Signal operation, shifting and scaling

Question

so I have a question regarding this continuous time signal:

$$y(t) = \int_{-\infty}^t x(2\tau) \, d\tau$$ Now the question was to find if this function was causal, so i proceeded to check the impulse response of $\delta(t)$ and $\delta(t-1)$, when i inserted these functions for $y(t)$ it appeared to be causal. Although, I was told I had made a mistake when inserting $\delta(t-1)$ for $x(2\tau)$. I did $\delta(2(\tau -1 ))$ I was told it should be $\delta(2\tau -1)$. The explanation i received only confused me more so, I am hoping for some clarification. Thankyou.

Answer 1

You cannot assume that the system can be characterized by a (one-dimensional) impulse response (i.e., by its response to the input signal $x(t)=\delta(t)$). This is only the case if the system is linear and time-invariant (LTI), but the given system is in fact time-varying.

But you don't need the impulse response to judge whether a system is causal or not. You just need to check if you need future input values to compute the current output signal or not. By substituting $u=2\tau$ your input-output relation becomes

$$y(t)=\frac12\int_{-\infty}^{2t}x(u)du\tag{1}$$

from which you can immediately see that for $t>0$ you need to know the future behavior of the input signal (up to $2t$) to be able to compute $y(t)$. So the system is not causal.

Answer 2

If you have a function $x(t)$, then $x(2\tau)$ means, in a sense, "the expression you get when you replace every $t$ with a $2 \tau$".

So, for example, if $x(t) = (t-1)^2 + t\sin(e^t)$, then $$ x(2 \tau) = ((2 \tau) - 1)^2 + (2 \tau) \sin(e^{2 \tau}) $$ Similarly, if $x(t) = \delta(t - 1)$, then $x(2 \tau) = \delta((2 \tau) - 1)$.