Simplify $y(t)=e^{-t}u(t) * \sum_{k=-\infty}^{\infty}\delta(t-3k)$ in the form of $y(t)=Ae^{-t}$ for $0\le t<3$ I am learning convolution, and this is one of my exercise problems. Let
$$y(t)=e^{-t}u(t) * \sum_{k=-\infty}^{\infty} \delta(t-3k)$$
Show that $y(t)=Ae^{-t}$ for $0\le t<3$, and determine the value of $A$.
Here, $u(t)$ is 1 if $t>0$ and 0 if $t<0$. (Undefined at $t=0$); $\delta(t)=0$ if $t\ne0$ and its integral over all $t$ on the real line is 1.
My (incorrect) solution:
If $0\le t<3$, then $\sum_{k=-\infty}^{\infty} \delta(t-3k)$ agrees at all points with $\delta(t)$, and $e^{-t}u(t)$ with $e^{-t}$, so
$$y(t)=e^{-t} * \delta(t)$$
Thus, $y(t)=e^{-t}$ and $A=1$.
Correct answer:
$$A = \frac{1}{1-e^{-3}}$$
Question:
I ask for a nice solution that leads to this answer, since my textbook only gives the answer. Also, I want to know which one of my assertions/premises are incorrect, since my final answer is wrong.
 A: Pretty sure you just assumed the sum meant it was already written as a convolution. This is the convolution of a sum with $e^{-t}u(t)$:
$y(t)=\int_{-\infty}^{\infty} (e^{-\tau}u(\tau)\sum_{k=-\infty}^{\infty} \delta(t - \tau -3k))d\tau$
$y(t)=\int_{-\infty}^{\infty} (e^{-(t-\tau)}u(t-\tau)\sum_{k=-\infty}^{\infty} \delta(\tau -3k)) d\tau$ by commutativity
$y(t)=e^{-t} \int_{-\infty}^{\infty} (e^{\tau}u(t-\tau)\sum_{k=-\infty}^{\infty}\delta(\tau -3k)) d\tau$
$y(t)=e^{-t} \int_{-\infty}^{t} e^{\tau} \sum_{k=-\infty}^{\infty}\delta(\tau -3k)d\tau$
$y(t)=e^{-t}  \int_{-\infty}^{t} \sum_{k=-\infty}^{\infty} e^{\tau} \delta(\tau -3k)d\tau$
$y(t)=e^{-t}  \int_{\infty}^{t} e^0 \delta(\tau) + e^{-3} \delta(\tau + 3) + e^{-6} \delta(\tau + 6 ) + ...d\tau$ for $0 < t < 3$
$y(t)=e^{-t} (e^0 + e^{-3} + e^{-6} + ...)$
This is an infinite geometric sum:
$A = \sum_{i=0}^{\infty} (e^{-3})^{i} = \frac{1}{1-e^{-3}}$
A: One way to restate @cmitch answer is to consider algebraic properties of convolution namely commutativity and the following
$$f(t)*\delta(t-a)=f(t-a)$$
And using these properties we have
$$e^{-t}u(t)*\delta(t-3k)=e^{-(t-3k)}u(t-3k)$$
Now for $0\leq t\lt 3$, $u(t)=1$ iff $-\infty\lt k\leq 0$ and $u(t-3k)=0$ otherwise.
So
$$y(t)=\sum_{k=-\infty}^0e^{-t}e^{3k}\cdot 1=e^{-t}\sum_{j=0}^{+\infty}e^{-3i}$$
And we have the expected result. There is nothing new in this proof because the convolution identity is just proved by manipulating the integrals.
