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Zorich
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In DTFT of a signal, the spectrum of a sequence is periodic with period $2\pi$ and all the information needed for derivation of the original signal from its spectrum is contained in $\pi <\omega <\pi$.

But , why do they integrate over $2\pi$ and not from $-\infty$ to $+\infty$ , as in continuous Fourier transform in inverse DTFT? After all, when you derive a transform of a signal, the inverse of it is not arbitrary and to recover the original signal, you can't integrate over an other interval because all the information is contained in there.

In DTFT of a signal, the spectrum of a sequence is periodic with period $2\pi$ and all the information needed for derivation of the original signal from its spectrum is contained in $\pi <\omega <\pi$.

But , why do they integrate over $2\pi$ and not from $-\infty$ to $+\infty$ , as in continuous Fourier transform in inverse DTFT?

In DTFT of a signal, the spectrum of a sequence is periodic with period $2\pi$ and all the information needed for derivation of the original signal from its spectrum is contained in $\pi <\omega <\pi$.

But , why do they integrate over $2\pi$ and not from $-\infty$ to $+\infty$ , as in continuous Fourier transform in inverse DTFT? After all, when you derive a transform of a signal, the inverse of it is not arbitrary and to recover the original signal, you can't integrate over an other interval because all the information is contained in there.

Source Link
Zorich
  • 171
  • 5

Why integrate over $2\pi$ in inverse DTFT?

In DTFT of a signal, the spectrum of a sequence is periodic with period $2\pi$ and all the information needed for derivation of the original signal from its spectrum is contained in $\pi <\omega <\pi$.

But , why do they integrate over $2\pi$ and not from $-\infty$ to $+\infty$ , as in continuous Fourier transform in inverse DTFT?