Transformada de Laplace - 2
Simone Ribeiro
1. Desenhe o gráfico das seguintes funções:
(a) u1(t)+2u3(t)−6u4(t). Resp 6.3 - Ex. 1
(b) f(t−π)uπ(t), onde f(t)=t2. Resp 6.3 - Ex. 3
(c) f(t)=(t−1)u1(t)−2(t−2)u2(t)+(t−3)u3(t). Resp 6.3 - Ex. 6
2. Ache a transformada de Laplace das seguintes funções:
(a) f(t) = (
0, t<2
(t−2)2, t≥2 . Resp 6.3 -Ex. 7
(b) f(t) =
0, t< π t−π, π≤t<2π
0, t≥2π
. Resp:
Resp 6.3 - Ex. 9
(c) f(t) = (t−3)u2(t)−(t−2)u3(t). Resp: Resp 6.3 - Ex. 11
(d) f(t) = (
0, t<1
t2−2t
+2, t≥1 Resp 6.3
-Ex. 8
(e) f(t)=u1(t)+2u3(t)−6u4(t). Resp 6.3
-Ex. 10.
3. Ache a transformada inversa das seguintes funções:
(a) F(s)= 3!
(s−2)4. Resp 6.3 - Ex. 13
(b) F(s)= 2(s−1)e −2s
s2−2s
+2. Resp 6.3 - Ex. 15
(c) F(s)= (s−2)e −s
s2−4s+3. Resp 6.3 - Ex. 17
(d) F(s)= e
−2s
s2+s−2. Resp 6.3 - Ex. 14
(e) F(s) = e −s
+e−2s−e−3s−e−4s
s . Resp 6.3
- Ex. 18
4. Suponha queF(s)=L{f(t)}existe paras>a≥0.
(a) Mostre que secé uma constante positiva, então
L{c f(t)}= 1
cF
s
c
, s>ca,
(b) Mostre que seké uma constante positiva, então
L−1{F(ks)}= 1
kf
t
k
,
(c) Mostre que seaebsão constantes coma>0, então
L−1{F(as+b)}= 1
ae
−bt/af t
a
,
Resp 6.3 - Ex. 19
5. Use o resultado do problema acima para calcular as seguintes transformadas in-versas:
(a) F(s)= 2nsn++11n!.
Resp 6.3 - Ex. 20
(b) F(s)= 9s2−112s+3.
Resp 6.3 - Ex. 22
(c) F(s)= 4s22+s+4s1+5.
Resp 6.3 - Ex. 21
(d) F(s)= e22se−−4s1. Resp 6.3 - Ex. 23
6. Ache a transformada de Laplace da função dada:
(a) f(t)= (
1, 0≤t<1 0, t≥1 Resp 6.3 - Ex. 24
(b) f(t)=
1, 0≤t<1 0, 1≤t<2 1, 2≤t<3 0, t≥3
Resp 6.3 - Ex. 25
(c) f(t)=1−u1(t)+· · ·+u2n(t)−u2n+1(t)= P2n+1
k=1 (−1)kuk(t).
Resp 6.3 - Ex. 26
(d) f(t) = 1 + P∞k=1(−1)kuk(t). Use
inte-gração termo a termo. Resp 6.3 - Ex. 27
7. Suponha f satisfaça f(t+T) = f(t) para todot ≥0 e para algum número positivo
T. Neste caso, f é dita periódica de períodoT. Mostre que
L{f(t)}=
RT
0 e
stf(t)dt
1−esT .
8. Use o resultado do problema acima para achar a transformada de Laplace das seguintes funções:
(a) f(t)= (
1, 0≤t<1 0, 1≤t<2; f(t+2)= f(t),
Resp 6.3 - Ex. 29
(b) f(t)=t,0≤t<1; f(t+1)= f(t),
Resp 6.3 - Ex. 31
(c) f(t)=sint,0≤t< π; f(t+π)= f(t),
9. Encontre a solução dos seguintes Problemas de Valor Inicial usando Transformada de Laplace.
(a) y′′
+y= f(t); y(0)=0, y′(0)=1, f(t)= (
1, 0≤t< π/2 0, π/2≤t<∞; Resp 6.4 - Ex. 1
(b) y′′2y′
+2y=h(t); y(0)=0, y′(0)=1,h(t)= (
1, π≤t<2π
0, 0≤t< πet≥2π; Resp 6.4 - Ex. 2
(c) y′′
+4y=sint−u2π(t) sin(t−2π); y(0)=0, y′(0)=0,
Resp 6.4 - Ex. 3 (d) y′′
+4y=sint+upi(t) sin(t−pi); y(0)=0,y′(0)=0,
Resp 6.4 - Ex. 4 (e) y′′
+3y′+2y=u2(t); y(0)=0, y′(0)=0,
Resp 6.4 - Ex. 6 (f) y′′
+y′+(5/4)y=t−uπ/2(t)(t−π/2);y(0)=0, y′(0)=0,
Resp 6.4 - Ex. 8 (g)
