Is there a time-domain proof of Nyquist sampling theorem?

For a continuous-time signal $x(t)$ that is bandlimited (in the baseband) to $[-W,W]$, the standard proof of Nyquist sampling theorem proceeds in the frequency domain by examining the Fourier transform $X_s(f)$ of the $x(t)$ sampled at rate $2W=1/T$, and then showing that one can re-construct $x(t)$ using the low-pass filter with $\operatorname{sinc}(t/T)$ as impulse response to input of samples modulated by a Dirac comb. The standard proof uses the Fourier transform $X(f)$ of the original signal $x(t)$, which is computed over an infinite time domain, thus implicitly using infinite number of samples.

I am wondering if there is an alternative proof of the sampling theorem that, given the same conditions on the signal $x(t)$ as in the standard proof (band-limited, continuous, etc) uses fixed number of samples $n$ (as opposed to infinite number in the standard version), and then proceeds to show that, if sampled at rate $2W$, the reconstruction of the signal $\hat{x}(t)$ gets "closer" to original signal $x(t)$ as one increases $n$ such that $$\lim_{n\rightarrow\infty}\|x(t)-\hat{x}(t)\|=0$$

for some metric $\|\cdot\|$ like $L_1$ or $L_2$.

I don't have the mathematical sophistication to prove this myself, but I feel that this might be in the literature somewhere... Perhaps this an extension of the standard proof somehow.

• I like to visualize the sampling theorem as how stroboscopic light can be used to measure the rpm of a rotating disk. – Abu Bakar Feb 9 '14 at 10:12

The key assumption is that the signal is band-limited. This is a frequency-domain assumption. Any sensible proof must go through the frequency domain.

Same goes for the proof being an approximation argument. Any real proof must be an approximation argument, including the one you alluded to, or its rigorous version.

Here is another Hilbert space argument that, in the end, gives us approximation in the topology of uniform convergence (much better than $L^1$ or $L^2$):

Let $H$ be the set of band-limited elements of $L^2(\mathbb{R})$ (no extra assumptions). By unitarity of the Fourier transform $\mathcal{F}$, $H$ is a Hilbert subspace of $L^2$. Since $L^2$ elements of compact support lie in $L^1$, the Fourier inversion theorem implies that elements of $H$ are in fact continuous almost everywhere. So band-limited assumption implies (for now) continuity and therefore sampling make sense.

$\mathcal{F}(H)$ has orthonormal basis $\{ e^{- 2 \pi i k\frac{\xi}{2T}} \}_{k \in \mathbb{Z}}$. So now it's natural to compute the Hilbert space expansion of $\hat{f}$ in this basis then apply $\mathcal{F}^{-1}$. By unitarity

$$\langle \hat{f}, e^{- 2 \pi i k\frac{\xi}{2T}} \rangle = \langle f, \delta_{\frac{k}{2T}}\rangle = f(\frac{k}{2T}).$$

Strictly speaking, one needs a rigged Hilbert space that includes distributions to make sense of inner products with delta functions but everything works out. On the other hand, the inverse Fourier transform of the basis $\{ e^{- 2 \pi i k\frac{\xi}{2T}} \}_{k \in \mathbb{Z}}$ are just shifts of the $\mbox{sinc}$ function. So we have that Shannon's sampling formula holds in the $L^2$-sense.

To strengthen the convergence, notice $L^2$-convergence (in the frequency domain) implies $L^1$-convergence by the band-limited assumption. By property of $\mathcal{F}^{-1}$, back in the time domain we have uniform convergence.

Since $\mbox{sinc}$ functions and its shifts are all smooth, we can actually conclude that a band-limited $L^2$ function is in fact smooth almost everywhere.

• Very nice answer, again! I didn't think about this Hilbert-space argument... – M.B.M. Jun 1 '13 at 5:54