IEDO Octave

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Aluno : Luís Henrique dos Santos 
Matrícula:16.2.8344 
 
NUMERO 1. 
𝑦′′(𝑡) + 2(𝐴1 + 1)𝑦′(𝑡) + (16 + (𝐴1 + 1)
2𝑦(𝑡) = 𝑔(𝑡); 𝑦(0) = 𝑦′(0) = 1 
𝑂𝑛𝑑𝑒 𝐴1 = 4 𝑒 𝑔(𝑡) = {
𝑡 − (𝐴1 + 1), 𝑡 ≥ 𝐴1 + 1
0, 𝑡 < 𝐴1 + 1
 
𝐸𝑛𝑡ã𝑜 𝑔(𝑡) = {
𝑡 − 5, 𝑡 ≥ 5
0, 𝑡 < 5
= 𝑢5(𝑡)(𝑡 − 5) 𝑒 
𝑦′′(𝑡) + 2(𝐴 + 1)𝑦′(𝑡) + (16 + (𝐴 + 1)2𝑦(𝑡) = 𝑢5(𝑡)(𝑡 − 5) 
𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑎 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑑𝑎 𝑒𝑚 𝑎𝑚𝑏𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 
𝑠2𝑌(𝑠) − 𝑠𝑦(0) − 𝑦′(0) + 10(𝑠𝑌(𝑠) − 𝑦(0)) + 41𝑌(𝑠) = 
𝑒−5𝑠
𝑠2
 
𝑌(𝑠)(𝑠2 + 10𝑠 + 41) − 𝑠 − 1 − 10 = 
𝑒−5𝑠
𝑠2
 
𝑌(𝑠)(𝑠2 + 10𝑠 + 41) = 
𝑒−5𝑠
𝑠2
 + 𝑠 + 11 
𝑌(𝑠)(𝑠2 + 10𝑠 + 41) = 
𝑒−5𝑠 + 𝑠3 + 11𝑠2
𝑠2
 
𝑌(𝑠) = 
𝑒−5𝑠 + 𝑠3 + 11𝑠2
𝑠2(𝑠2 + 10𝑠 + 41)
 𝑠𝑒 𝐻(𝑠) = 
1
𝑠2(𝑠2 + 10𝑠 + 41)
 
𝐸𝑛𝑡ã𝑜 𝑌(𝑠) = (𝑒−5𝑠 + 𝑠3 + 11𝑠2)𝐻(𝑠) = 𝑒−5𝑠𝐻(𝑠) + 𝑠3𝐻(𝑠) + 11𝑠2𝐻(𝑠) 
 
𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑛𝑑𝑜 𝑎 𝑖𝑛𝑣𝑒𝑟𝑠𝑎 𝑑𝑒 𝐻(𝑠) = ℎ(𝑡) 𝑝𝑜𝑟 𝑓𝑟𝑎çõ𝑒𝑠 𝑝𝑎𝑟𝑐𝑖𝑎𝑖𝑠: 
𝐻(𝑠) = 
1
𝑠2(𝑠2 + 10𝑠 + 41)
= 
𝐴
𝑠
+
𝐵
𝑠2
+
𝐶𝑠 + 𝐷
𝑠2 + 10𝑠 + 41
 
 𝐴𝑠(𝑠2 + 10𝑠 + 41) + 𝐵(𝑠2 + 10𝑠 + 41) + (𝐶𝑠 + 𝐷)𝑠2 = 1 
𝑠3(𝐴 + 𝐶) + 𝑠2(10𝐴 + 𝐵 + 𝐷) + 𝑠(41𝐴 + 10𝐵) + 41𝐵 = 1 
𝐵 = 
1
41
 𝐸𝑛𝑡ã𝑜 𝑐𝑜𝑚𝑜 41𝐴 = −10𝐵, 41𝐴 = −
10
41
 𝑒 𝐴 = −
10
412
 
𝐶𝑜𝑚𝑜 𝐴 = −𝐶, 𝐶 = 
10
412
 𝑒 𝑝𝑜𝑟 𝑓𝑖𝑚, 𝑐𝑜𝑚𝑜 𝐷 = −10𝐴 − 𝐵 𝐷 = 
102
412
 − 
1
41
= 
59
412
 
𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜 ℎ(𝑡) = −
10
412
 + 
𝑡
41
 +
10𝑠
412
 +
59
412
(𝑠 + 5)2 + 42
 = −
10
412
 + 
𝑡
41
 +
1
422
10𝑠 + 59
(𝑠 + 5)2 + 42
= 
 = −
10
412
 + 
𝑡
41
 +
10
422
𝑠 + 
59
10
(𝑠 + 5)2 + 42
= 
= −
10
412
 + 
𝑡
41
 +
10
422
(
𝑠 + 5
(𝑠 + 5)2 + 42
 + 
9
10
(𝑠 + 5)2 + 42
) = 
 = −
10
412
 + 
𝑡
41
 +
10
422
(𝑒−5𝑡cos4𝑡 +
9
40
4
(𝑠 + 5)2 + 42
) = 
 
= −
10
412
 + 
𝑡
41
 +
10
422
(𝑒−5𝑡cos4𝑡 +
9
40
𝑒−5𝑡𝑠𝑒𝑛4𝑡) 
 
𝑆𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒 𝓛−1{𝑌(𝑠)} = 𝑦(𝑡) 𝑞𝑢𝑒𝑟𝑒𝑚𝑜𝑠 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 
𝑦(𝑡) = 𝓛−1{𝑒−5𝑠𝐻(𝑠) + 𝑠3𝐻(𝑠) + 11𝑠2𝐻(𝑠)} 
 
𝐷𝑖𝑟𝑒𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑑𝑎 𝑡𝑎𝑏𝑒𝑙𝑎 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝓛{𝑒−5𝐻(𝑠)} = 𝑢5(𝑡)ℎ(𝑡 − 5) = 
 = 𝑢5(𝑡) (−
10
412
 + 
𝑡 − 5
41
 + 
10
412
(𝑒−5(𝑡−5)cos4(𝑡 − 5) +
9
40
𝑒−5(𝑡−5)𝑠𝑒𝑛4(𝑡 − 5))) (1) 
𝑆𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒 𝓛−1{𝑌(𝑠)} = 𝑦(𝑡) 
11𝓛−1{𝑠2𝐻(𝑠)} = ? 
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑛𝑑𝑜 𝑝𝑒𝑙𝑎 𝑡𝑎𝑏𝑒𝑙𝑎 𝑞𝑢𝑒 𝓛{ℎ′′(𝑡)} = 𝑠2𝐻(𝑠) − 𝑠ℎ(0) − ℎ′(0) 
𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑠2𝐻(𝑠) = 𝓛{ℎ′′(𝑡)} + 𝑠ℎ(0) + ℎ′(0) 
𝐸𝑛𝑡ã𝑜 𝓛−1{ 𝑠2𝐻(𝑠)} = ℎ′′(𝑡) + 𝓛−1{𝑠ℎ(0)} + 𝓛−1{ℎ′(0)} 
ℎ(0) = −
10
412
+ 
10
412
(1) = 0 
𝑆𝑒 ℎ′(𝑡 ) = 
1
41
+
10
412
(−4𝑒−5𝑡𝑠𝑒𝑛4𝑡 − 5𝑒−5𝑡cos4𝑡 +
9
40
(4𝑒−5𝑡cos4𝑡 − 5𝑒−5𝑡𝑠𝑒𝑛4𝑡)) 
 
ℎ′(0) = 
1
41
+
10
412
(−5 + 
9
40
4) = 
1
41
+
10
412
(−
41
10
) = 0 
 
ℎ′′(𝑡) = 
10
412
(−16𝑒−5𝑡cos4𝑡 + 20𝑒−5𝑡𝑠𝑒𝑛4𝑡 + 20𝑒−5𝑡𝑠𝑒𝑛4𝑡 + 25𝑒−5𝑡cos4𝑡 
+
9
40
(−16𝑒−5𝑡𝑠𝑒𝑛4𝑡 − 20𝑒−5𝑡cos4𝑡 − 20𝑒−5𝑡cos4𝑡 + 25𝑒−5𝑡𝑠𝑒𝑛4𝑡)) = 
 = 
10
412
(9𝑒−5𝑡cos4𝑡 + 40𝑒−5𝑡𝑠𝑒𝑛4𝑡 + 
9
40
(9𝑒−5𝑡𝑠𝑒𝑛4𝑡 − 40𝑒−5𝑡cos4𝑡)) 
𝐶𝑜𝑚𝑜 𝓛−1{0} = 0, 11𝓛−1{ 𝑠2𝐻(𝑠)} = 11ℎ′′(𝑡) = 
 = 
110
412
(9𝑒−5𝑡cos4𝑡 + 40𝑒−5𝑡𝑠𝑒𝑛4𝑡 + 
9
40
(9𝑒−5𝑡𝑠𝑒𝑛4𝑡 − 40𝑒−5𝑡cos4𝑡)) (2) 
 
