# Convolution with Heaviside function (integration)

To clarify notation, I use $u_n = 1$ when $x>n$, and $0$ otherwise.

I am having troubles with the following convolution/integration:

$u_2(t) \ast sin(\sqrt{2}t) = \int^t_0u_2(\tau) \cdot sin(\sqrt{2}(t-\tau))\ d\tau$.

At first I thought of splitting the integral up so that I can make the Heaviside function some definitive value (0 or 1) on an interval, but I do not know how that would work since $t$ has no specific value. That leads me to think perhaps my problem is that I am not very good at integration.

Any hints or tips will be appreciated.

• Well if $t\leq 2$ then the integral is zero, while if $t>2$ it is $\int_2^t{\sin(\sqrt{2}(t-\tau)d\tau}=\sqrt{2}(\cos(\sqrt{2}(t-2))-1)$
– RTJ
Commented Apr 8, 2015 at 19:40
Since $u_2(\tau)=0$ $\forall \tau\in(-\infty,2)$ and $u_2(\tau)=1$ $\forall \tau\in[2,\infty)$ the integral can be written as $$\int_0^t{u_2(\tau)\sin(\sqrt{2}(t-\tau))d\tau}=\bigg\{\array{0\:,\qquad t\leq 2\\ \int_2^t{\sin(\sqrt{2}(t-\tau))d\tau},\quad t>2}$$ The last integral can be calculated by a simple change of variables $w=\sqrt{2}(t-\tau)$ $$\int_2^t{\sin(\sqrt{2}(t-\tau))d\tau}=\sqrt{2}\int_0^{\sqrt{2}(t-2)}{\sin(w)dw}=\sqrt{2}\left[1-\cos(\sqrt{2}(t-2))\right]$$
• Thank you. So essentially it is equal to $u_2(t) \cdot \int^t_2 sin(\sqrt{2}(t- \tau ))d\tau$. Thank you once again!