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 $\{\exp\left(\alpha\cdot i 2\pi k /N\right)\}_{k=0}^{N-1}$, where $\alpha\in(0,1)$ 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 ${X(t)}{t\in \mathbb{R}}$, as in $x_k=X(kT_s)$ for $k=0,\ldots, N-1$, where $T_s\in \mathbb{R}{++}$ is the sampling period. If the continuous-time signal has a compactly supported Fourier transform, than by Shannon's sampling theorem (http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem) $X$ is determined by the infinite sequence ${X(kT_s)}_{k\in \mathbb{Z}}$ 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

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 $\{\exp\left(\alpha\cdot i 2\pi k /N\right)\}_{k=0}^{N-1}$, where $\alpha\in(0,1)$ 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 $\{X(t)\}_{t\in \mathbb{R}}$, as in $x_k=X(kT_s)$ for $k=0,\ldots, N-1$, where $T_s\in \mathbb{R}_{++}$ 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 $\{X(kT_s)\}_{k\in \mathbb{Z}}$ 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

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 $\{\exp\left(\alpha\cdot i 2\pi k /N\right)\}_{k=0}^{N-1}$, where $\alpha\in(0,1)$ 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 ${X(t)}{t\in \mathbb{R}}$, as in $x_k=X(kT_s)$ for $k=0,\ldots, N-1$, where $T_s\in \mathbb{R}{++}$ is the sampling period. If the continuous-time signal has a compactly supported Fourier transform, than by Shannon's sampling theorem (http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem) $X$ is determined by the infinite sequence ${X(kT_s)}_{k\in \mathbb{Z}}$ for large 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

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 $\{\exp\left(\alpha\cdot i 2\pi k /N\right)\}_{k=0}^{N-1}$, where $\alpha\in(0,1)$ 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 ${X(t)}{t\in \mathbb{R}}$, as in $x_k=X(kT_s)$ for $k=0,\ldots, N-1$, where $T_s\in \mathbb{R}{++}$ is the sampling period. If the continuous-time signal has a compactly supported Fourier transform, than by Shannon's sampling theorem (http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem) $X$ is determined by the infinite sequence ${X(kT_s)}_{k\in \mathbb{Z}}$ 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