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FACULTAD DE INGENIERÍA
ESCUELA PROFECIONAL DE INGENIERIA CIVIL Y AMBIENTAL

Tema:
Transformadas de Laplace

Estudiante Fustamante Fustamante Jhonatan
Távara Cancino Víctor
Tejada Chapoñan Dennis
López Mestansa Enrique

Escuela Profesional: IngenieríaCivil

Curso: Dinámica

Profesor:
Vera Lázaro Alejandro Segundo

TRANSFORMADAS DE LAPLACE

1. l[et+5]
l[et+5]= ∫e-stet+5dt
l[et+5]= ∫e-st+t+5dt
l[et+5]= ∫e-ts-1+5dt
l[et+5]=-e-ts-1+5s-1
l[et+5]= e5s-1

2. l[e-3t+4]
l[e-3t+4]=∫e-ste-3t+4dt
l[e-3t+4]=∫e-st-3t+4dtl[e-3t+4]=∫e-t(s+3)+4dt
l[e-3t+4]=-ets+3+4s+3
l[e-3t+4]=e4s+3

3. l[te3t]
l[te3t]=n!s-an+1
l[te3t]=1!s-31+1
l[te3t]=1s-32
l[te3t]=1s2-6s+9

4. l[2e-tsent]
l[2e-tsent]=ws+a2+w2
l[2e-tsent]=1s-12+12
l[2e-tsent]=1s2-2s

5. l[4t6]
l[4t6]=4l[t6]=n!sn+1
l[4t6]=4(6!)s6+1
l[4t6]=2880s7

6. l[3t2+8t+5]
l[3t2+8t+5]=3l[t2]+8l[t]+l[5]
l[3t2+8t+5]= 3(2!)s2+1+8(1!)s1+1+5sl[3t2+8t+5]= 6s3+8s2+5s

7. l[-4t2+16t+9]
l[-4t2+16t+9]= -4l[t2]+16l[t]+l[9]
l[-4t2+16t+9]=-4(2!)s2+1+16(1!)s1+1+9s
l[-4t2+16t+9]=-8s3+16s2+9s

8. l[(t+2)3]
l[(t+2)3]=l[t3+6t2+12t+8]
l[t3+6t2+12t+8]=l[t3]+6l[t2]+12l[t]+l[8]
l[t3+6t2+12t+8]=(3!)s3+1+6(2!)s2+1+12(1!)s1+1+8s
l[t3+6t2+12t+8]=6s4+12s3+12s2+8s

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9. l[(1+e3t)2]
l[(1+e3t)2]=1+2e3t+e6tl[(1+e3t)2]=l[1]+2l[e3t]+l[e6t]
l[(1+e3t)2]= 1s+2s-3+1s-6

10. l[5t2-6sen3]
l[5t2-6sen3]= 5l[t2]-6l[sen3t]
l[5t2-6sen3]= 5(2!)s2+1-6(3)s2+32
l[5t2-6sen3]= 10s3-18s2+9

11. l[cos3t+sen4t]
l[cos3t+sen4t]= l[cos3t]+l[sen4t]
l[cos3t+sen4t]= ss2+32+4s2+42
l[cos3t+sen4t]= ss2+9+4s2+16

12. l[e-2tsenh3t]
l[e-2tsenh3t]= b(s-a)2-b2
l[e-2tsenh3t]= 3(s+2)2-32
l[e-2tsenh3t]= 3(s)2+4s-5

13. l[e-tcosht]l[e-tcosht]= s-a(s-a)2-b2
l[e-tcosht]= s+1(s+1)2-12
l[e-tcosht]= s+1s2+2s


14. l[cos2t]
l[cos2t]= s2+2k2s(s2+4k2)
l[cos2t]= s2+2(1)2s(s2+4(1)2)
l[cos2t]= s2+2s(s2+4)

15. [cos2tcos4t]
cos2tcos4t= s3(s4+4k4)
cos2tcos2t(2)= s3(s4+4k4)
cos2tcos2t(2)= s3(s4+4(2)4)
cos2tcos2t(2)= s3(s4+64)

TRANSFORMADAS DE LAPLACE INVERSA

I.- Hallar la l-1

a)l-12s2+5s-4s3+s2-2s
l-12s2+5s-4s3+s2-2s= l-12s2+5s-4ss+2(s-1)
2s2+5s-4ss+2(s-1) = As+B(s+2)+C(s-1)
2s2+5s-4ss+2(s-1) = As+2s-1+Bss-1+Cs(s+2)ss+2(s-1)

(2s2+5s-4) = As+2s-1+Bss-1+Cs(s+2)

s=0
-4 = A0+20-1+B00-1+C0(0+2)
A=2

s=-2
-6 = A-2+2-2-1+B-2-2-1+C-2(-2+2)
B=-1

s=1
3 = A1+21-1+B11-1+C1(1+2)
C=1

l-12s2+5s-4s3+s2-2s= l-12s-1(s+2)+1(s-1)
l-12s2+5s-4s3+s2-2s=l-12s-l-11s+2+l-11s-1
l-12s2+5s-4s3+s2-2s=2-e-2t+et

b) l-12s-6s2-1
l-12s-6s2-1 = l-12s-6s+1(s-1)
2s-6s+1(s-1) = As+1+Bs-1
2s-6s+1(s-1) = As-1+B(s+1)s+1(s-1)

(2s-6) = As-1+B(s+1)

s=-1
-8 = A-1-1+B-1+1
A=4

s=1
-4 = A1-1+B1+1
B=-2

l-12s-6s+1(s-1)= l-14s+1-2S-1l-12s-6s+1(s-1)= l-14s+1-l-12s-1
l-12s-6s+1(s-1)=4e-t-32et

c) l-1s+12s2+4S
l-1s+12s2+4S = l-1s+12ss+4
s+12ss+4 = As+Bs+4
s+12ss+4 = As+4+B(s)s(s+4)

(s+12) = As+4+B(s)

s=0
12 = A0+4+B(0)
A=3

s=-4
8 = A-4+4+B(8)
B=1

l-1s+12ss+4 = l-13s-1S+4
l-1s+12ss+4 = l-13s-l-11s+4l-1s+12ss+4 = 3-e-4t
d) l-11(s+3)6
l-11(s+3)6 = l-11(s+3)65!5!
l-11(s+3)6 = 1!5!l-11(s+3)5+15!5!
l-11(s+3)6 = 1120(t5e-3t)

II. - Hallar l-1para funciones parciales

a) s+3s+1(s-3)
s+3s+1(s-3)= As+1+Bs-3
s+3s+1(s-3)= AS-3+B(S+1)s+1(S-3)

s+3=AS-3+B(S+1)

s=-1 s=3
-1+3=A-1-3+B(-1+1)...