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The strength of the delta function by definition is infinity, so how come in some questions a number is assigned to the strength of the impulse function? Although these numbers are not necessarily big.

Ethan Bolker
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Jack
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2 Answers2

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A "delta function" is not really a function, it's what is called a "generalized function", or distribution. So, when you see people write $\delta(0) = \infty$, this is just a "formality", and a very misleading one at that. The only correct way to think about a delta function is under an integral sign with another ("normal") function $f(x)$:

$$ \int \delta(x)f(x)dx $$ This expression makes perfect sense, and as long as $f(x)$ is "nice enough" (e.g. continuous at 0), it will evaluate to $f(0)$. If someone wants to "scale" a delta function, they will usually write $a\delta(x)$, but again, this isn't a function, so it doesn't make sense to write

$$ a\delta(x) = \left\{\begin{array}{cc} 0 & x\neq 0 \\ a\cdot \infty & x = 0 \end{array}\right. $$ so the only way to make sense of $a\delta$ is under an integral sign:

$$ \int a\delta(x)f(x)dx = a\int \delta(x)f(x)dx = af(0) $$

icurays1
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  • Thank you, I read that aδ(x), means that a is the area under the delta function which is its strength too. It does not makes sense if we look at it as a function, so it is just a definition of a distribution, meaning the area under the distribution is always equal to their strengths? I read the wiki link but I find it a bit confusing. – Jack Apr 07 '16 at 19:47
  • Yes, the correct way to think about the "strength" of the delta function is its area. – icurays1 Apr 07 '16 at 20:04
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The Dirac function is not really a function, rather a distribution representing a point mass or impulse. It is helpful to view the Dirac function as a limit of functions $$ \delta_a(t) = \left\{ \begin{array}{cl} \dfrac{1}{2a}, \ \ & |t| < a \\[10pt] 0, \ \ & \text{otherwise} \end{array} \right. , $$ as $a \to 0^+.$ Notice that $\int^\infty_{-\infty} \delta_a (t) \ dt = 1$ regardless of the value of $a.$ If we scaled by positive constant, say $ m \delta_a(t),$ then we would increasing the "impulse".

matt biesecker
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  • Isn't the strength the area under the distribution then? In the limit of function that you wrote , does it account for area? – Jack Apr 07 '16 at 19:21