Use convolution integral to find step response of a system

[TEEE]

Homework Statement

An electrical network has the unit-impulse response :h(t)=3t⋅e^-4t .If a unit voltage step is applied to the network, use the convolution integral to work out the value of the output after 0.25 seconds.

Homework Equations

Convolution integral: y(t)=f(t)*h(t)
Unit step function u(t) = 1 when t>=0, 0 when t<0

The Attempt at a Solution

This is a problem on the problem sheet of my circuit analysis course. The lecturer also gives a solution which is not complete:
v(τ)=∫^∞₀ 1⋅3(τ-t)e^(-4(τ-t)) dt
v(0.25) = 4.95×10^-2V
I'm not sure how to proceed from the above equation to get the final answer. I have tried solving using convolution:y(t) =∫^∞₀h(τ)u(t-τ)dτ or y(t)=∫^∞₀u(τ)h(t-τ)dτ,but get something like 0 or -0.192. I guess I dealt with the u(τ) or u(t-τ) in the integral wrong as I'm not sure how to deal with it in an integral. I know that u(t) might affect the limit of the integral but I'm not sure how.(p.s.the∫ part above means integrate from ∞ to 0, if that causes any confusion)
Any help or advise would be really appreciated as I've been bothered by this for quite some time.Thanks!

 

[STEMucator]

Let ##h(t) = 3t e^{-4t}## and ##u(t)## be the unit step function, then:

$$(h * u)(t) = \int_0^{\infty} h(x) u(t-x) \space dx = 3 \int_0^{0.25} x e^{-4x} \space dx$$

Integration by parts will yield the answer.

 

[TEEE]

Let ##h(t) = 3t e^{-4t}## and ##u(t)## be the unit step function, then:

$$(h * u)(t) = \int_0^{\infty} h(x) u(t-x) \space dx = 3 \int_0^{0.25} x e^{-4x} \space dx$$

Integration by parts will yield the answer.

Thank you for your reply, it helped me find out where I went wrong in my approach to the solution. I changed the limit to infinity to 0.25 instead of 0.25 to 0 when evaluating (h*u)(t). Thanks again much appreciated.

 

[rude man]

Actually, the complete convolution integral limits are -∞ to +∞. But the lower limit changes to zero because your h(τ) = 0, τ < 0, and the upper limit changes to t since U(t-τ) = 0 for τ > t. So then v(t) = 3∫₀^t τ e^-4τ dτ
and you can let t = 0.25 for your particular problem after performing the definite integral.