Given $x(t)=\delta\left(t-\dfrac{\pi}{2}\right)$
Is it correct to say that $|x(t)|=\delta\left(t-\dfrac{\pi}{2}\right)$ ?
Or can't we just apply magnitude to Delta Function? Thanks in advance.
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$\begingroup$ Since $\delta_{t_0}$ is a positive-definite distribution, you could say that $|\delta_{t_0}| = \delta_{t_0}$, yes. $\endgroup$– m93aFeb 7, 2021 at 18:47
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1 Answer
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As opposed to what phisicist think, the Dirac delta is not a function. The Dirac delta is a measure. That is it is defined on a $\sigma$-algebra.
In simple terms $\delta_0(I)=1$ if and only if $0\in I$ and $\delta_0(I)=0$ otherwise, for some class of subsets $I$ of $\mathbb{R}$.
So your question becomes: can we give meaning to the absolute value/magnitude of a measure?
The answer is yes and is the same idea that is behind the absolute value of a function: if $f^+$ and $f^-$ are two positive functions and $$f=f^+-f^-$$ then $$|f|=f^++f^-.$$ Any function can be written as the difference of a positive and negative part.
Less trivially any finite measure can be written as the difference of two finite positive measures: $\mu=\mu^+-\mu^-$. Then what you speak of is the total variation measure of a measure (a little bit of abuse): $$|\mu|=\mu^++\mu^-.$$ As such the total variation measure of a positive measure such as the Dirac delta is just the Dirac delta itself: $|\delta_0|=\delta_0$.
