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This is basically a communication engineering and signal processing question. However, since this question involves mathematics, I was adviced by the members of Electrical Engineering Stack exchange to post it in this forum. So if anybody is involved in this field and have concepts pertaining to unit step and unit impulse functions, they may solve this problem.

It would be very helpful if someone could provide the intuitive reasoning and mathematical proof for:

$$\lim \limits_{a \to 0} \frac{1}{a}[u(\frac{t}{a}+\frac{1}{2})-u(\frac{t}{a}-\frac{1}{2})]= \delta(t) $$

where,

u(t)=Unit step function.

$\delta(t)$ is the unit impulse function.

Thank you.

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    $\begingroup$ $\delta$ $\delta$ $\endgroup$ – Hurkyl Oct 16 '17 at 14:09
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Look at the two cases:

$t\neq0$:

If $|a| < |t|$ then $|\frac{t}{a}| > 1$ so both $u(\frac{t}{a}+\frac{1}{2})$ and $u(\frac{t}{a}-\frac{1}{2})$ have the same sign.

Because either $\frac{t}{a} < -1$ then we get $\frac{t}{a}+\frac{1}{2} < -1 + 0.5 = -0.5 < 0 $ and $\frac{t}{a}-\frac{1}{2} < -1 - 0.5 = -1.5 < 0 $. $\quad$ So both arguments of the unit step function are negative which leads to: $u(\frac{t}{a}+\frac{1}{2})-u(\frac{t}{a}-\frac{1}{2}) = -1 - (-1) = 0$

or $\frac{t}{a} > 1$ then we get $\frac{t}{a}+\frac{1}{2} > 1 + 0.5 = 1.5 > 0 $ and $\frac{t}{a}-\frac{1}{2} > 1 - 0.5 = 0.5 > 0 $. $\quad$ So both arguments of the unit step function are positive which leads to: $u(\frac{t}{a}+\frac{1}{2})-u(\frac{t}{a}-\frac{1}{2}) = 1 - 1 = 0$

So $u(\frac{t}{a}+\frac{1}{2}) - u(\frac{t}{a}-\frac{1}{2}) = 0$ which makes the series constant $0$ from that point on and the limit being $0$.

$t=0$

$\frac{1}{a}[u(\frac{0}{a}+\frac{1}{2}) - u(\frac{0}{a}-\frac{1}{2})] = \frac{1}{a}[1- (-1)] = 2*\frac{1}{a}$

which should be a well known series that diverges to infinity.

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  • $\begingroup$ Sir, I have understood your explanation. Could you please elaborate this point a bit more: If $|a| < |t|$ then $|\frac{t}{a}| > 1$ so both $u(\frac{t}{a}+\frac{1}{2})$ and $u(\frac{t}{a}-\frac{1}{2})$ have the same sign. $\endgroup$ – Soumee Oct 16 '17 at 13:37
  • $\begingroup$ Sir, could it mean that when the magnitude of a is so less than t, that it makes t/a very large value(in magnitude), the addition or subtraction of 0.5 makes both $u(\frac{t}{a}$ and $\frac{1}{2})$ fall on the same side (either +ve or -ve) of x axis? $\endgroup$ – Soumee Oct 16 '17 at 13:58
  • $\begingroup$ Yes exactly.... $\endgroup$ – mathdotrandom Oct 16 '17 at 14:06
  • $\begingroup$ Sir, then what is the significance of 1/2 here? Can it be any other fraction? $\endgroup$ – Soumee Oct 16 '17 at 14:11
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    $\begingroup$ Yes it would be the same result for every other numbers. We need one positive and one negative number in the second case though. $\endgroup$ – mathdotrandom Oct 16 '17 at 14:15

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