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Looking for a solution to the following problem:

A signal $x(t)$ is band limited to $B$ Hz, and sampled above the Nyquist rate, with corresponding $f_s = 1/T$. If the sampled signal is given by

$$ y(n) = x(nT), $$

describe how to compute the phase-shifted sampled signal

$$ z(n) = x(nT + T/2), $$

from the samples $y(n)$ without first reconstructing $x(t)$ from $y(n)$ and resampling.

Have not been able to find one on the web.

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2 Answers 2

Because the sampling frequency is above the Nyquist rate, the original signal can be written in terms of its samples $y(n)$:

$$x(t)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(t-mT)/T]}{\pi(t-mT)/T}\tag{1}$$

By setting $t=nT+T/2$ you get from (1)

$$z(n)=\sum_{m=-\infty}^{\infty}y(m)\frac{\sin[\pi(n-m)+\pi/2]}{\pi(n-m)+\pi/2}$$

and since

$$\sin[\pi(n-m)+\pi/2]=(-1)^{n-m}$$ you finally get

$$z(n)=\sum_{m=-\infty}^{\infty}y(m)\frac{(-1)^{(n-m)}}{\pi(n-m)+\pi/2}$$

So for computing $z(n)$ you do not need to reconstruct $x(t)$ but you need to compute a sum over all values $y(m)$. In practice a few samples $y(m)$ around the value $n$ will be sufficient because the weighting factors in the sum decrease with increasing distance $|n-m|$.

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Yet another possibility is to zero-pad the signal sample vector, FFT it, rotate the phase of each FFT result bin linearly with the bin index, and IFFT a time shifted result.

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