Resampling of DTFT

Question

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?

Answer 1

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.

Answer 2

First the FFT of $$x[n]=[1, 1, 1, 1]$$ and its inverse for $N=2^3$

where red dot indicates sample location.

Now shift samples as given by you,

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