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Sign function property proof

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  1. Jun 16, 2016 #1
    1. The problem statement, all variables and given/known data
    The signum function is defined by$$sgn(t)=\left\{\begin{matrix}-1, \ t<0\\0, \ t=0 \\ 1, \ t>0 \end{matrix}\right.$$It has derivative$$\frac{d}{dt} sign(t) = 2 \delta(t)$$Use this result to show that ##j2\pi \nu S(\nu)=2,## and give an argument why ##S(0)=0.## Where ##S(\nu)## denotes the Fourier transform of the signum function

    3. The attempt at a solution

    I know that the Fourier transform of the sign function is:$$S(\nu)=\frac{1}{j\pi \nu}.$$
    If we substitute this into ##j2\pi \nu S(\nu),## we get ##2## as expected. But the question wants us to show this result without deriving an expression for the Fourier transform of ##sgn(t).## How can we do this?

    Furthermore, if we substitute ##0## in ##S(\nu)## we get ##S(0)=1/j\pi (0) = \infty.## So what does the question mean by giving an argument that ##S(0)=0##?
     
  2. jcsd
  3. Jun 16, 2016 #2

    Charles Link

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    If the F.T. of ## f(t) ## is ## F(\nu) ##, isn't F.T. ## f'(t)=j \nu F(\nu) ## ? (with possibly a factor of ## 2\pi ## depending on your definition of F.T. and your frequency variable.) And you can readily compute the F.T. of the delta function..I don't have a good answer for the S(0)=0 yet...editing...For S(0), it appears the function ## S(\nu) ## goes symmetrically from minus infinity to plus infinity near ## \nu=0 ##...
     
    Last edited: Jun 16, 2016
  4. Jun 16, 2016 #3

    mfb

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    Don't trust a given formula for S(v) here. You can derive S(0) yourself, and the result should be zero.
     
  5. Jun 16, 2016 #4
    How would you derive ##S(0)##? Did you use the sifting property ##S(0)=\int^\infty_{-\infty} \delta (t) \frac{1}{j \pi \nu} d\nu## or some other method?
     
  6. Jun 16, 2016 #5

    Charles Link

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    Why not use ## S(\nu)=\int sgn(t)exp(-i 2\pi \nu t) dt ## and set ## \nu=0 ##. The function is odd so the integral is 0. See also my post #2.
     
  7. Jun 16, 2016 #6
    So using the property ##\frac{d f(t)}{dt} \leftrightarrow j 2 \pi \nu F(\nu),## we will have:

    $$2 \delta (t) = j 2 \pi \nu S(\nu).$$

    But according to the question this should be equal to just ##2.## How do I make the Dirac delta disappear?
     
  8. Jun 16, 2016 #7

    Charles Link

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    Simple: ## 2\pi j \nu S(\nu)=F.T.( 2 \delta(t)) ##. The F.T. of the delta function is 1.
     
  9. Jun 16, 2016 #8
    Thank you, that makes perfect sense now.
     
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