I have two functions in a convolution problem, and I'm struggling to wrap my head conceptually around what I'm supposed to do. The functions are: \begin{align*} h(t) &= \delta(t-1) + \delta(t-3)\\ x(t) &= \delta(t-3) - 2\delta(t-4), \end{align*}
and I want to find $y(t) = x(t)*h(t).$
On
The spirit of the thing is that $\delta*f=f$ for almost any $f$. Similarly, letting $T_af(x)=f(x+a)$ be the translation operator, $(T_a\delta)*f=T_af$.
It does require something (there are several choices) to believe/prove/know that $\delta*\delta=\delta$... but, granting this, $(T_a\delta)*(T_b\delta)=T_{a+b}\delta$.
The convolution is defined as \begin{align*} x(t) * h(t) &=\int_{-\infty}^{\infty}x(\tau)\,h(t-\tau)\,d\tau\\ &=\int_{-\infty}^{\infty}[\delta(\tau-3) - 2\delta(\tau-4)]\,[\delta(t-\tau-1) + \delta(t-\tau-3)]\,d\tau\\ &=\int_{-\infty}^{\infty}\delta(\tau-3)\,\delta(t-\tau-1)\,d\tau\\ &+\int_{-\infty}^{\infty}\delta(\tau-3)\,\delta(t-\tau-3)\,d\tau\\ &-2\int_{-\infty}^{\infty}\delta(\tau-4)\,\delta(t-\tau-1)\,d\tau\\ &-2\int_{-\infty}^{\infty}\delta(\tau-4)\,\delta(t-\tau-3)\,d\tau. \end{align*} Now the way the Dirac delta function works, it "picks out" values under the integral sign. The fact that the other function is a Dirac delta function doesn't change this. So we obtain: \begin{align*} x(t) * h(t) &=\delta(t-3-1)+\delta(t-3-3)-2\delta(t-4-1)-2\delta(t-7-3)\\ &=\delta(t-4)+\delta(t-6)-2\delta(t-5)-2\delta(t-10). \end{align*}