I need to determine if the following system is time-invariant or not, and I'm a bit unsure about it. $$y(t)= \int_{-\infty}^{t-2} \tau \cdot x(2\tau)d\tau $$
Asked
Active
Viewed 779 times
1
-
Have you checked if $y(t+\delta t) = \int_{-\infty}^{t- 2} \tau\cdot x(2\tau + \delta t)d\tau$? – user7530 Apr 03 '14 at 15:31
-
@copper.hat Hmm it is? How so? – user7530 Apr 03 '14 at 15:36
-
I don't think so? The system is time-invariant if you can compute $y(t+\delta t)$ by replacing $x(s)$ by $x(s+\delta t)$ at every point in the calculation of $y(t)$ – user7530 Apr 03 '14 at 15:42
-
@copper.hat $x'(2a) = x(2a+\delta t)$... $\delta t$ has no dependence on $a$. – user7530 Apr 03 '14 at 15:44
-
@copper.hat A system $y(x(s), t)$ is time-invariant if, for every shift $\delta t\in \mathbb{R}$, $$y(x(s+\delta t), t) = y(x(s), t+\delta t).$$ – user7530 Apr 03 '14 at 15:54
-
@user7530: I am removing my embarrassing comments above... – copper.hat Apr 03 '14 at 16:01
2 Answers
0
The system is not time-invariant. Consider the response to the input signal $x_1(t)=x(t-t_0)$:
$$y_1(t)=\int_{-\infty}^{t-2}\tau x_1(2\tau)d\tau=\int_{-\infty}^{t-2}\tau x(2\tau-t_0)d\tau=\int_{-\infty}^{t-2-t_0/2}(\tau+t_0/2) x(2\tau)d\tau \neq y(t-t_0)$$
Matt L.
- 10,636
0
Let $x=1_{ [0,2] }$, then for $t > 100$, we have $y_{x}(t) = \int_{-\infty}^{98} \tau 1_{ [0,2] }(2 \tau) d \tau = \int_0^1 \tau d \tau = {1 \over 2}$.
If we let $x' = 1_{ [2,4] } $ (note that $x'(t) = x(t-2)$), then $y_{x'}(t) = \int_{-\infty}^{98} \tau 1_{ [2,4] }(2 \tau) d \tau = \int_1^2 \tau d \tau = {3 \over 2}$.
Hence the system is not time invariant.
If it was, we would have $y_{x'}(t) = y_x (t-2)$.
copper.hat
- 172,524