Derivative of sawtooth - interpretation of result

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Gennaro Arguzzi 2017-7-3

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编辑： David Goodmanson 2017-7-4

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Hi everyone. I derived a sawtooth function:

with the following code:

syms t
x=t*(heaviside(t)-heaviside(t-3));
y=diff(x);

The result is: y = heaviside(t) - heaviside(t - 3) - t*(dirac(t - 3) - dirac(t)). I don't understand why the delta Dirac is multiplied per t. I know that x(t)*delta(t-3)=x(0)*delta(t-3), but x(0)=0 thus, for matlab x(0)*delta(t-3)=0.

Thank you very much.

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Answer 1

David Goodmanson 2017-7-3

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编辑：David Goodmanson 2017-7-3

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Hello Gennaro,

This is just the formula for the derivative of a product,

t * ( heaviside(t) -heaviside(t-3) )
where
d/dt heaviside(t) = delta(t)  and   d/dt heaviside(t-3) = delta(t-3).

And as a distribution,

     Integral x(t)*delta(t-3) dt = x(3)
  and (assuming limits of integration include t=3)
     Integral x(3)*delta(t-3) dt = x(3) * Integral delta(t-3) dt = x(3)
  so effectively
     (x(t)-x(3))*delta(t-3) = 0 
  not 
     (x(t)-x(0))*delta(t-3) = 0

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Gennaro Arguzzi 2017-7-3

编辑：Gennaro Arguzzi 2017-7-3

Hi @David Goodmanson, I don't understand the graphic meaning of t*delta(t-3). When my professor evaluate the derivative, he wrote only the term d(t)/dt*delta(t-3)=1*delta(t-3). The graphic meaning of delta(t-3) is clear, but t*delta(t-3) is unclear for me. PS: I don't know the distributions. Thank you very much.

David Goodmanson 2017-7-3

编辑：David Goodmanson 2017-7-4

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Hello Gennaro, I will try to give a better version of what is in the answer. The delta "function" delta(t) is a kind of spike that has a nonzero value only at t=0, and the total area under its curve = 1. The translated function delta(t-3) is nonzero only when its argument is zero, so at t=3.

For the product t*delta(t-3), t=3 is the only value that matters and gives something nonzero, so you can substitute and effectively

t*delta(t-3) = 3*delta(t-3)

You can also look at what happens when you integrate across the location of the delta function.

    fundamental definition:
       Integral g(t)*delta(t-t0) dt = g(t0)   for any g(t)
    so if g(t) = t,
       Integral t*delta(t-3) dt = 3
    but
       Integral 3*delta(t-3) dt = 3    (delta function has area 1)
    so effectively
       t*delta(t-3) = 3*delta(t-3)