Invertiblity of LTI System

Can any one please explain me how the convolution of the following leads to delta(t).

To convolve these to function i considered delta(t+T) as x(t) and

delta(t-T) as the response. Then these two function overlaps only at t=-T and then the convolution should result in only delta(t+T). Then why delta(t) is the answer.

Your reasoning can bring you to the answer, but it's not spot on. Consider: $$x(t) = \delta(t)$$ $$h(t) = \delta(t-\tau)$$ $$y(t) = x(t)*h(t) = \delta(t-\tau)$$

System output to an impulse at $t = 0$ is another impulse at $t = \tau$ (a time delay of $\tau$). If the system is LTI, then the following holds true:

$$x(t) = \delta(t+\tau)$$ $$h(t) = \delta(t-\tau)$$ $$y(t) = x(t)*h(t) = \delta(t)$$

Because the impulse now happens at $t = -\tau$, system output now happens at $t = 0$.

• i got it. Thank you so much. You explained really well Commented Sep 9, 2016 at 19:09

Treat convolution as: fold, slide, multiply, add. If the impulse at $t=\tau$ is folded (about the vertical axis) it lies exactly coincident with the impulse at $t=-\tau$, and this initial position corresponds to $t=0$. The product of the two impulses is the unit impulse.

Subsequently sliding, ie for all other values of $t$, the multiplication of the two impulses is zero.