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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.
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 $|n-m|$.
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.
