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

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

$= \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


1 Answer 1


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 $y(t-d)$.

  • $\begingroup$ So basically if the upper limit will become $x(t)$ and then take inputs then the system-2 will become Time-Invariant $\endgroup$
    – fpsshubham
    Commented Oct 4, 2017 at 12:10
  • $\begingroup$ You mean this : $$y(t)=\mathcal{T} \{ x(t) \} = \int_{-\infty}^{x(t)} x(\tau) d\tau $$ ? $\endgroup$
    – Fat32
    Commented Oct 4, 2017 at 12:49
  • $\begingroup$ Yes thats what i am sayin $\endgroup$
    – fpsshubham
    Commented Oct 4, 2017 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.