What is the derivative (in the engineer's or distributional sense) of the causal function $f(t)\theta(t)$, where $\theta$ is the Heaviside unit step function?

I've seen the formula $f'(t)\theta(t)+f(0)\delta(t)$, where $\delta$ is Dirac's delta. This looks like a kind of "product rule": differentiating the product gives $f'\theta+f\theta'$, but $\theta'$ is $\delta$, and $\color{blue}{f(t)\delta(t)=f(0)\delta(t)}$.

If this is right, I don't understand the following argument, from the solutions manual to Oppenheim and Wilsky's Signals and Systems. The solutions manual says the derivative of the function $2e^{-3t}\theta(t-1)$ is


It's the second term I don't understand. Using the "product rule" heuristic, the second term should be $2e^{-3t}\delta(t-1)$, which using the blue formula above gives $\color{red}{2e^{-3}}$ times the delayed delta function, not just twice the delayed delta function.

Is the solutions manual wrong?

  • $\begingroup$ One cannot differentiate the Heaviside step function in the classical sense (since $\theta\notin C^{1}(\mathbb{R})$), thus one must opt to deduce the distributional derivative. I presume that you define $$\theta(t)=\begin{cases}1,&t>0, \\ 0,& t\le 0.\end{cases}$$ Thus, for all test functions $\phi\in\mathcal{D}(\mathbb{R})$, one can compute, for $f\in C^{\infty}(\mathbb{R})$, $$\begin{aligned} \langle (f\theta)',\phi\rangle&=-\langle f\theta,\phi'\rangle \\ &=-\langle\theta,f\phi'\rangle \\ &=-\int_{0}^{\infty}f(t)\phi'(t)\,dt\end{aligned}$$ $\endgroup$ – Jason Born Feb 28 '17 at 17:39
  • $\begingroup$ @user3482534: yes, I'm working in engineer's notation, where it is customary to give calculations and formulas like I did. Can you translate your comment into a more direct answer to the question about whether the solutions manual is wrong? $\endgroup$ – symplectomorphic Feb 28 '17 at 17:42

You are right.

Indeed, $$\langle (e^{-3t}\theta(t-1))',\phi\rangle = -\langle e^{-3t}\theta(t-1),\phi'\rangle = -\int_1^\infty e^{-3t}\phi'(t)dt$$ $$=-e^{-3t}\phi(t)\big|_1^{\infty}-3\int_1^\infty e^{-3t}\phi(t)dt = e^{-3}\phi(1) - 3 \langle e^{-3t}\theta(t-1),\phi \rangle = e^{-3}\langle\delta_1,\phi\rangle - 3 \langle e^{-3t}\theta(t-1),\phi \rangle, $$ hence we can conclude that

$$(e^{-3t}\theta(t-1))' =- 3 e^{-3t}\theta(t-1) +e^{-3} \delta_1. $$

I think the authors of the manual forgot that they were working with $\theta(t-1)$, not just $\theta(t)$.


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