10. Ache a solução dos seguintes problemas de valor inicial usando Transformada de Laplace.
(a) y′′
+2y′+2y=δ(t−π); y(0)=1, y′(0)=0,
Resp 6.5 - Ex. 1 (b) y′′
+4y=δ(t−π)−δ(t−2π);y(0)=0, y′(0)=0,
Resp 6.5 - Ex. 2 (c) y′′
+3y′+2y=δ(t−5)+u10(t); y(0)=0,y′(0)=1/2,
Resp 6.5 - Ex. 3 (d) y′′−y
=−20δ(t−3); y(0)=1, y′(0)=0,
Resp 6.5 - Ex. 4 (e) y′′
+2y′+3y=sint+δ(t−3π); y(0)=0, y′(0)=1/2,
Resp 6.5 - Ex. 5
11. Mostre as propriedades comutativa, distributiva e associativa da integral de con-volução.
(a) f ∗g= g∗ f.
(b) f ∗(g1+g2)= f ∗g1+ f ∗g2.
(c) f ∗(g∗h)=(f ∗g)∗h.
(a) f(t)= Rt
0(t−τ)
2cos 2τdτ. Resp: 4
(b) f(t)=R0te−(t−τ)sinτdτ. Resp 6.6 - Ex. 5
(c) f(t)=R0t(t−τ)eτdτ. Resp 6.6 - Ex. 6
(d) f(t)=R0tsin(t−τ) cosτdτ. Resp 6.6 - Ex. 7
13. Ache a transformada inversa das seguintes funções usando o Teorema da Con-volução:
(a) F(s)= 1
s4(s2+1). Resp 6.6 - Ex. 8
(b) F(s)= s
(s+1)(s2+4). Resp 6.6 - Ex. 9
(c) F(s)= 1
(s+1)2(s2+4). Resp 6.6 - Ex. 10
(d) F(s)= G(s)
s2+1. Resp 6.6 - Ex. 11
14. Expresse a solução dos seguintes Problemas de Valor Inicial em termos de uma integral da convolução.
(a) y′′
+ω2y= g(t); y(0)=0, y′(0)=1,
Resp 6.6 - Ex. 12 (b) y′′
+2y′+2y=sinαt; y(0)=0, y′(0)=0,
Resp 6.6 - Ex. 13
(c) y′′
+y′+(5/4)y=1−uπ(t); y(0)=1, y′(0)=−1,
Resp 6.6 - Ex. 15 (d) y′′
+3y′+2y=cosαt;y(0)=1, y′(0)=0,
17. = − +3 18. =cosh 19. y=cos√2t
20. y=(ω2−4)−1[(ω2−5)cosωt+cos 2t] 21. y=15(cost−2 sint+4etcost−2etsint)
22. y=15(e−t−etcost+7etsint) 23. y=2e−t+te−t+2t2e−t
24. Y(s)= s
s2+4+
1−e−πs
s(s2+4) 25. Y(s)=
1
s2(s2+1)−
e−s(s+1)
s2(s2+1)
26. Y(s)=(1−e−s)/s2(s2+4) 29. 1/(s−a)2
30. 2b(3s2−b2)/(s2+b2)3 31. n!/sn+1
32. n!/(s−a)n+1 33. 2b(s−a)/[(s−a)2+b2]2 34. [(s−a)2−b2]/[(s
−a)2+b2]2
36. (a)Y�+s2Y =s (b)s2Y��+2sY�−[s2+α(α+1)]Y = −1
Section 6.3, page 314
7. F(s)=2e−s/s3 8. F(s)=e−s(s2+2)/s3
9. F(s)=e
−πs
s2 − e−2πs
s2 (1+πs) 10. F(s)=
1
s(e −s
+2e−3s−6e−4s)
11. F(s)=s−2[(1−s)e−2s−(1+s)e−3s] 12. F(s)=(1−e−s)/s2
13. f(t)=t3e2t 14. f(t)= 13u2(t)[et−2−e−2(t−2)] 15. f(t)=2u2(t)et−2cos
(t−2) 16. f(t)=u2(t)sinh 2(t−2)
17. f(t)=u1(t)e2(t−1)cosh(t−1) 18. f(t)=u1(t)+u2(t)−u3(t)−u4(t)
20. f(t)=2(2t)n 21. f(t)= 12e−t/2cost
22. f(t)=16et/3(e2t/3−1) 23. f(t)= 12et/2u2(t/2)
24. F(s)=s−1(1−e−s), s>0 25. F(s)=s−1(1
−e−s+e−2s−e−3s), s>0
26. F(s)=1
s[1−e −s
+ · · · +e−2ns−e−(2n+1)s]= 1−e
−(2n+2)s
s(1+e−s) , s>0
27. F(s)=1
s ∞
�
n=0
(−1)ne−ns
= 1/s
1+e−s, s>0
29. L{f(t)} = 1/s
1+e−s, s>0 30. L{f(t)} =
1−e−s
s(1+e−s), s>0
31. L{f(t)} =1−(1+s)e
−s
s2(1−e−s) , s>0 32.
