Convolution of unit step function with itself

Question

How does the convolution of the unit step function with itself compute?

Convolution integral I am referring to

I appreciate the response

Answer 1

Let $$f(t) = (h*h) = \int_{-\infty}^\infty h(t-\lambda)h(\lambda)d\lambda $$ For $\lambda<0$ we have $h(\lambda) = 0$. So $$f(t) = \int_{0}^\infty h(t-\lambda)d\lambda$$ Here $\lambda$ changes from $0$ to $\infty$ and $t$ goes from $-\infty$ to $\infty$. If $t\lt0$ then $f(t) =0$ and if $t \gt0$ then $$f(t) = \int_{0}^t 1d\lambda = t$$ So the answer is $f(t) = h(t)\times t$.