1
$\begingroup$

Prove: $$ \left. \int_{-\infty}^{\infty}\frac{d\delta(t)}{dt}\phi(t)dt=\frac{-d\phi(t)}{dt}\right|_{t=0} $$

$\delta(t)$ is impulse signal and other one is any general signal.

  • Can anybody do the proof for me?
  • Or at-least how to proceed further?
$\endgroup$
4
  • $\begingroup$ Integrate by parts. $\endgroup$ – Jazzmaniac Feb 13 '17 at 12:27
  • 3
    $\begingroup$ Shouldn't this be asked in math SE? $\endgroup$ – Tendero Feb 13 '17 at 13:32
  • $\begingroup$ Which of $f$ and $\phi$ is the impulse? $\endgroup$ – Dilip Sarwate Feb 13 '17 at 14:23
  • 4
    $\begingroup$ This question needs some editing. You didn't explain what $f(t)$ and $\phi(t)$ are. $\endgroup$ – Matt L. Feb 13 '17 at 17:21
2
$\begingroup$

For any pair of functions $u(t)$ and $v(t)$ we have that

$$\frac{d}{dt}(u(t)v(t))=v(t)\frac{du(t)}{dt}+u(t)\frac{dv(t)}{dt}$$

If we integrate on both sides of the equation we get

$$\int\frac{d}{dt}(u(t)v(t)) \ dt=\int v(t)\frac{du(t)}{dt} \ dt+\int u(t)\frac{dv(t)}{dt} \ dt$$

In this case, $u(t)=\delta(t)$ and $v(t)=\phi(t)$, so

$$\int\frac{d}{dt}(\delta(t)\phi(t)) \ dt=\int \phi(t)\frac{d\delta(t)}{dt} \ dt+\int \delta(t)\frac{d\phi(t)}{dt} \ dt$$

The left side of the equation can be solved easily as

$$\int\frac{d}{dt}(\delta(t)\phi(t)) \ dt = \frac{d}{dt}\int\delta(t)\phi(t) \ dt=\frac{d}{dt}\phi(0)=0$$

as $\phi(0)$ is a constant. We also know that the delta function has the following property

$$\int \delta(t)\frac{d\phi(t)}{dt} \ dt = \left.\frac{d\phi(t)}{dt}\right|_{t=0}$$

So going back to the first equation and replacing with these results:

$$0=\int \phi(t)\frac{d\delta(t)}{dt} \ dt+\left.\frac{d\phi(t)}{dt}\right|_{t=0}$$

And that's the proof.

$\endgroup$
2
  • $\begingroup$ Nice effort! You could also begin with $\int{\phi(t) \delta(t) dt} = \phi(0)$ and then proceed as $\int{\phi(t)' \delta(t) dt} + \int{\phi(t) \delta(t)' dt}= 0$ and simply arrive at the conclusion as $\phi(0)' + \int{\phi(t) \delta(t)' dt}= 0$. for a much simpler layout. Note that when you work with generalized functions such as the $\delta(t)$ all these classical sense operators of differentiation, integration, and limit are just virtually acting on the impulse function. $\endgroup$ – Fat32 Feb 14 '17 at 21:20
  • $\begingroup$ @Fat32 Thanks! Yes, that may have been simpler. I wanted to start from the very beginning anyway, just in case the OP couldn't work out the missing steps by his own. Apart from that, I must admit I'm not that familiar with distribution theory, so I followed the steps that helped me understand it when I studied this. $\endgroup$ – Tendero Feb 14 '17 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.