s s
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l s 2 w 1 e C C Y ( s) 0 5 5 31 42 EI s s s s
aplicamo aplicamoss L1 l s 2 w 1 e C C L1 Y (s ) L1 0 5 5 31 42 s s s EI s C1x 2 C2 x 3 w0 x 4 ( x l / 2)4 ( x l / 2) y( x) 4! 3! EI 4! 2!
y( x)
w0
24 EI
x
4
( x l / 2) ( x l / 2) 4
C1 x 2
2
C2 x 3
6
0 0 x l / 2 ( x l / 2) x l / 2 0 w0 4 C1 x 2 C2 x3 4 24 EI x ( x l / 2) ( x l / 2) 2 6 ( )
0 x l / 2
( x l / 2) toma toma el valor valor de 0
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10. Una viga uniforme de longitud L sostiene una carga concentrada w0 en x Está empotrada en su extremo izquierdo y libre en el derecho. Emplee la transformada de Laplace para determinar la flexión y x partiendo de
EI
d4y
1 w x L, 0 4 2 dx
Donde y 0 0, y ' 0 0, y '' L 0, y ''' L 0 11. Resuelva problema
la ecuación diferencial del 10, con las condiciones
w0
0, y ' 0 0, 0, y L 0, y ' L 0 . y 0 0, En este caso, la viga esta empotrada en ambos extremos
x L y
1 L. 2
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AHORA HALLAMOS LAS CONSTANTES C1 y C2,UTILIZANDO LAS CONDICIONES DE FRONTRERA y(L)=0, y'(L)=0 , EN LA FUNCION EN
l 2
x l
l ( x ) W 2 Y ( x) O EI 6 Y (l ) 0 WOl
24 EI
C 1
3
C1
x2
2
C2
x3
6
C2l
0 3 WOl
C2l 3 24 EI
C 1
PARA CONSEGUIR LA SEGUNDA ECUACION DERIVAMOS LA
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Y ( x)
3WOl x2 8 EI 2
5WO x3 4EI 6
l ( x ) W 2 Y ( x) O 6 EI
,0 x
l
2
3
3WOl x2
5WO x3 l , xl 8EI 2 4EI 6 2
Sistemas de Ecuaciones Lineales
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Resuelva el sistema si R1 10, R2
120, i3 0 0 y E t 0,
0 t 2
(b) calcule la corriente i1 t Solución:
5, L 1h, C 0.2 f , i2 0 0 ,
t 2
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LD( Dq3 i2 ) R1i2 E R1Dq3
LD2 q3 LDi2 R1 i2 E ..........8 '
1 q3 R1i2 0..............8'' C
De 3 LDi1
R2 i3
1 q3 C
E ...............9
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( s 10) I sI 120 (1 e2 s ) 2 3 s 10 sI2 (5s 5) I 3 0 Resolviendo por determinantes:
120(1 e2 s ) s 0
s
5 s 5
1 e2 s (5 s 5)120( )
120s 120
(1 (120s 120)e2 s
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145e5/3t 180e2t 36 i2 (t ) 144e 5/ 3(t 2) 180e 2(t 2) 145e5/3t 180e2t 14
; 0 t 2
;t
2
Hallando i3(t) Reemplazando
d (145e5/ 3t
i2 (t ) en una de la primera ecuación del sistema inicial.
180e2t 36) d
di3 d
10(145e5/3t 180e2t 36) 120
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1
4
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d 2q
dq
1
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21. Una viga uniforme en voladizo, de longitud L , está empotrada en su extremo
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