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Question Number 129446 by Eric002 last updated on 15/Jan/21

f i n d t h e f o u r i e r t r a n s f o r m o f s i g n u m

f u n c t i o n

Answered by Olaf last updated on 16/Jan/21

sgn (x) = {\begin{cases} - 1, x < 1 \\ 0, x = 0 \\ + 1, x > 1 \end{cases}

\hat{sgn} (f) = \lim_{T \to \infty, α \to 0} {\int_{- \frac{T}{2}}^{+ \frac{T}{2}} {sgn}_{α} (t) e^{- j 2 π f t} d t}

with {sgn}_{α} (t) = {\begin{cases} - e^{α t}, t < 0 \\ + e^{α t}, t > 0 \end{cases}

\hat{sgn} (f) = \lim_{T \to \infty, α \to 0} {- \int_{- \frac{T}{2}}^{+ 0} e^{- j t (2 π f + j α)} d t

+ \int_{0}^{+ \frac{T}{2}} e^{- j t (2 π f - j α)} d t}

\hat{sgn} (f) = \lim_{T \to \infty, α \to 0} {\frac{1}{- α + j 2 π f} + \frac{1}{α + j 2 π f}

+ \frac{e^{- α \frac{T}{2}} e^{j π f t}}{α - j 2 π f} + \frac{e^{- α \frac{T}{2}} e^{- j 2 π f}}{α + j 2 π f}}

Finally, by imposing α T \to \infty :

\hat{sgn} (f) = \lim_{α \to 0} {\frac{j 4 π f}{- α^{2} - 4 π^{2} f^{2}}} = \frac{1}{j π f}

\hat{sgn} (f) = = \frac{1}{j π f}, f is the frequency [Hertz]

or \hat{sgn} (ω) = \frac{1}{\sqrt{2 π}} . \frac{1}{j π \frac{ω}{2 π}} = \sqrt{\frac{2}{π}} . \frac{1}{j ω}

(ω is the p u l s a t i o n [rad . s^{- 1}])

Commented by Eric002 last updated on 16/Jan/21

w e l l d o n e s i r

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