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

enter image description here

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.

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2 Answers 2

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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\$.

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  • \$\begingroup\$ i got it. Thank you so much. You explained really well \$\endgroup\$
    – Mayank Pal
    Sep 9, 2016 at 19:09
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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.

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