Derivative of sawtooth - interpretation of result

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Gennaro Arguzzi
Gennaro Arguzzi 2017-7-3
编辑： David Goodmanson 2017-7-4
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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回答（1 个）

David Goodmanson
David Goodmanson 2017-7-3
编辑：David Goodmanson 2017-7-3
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
  2 个评论
Gennaro Arguzzi
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
David Goodmanson 2017-7-3
编辑：David Goodmanson 2017-7-4
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)
If your prof wrote that effectively
t*delta(t-3) = delta(t-3)
I would say he or she made a mistake.

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