# DFT of signal with frequency-dependent time shift

I have a signal $s[n]$, and I want to compute the DFT of it with a frequency selective shift (computing all $k$ here but I really only care about $k > 0$)

$$S[k] = \sum_{n=0}^{N-1} s[n-\dfrac{N}{2k}] e^{(-j2\pi nk/N)}\space\space\space -N/2 \le k \lt N/2$$

Following the derivation of the general shift theorem here, I can make the change of variables $m = n-\dfrac{N}{2k}$:

$$S[k] = \sum_{m=-N/2k}^{N-1-N/2k} s[m] e^{(-j2\pi (m+N/2k)k/N)}$$

Carrying through the product in the exponential term:

$$S[k] = \sum_{m=-N/2k}^{N-1-N/2k} s[m] e^{(-j2\pi mk/N)} e^{-j\pi}$$

And realizing $e^{-j\pi} = -1$:

$$S[k] = -\sum_{m=-N/2k}^{N-1-N/2k} s[m] e^{(-j2\pi mk/N)}$$

Which, to my surprise, indicates that the shift drops out, though now I'm summing over shifted intervals. I believe that's just another phase adjustment that has to be made though, given $s[n]$ is periodic.

I have a hard time believing this, I would have expected something like a quadratic phase shift in frequency to result. Can anyone point to where I've made a mistake?

• How do you guarantee that the shift $N/2k$ is integer? And, btw, $e^{-j\pi}=-1$ (not $1$). – Matt L. Aug 23 '17 at 21:19
• I don't guarantee it, I'm hand waving a little bit, but I believe the new summation limits drop out (at least that's what the shift theorem derivation does), and the rest is handled by the continuous phase shift terms. Good eye on the $e^{-j\pi}$ – gct Aug 23 '17 at 21:25
• Well, IF for a given $k$ it happens that $N/2k$ is integer, then your result is correct (just replace $\Delta$ by $N/2k$ in the formula you linked to). If $N/2k$ is not an integer, the result doesn't make sense. – Matt L. Aug 23 '17 at 21:47
• I can find the value at any time point by interpolating if necessary, so I don't think the non-integral values matter. – gct Aug 24 '17 at 0:16