Compute the following Laplace transforms by hand using the definition of a Laplace trans-form (not using a table, although you can check your answers with a table):(a) Delta function: f(t) = 8(t)Note: the Laplace transform integral is from just before 0 (what you called 0 incalculus) to infinity. So, integrate from 0 to ∞.0,(b) Heaviside function: ƒ(z) = { 1 1,(c) Time-shifted delta function: f(t) = 8(t-a).t < 0t≥0

The Laplace transform of the function f(t) is defined as Lft=∫0∞e-stf(t)dt. The transform transforms the function of variable t into another function of variable s. In the first part of the problem, we have to find the Laplace transform of the function ft=δt. In the second part of the problem, we have to find fx=0 t<01 t≥0. In the final part, we have to find the transform of the shifted delta function ft=δt-a.(a) A delta function is defined as δt-a=0 for t≠a ∞ for t=a. The fundamental property of the delta function is it satisfies the condition ∫-∞∞f(t)δ(t-a)dt=f(a). To find the transform, use the definition Lft=∫0∞e-stf(t)dt and then use the fundamental property of the delta function. The point t=0 lies inside the region 0- to ∞. Lft=∫0-∞e-stδtdt=e-s·0=e0=1 Thus the transform of ft=δt is 1.(b) According to the definition of the Heaviside function, f(x)=0 for t<0 1 for t≥0. To find the Laplace transform use the definition Lfx=∫0∞e-sxf(x)dx. Since the function is defined separately for the region t<0 and t≥0, for calculating the Laplace transform, choose fx=1. Lfx=∫0∞e-sxfxdx=∫0∞1·e-sxdx=-e-sxs0∞=limx→∞-e-sxs--e-s·0s=0+1s=1s Thus the Laplace transfrom of f(x)=0 for t<0 1 for t≥0is 1s(c) A delta function is defined as δt-a=0 for t≠a ∞ for t=a. The fundamental property of the delta function is it satisfies the condition ∫-∞∞f(t)δ(t-a)dt=f(a). To find the transform, use the definition Lft=∫0∞e-stf(t)dt and then use the fundamental property of the delta function. The point t=a lies inside the region 0 to ∞. Lft=∫0∞e-stδt-adt=e-sa Thus the transform of ft=δt-a is e-sa.