I am wondering what the convolution of these 2 functions is equal:
- $$\int_{-\infty}^{\infty}y\left(t\right)x\left(\alpha\left(t-\tau\right)\right)$$
- $$ \int_{-\infty}^{\infty}y\left(t\right)x\left(\alpha t-\tau\right) $$
I tried to use some convolution properties to check which of them are correct but it didn't come to any results.
thanks, Adam.
Your definition of convolution appears to be incorrect.
Convolution is defined as:
$$(f*g)(t) = \int_{-\infty}^\infty f(\tau)g(t-\tau) d\tau $$
Rewriting your expression 1 to what I think you meant:
$$\begin{align*} \int_{-\infty}^\infty y(\tau)x(\alpha(t-\tau)) d\tau \\ \end{align*}$$
This is just $y(t)$ convolved with a scaled version of $x(t)$. If we let $u(t) = x(\alpha t)$ then
$$\begin{align*} \int_{-\infty}^\infty y(\tau)x(\alpha(t-\tau)) d\tau &= (y*u)(t)\\ \end{align*}$$
Rewriting your expression 2 to what I think you meant:
$$\begin{align*} \int_{-\infty}^\infty y(\tau)x(\alpha t-\tau) d\tau \\ \end{align*}$$
This is just a scaled version of the convolution of $y(t)$ and $x(t)$:
$$\begin{align*} \int_{-\infty}^\infty y(\tau)x(\alpha t-\tau) d\tau &= \int_{-\infty}^\infty y(\alpha u)x(\alpha (t-u)) \alpha du\\ \\ &= \alpha\cdot(y*x)(\alpha t)\\ \end{align*}$$
To summarize:
Expression 1 has one operand scaled in time and the other is not.
Expression 2 has the result scaled in time and magnitude, or equivalently both operands are scaled in time and magnitude.
In the general case, where $u(t) = x(\alpha t)$:
$$(y*u)(t) \ne \alpha\cdot(y*x)(\alpha t) $$