 𝐸 𝓛−1{𝑠3𝐻(𝑠)} = ? 
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑛𝑑𝑜 𝑝𝑒𝑙𝑎 𝑡𝑎𝑏𝑒𝑙𝑎 𝑞𝑢𝑒 𝓛{ℎ′′′(𝑡)} = 𝑠3𝐻(𝑠) − 𝑠2ℎ(0) − 𝑠ℎ′(0) − ℎ′′(0) 
𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑠3𝐻(𝑠) = 𝓛{ℎ′′′(𝑡)} + 𝑠2ℎ(0) + 𝑠ℎ′(0) + ℎ′′(0) 
𝐸𝑛𝑡ã𝑜 𝓛−1{ 𝑠3𝐻(𝑠)} = ℎ′′′(𝑡) + 𝓛−1{𝑠2ℎ(0)} + 𝓛−1{𝑠ℎ′(0)} + 𝓛−1{ℎ′′(0)} 
ℎ′′(0) = 
10
412
(9 +
9
40
(−40)) = 0 
𝐴𝑠𝑠𝑖𝑚 𝓛−1{ 𝑠3𝐻(𝑠)} = ℎ′′′(𝑡) 
ℎ′′′(𝑡) = 
10
412
(−36𝑒−5𝑡𝑠𝑒𝑛4𝑡 − 45𝑒−5𝑡cos4𝑡 + 160𝑒−5𝑡cos4𝑡 − 20𝑒−5𝑡𝑠𝑒𝑛4𝑡 
+ 
9
40
(36𝑒−5𝑡cos4𝑡 − 45𝑒−5𝑡𝑠𝑒𝑛4𝑡 + 160𝑒−5𝑡𝑠𝑒𝑛4𝑡 + 200𝑒−5𝑡cos4𝑡)) = 
 = 
10
412
(−236𝑒−5𝑡𝑠𝑒𝑛4𝑡 + 115𝑒−5𝑡cos4𝑡 +
9
40
(236𝑒−5𝑡cos4𝑡 + 115𝑒−5𝑡𝑠𝑒𝑛4𝑡)) (3) 
𝑦(𝑡) = (1) + (2) + (3) 
Com manipulações algébricas pode-se chegar à mesma resposta da questão 3 para o que não é g(t): 
𝑇𝑜𝑚𝑎𝑛𝑑𝑜 𝑒−5𝑡𝑠𝑒𝑛(4𝑡) = 𝑆 𝑒 𝑒−5𝑡 cos(4𝑡) = 𝐶 
𝑃𝑜𝑑𝑒𝑚𝑜𝑠 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑟 2 
110
412
(9𝐶 + 40𝑆 +
9
40
(9𝑆 − 40𝐶)) = 
110
412
(40𝑆 +
81𝑆
40
) =
110
412
(
1681𝑆
40
) = 
=
11
4
𝑒−5𝑡𝑠𝑒𝑛(4𝑡) 
𝐸 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑖𝑐𝑎𝑟 3 
10
412
(−236𝑆 + 115𝐶 +
9
40
(236𝐶 + 115𝑆)) =
10
412
(−236𝑆 + 115𝐶 +
2124𝐶
40
+
1035𝐶
40
) = 
=
10
412
(
6724𝐶
40
−
8405𝐶
40
) = 𝑒−5𝑡cos(4𝑡) −
5
4
𝑒−5𝑡𝑠𝑒𝑛(4𝑡) 
 
𝑦(𝑡) = 𝑢5(𝑡) (−
10
412
 + 
𝑡 − 5
41
 + 
10
412
(𝑒−5(𝑡−5)cos4(𝑡 − 5) +
9
40
𝑒−5(𝑡−5)𝑠𝑒𝑛4(𝑡 − 5))) + 
+𝑒−5𝑡cos4𝑡 +
3
2
𝑒−5𝑡𝑠𝑒𝑛4𝑡 
 
