Recent content by mechGTO

ahh, got it. thanks a lot. (sleep deprivation makes you miss things)

Yes, I do. Does that mean my attempt was incorrect?

I would say g(t-τ) would be (t-τ)2sin(t-τ)

\int_0^t (t-\tau)cos(2(t-\tau))e^{-4\tau} d\tau it looks as if f(t) = t and g(t) = cos(2*f(t)) Would it make sense to do a simple substitution and end up doing the integral \int_0^t u\cdot cos(2u) du giving \frac{1}{2} u\cdot sin(2u) + cos(2u) \frac{1}{2} t\cdot sin(2t) + cos(2t)...

In that problem, it would be setup so that f(t) = t^2 and g(t) = sin(t) making the convolution integral ∫ from 0 to t: t2*sin(t-τ) dτ and then integrated using trig identities

1. I'm not sure because I've never seen (t-tau) twice in one of these. 2. Laplace(f*g) = F(s)G(s)

Homework Statement Determine the Laplace Transform of ∫(from 0 to t) (t-τ)cos(2(t-τ))e-4τ dτ using Laplace Transform tables. Homework Equations I know the basic convolution theorem is (f*g)(t) = ∫f(τ)g(t-τ)dτ The Attempt at a Solution I'm not sure if this is double convolution...