Help me get a Fourier transform of sinc

can anyone help to get the Fourier transform of
sinc(sq) (1000*pi*t) ??? (sinc square) in "w" domain .....

fourier sinc

hi murat,
You can either use the Fourier Transform pairs to get it, or you'll have to use the residue theorem from complex analysis to solve it using the Residue Theorem

sinc fourier

You have it.

sinc function fourier

The fourier transform of a triangular function is sinc function.Then by symmetry of FT pairs,FT of sinc square will be triangular function.

sinc squared

The fourier transform of a sinc function is not a triangular function.

THe fourier transform of a sinc is a box shaped function. Rectangular.

https://en.wikipedia.org/wiki/Sinc_function

Hope that helps.

sinc function fourier transform pair

hi we know that
F \[\frac{\sin(Wt)}{\pi t}\] is \[rect(\frac{\omega}{2W})\]

and also we know that multiplication in time domain is convolution in frequency domian. i.e.,

\[x(t)\times h(t) \longarrow \frac{1}{2 \pi}X(j\omega)*H(j\omega)\]

using \[h(t) = x(t)\] we get

\[x^2(t) \longarrow \frac{1}{2\pi}X(j\omega)*X(j\omega)\]

For us \[x(t) = \frac{\sin(1000 \pit)}{\pi t}\]

Thus the FT of \[\frac{\sin^{2}(Wt)}{(\pi t)^{2}}\] turns out to be \[1000 tri(\frac{\omega}{2000 pi})\]. Hope this helps you

thnx

purna!

Added after 2 minutes:

hi we know that
F.T. of \[\frac{\sin(Wt)}{\pi t}\] is \[rect(\frac{\omega}{2W})\]

and also we know that multiplication in time domain is convolution in frequency domian. i.e.,

\[x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)\]

using \[h(t) = x(t)\] we get

\[x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)\]

For us \[x(t) = \frac{\sin(1000 \pit)}{\pi t}\]

Thus the FT of \[\frac{\sin^{2}(Wt)}{(\pi t)^{2}}\] turns out to be \[1000 tri(\frac{\omega}{2000 \pi})\]. Hope this helps you

thnx

purna!

Added after 1 minutes:

hi we know that
F.T. of \[\frac{\sin(Wt)}{\pi t}\] is \[rect(\frac{\omega}{2W})\]

and also we know that multiplication in time domain is convolution in frequency domian. i.e.,

\[x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)\]

using \[h(t) = x(t)\] we get

\[x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)\]

For us \[x(t) = \frac{\sin(1000 \pit)}{\pi t}\]

Thus the FT of \[\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}\] turns out to be \[1000 tri(\frac{\omega}{2000 \pi})\]. Hope this helps you

thnx

purna!

Added after 30 seconds:

hi we know that
F.T. of \[\frac{\sin(Wt)}{\pi t}\] is \[rect(\frac{\omega}{2W})\]

and also we know that multiplication in time domain is convolution in frequency domian. i.e.,

\[x(t)\times h(t) ---> \frac{1}{2 \pi}X(j\omega)*H(j\omega)\]

using \[h(t) = x(t)\] we get

\[x^2(t) ---> \frac{1}{2\pi}X(j\omega)*X(j\omega)\]

For us \[x(t) = \frac{\sin(1000 \pi t)}{\pi t}\]

Thus the FT of \[\frac{\sin^{2}(1000 \pi t)}{(\pi t)^{2}}\] turns out to be \[1000 tri(\frac{\omega}{2000 \pi})\]. Hope this helps you

thnx

purna!

sinc square

basically,the fourier transform of triangular function is sinc(sq),if the tr fn range is of A to B(time fn) and amplitude is C then the fourier transform will be (B-A)*c*sinc(sq)((B-A)f)

Hi, I think you can find the anwser here
https://eagle.lamost.org/?p=38679

It showes every steps clearly.:smile:

It show the Fourier Transform of sinc(x) function --> a rectangular pulse.

But the question is the FT of sinc(x)^2 --> a triangular pulse.

The easiest way is simply enter Fourier(sinc(x)^2) at WolframAlpha