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Derivative of a unit step function

  1. Mar 9, 2013 #1
    So I'm busy struggling with some worked examples in my signals class. I understand the theory from the notes and textbook but I cannot seem to apply them to proper examples.
    We are asked to find the derivative of g(t) = (1-e^(-t))*u(t) where u(t) is a unit step function.

    I know the derivative of u(t) is the delta function, d(x). So when I try solve the derivative I use the chain rule and get:
    g'(t) = e^(-t)*u(t) + (1-^e(-t))*d(x)
    However I get stuck at this point and not sure where to go from here.

    Thanks in advance
     
  2. jcsd
  3. Mar 9, 2013 #2

    vela

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    That should be d(t), not d(x).

    Use the fact that f(t)δ(t) = f(0)δ(t).
     
  4. Mar 9, 2013 #3

    Ray Vickson

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    You do not need the δ-function, since g(t) is continuous at t = 0 (no jump discontinuity). However, the left-hand and right-hand derivatives are different, so the answer will have a u(t) in it.
     
  5. Mar 10, 2013 #4

    tiny-tim

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    Welcome to PF!

    Hi Crabbz! Welcome to PF! :smile:

    (have a δ from the "Quick Symbols" box next to the Reply box :wink:)
    (as vela :smile: says, of course that should be δ(t) :wink:)

    Technically, it's not the delta function, it's the dirac delta, which isn't a function at all, it's a "generalised function", or "distribution", which really only works inside an integral.

    (unfortunately, they're both written δ ! :rolleyes:)

    In particular, dirac δ(0) is not 1, it's actually ∞ (just look at the graph … how else can you get its total integral to equal 1 ?! :biggrin:)
    Let's take an easier example, f(x) = (2x + 3)u(x) …

    just by looking. that's the negative x-axis, then a line of slope 2 starting at (0,3): so its derivative is obviously 0 for x < 0, and 2 for x > 0, and there's a discontinuity at x = 0.

    Now let's apply the chain rule, instead of just looking:

    f'(x) = 2u(x) + (2x + 3)u'(x)

    for x ≠ 0, that gives us f'(x) = 2u(x) + 0 (which is correct for x≠ 0)

    for x = 0, it doesn't really tell us anything, since u'(0) = ∞​

    The way to interpret f'(x) = 2u(x) + (2x + 3)u'(x) at x = 0 is to look at a small interval (h,k), with k > 0, and integrate:

    ∫ f'(x) dx = ∫ 2u(x) dx + ∫ (2x + 3)u'(x) dx

    the LHS is f(k) - f(h), ie 2(k - h) if h ≥ 0, and 2k + 3 if h < 0

    the first term on the RHS is 2(k - h) if h ≥ 0, but only 2k if h < 0

    the second term on the RHS is (2*0 + 3)*0 = 0 if h ≥ 0, and (2*0 + 3)*1 = 3 if h < 0

    so it all adds up! :smile:
     
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