Request
Please check my work. I am not certain how to calculate the convolution of the unit step function.
Given:
Find the convolution of $f(t)=t$ and $g(t)=u(t-1)$.
$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$
My Solution:
When $0<t<1$, $u(t-1)=0$ and when $t>1$, $u(t-1)=1$. We shall solve for $t>1$.
$$h(t)=t*1=\int_1^t 1\cdot (t-\tau) d\tau=\frac{t^2}{2}-\frac{1}{2}$$
Answer
When $0<t<1$, $h(t)=0$
When $t>1$, $h(t)=\frac{t^2}{2}-\frac{1}{2}$
Question
Is this correct?
Best Answer
Your solution isn't correct. The integral is \begin{align} \int_1^t(t - \tau)d\tau &= t\tau - \tau^2/2\Bigr|_1^t\\ &=t^2-t^2/2-t+1/2=\frac{t^2}{2}-t+\frac{1}{2}\\ &= \frac{(t-1)^2}{2} \end{align} You marked this post as a Laplace transform. Are we supposed to also take the transform of the convolution?