Monday, 18 January 2016

Proof of time-invariance of continuous-time system


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.


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