Prove that the following system is invertible.

$$y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{3t} x(\tau) \,\mathrm d \tau$$

Answer: yes, the system is invertible. I need some hint here, not the full solution.

  • $\begingroup$ i have tried Differentiation but there are limits that giving me problems. $\endgroup$ – fpsshubham Oct 4 '17 at 13:22
  • $\begingroup$ That's right, so basicaly you have a problem with differentiating an integral with respect to its variable limits ? $\endgroup$ – Fat32 Oct 4 '17 at 13:23
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    $\begingroup$ Yes that's right. I think i have to revise Calculus little bit $\endgroup$ – fpsshubham Oct 4 '17 at 13:26

In this problem of finding the inverse system (if it exists) its intuitive to try differentiating the integral as the system input/output is given by:

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

Before differentiating te integral however, I would like to make this little change which is quite clear I assume: $$y(t/3) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{t} x(\tau) d\tau$$

Then I differentiate both sides of the equality: $$ \frac{d}{dt} \left( y(t/3) \right) = \frac{d}{dt} \int_{-\infty}^{t} x(\tau) d\tau $$

Which proceeds as: $$ \frac{1}{3} y'(t/3) = x(t) $$ which yields the inverse system.

Note that differentiation of an integral with variable limits is known as the Leibnitz Rule and can be summarized as below:

given $$F(x) = \int_{\alpha(x)}^{\beta(x)} g(x,t) dt$$ then $$ \frac{d}{dx} F(x) = \frac{d}{dx} \int_{\alpha(x)}^{\beta(x)} g(x,t) dt $$ $$ F'(x) = g(x,\beta(x)) \beta'(x) - g(x,\alpha(x)) \alpha'(x) + \int_{\alpha(x)}^{\beta(x)} \frac{d}{dx} g(x,t) dt $$

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    $\begingroup$ Dude why are you so Intelligent i mean you have answer to every question $\endgroup$ – fpsshubham Oct 4 '17 at 14:05

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