 
NUMERO 2. 
Código utilizado no SciLab, o Octave não queria plotar meus gráficos. 
u5 = zeros(1,100) 
 
for i = 1:101 
 if(i<5) 
 u5(i) = 0; 
 else 
 u5(i) = 1; 
 end 
end 
 
res = zeros(1,101); 
t = 0; 
for a = 1:101 
 res(a) = u5(a)*(-10/(41^2) + (t-5)/41 + (10/(41^2))*(exp(-5*t + 25))*cos(4*t - 20) + (-
9/40)*(exp(-5*t + 25))*sin(4*t - 20)) + (110/(41^2))*(9*exp(-5*t)*cos(4*t) + 40*exp(-
5*t)*sin(4*t) +(9/40)*(9*exp(-5*t)*sin(4*t) - 40*exp(-5*t)*cos(4*t))) + (10/(41^2))*(115*exp(-
5*t)*cos(4*t) -236*exp(-5*t)*sin(4*t) +(9/40)*(115*exp(-5*t)*sin(4*t) +236*exp(-5*t)*cos(4*t))); 
 t = t + 0.015; //Apenas para mostrar as curvas mais suaves 
end 
 
plot(res) 
 
 
Após simplificações podemos escrever 
 
 
u5 = zeros(1,100) 
 
for i = 1:101 
 if(i<5) 
 u5(i) = 0; 
 else 
 u5(i) = 1; 
 end 
end 
 
res = zeros(1,101); 
t = 0; 
for a = 1:101 
 res(a) = u5(a)*(-10/(41^2) + (t-5)/41 + (10/(41^2))*(exp(-5*t + 25))*cos(4*t - 20) + (-
9/40)*(exp(-5*t + 25))*sin(4*t - 20)) + exp(-5*t)*cos(4*t) + (3/2)*exp(-5*t)*sin(4*t) 
 t = t + 0.015; //Apenas para mostrar as curvas mais suaves 
end 
 
plot(res) 
 
 
 
Destaque da descontinuidade em 5 assim como era g(t). 
 
 
 
NUMERO 3. 
𝑦′′(𝑡) + 10′(𝑡) + 41𝑦(𝑡) = 𝑔(𝑡) 
𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑎 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑑𝑎 𝑒𝑚 𝑎𝑚𝑏𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 
𝑠2𝑌(𝑠) − 𝑠𝑦(0) − 𝑦′(0) + 10(𝑠𝑌(𝑠) − 𝑦(0)) + 41𝑌(𝑠) = 𝐺(𝑠) 
𝑌(𝑠)(𝑠2 + 10𝑠 + 41) − 𝑠 − 1 − 10 = 𝐺(𝑠) 
𝑌(𝑠)(𝑠2 + 10𝑠 + 41) = 𝐺(𝑠) + 𝑠 + 11 
𝑌(𝑠) = 
 𝐺(𝑠) + 𝑠 + 11
(𝑠2 + 10𝑠 + 41)
= 
= 
 𝐺(𝑠)
(𝑠2 + 10𝑠 + 41)
+ 
 𝑠 
(𝑠2 + 10𝑠 + 41)
+ 
 11
(𝑠2 + 10𝑠 + 41)
 
 (4) (5) (6) 
 
 𝑆𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒 𝓛{𝐺(𝑠)} = 𝑔(𝑡) 
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑑𝑜 4: 𝐺(𝑠)
1
𝑠2 + 10𝑠 + 41
=
𝐺(𝑠)
4
4
(𝑠 + 5)2 + 42
 
𝓛−1{𝑌(𝑠)} = 𝑦(𝑡) =
1
4
(𝑔(𝑡) ∗ 𝑠𝑒𝑛4𝑡) =
1
4
∫ 𝑔(𝑡 − 𝓣)𝑠𝑒𝑛4𝓣𝑑𝓣
𝑡
0
 
 
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑜 5: 
 𝑠 
(𝑠2 + 10𝑠 + 41)
=
𝑠 + 5
(𝑠 + 5)2 + 42
−
5
(𝑠 + 5)2 + 42
=
𝑠 + 5
(𝑠 + 5)2 + 42
−
5
4
4
(𝑠 + 5)2 + 42
= 𝑒−5𝑡cos4𝑡 −
5
4
𝑒−5𝑡𝑠𝑒𝑛(4𝑡) 
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑜 6: 
11
𝑠2 + 10𝑠 + 41
= 
11
(𝑠 + 5)2 + 42
=11
4
4
(𝑠 + 5)2 + 42
=
11
4
𝑒−5𝑡𝑠𝑒𝑛4𝑡 
𝑦(𝑡) = (4) + (5) + (6) 
𝑦(𝑡) =
1
4
∫ 𝑔(𝑡 − 𝓣)𝑠𝑒𝑛4𝓣𝑑𝓣
𝑡
0
+ 𝑒−5𝑡cos4𝑡 +
3
2
𝑒−5𝑡𝑠𝑒𝑛4𝑡