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Delta Function Graphs

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data

    For each of these sketch and provide a formula for the function (i.e. in terms of ##u(t)##, ##\delta(t)##) and its derivative and anti-derivative. Denote the ##\delta## function with a vertical arrow of length 1.

    (a) ##f(t)=\frac{|t|}{t}##

    (b) ##f(t)=u(t) exp(-t)##

    2. Relevant equations

    $$u'(t)=\delta(t)$$

    $$\int^t_{-\infty} \delta (\tau) d \tau = u(t)$$

    3. The attempt at a solution

    (a) ##f(t)=\frac{|t|}{t}## in terms of the Heaviside unit step can be written as:

    $$f(t)=2 u(t)-1$$

    Using the property stated above the derivative becomes:

    $$f'(t)=2 \delta(t)$$

    But shouldn't it be zero? Because when I take the "normal" derivative of ##f(t)=\frac{|t|}{t}## it becomes 0 (derivative of -1 for t<0 an 1 for t>0).

    Here is the sketch of f(t) and the two possible versions of f'(t):

    sketch1.jpg

    For the antiderivative, if we integrate the original function we will have ##\int f(t) dt=|x| + c##. But I am not sure how to represent it in terms of ##\delta## because we do not know the indefinite integral of u(t). :confused:

    (b) So here I made a graph of this function and its derivative:

    sketch2.jpg

    From the graph for t<0 the derivative of zero is just 0. For t>0 derivative of e-t is -e-t (this is also equal to this function's anti-derivative).

    But now if we differentiate this in terms of delta using the product rule, we get ##f'(t)=-u(t)e^{-t}+e^{-t} \delta (t)##. This is very different from the previous answer. And how can I represent ##e^{-t} \delta (t)## on the graph?

    I have noticed that in some websites they draw the delta function as just being scaled by the size of the jump discontinuity, so in this case:

    sketch3.jpg

    So I am not sure which one of these solutions is correct. Any help is greatly appreciated.
     
  2. jcsd
  3. Mar 9, 2016 #2

    BvU

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    (a) You're doing fine. There is no indefiniti integral, but the exercise only asks for the antiderivative.

    (b) something goes wrong at t=0: I see the function jump up but your derivative stays below 0. In fact you can use the rule for differentiation of a product here.

    That is exactly what the exercise composer means with
     
  4. Mar 9, 2016 #3
    Thank you very much for the response.

    But when I differentiate, where do I exactly need to introduce the ##\delta(t)## into the equation for ##f'(t)##? :confused:

    For t<0 we have (by product rule):

    $$f'(t)= u'(t) e^{-t} + - e^{-t} u(t) = u'(t) e^{-t}$$

    Do I need to rewrite this as ##\delta(t) e^{-t}## or just write it as zero (since the u(t) is constant on ##(-\infty, 0)##)?

    Likewise for t<0:

    $$f'(t)= u'(t) e^{-t} + - e^{-t} u(t) = u'(t) e^{-t} - e^{-t}=0$$

    It is also possible to introduce delta here as: ##f'(t) = \delta(t) e^{-t}##.

    Beut if I have a separate case for t=0, the derivative becomes ##\delta(0)e^{-0} - e^{-0} u(0)##. But I am not sure how to evaluate this since the u(t) is infinitely steep there, and does not have a particular value.

    So I think this would be how the graph looks like for the derivative:

    sketch4.jpg

    Is that correct, or do I need to scale the (t) somehow?

    Do we not need to use (t) or something else in the expression for the antiderivative? Because the graph looks exactly the same as the one for the derivative (and we have a discontinuity at 0):

    image.jpg

    For t<0, ##f(t)=0 \implies F(t)= \int 0 = 0 + c##, and for t>0 ##f(t)= e^{-t} \implies F(t) = - e^{-t}##.
     
