Questão - Transformada de Laplace - 2023-2

1) Usando a Transformada de Laplace, resolva os seguintes problemas: a) y_0''(T) + 16 y_0(T) = 1 y_0(0) = 1 ; y_0'(0) = 2 b) y_0''(T) - y_0'(T) = e^T cos(T) ; y_0(0) = y_0'(0) = 0 c) y_0'(T) + 2 y_0(T) = f(T) ; f(T) = { T 0<=T<1 ; 0 T>=1 } ; y_0(0)=0 a) y'' + y + 16y = 1 L(y'' + 16y) = L(1) s^2Y - sy(0) - y'(0) + 16Y = 1/s Y = 1/4 [s^2+16] s(s^2+16) -> s^2 / s Como int[f(t)] delta = s * L[g(t)/s] Int[seno(u) du = s^1 / z(s^2+16)/s Int[sen(4u) du = y(t) y(t) = 1/4 [1 - cos 4t] -> y(t) = 1/16 (1 - cos 4t) = sen 2yr/8 b) y'' - y' = e^tcosT (s^2-5)Y = 5-1 -> Y = 5-1 (s-1)^2+1 A/B/s + C/s^2+D/s s[(s)(s-y^2-1)]/s(s-1) A(s-1)(s^2+5) + D -(s^2-2s+6) + C(s+0)(s^2-s) = 5-1 A(s^3-2s^2-2) - D(3-2s*B + 3sB +(s^3sC-2C^2 + D's2)D's = 5-1 s^3(A+B+C) + 5s(-3D-2B -C+D) + 6(2B+D) - 29A = 5-1 A=1/2 B=0 -> Y = 1 + 1 - S C.=-1/12 D=1 y(t) = 1/2 + 1/2senT - ecosT - esenT) gy(t) = 1/2 (esen(T - ecosT + 1) 3) f(t) = t[u_0(T) - u(T,4)] - 10(u_1(T)) - t[u(T)-u_0(t)] Y' + 2y = t(10-10) (s+2)Y - 10 / s^2 / s V = V_0 - U_0 / s(5^2+2) / s(5^2+2) Y = V_0 - V_1 / s (5^2 / s(5^2 (\/s/Vref)/; 5^2 - Y ( ,-d+ +C = [ \-/ . ( -- [s + s ; + 4^ (3 - B+C = A (5-0-3s) -|B (s \ 4) + Cst = 3t2s // 5 // 5 ^5+2 s^2( A - c) + 5(2A+B ) +2B = 1 B= 1/2 A= -1/4 C=1/4 \ 1 / / ] n Y / ( (\) 2^(s + t ) [ -> V_0 / -1 + 1 + / s, where / (u_1(s) 2 +\ / ^ s ] = s) , y(t ) => (/t ), 5 ),\ / \ f / / -1+ s*\t + \e-2t /2 2/ s2 \ \