Question

1. Sketch a graph of the following function: f(t)-5(H(t) -2H(t-1)+ H(t-2)) 2. Use the definiton of the Laplace Transform to show that -5s 3. Find the inverse Laplace transform of the following: (a) T(s)-( (b) Y(s)=式 ) (劫 +3s 4. Solve the the following initial value problem: Sketch the function, f(), and the solution vit) on the same graph. 5. Solve the the following initial value problem: y" + 144y = δ(t); y(0) = 0, ช่ (0) =0

Answer 1

H(t-a) = {0 1 if if t < a t > a then f(t) = 5H(t)-10H(t-1) + 5H(t-2) take t < 0 implies t < 1 and t < 2 Then f(t) = 50-10.0 + 5.0 = 0 take t greaterthanorequalto 0 and t < 1 implies t elementof [0,1] implies f(t) = 5-10.0 + 5.0 = 5 Take t elementof [1,2] then f(t) = 5-10 + 0 = -5 t elementof [2, infinity] then f(t) = 5-10 + 5 = (-) so, H(t-5) = {0 1 if if t,5 t greaterthanorequalto 5 alpha {H(t-5)^2 = integral^infinity_0 e^-st H(t-5) dt = integral^5_0 e^st o dt + integral^infinity_5 e^-st. 1 dt = 0 + integral^infinity_5 e^-st dt = [e^-st/-s]^infinity_5 = 0 + e^5s /s = [e^-5s/s] because l^-ts rightarrow 0 for t rightarrow infinity and s > 0