Calculating fourier transform at any frequency

If what you need is a simple practical method to do the interpolation,
then just multiply the "time domain" samples by the linear-phase
signal ${expleft(alphacdot i 2pi k /Nright)}_{k=0}^{N-1}$, where
$alphain(0,1)$ is the sub-sample shift in the "frequency domain."
(I'm not sure if an additional constant of absolute value 1 is
required here). As
already noted in previous answers, the motivation for this is the
assumption that your samples vector ${x_k}$ comes from sampling some
continuous-time signal ${X(t)}_{tin mathbb{R}}$, as in
$x_k=X(kT_s)$ for $k=0,ldots, N-1$, where $T_sin mathbb{R}_{++}$ is
the sampling period. If the continuous-time signal has a compactly
supported Fourier transform, than by Shannon's sampling theorem $X$ is determined by the infinite sequence ${X(kT_s)}_{kin mathbb{Z}}$ for small enough $T_s$. Since we only have $N$ entries
of the infinite sequence, we loose some information on $X$. But if
$X$ decays rapidly (rather than having a compact support), then at
least intuitively we don't loose much (I know that this is a dangerous
and imprecise statement :) ).

For example, try the following Matlab lines:

ttt=-32:31;

x_time=exp(-ttt.^2/10);

figure;plot(x_time);

x_freq = fft(x_time);

figure;plot(fftshift(abs(x_freq)));

figure

for alpha=-4:0.5:4

x_time_2 = x_time .* exp(2 * pi * i * (0:63)/64 * alpha);

x_freq=fft(x_time_2);

plot(fftshift(abs(x_freq)));

axis([26,37,0,10]);

grid on;

pause(1);

end

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