reconstruction through sinc interpolation

Question

I have a discrete-time signal $x_k = \sum_l a_l g(kT - l(T+\Delta T))$ where $g(t) = \frac {\sin(\pi t/(T+\Delta T))}{\pi t/(T+\Delta T)}$.

Since the signal $x$ has been sampled at rate $1/T>2 \frac {1}{2(T+\Delta T)}$ where $\Delta T >0$ it follows from the sampling theorem that we should be able to reconstruct the discrete samples $a_l$'s from $x_k$'s.

How to do that is my question.

I write: $y_k = \sum_l x_l f(l,k)$ and I want to find $f(k,l)$ such that $y_k = a_k$ for all $k$.

Therefore: \begin{align} y_k & = \sum_l \sum_i a_i g(lT-i(T+\Delta T)) f(k,l) \\ & =\sum_i a_i \sum_l g(lT-i(T+\Delta T)) f(k,l) =a_k \end{align}

Therefore: $\hspace{8 mm}$$\sum_l g(lT-i(T+\Delta T)) f(k,l) =\delta (k-i)$

Therefore: $\hspace{8 mm}$$\sum_l g(lT-k(T+\Delta T)) f(k,l) =1$,

which gives: $f(k,l)=\operatorname{sinc}(k(T+\Delta T) - lT).$

Therefore, I think $y_k = a_k = \sum_l x_l \operatorname{sinc}(k(T+\Delta T)-lT)$.

Do you agree? I implement the above in MATLAB and I do not get the expected output, so I don't know if my reasoning is correct.

Thank you.

Answer 1

The sampling theorem is just a special case of Poisson summation where you look at the product of $\widehat{f}(\xi)$ with $\text{sinc}(\xi)$ and invert the Fourier transform (obviously we assume $f\in L^1$ as well), noting that it's the Dirichlet kernel that is what gives finite, symmetric sums.

The specific sampling points are given by $t_k={k\over\Omega}$, so we know

$$\widehat{f}_s(\xi)=\sum_{k\in\Bbb Z}\widehat{f}\left(\xi-k\Omega\right)$$

now using Poisson summation, we have that this is

$$\sum_{k\in\Bbb Z}{1\over\Omega} f\left({k\over \Omega}\right)e^{-2\pi in \xi/\Omega}$$

Now simply invert the Fourier transform on both sides (the theorem presumes $f\in L^1$ so that the inverse transform is still given by the integral identity)

$$f(t)=\int_{\Bbb R}\sum_{n\in\Bbb Z}{1\over \Omega} f\left({n\over \Omega}\right)e^{-2\pi in \xi/\Omega}e^{2\pi i t\xi}\,d\xi$$

Your sample values are the $f\left({n\over \Omega}\right)$, so those are the numbers you sampled, $t_k={k\over\Omega}$.

Your $\Omega = T+\Delta T$ is greater than the band-limit, so that the inversion recovers the answer pointwise, as you noted.

Now to compute you just need to actually do the integral, which you should know already produces

$$f(t)=\sum_{k\in\Bbb Z}t_k\text{sinc}\left(t\Omega-k\right)$$

You'll note the difference is that in your example you get a $T$ next to the $k$, where there should be none, (recall $\Omega = T+\Delta T$).

I'm not sure what you're doing with these $y_k$, but if you want a specific value, you can pick a fixed $t$ to get such a $y_k$.