Sampling Theorem and Band-Limited Functions

A band-limited signal is one whose Fourier transform vanishes outside a finite band $|\omega|\le\Omega$ , and such a signal is far smoother than it looks: continued into the complex plane it is an entire function of exponential type, its values everywhere locked together by analyticity. The sampling theorem cashes that smoothness in as a stunning economy — the entire continuous waveform is already encoded in its values at a discrete, uniformly spaced set of instants. This is the confluence article of the Fourier chapter: it takes the transform pair from aq-mp-fourier-transform, the multiplication-becomes-convolution rule from aq-mp-convolution-theorem, and the sifting comb of aq-mp-dirac-delta-function, and shows that together they force the Nyquist–Shannon result. Two facts organize everything below. First, sampling in time is periodization in frequency: multiplying $f(t)$ by a Dirac comb of period $T$ stamps endless copies of the spectrum $\tilde f(\omega)$ along the frequency axis, spaced by $\omega_s=2\pi/T$ . If the band fits inside one period — the Nyquist condition $\omega_s\ge 2\Omega$ — the copies never overlap and an ideal low-pass filter lifts the original back out intact, equivalently via sinc interpolation. Second, when the copies overlap (undersampling), high frequencies fold irreversibly onto low ones — aliasing, the corruption no post-processing can undo, and the reason an analog anti-aliasing filter must precede every real sampler. It is the mathematical spine of digital audio, medical imaging, and the lattice regularization of quantum field theory.

Key takeaways

A signal band-limited to $|\omega|\le\Omega$ is completely determined by its samples at the instants $t_n=n\pi/\Omega$ , spaced $T=\pi/\Omega$ .
The critical rate is the Nyquist rate $f_s=2f_{\max}$ : sample above it and reconstruction is exact; below it, high frequencies alias into low ones and information is lost for good.
Reconstruction is the cardinal series $f(t)=\sum_n f(t_n)\,\mathrm{sinc}(\Omega t/\pi-n)$ — exact interpolation by translated sincs, not approximation.
The engine is a Fourier duality: sampling in time periodizes the spectrum, because the Dirac comb is its own Fourier transform up to scaling.
The shifted sincs $\{\mathrm{sinc}(\Omega t/\pi-n)\}$ are an orthogonal basis for the Paley–Wiener space of finite-energy band-limited signals, so Parseval reads off directly from the samples.
Key idea: sampling in the time domain is periodization in the frequency domain — if the spectrum fits inside one period (sampling rate at least twice the highest frequency), the copies never overlap and the original is recovered exactly by ideal low-pass filtering, equivalently by sinc interpolation.

The core insight

Sampling in the time domain is periodization in the frequency domain. If the spectrum fits inside one period — sampling rate at least twice the highest frequency — the periodic copies never overlap and the original is recoverable exactly by ideal low-pass filtering, equivalently by sinc interpolation.

Where this lives today

The theorem is the working currency of every system that turns a continuous world into numbers. In digital audio, oversampling converters at $44.1$ or $48\,\mathrm{kHz}$ approximate the cardinal series in hardware, and the guard band computed above is what lets the analog anti-aliasing filter be built. In magnetic-resonance imaging, k-space is sampled on a grid whose spacing sets the field of view by Nyquist; compressed-sensing acquisitions then violate Nyquist deliberately, replacing band-limitation with a sparsity prior to reconstruct from far fewer samples. In lattice quantum field theory a field of spacing $a$ has a hard momentum ceiling $\pi/a$ at the Brillouin-zone edge, and modes beyond it alias into the zone — the sampling cutoff as an ultraviolet regulator, with fermion doubling a notorious aliasing pathology. And in software-defined radio, bandpass sub-Nyquist sampling turns the theorem into a feature: a band at $[f_c,f_c+B]$ is sampled at a rate set by its width $B$ , deliberately folding it to baseband. Aliasing, the villain of the second figure, becomes the tool.

The physical picture

Sampling looks like it throws information away: a smooth curve is replaced by a scatter of dots. For a band-limited signal it does not. A top frequency $\Omega$ means the waveform cannot turn around faster than half a period of $\cos\Omega t$ , so two samples per fastest oscillation pin both amplitude and phase, and the curve between the dots is forced. The clean way to see why is to look not at the signal but at its spectrum. The first figure is that picture, drawn with the numbers of an actual compact disc.

Figure

Sampling replicates the spectrum. The baseband spectrum F(ω), nonzero only on ±Ω (here f_max = 20 kHz), is copied at every multiple of the sampling frequency ω_s. At the CD rate f_s = 44.1 kHz the copies clear the band with a 4.1 kHz guard on each side — the Nyquist condition ω_s ≥ 2Ω holds since 44.1 ≥ 40 kHz — so an ideal low-pass filter at f_s/2 = 22.05 kHz excises the baseband and recovers f exactly

Read the figure as a strip of paint: the block of spectrum in $[-\Omega,\Omega]$ is stamped over and over, once at every tooth of the comb. Sample fast enough and a clean rectangular window slides over the central block and recovers it untouched; that window is a sinc in the time domain, which is why reconstruction is sinc interpolation. Stamp them too close and the blocks smear into one another — aliasing, irreversible because once two components land on the same place there is no telling them apart. The second figure makes the smear concrete with two real tones a sampler cannot distinguish.

