I am studying Signals and Systems.

The textbook told me $\delta'(t)$ has the following properties.

$1$. $x(t)\delta'(t-t_0)=x(t_0)\delta'(t-t_0)-x'(t_0)\delta(t-t_0)$

$2$. $\displaystyle\int_{-\infty}^t\delta'(\tau-t_0)d\tau=\delta(t-t_0)$

$3$. $\displaystyle\delta'(at+b)=\frac{1}{|a|}\delta'\left(t+\frac ba\right)$

Please prove these three equations mathematically.

Thank you.

  • $\begingroup$ using which formalism? Theory of distributions? $\endgroup$ – Nikolay Gromov Sep 21 '15 at 12:29

In the theory of distributions one defines $$ D_{\delta(t-t_0)}[f]=f(t_0)\;\;,\;\;D_{\delta'(t-t_0)}[f]=-D_{\delta(t-t_0)}[f'] $$ using these definitions, we think about $x(t)\delta'(t-t_0)$ as another distribution, such that formally $$ D_{x(t)\delta'(t-t_0)}[f]=D_{\delta'(t-t_0)}[x(t)f(t)]=-x'(t_0)f(x_0)-x(t_0)f'(t_0) $$ which matches the r.h.s. of 1

For 3) one simply uses the definition $$ D_{\delta'(a t+b)}(f(t))=-D_{\delta(a t+b)}(f'(t)) $$ and then the propery of $\delta(a t+b)=1/|a| \delta(t+b/a)$. So in fact your 3. is incorrect. It should be $$\delta'(a t+b)={\rm sign}(a) \delta'(t+b/a)$$

2) is just a notation, depending on its interpretation there is nothing to prove

  • $\begingroup$ Eh...2 is applying $\delta'(t-t_0)$ to the function $1_{(-\infty,t]}$. So it's not entirely trivial. $\endgroup$ – Ian Sep 21 '15 at 12:50
  • $\begingroup$ Most definitions of the distributions assume that the space of the probe functions is at least differentiable, this is not the case for the step function, so that's really just a notation. As a physicist I am perfectly OK with this notation... $\endgroup$ – Nikolay Gromov Sep 21 '15 at 12:52
  • $\begingroup$ For general distributions, the test functions are smooth and compactly supported. But for "nice" distributions, we can make extensions. For example, the Dirac delta extends nicely to all continuous functions. In this case you're right, I was a bit hasty, since the result of applying $\delta'$ to the step function can only be understood as a distribution, not a number (and thus really shouldn't be understood as being anything at all). $\endgroup$ – Ian Sep 21 '15 at 12:57
  • $\begingroup$ with the step function there is even a problem with delta function, as the step function is not really well defined at the origin... It is safe to assume $\Theta(0)=1/2$ which gives the right result in most of the reasonable cases. $\endgroup$ – Nikolay Gromov Sep 21 '15 at 13:01
  • 1
    $\begingroup$ that's why I mentioned "at least differentiable" which is the maximal extension one could allow for delta' $\endgroup$ – Nikolay Gromov Sep 21 '15 at 13:04

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