Heaviside step function with function as argument

Question

Is the following computations correct?

Can the Heaviside step function have an arbitrary function as argument?

It seams reasonable and leads to the correct/same answers, but I have not been able to find any source of information about it.

Does anyone know where I can find information about the Heaviside step function with a function as argument?

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$f(t) := |t| = H(t)t+\left((1-H(t)\right)(-t)=\left[2H(t)-1\right]t$

$g(t) := \sin(t)$

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$f^\prime(t) = 2H(t)-1 + 2t\delta(t)= 2H(t)-1 $

$g^\prime(t) = \cos(t)$

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$f^{\prime\prime}(t) = 2\delta(t) $

$g^{\prime\prime}(t) = -\sin(t)$

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$h(t) := |\sin(t)|= f\left[(g(t)\right]= \left[2H(\sin(t))-1\right]\sin(t)$

$h^\prime(t) = f^\prime\left[g(t)\right] g^\prime(t) = \left[2H\left[\sin(t)\right]-1\right]\cos(t)$

$h^{\prime\prime}(t) = f^{\prime\prime}\left[g(t)\right] \left[g^\prime(t)\right]^2 + f^\prime\left[g(t)\right] g^{\prime\prime}(t) = \\ = 2\delta\left[\sin(t)\right]\cos^2(t) + \left[2H\left[\sin(t)\right]-1\right]|\left[-\sin(t)\right]=\\ = 2\delta\left[\sin(t)\right]\cos^2(t) -h(t) =\\ = \{\sin(t)=0 \Leftrightarrow t=\pm n\pi,\; n\in\mathbb{Z}\Leftrightarrow \cos^2(t)=1\}=\\ =2\delta\left[\sin(t)\right]-|\sin(t)|$