Lista B2 - Vinícius Santos

Nome: Vinícius de Souza Santos\nRa: B2300803\nDisciplina: PDS27 - Tratamento Digital do Sinal\nProfessor: Anderson Betuel\nSemestre: Segundo Bimestre\n\n1) Calculo a Transformada de Laplace das funções a seguir.\n\ng(t) = 3t\nU: s = j\nN: s = e^{-s}\ndu: du = dt\n\n3t^2\n\ne^{-s}\int_0^5 e^{-st} dt\n\nI - e^{-s} I\n\n-e^{-st}\n\nd(t)\n\n\n\n-3\n\n\n\n0\n\n0 - 0\n\n\n\n+ 2(0).0 + 2(2)\n\n\n\n\n\n= 3.2/λ^3\n\n d) 4. cos(5t)\nU: U = cos(5t)\nN: N = -e^{-st}\ndu: du = -5 sin(5t) dt\n\n4. ∫_0^∞ e^{-st} cos(5t) dt = (cos(5t)). e^{-st}\n\n\ne^{-st} ∫ (cos(5t)) dt\n\n\ndu: du = -5 sin(5t) dt\n\n\n\n\n\n\n0\n\n5/\lambda^2\n\n\n0 - 5/\lambda^2 + ∫ (sin(5t) dt)\n\n\ne^{-st} λ+5/2\ne^{-2s}\n\n\n 2) 5. sinh(4t)\n5. 5sinh(4t)\n\n\n(1 - e^(-4t)) = 2{ 1/(n^2 + 16)}\n\n-(1/8(n^4 - (n+4) + 4. y - y\n\n4-4\n\n2 = 20\nλ^2 + 16 = n^2 + 16\n\n\ne^{-5t}\n\n∫ e^{-5t} dt = c e^{-5t}\ndS = 4/λ = (4/8)\n\n\ne^{-st} ∫ dt\n\n0\n\n An(m(Y)) e^{-M} cos(4t) dt\n\nu2: cos(wt)\n\nu1 = sin(wt) - e^{-Mt} e^{-M} e^{16} ∫ e^{-st} D1(m(Y)) dt\n\nA = \\ A = \sin(m(Y)) e^{-4t} e^N 16 . A\n\nA ± 1 . A = \sin(m(Y)) e^{-Mt} e^N(16-DL)\n\nA(1) + 1 . A = (λ2 + λ1) - (λ2 + λ1) e^{-M} e \n\nA = 1/λ2 + 16\n\nD = 0/2 + 16\n\nD(sin(4t)) = 4/λ2 + 16\n\ns(λ e^{8t} F(n0)) = 4. (C + 1)\n1 - 2 - 28 + 16 y(3, 5, e^{-5t}) = (-1)^2 d^2/dt^2 / d^2;\n\nD1 = 2/5 e^{-5t}\nF(1) = 2.8 e^{-5t}\n\n∫(0 to ∞) e^{-st} e . e - e^{(k + 5)t} 1/(λ + 5)\ne^{-1}\n\n(-1) ^2 d^2/(t - 1) = (-1/λ + 5)^{2}[(λ + 5)^2 - 2/(λ + 5)]\n\nD^2 = 2/4 -1/(2 + 5)\n\nD^3 = (2 + 5)^{3}\n\n3.2/(x + 5)^3;\n\nF(k) = e^{(5t)} 3^3 e^{-3} - 3/2 [e^{2t} e - . ] \n\n3 e^{-2} + e^n\n\n3 e^{-2} – 3 = (t + 5)^3 - 2[1 - e^{-3t}]\n\n1 { (1 - 1) = . + 5)(n-7)} e^{-6} dt\n1. - λ2/2 + λ4\n\n(6/λt) + 3/4 + (6 /λ) + λ\n\n∫1 4 / 6 ^ 2 + 2 4 λ = (λ + 6 + 1) λ \n\n= λ / λ - 1\n\n-3 e^{-4}/ λ - 1\n\ne^{-t} U(t - 4)\n\ns2 b(t - e). N(t) = e^{-mt} b6(t + c1)\n\ne^{-mt} U(xc) = e^{-4} e6/m(yt-t) = .e = 1/(1 - λ)\n\n- 3 e^{-t} λ = λ 3 / λ = 1 - E3s. 02) Calcule a transformada inversa.\n\na) g(t) = 3 / (s - 3)\n\n3\n--------\nA1\nA\nK1 - λ. 3\n\n| λ = 0 3\n\nθ(g(s)) 3\n = 3 / (λ^2 + 9)\n\nd2 c os (αt) = a\n~ d2 (sin (3t)) : 3\n λ^2 + 2\n\nβλ = λs\n\nG(λ) G(s) = 4. cos (s.t)\nλ^2 + 25\n\n\\\nd ---------\n s2 + 9\n\nd(g(t)) 3\n\nλ^2 - 2λ + 16\n\nµ(n) = 4\n\n\n\n\\\ns[ F(ω).\n- λ.e^(-αt) − sθ( n − t ) = e^(-αt) 01) g(s) = (s^2 + 1)/ (s + 3) . (s + 1) . (s - 2)\n\nA1 A2 A3\n---------\n (s + 3) (s + 1) (s - 2)\n\nA1 = -40/ (s^2 + 180 + 5)\n (n/6) .(s - 0)\n | n = -3 - 10\n\nA2 = A1(s^2 + 180 + 5)↓\n----------\n (s + 3)(s + 1)(s - 2)\n\nn = -1 = 2\n\nA3 = A2(s^2 + 180 + 5)\n| s + 3 | (n)^1\n 45 3\n\n4 + 2 + 3\n\nn + 1 + 2\n\n4 . e^(3t) + 2. e^(-2t) + 3.c\n\nf) G(λ) = 2λ^2 + λ - 2\n\nd(1 + 2) G(1 )\n\nd (n + 1) 3\n\nA1 = G + 3\n\nA2 = 2. 12\n\ns + 1 |(λ)! | c |\n g) θ(λ) = λ - 3 . (λ - 3)\n λ^2 - 6λ - 7 (n + 1).(n - 7)\n\nA + B\n\n= A.(θ(λ)) . (λ - 3).\n A.(θ(λ)) . (λ - 3) | |\n\ndiv | λ^2 - 7 |\n | λ = -8 . 1/ 2\n\nB = _D/ + 1 / λ - 3\n\nd | 4 - 1 / (λ + 1)(n - 7) . (2) | λ = -3 - 2 \n\n1/ 2 | 2 | = 1/ 1\n n + 1 (3)(7)\n λ(2)(2)n\n\n({e^(-t) + e^t) e^(-t)}\n\nλ\n= {\n 2 e^t - 2 e^(-t) − e^t − t 2 } d) y''(t) + y(t) = cos(2.t) + log(q(t)) = 3, y'(0) = 4\n\nn² F(n) - Λ(n - 1) + F(n) = Λ(n)\nF(n) (n + 1) - Λ(n - 1) = Λ(n + 1)\nF(n) : Λ(n + Λ(t)) = n³ + n² + 1 + n³ + n + 1)\n(n² + 1), (Λ + 1)\n\nA + B = Λ(n + 1) + Λ(n)\n (n² + 1) (n³ + 1) (n² + 1), (n + 1)²\n\ncos(1) + Λ(n + 1)