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I understand that the dirac delta function basically describes a pulse of area one, if the pulse is very narrow, the height will be infinity. However, Im confused because sometimes, people consider the dirac delta function as being one at t=0 (AKA the unit impulse function) rather than infinity, so when is it considered to be one rather than infinity?

P.S. In MATLAB the dirac delta function is infinity at t=0

S.s.
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  • They are probably referring to the dirac measure, which places a mass of $1$ at the singleton $0$. – Daniel Xiang Aug 19 '18 at 22:33
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    Don't confuse the Dirac Delta with the Kronecker Delta https://en.wikipedia.org/wiki/Kronecker_delta . The Dirac delta is used for continuous time; the Kronecker delta for discrete time. – Andy Walls Aug 19 '18 at 23:05
  • I know the difference between the two, what Im talking about is continuous time, sometimes they take the dirac delta as one and sometimes as infinity. – S.s. Aug 19 '18 at 23:18
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    The only thing I can think of is that given some metric $d(x,y)$ then $\frac{d(x,y)}{d(x,y)+1}$ is also a metric with $\infty = 1$. – CyclotomicField Aug 19 '18 at 23:35
  • The Dirac delta function, $\delta(t)$, has no meaning by itself, but is rather defined as $\int_{-\infty}^{\infty} \delta(t) \mathrm{d}t = 1$. As Andy Walls pointed out, this is different from the Kronecker Delta function that is used in Digital signal processing which is defined as $\delta(n) = \left{ \begin{array}{rl} 1 & \text{if } n = 0 \ 0 & \text{if } n \neq 0 \end{array} \right. $ – Winter Soldier Aug 20 '18 at 01:21
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    The convention $\delta(0)=1$ is used in some books about math for engineers. Just ignore them. The correct intuition is indeed that the "value" at zero is infinite and not 1. – Abdelmalek Abdesselam Aug 20 '18 at 14:56
  • Thats where I get confused, Im an engineer and the books I read sometimes take 1 for the magnitude of the dirac delta. – S.s. Aug 21 '18 at 01:57

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As you say, the "value" of the delta "function" is infinite at $0$ and the "area under it" is $1$. I am not aware of any time we take the value to be $1$. The unit impulse refers to the total impulse delivered, which is the area under the force-time curve.

Ross Millikan
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  • So say, if we take the derivative of the unit step function, then we get a dirac delta with a magnitud of 1 at zero? – S.s. Aug 20 '18 at 01:40
  • Yes, that is true. You just have to remember that the magnitude of a delta is not the value at $0$ but the integral across $0$. – Ross Millikan Aug 20 '18 at 02:15