system delay: $x(2t-t_o) \,\,or \,\, x(2t-2t_o)$?

Question

I have a system as this: $x(t)--->system-->y(t) = x(2t)$

If i delay the output y(t), what do i get $y(t) = x(2t-t_o)\,\, or\,\, y(t) = x(2t-2t_o)$? What is the correct expression and why?

Answer 1

The answer follows by simply noting that an arbitrary signal $z(t)$ delayed by $t_0$ is written as $z(t-t_0)$, that is, replace the variable $t$ appearing in the argument of $z(\cdot)$ by $t-t_0$. In your case, the signal $z(t)=x(2t)$. Therefore, the output of the delay equals $z(t-t_0)=x(2(t-t_0))=x(2t-2t_0)$.

Therefore, the signal $x(2t-2t_0)$ is generated from $x(t)$ by the following two steps (in that order):

1. The signal $x(t)$ is "time compressed" by a factor of $2$, resulting in $x(2t)$;
2. The compressed signal is delayed by $t_0$ resulting in $x(2(t-t_0))$.

It may be useful to see what happens when you reverse the order:

1. The signal $x(t)$ is delayed by $t_0$, resulting in $x(t-t_0)$;
2. The delayed signal is then time compressed by a factor of $2$, resulting in $x(2t-t_0)$.

Of course, the signal you get when you reverse the order of operations is different and this is captured by the notation.