Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac function derivatives?

  1. Jul 3, 2007 #1

    I have a system from which i want to compute a expression for a time domain impulse response. The expressions for modulus and phase of the system is quite complicated and I'm using maple in order to do the inverse transforming.

    now, maple tells me the inverse transform is an expression involving derivatives of the dirac delta function, like this:

    h(t) = exp( c0 ) * ( c1 * dirac(t) + c2 * dirac(2,t) + c3 * dirac(4,t) )

    (where c0 c1 c2 and c3 are just numbers if I plug in some values for my variables. )

    now my question, how do I interpret dirac(2,t) and dirac(4,t) ? the maple help says they represent 2nd and 4th order derivates of the delta function...? dirac(t) is trivial but I really don't understand what the nth order derivative of the dirac function is supposed to mean here... help!
  2. jcsd
  3. Jul 3, 2007 #2


    User Avatar
    Science Advisor

    Fred, I'm not really sure about dirac derivatives either but they sound pretty nasty. My hunch is that their presence in h(t) implies that your system is non-casual (output depends on derivatives of input). You can't actually build such a system, not in an ideal sense anyway.

    Like all distributions they only really make sense when inside an integral, like a convolution-integral or integral-transform of some kind.

    I was just scratching around with delta(1,t) now and I worked out the following result which I think underlies what these nasties are doing in your impulse response :

    [tex]\int f(\lambda) \, \delta(1,x-\lambda) \, d \lambda = f'(x)[/tex]

    Here the integration is over any interval containing x. (BTW I worked this out by applying first principles differentiation to the delta function and exchanging the order of limits and integration).

    I assume you would get something similar but involving f''(x) and f''''(x) for convolution integrals with delta(2,x) and delta(4,x), that's how I came to the conclusion that your system is non-causal.
    Last edited: Jul 3, 2007
  4. Jul 3, 2007 #3


    User Avatar
    Science Advisor

  5. Jul 3, 2007 #4
    The "Dirac Delta function" is not really a function. It is referred to as a "Generalized Function". However, these things are not mentioned in engineering and physics classes which is why most students have not heard of them. The idea is the a "Generalized Function" is a sequence of regular functions. And we view this Delta function as its limits. Then the ideas of differentiability are extended (generalized).
  6. Jul 3, 2007 #5

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I get a sign difference from what uart has.


    [tex]\int^{\infty}_{-\infty} f \left( x \right) \delta' \left( x \right) dx,[/tex]

    and assume that standard integration by parts still works. Take [itex]u = f \left( x \right)[/itex] and [itex]dv = \delta' \left( x \right) dx.[/itex] Then,

    [tex]\int^{\infty}_{-\infty} f \left( x \right) \delta' \left( x \right) dx = \left[ f \left( x \right) \delta \left( x \right) \right]^{\infty}_{-\infty} - \int^{\infty}_{-\infty} f' \left( x \right) \delta \left( x \right) dx.[/tex]

    The test function [itex]f \left( x \right)[/itex] dies at [itex]\pm \infty[/itex], so only the second term survives, wich gives [itex]- f' \left( 0 \right)[/itex].


    [tex]\int^{\infty}_{-\infty} f \left( x \right) \delta^{(n)} \left( x \right) dx = \left(-1 \right)^n f^{(n)} \left( 0\right). [/tex]
    Last edited: Jul 3, 2007
  7. Jul 3, 2007 #6


    User Avatar
    Science Advisor

    No George you got exactly the same as what I got. In my first "scratchings" I got the negative sign exactly as you have posted. Though I proceded a different way I got the same result (and later the link provided by HallsOI confirmed that this negative sign is indeed correct there).

    But since the impulse response h(t) is normally used within a convolution integral I rewrote the integral in that form, that is :

    [tex]\int h(x-\lambda) \, f(\lambda) d \lambda[/tex]

    and the negative signs cancelled.

    EDIT: I don't mean that the whole (-1)^n vanishes, I only did it for the case of n=1. All I meant is that when you put it in a convolution integral you get one extra negative sign. The general case would be :

    [tex] \int f(\lambda) \, \delta(n,x-\lambda) \, d\lambda = (-1)^{n+1} f^{(n)}(x) [/tex]
    Last edited: Jul 3, 2007
  8. Jul 3, 2007 #7

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right; I scanned your post too quickly and missed the negative sign in front of the integration variable. :grumpy:
  9. Jun 18, 2008 #8
    Hi uart. Could you explain why such a case would make the system non-causal? Im not sure how the derivative of the input would violate causality. (All you would have to do is make find the derrivative as (x-xo)/(t-to), where you use x as the current value, and x0,t0 as the value a time step backwards. No?

    Also, im have the same question as the OP, but im using matlab. So, if I had [tex]\frac{d}{dt} \delta [/tex]

    What would the ans. be? It seems that the derivative of the impulse should give itself. It has no slope, then a big spike. So you'd expect the result to look exactly the same.
  10. Mar 30, 2010 #9
    Right off the bat check which derivative you are going wrt to. Usually the $\delta$ function is written in terms of spatial variables, not time. If this is the case and all you need is a time derivative then the answer is trivial.

    The following is in response to the last couple sentences in your post: You are attempting to qualitatively take the derivative of something that seems like a function; it is quite bold to expect that the result of such vague computations will hit on the correct idea. Keep in mind that all of these are generalized functions; furthermore, when considered as a typical function, they are a.e. equal to the 0 function and thus as far as integration can tell us, equal to the 0 function. This conclusion is neither interesting nor useful. Hence, the use of generalized functions.

    There is a bit of mathematical theory here you must understand before trying to obtain an intuitive feel for the situation. A generalized function in this sense is an operator defined on function spaces that acts on functions by a "recipe" involving integrals. You should research this topic on your own (for instance, mathworld is a good starting place) if you want to understand the situation any better.

    BTW, there is one part of your intuition that didn't suffer from faulty logic: you are correct about the behavior of these functionals "away from zero"; I am giving credit where credit is due.

    Sorry for reviving such an old post but this is the first thing in various google searches while trying to research derivatives of delta functions, so this post is more of a public service announcement to people coming from wherever.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook