# Sign function property proof

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

## 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##?

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 ##...

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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.

mfb said:
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?

roam said:
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.

mfb and 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?

roam said:
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.