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Help me get a Fourier transform of sinc

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murat_mc

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

elmolla

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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
 

neils_arm_strong

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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.
 

purnapragna

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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!
 
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maxwell_28

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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)
 

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