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I'm trying to figure out whether the input $x(t)$ and output $y(t)$ is time invariant. I was able to solve the other questions but I ran into a problem with this one. I'm bad at editing these so I posted it as a picture

$$y(t)=\int_{t}^{t+1}x(\tau-\alpha)d\tau$$

where $\alpha$ is a constant.

This is how I usually solve these. I'm posting an example from my text book

Textbook examples

I just don't know how to do this question cause it has two terms, and it's not the t variable, and there's a constant. I'd really appreciate the help please.

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  • $\begingroup$ Look how I edited the question. It would be better to learn some Latex... $\endgroup$ – msm Sep 28 '16 at 2:55
  • $\begingroup$ Latex? I haven't really heard of that before, I'll look into that, thank you $\endgroup$ – Drew U Sep 28 '16 at 3:13
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The output to $x_1(t)$ is $$y_1(t)=\int_{t}^{t+1}x_1(\tau-\alpha)d\tau$$ Now select $x_2(t)=x_1(t-t_0)$ $$y_2(t)=\int_{t}^{t+1}x_1(\tau-t_0-\alpha)d\tau$$ Choose $\tau'=\tau-t_0\Rightarrow d\tau=d\tau'.$ $$y_2(t)=\int_{t-t_0}^{t-t_0+1}x_1(\tau'-\alpha)d\tau'=y_1(t-t_0)$$

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  • $\begingroup$ I get it, you replaced the τ with τ-t0. I thought I'd have to figure out a way to substitute it with a variable. So this is time invariant. The only thing I didn't get was why you didn't replace t with t-t0 in the first y2(t) expression like you did in the last one. Was it just not necessary? $\endgroup$ – Drew U Sep 28 '16 at 3:19
  • $\begingroup$ I just calculated the output the shifted input in the first one, according to the definition of the system. $\endgroup$ – msm Sep 28 '16 at 3:21
  • $\begingroup$ Oh, okay. I get it great now, thanks $\endgroup$ – Drew U Sep 28 '16 at 3:23

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