What is the result of $x(at) * δ(t-k)$

Question

It seems rational that $x(at) * δ(t-k) = x(at - ak)$. I tried to prove this: Let $x_1(t) = x(at)$ then $x_1(t) * δ(t-k) = x_1(t-k) = x(at - ak)$. But I thought this as well: $$x(at) * δ(t-k) = x(at) * δ\left(\frac{at-ak} a \right) = a(x(at) * δ(at-ak)) = ax(at-ak)$$ Well I know that this may sound silly but I can't find where I'm wrong.

Answer 1

Dirac Delta is not zero at $t = k$ hence

$$x(at)\delta(t-k) = x(ak)$$

By the general distribution theory

$$f(x)\delta(x-a) = f(a)$$

Convolution

When it's about convolution, a simple rule reads:

\begin{align} f(t)*\delta(t-a)=\int_0^t f(t-s)\delta(s-a)\,ds&=\begin{cases} 0, &t<a,\\ f(t-a) &t\ge a,\end{cases}=f(t-a)\theta(t-a), \end{align} where $\theta(t)$ denotes the unit step function.

Can you solve it for your function?

The fastest solution

Call $t-k = p$ in order to have

$$x(a(p+k))*\delta(p) = x(a(\tau + k))$$

In which we made use of the convolution identity

$$F[x\pm a]*\delta(x) = F[\tau \pm a]$$

Answer 2

The way you "prove" $x(at)*\delta(t-k)=x(at-kt)$ is not correct. Note that, starting from the signal $x(t)$, two operations are performed in series:

1. A time scaling operation to the signal $x(t)$, resulting in the signal $y(t)=x(at)$
2. A shift of the signal $y(t)$ by $k$, resulting in the signal $z(t)=y(t)*\delta(t-k)=y(t-k)=x(a(t-k))=x(at-ak)$.

The second equation in the "alternative" proof does not hold.