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I have a constant digital signal that is 1 for every sample and of length 4. 4 point DFT coefficients are $$[4,0,0,0]^T$$ Obviously. I wonder, if I resample the DTFT such that samples are taken at $$w=[\pi/4, 3\pi/4,5\pi/4,7\pi/4]^T $$ If I take the inverse DFT, can i find the same signal as before? If not, what is the effect of shifting the samples?

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The DTFT of a length $N$ sequence $x[n]$ is given by

$$X(e^{j\omega})=\sum_{n=0}^{N-1}x[n]e^{-jn\omega}\tag{1}$$

Sampling the DTFT at frequencies $\omega_k$ with

$$\omega_k=\frac{\pi}{4}(2k+1),\qquad k=\{0,1,2,3\}\tag{2}$$

and with $N=4$ we obtain from $(1)$ and $(2)$

$$\begin{align}X(e^{j\omega_k})&=\sum_{n=0}^{3}x[n]e^{-jn\omega_k}\\&=\sum_{n=0}^{3}x[n]e^{-jn\pi (2k+1)/4}\\&=\sum_{n=0}^{3}x[n]e^{-jn\frac{\pi}{4} }e^{-j2\pi nk/4}\tag{3}\end{align}$$

Comparing $(3)$ with the formula for a length $N=4$ DFT

$$\tilde{X}[k]=\sum_{n=0}^3\tilde{x}[n]e^{-j2\pi nk/4}\tag{4}$$

we see that an inverse DFT of $(3)$ results in a sequence

$$\tilde{x}[n]=x[n]e^{-jn\frac{\pi}{4} }\tag{5}$$

I.e., shifting the spectrum corresponds to a modulation of the original sequence.

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  • $\begingroup$ Great answer. I would like to add 1 more question: How do I reconstruct the dtft in $w$ by using these shifted samples? I guess I should use the sinc interpolation, but how do the interpolant functions change with the changing sampling shift? $\endgroup$ – strahd Feb 2 at 13:13
  • $\begingroup$ @strahd: Is this related to this question of yours? I'm actually not completely sure what you're asking and what exactly you're trying to do. Maybe you can edit that other question to clarify what exactly it is that you're trying to do. $\endgroup$ – Matt L. Feb 2 at 18:04
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First the FFT of $$x[n]=[1, 1, 1, 1]$$ and its inverse for $N=2^3$

enter image description here

where red dot indicates sample location.

Now shift samples as given by you,

enter image description here

I may be totally wrong but keen to learn my mistakes. Nevertheless, the $\left|x[n]\right|$

enter image description here

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  • $\begingroup$ I am sorry but this teems wrong :/ $\endgroup$ – strahd Jan 31 at 20:56
  • $\begingroup$ I am working on it. Please give me bit more time... $\endgroup$ – jomegaA Jan 31 at 21:24
  • $\begingroup$ Ah ok I am sorry did not mean to bother you $\endgroup$ – strahd Jan 31 at 21:26

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