L{f(t)} = 1+e
−πs
(1+s2)(1−e−πs), s>0
33. (a)L{f(t)} =s−1(1
−e−s), s>0 (b)L{g(t)} =s−2(1
−e−s), s>0 (c)L{h(t)} =s−2(1
−e−s)2, s>0
34. (b)L{p(t)} = 1−e
−s
s2(1
+e−s), s>0
Section 6.4, page 321
1. y=1−cost+sint−uπ/2(t)(1−sint)
2. y=e−tsint+1
2uπ(t)[1+e
−(t−π )cost+e−(t−π )sint] −12u2π(t)[1−e−
(t−2π )cost
−e−(t−2π )sint] 3. y=1
6[1−u2π(t)](2 sint−sin 2t) 4. y=1
6(2 sint−sin 2t)− 1
6uπ(t)(2 sint+sin 2t) 5. y=12+12e−2t−e−t−u10(t)[12+12e−
2(t−10)
−e−(t−10)]
See SSM for detailed solutions to 17, 20, 22, 24
27b, 30, 32
36a, 38ab
2, 4, 8
14, 21, 22, 27
28, 30
1
Answers to Problems 705
6. y=e−t−e−2t+u2(t)[12−e−(t−2)+21e−2(t−2)] 7. y=cost+u3π(t)[1−cos(t−3π )]
8. y=h(t)−uπ/2(t)h(t−π/2), h(t)= 254(−4+5t+4e−t/2cost−3e−t/2sint)
9. y= 12sint+12t−12u6(t)[t−6−sin(t−6)]
10. y=h(t)+uπ(t)h(t−π ), h(t)=174[−4 cost+sint+4e−t/2cost+e−t/2sint] 11. y=uπ(t)[14−14cos(2t−2π )]−u3π(t)[
1 4−
1
4cos(2t−6π )] 12. y=u1(t)h(t−1)−u2(t)h(t−2), h(t)= −1+(cost+cosht)/2 13. y=h(t)−uπ(t)h(t−π ), h(t)=(3−4 cost+cos 2t)/12 14. f(t)=[ut
0(t)(t−t0)−ut0+k(t)(t−t0−k)](h/k)
15. g(t)=[ut
0(t)(t−t0)−2ut0+k(t)(t−t0−k)+ut0+2k(t)(t−t0−2k)](h/k)
16. (b)u(t)=4ku3/2(t)h(t−32)−4ku5/2(t)h(t−52),
h(t)= 14−(√7/84)e−t/8sin(3√7t/8)−14e−t/8cos(3√7t/8)
(d)k=2.51 (e)τ =25.6773 17. (a)k=5
(b)y=[u5(t)h(t−5)−u5
+k(t)h(t−5−k)]/k, h(t)=
1 4t−
1 8sin 2t 18. (b) fk(t)=[u4
−k(t)−u4+k(t)]/2k;
y=[u4
−k(t)h(t−4+k)−u4+k(t)h(t−4−k)]/2k,
h(t)= 14−14e−t/6cos(√143t/6)−(√143/572)e−t/6sin(√143t/6)
19. (b)y=1−cost+2
n
�
k=1
(−1)kukπ(t)[1−cos(t−kπ )]
21. (b)y=1−cost+
n
�
k=1
(−1)ku
kπ(t)[1−cos(t−kπ )]
23. (a)y=1−cost+2
n
�
k=1
(−1)ku11k/4(t)[1−cos(t−11k/4)]
Section 6.5, page 328
1. y=e−tcost+e−tsint+uπ(t)e−(t−π )sin(t−π )
2. y= 12uπ(t)sin 2(t−π )−12u2π(t)sin 2(t−2π )
3. y= −1 2e
−2t
+1 2e
−t
+u5(t)[−e−2(t−5)+e−(t−5)]+u10(t)[1 2+
1 2e
−2(t−10)
−e−(t−10)] 4. y=cosh(t)−20u3(t)sinh(t−3)
5. y= 14sint−14cost+14e−tcos√2t+(1/√2)u3π(t)e−(t−3π )sin√2(t−3π )
6. y= 12cos 2t+12u4π(t)sin 2(t−4π )
7. y=sint+u2π(t)sin(t−2π )
8. y=uπ/4(t)sin 2(t−π/4)
9. y=uπ/2(t)[1−cos(t−π/2)]+3u3π/2(t)sin(t−3π/2)−u2π(t)[1−cos(t−2π )] 10. y=(1/√31)uπ/6(t)exp[−14(t−π/6)] sin(
