$x(2t) * \delta(5t+5)$

Question

How can i compute such convolution? $$x(2t) * \delta(5t+5)$$ I know that $$x(t) * \delta(t) = x(t)$$ but how I tackle such one? Thanks

Answer 1

A fundamental formula (see for example here) is:

\begin{equation} \delta(f(x))=\textstyle \sum_i\frac{\delta(x-a_i)}{\lvert f^\prime(a_i)\rvert}. \end{equation}

where the $a_i$ are the roots of $f$. Have you seen it ?

Here with $f(t)=5t+5$, you can use it to convert:

$$\delta(5t+5)=\frac15 \delta(t+1)=\frac15 \delta_{-1}(t)$$

Convolving with $\delta_a$ amounts to a shift of $a$ units. Once again, did you know this important result ?

Therefore, your final result is the translation of $x(2t)$ on the left of one unit followed by multiplication by $\frac15$.

Said otherwise, the result is:

$$\frac15 x(2(t+1))$$

(Thanks to Lutz Lehmann who pointed an error...)

Answer 2

Hint: \begin{align} &\int_\mathbb{R} x(2(t-s))\,\delta(5s+5)\,ds \\ &\stackrel{u=5s+5}{=}\frac15\int_\mathbb{R} x(2(t-\frac15(u-5)))\delta(u)\,du \end{align}