Derivative of unit step function

  • Thread starter Will
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  • #1
Will

Main Question or Discussion Point

[SOLVED] Derivative of unit step function

How does one do this, for example x= e^(-3t)u(t-4); how do you get x' ??
 

Answers and Replies

  • #2
enigma
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Do laplace transforms on it.
 
  • #3
Hurkyl
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Write out the definition of the unit step function and it might be easier to see.
 
  • #4
Will
I think I got it now. I used the property L{f'}(s) = sL{f}(s) - f(0)
Is that correct?
 
  • #5
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The Fourier Transform

can also be used. It can be used for many unbounded functions.
 
  • #6
Hurkyl
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You could just differentiate it directly.

x(t) = e^(-3t)u(t-4)

is equivalent to:

Code:
x(t) =  e^(-3t)   (for t > 4)
           0      (for t < 4)
with x(4) depending on the precise definition of u.

Differentiating on each piece gives:

Code:
x'(t) = (-3) e^(-3t)   (for t > 4)
          0            (for t < 4)
And x'(4) does not exist because x(t) is discontinuous at t = 4

IOW:

x'(t) = (-3) e^(-3t) u(t - 4) for t [x=] 4
 
Last edited:
  • #7
ahrkron
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Sometimes you can safely assume the derivative of a step to be a delta function (for instance, when you integrate a delta, you get a step).

They need to be used as distributions, and there may be some requirements on the functions you use along with them (integrability, continuity,...).

I'm sorry I don't remember much about it.
 

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