I have a system with the following input/output relation:
$$ y(t)=x(-t) $$
and I want to prove (not graphically/draw) that its not TI (time invariant).
I tried to write down $y(t-T)$ and compare it to the response to $x(t-T)$ but I get both terms equal which shouldn't be the case.
Thanks.
Answer
Let $y_1(t)$ be the response to the signal $x_1(t)$:
$$y_1(t)=x_1(-t)\tag{1}$$
Now let $x_2(t)$ be a shifted version of $x_1(t)$:
$$x_2(t)=x_1(t-T)\tag{2}$$
The response to $x_2(t)$ is
$$y_2(t)=x_2(-t)=x_1(-t-T)\tag{3}$$
If the system were time-invariant, its response to $x_2(t)$ should be a shifted version of its response to $x_1(t)$:
$$y_2(t)=y_1(t-T)\tag{4}$$
However, from (1) we have $y_1(t-T)=x_1(-(t-T))=x_1(-t+T)$. Comparing this to (3) we see that (4) is not satisfied, and, consequently, the system is not time-invariant. Now that you've seen the proof, try drawing all the signals in order to gain a better understanding.
No comments:
Post a Comment