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$S_1:x(t)\longrightarrow y(t) = \int_{-\infty}^{3t}{x(\tau)\,d\tau}$

$x(t-t_0)\longrightarrow y_1(t) = \int_{-\infty}^{3t}{x(\tau-t_0)\,d\tau}$

$= \int_{-\infty}^{3t-t_0}{x(z)dz} \neq y(t-t_0) $ (Time Variant)


My teacher showed my the Above relation, shows that the system is Time Variant but if i change its upper limit to $3t-3t_0$ i.e $\int_{-\infty}^{3t-3t_0}{x(z)dz}$ then the system becomes Time-Invariant, my teacher told me this and now i am confused can someone tells me why upper limit $3t-3t_0$ this makes it time-Invariant

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The system-1 whose input/output equation is $$ y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{t} x(\tau) d\tau $$ is time-invariant.

Whereas the system-2 $$ y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{3t} x(\tau) d\tau $$ is time-varying.

On the other hand the system-3 $$ y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{3t-3t_0} x(\tau) d\tau $$ is still time-varying for any (fixed) value of $t_0$.

I presume that your teacher have tried to underline the fact that what makes the system-2 time-varying is the fact that $y(t-t_0) \neq \mathcal{T}\{x(t-t_0)\}$ which becomes as the equation shows:

$$y(t-t_0) = \int_{-\infty}^{3(t-t_0)} x(\tau) d\tau = \int_{-\infty}^{3t-3t_0} x(\tau) d\tau $$

and $$\mathcal{T}\{x(t-t_0)\} = \int_{-\infty}^{3t} x(\tau-t_0) d\tau = \int_{-\infty}^{3t-t_0} x(\tau) d\tau $$

These two would be equal if th upper limit of the second equation would be $3t-3 t_0$. That's what your teacher told. Since they are not equal therefore the system-2 is time-varying.

Note that artifically making the limit as $3t-3t_0$ is quite meaningless, as the fixed value of $t_0$ has nothing to do with the variable amount of shifts that occurs in $x(t-d)$ or $y(t-d)$.

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  • $\begingroup$ So basically if the upper limit will become $x(t)$ and then take inputs then the system-2 will become Time-Invariant $\endgroup$ – fpsshubham Oct 4 '17 at 12:10
  • $\begingroup$ You mean this : $$y(t)=\mathcal{T} \{ x(t) \} = \int_{-\infty}^{x(t)} x(\tau) d\tau $$ ? $\endgroup$ – Fat32 Oct 4 '17 at 12:49
  • $\begingroup$ Yes thats what i am sayin $\endgroup$ – fpsshubham Oct 4 '17 at 12:53

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