    Last edited: Mar 9, 2016
  5. Mar 10, 2016 #4

    BvU

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    $$
    f'(t)= u'(t) e^{-t} - e^{-t} u(t) $$And $$u'(t) e^{-t} = \delta(t) e^{-t} = \delta(t) \ \ ! $$It's so simple you would almost overlook it :smile: !

    derivative plot looks good to me now; no scaling needed.

    However, for the antiderivative: if I differentiate that, do I get f(t) ?
     
  6. Mar 10, 2016 #5
    Thank you. No, I think we need to multiply it by the Heaviside step function: ##F(t)= -e^{-t} u(t)##.

    But is the graph right for F(t)?

    Would it be correct to write the following?

    For t<0, $$f'(t)= \delta(t)$$

    And for t>1, $$f'(t)= \delta (t) - e^{-t}$$

    Whenever we have t=0, in both cases ##f'(0)=\infty##.
     
  7. Mar 10, 2016 #6

    BvU

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    for t < 0 and for t > 0 ##\ \ \delta(t) =0\quad ## so yes, it's correct.

    However, you should understand that you can't write "for t = 0 ##f'(t)=\infty##"

    You can only think that :smile: -- and it's exactly to deal with that kind of situations that the ##\delta## 'function' is so useful: it allows us to write $$
    f'(t)= \delta(t) - u(t)\;e^{-t} \quad\forall t \qquad $$


    Coming back to the graph of antiderivative: If I differentiate that, I get a ##-\delta## that isn't in f(t). so: no, the graph can't be right !
     
  8. Mar 10, 2016 #7
    Would it then be correct to add a ##+\delta## to the graph of F(t) at zero?

    Is the actual formula of ##F(t)=-e^{-t} u(t)## for the antiderivative correct? Because clearly the integral is ##-e^{-t}## on the RHS of the graph and zero to the left.
     
  9. Mar 10, 2016 #8

    BvU

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    Expanding on
    The (pretty extensive -- worth investigating) Wiki lemma calls this a heuristic characterization ("can be loosely thought of"). For physicists this is usually good enough and we treat it (with proper caution) as an ordinary function.
     
  10. Mar 10, 2016 #9

    BvU

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    As I try to bring across: no. Check for yourself and do the differentiation !
     
  11. Mar 10, 2016 #10
    Yes, we don't get the original function when differentiating using the product rule (we want the derivative of F(t) to be ##u(t)e^{-t}##). So what is the mistake? :confused:

    As I said in the last post when we integrate the original function we get ##-e^{-t}## on the right, and zero on the negative side. The only way I can think of representing that is ##-e^{-t} u(t)##, but when we differentiate that we get ##-e^{-t} u'(t) + u(t) e^{-t} = f'(t)##.

    EDIT: we can rewrite ##-e^{-t} u'(t) + u(t) e^{-t} = -e^{-t} \delta (t) + u(t) e^{-t}## which is equal to ##u(t)e^{-t} = f(t)## everywhere for ##t \neq 0##. But I'm not sure if that's correct...
     
  12. Mar 10, 2016 #11

    BvU

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    As a candidate for the antiderivative $$
    F(t)=-e^{-t} u(t)
    $$ is almost good. We are free to add a constant to the function. But you need something to get rid of the delta function when you differentiate F. Now (hint, hint), what yields a ##\delta## 'function' when differentiated ?
     
  13. Mar 10, 2016 #12
    Thank you very much for the massive hint. So for ##F(t)=-u(t)exp(-t) +u(t)## this is how the plot looks like in Matlab:

    sk2.jpg
    Hopefully it's correct now.
     
  14. Mar 10, 2016 #13

    BvU

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    Well done. Indeed, the delta function is gone when we differentiate F. In other words: the discontinuity in F itself is gone.

    A good and useful exercise. After working with these things for a while you take a lot for granted and don't think twice what it actually means.
     
  15. Mar 10, 2016 #14
    Thank you very much for the help. :oldsmile:
     
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