Sign function property proof

[roam]

Homework Statement

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

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 $S(0)=0$?

 

[Charles Link]

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 $\nu=0$...

 

[mfb]

Don't trust a given formula for S(v) here. You can derive S(0) yourself, and the result should be zero.

 

[roam]

Don't trust a given formula for S(v) here. You can derive S(0) yourself, and the result should be zero.

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?

 

[Charles Link]

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?

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.

 

[roam]

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?

 

[Charles Link]

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?

Simple: $2\pi j \nu S(\nu)=F.T.( 2 \delta(t))$. The F.T. of the delta function is 1.

 

[roam]

Thank you, that makes perfect sense now.