√
31/4)(t−π/6)
11. y= 1 5cost+
2 5sint−
1 5e−
tcost−3 5e−
tsint+u
π/2(t)e−
(t−π/2)sin(t−π/2) 12. y=u1(t)[sinh(t−1)−sin(t−1)]/2
13. (a)−e−T/4δ(t−5−T), T=8π/√15 14. (a)y=(4/√15)u1(t)e−(t−1)/4sin(√15/4)(t−1)
(b)t1∼=2.3613, y1∼=0.71153
(c)y=(8√7/21)u1(t)e−(t−1)/8sin(3√7/8) (t−1); t1∼=2.4569, y1∼=0.83351 (d)t1=1+π/2=∼2.5708, y1=1
15. (a)k1∼=2.8108 (b)k1∼=2.3995 (c)k1=2 16. (a)φ(t,k)=[u4
−k(t)h(t−4+k)−u4+k(t)h(t−4−k)]/2k, h(t)=1−cost
(b)u4(t)sin(t−4) (c) Yes
See SSM for detailed solutions to 8, 10, 16bc
19abd 20, 20abc
1, 3, 5, 7
10
17. (b)y=
k=1
ukπ(t)sin(t−kπ ) 18. (b)y=
k=1
(−1) ukπ(t)sin(t−kπ )
19. (b)y=
20 �
k=1
ukπ/2(t)sin(t−kπ/2)
20. (b)y=
20 �
k=1
(−1)k+1ukπ/2(t)sin(t−kπ/2)
21. (b)y=
15 �
k=1
u(2k
−1)π(t)sin[t−(2k−1)π]
22. (b)y=
40 �
k=1
(−1)k+1u11k/4(t)sin(t−11k/4)
23. (b)y=√20 399
20 �
k=1
(−1)k+1ukπ(t)e−(t−kπ )/20sin[√399(t−kπ )/20]
24. (b)y=√20 399
15 �
k=1
u(2k −1)π(t)e
−[t−(2k−1)π]/20sin
{√399[t−(2k−1)π]/20}
Section 6.6, page 335
3. sint∗sint=12(sint−tcost)is negative whent=2π, for example. 4. F(s)=2/s2(s2+4) 5. F(s)=1/(s+1)(s2+1)
6. F(s)=1/s2(s−1) 7. F(s)=s/(s2+1)2
8. f(t)=16
�
t0 (t−τ )
3sinτdτ 9. f(t)
=
�
0te−(t−τ )cos 2τdτ10. f(t)=12
�
t0 (t−τ )e
−(t−τ )sin 2τdτ 11. f(t)
=
�
0tsin(t−τ )g(τ )dτ12. y= 1
ωsinωt+
1
ω
�
t0 sinω(t−τ )g(τ )dτ13. y=
�
t0 e
−(t−τ )sin(t
−τ )sinατdτ
14. y=18
�
t0 e
−(t−τ )/2sin 2(t
−τ )g(τ )dτ
15. y=e−t/2cost−12e−t/2sint+
�
t0 e
−(t−τ )/2sin(t
−τ )[1−uπ(τ )]dτ
16. y=2e−2t+te−2t+
�
t0 (t−τ )e
−2(t−τ )g(τ )dτ
17. y=2e−t−e−2t+
�
t0 [e −(t−τ )
−e−2(t−τ )] cosατdτ
18. y=12
�
t0 [sinh(t−τ )−sin(t−τ )]g(τ )dτ 19. y=43cost−13cos 2t+16
�
t0 [2 sin(t−τ )−sin 2(t−τ )]g(τ )dτ 20. �(s)= F(s)
1+K(s)
21. (c)φ(t)=13(4 sin 2t−2 sint) (d)u(t)=13(2 sint−sin 2t)
C H A P T E R 7 Section 7.1, page 344
1. x1�=x2, x2�= −2x1−0.5x2
2. x1�=x2, x2�= −2x1−0.5x2+3 sint
3. x1�=x2, x2�= −(1−0.25t−2)x1−t−1x2
4. x1�=x2, x2�=x3, x3�=x4, x4�=x1
5. x�
1=x2, x2�= −q(t)x1−p(t)x2+g(t); x1(0)=u0, x2(0)=u�0 6. y1�=y2, y2�= −(k1+k2)y1/m1+k2y3/m1+F1(t)/m1,
y3�=y4, y4�=k2y1/m2−(k2+k3)y3/m2+F2(t)/m2
See SSM for detailed solutions to 17b
1c
4
8, 13, 15, 17
20
2, 4, 5 21b