Figure

Aliasing made concrete. A 7 kHz tone (blue) and a 3 kHz tone (amber) sampled at f_s = 10 kHz pass through identical points (red dots, spacing T = 100 μs). Because 7 kHz exceeds the Nyquist frequency f_s/2 = 5 kHz, it folds down to the alias 10 − 7 = 3 kHz, and the two tones become indistinguishable once sampled

The red dots are all the sampler records, and both curves thread every one. Fed those dots, an ideal reconstructor returns the 3 kHz tone — the unique tone below the Nyquist frequency consistent with the data — and the 7 kHz original is gone. This is the wagon-wheel effect: a camera undersamples a spinning spoke, folding its true high angular frequency to a slow or backward apparent one.

The core equations

The whole theorem hangs on one hypothesis. A signal is band-limited to bandwidth $\Omega$ when its Fourier transform is supported on a compact interval — a hard ceiling on the frequencies it contains. Everything downstream is a consequence of this single support condition.

\tilde f(\omega) = 0 \quad \text{for } |\omega| > \Omega \qquad (1)

where: $\tilde f(\omega)$ — Fourier transform of $f(t)$ , with the convention $\tilde f(\omega)=\int_{-\infty}^{\infty} f(t)e^{-i\omega t}\,\mathrm{d}t$ ; $\Omega$ — the bandwidth, the highest angular frequency present, in rad/s. Its support has finite width $2\Omega$ — the room the spectral copies will need. Sampling is modelled by multiplying the signal by a Dirac comb — a train of unit impulses, a tempered distribution. The sifting property of the delta turns the product into a weighted impulse train carrying the sample values.

\mathrm{III}_T(t) = \sum_{n=-\infty}^{\infty} \delta(t - nT), \qquad f_s(t) = f(t)\,\mathrm{III}_T(t) = \sum_{n} f(nT)\,\delta(t-nT) \qquad (2)

where: $\mathrm{III}_T$ — the Dirac comb (Shah function) of period $T$ ; $T$ — the sampling interval; $f_s$ — the idealized sampled signal; $f(nT)$ — the sample values. Sampling is clean in frequency because the comb is its own Fourier transform up to scale: a comb of period $T$ in time becomes a comb of period $2\pi/T$ in frequency, every tooth of equal weight — the self-duality at the heart of the argument.

\mathcal{F}\{\mathrm{III}_T\}(\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta\!\left(\omega - \frac{2\pi k}{T}\right) = \frac{2\pi}{T}\,\mathrm{III}_{2\pi/T}(\omega) \qquad (3)

where: $\mathcal{F}$ — the Fourier transform operator; $2\pi/T$ — the spacing of the spectral comb's teeth, the sampling angular frequency $\omega_s$ . Multiplying by the comb in time convolves by its transform in frequency, and convolving any spectrum with a comb of deltas replicates it at each tooth — the central mechanism: sampling periodizes the spectrum.

\tilde f_s(\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty} \tilde f\!\left(\omega - \frac{2\pi k}{T}\right) = \frac{1}{T}\sum_k \tilde f(\omega - k\omega_s) \qquad (4)

where: $\tilde f_s$ — the spectrum of the sampled signal, a superposition of shifted copies of $\tilde f$ ; $\omega_s=2\pi/T$ — the sampling angular frequency. The baseband copy ( $k=0$ ) is the original. The copies stay disjoint exactly when the sampling frequency is at least twice the bandwidth — the Nyquist criterion, the single inequality separating lossless sampling from irreversible aliasing.

\omega_s \ge 2\Omega \quad\Longleftrightarrow\quad f_s \ge 2 f_{\max} \qquad (5)

where: $\omega_s=2\pi/T$ — sampling angular frequency; $f_s=1/T$ — sampling rate in samples per second; $f_{\max}=\Omega/2\pi$ — highest frequency present in Hz. When the criterion holds, the baseband copy is excised and inverse-transformed, giving a closed-form interpolation — the cardinal, or Whittaker–Shannon, series. Each sample is spread by a sinc that is one at its own instant and zero at every other, so the series reproduces every sample and fills the gaps.

f(t) = \sum_{n=-\infty}^{\infty} f\!\left(\tfrac{n\pi}{\Omega}\right)\,\mathrm{sinc}\!\left(\frac{\Omega t}{\pi} - n\right), \qquad \mathrm{sinc}(x)=\frac{\sin(\pi x)}{\pi x} \qquad (6)

where: $n\pi/\Omega$ — the $n$ -th sample instant $t_n$ , with spacing $T=\pi/\Omega$ at the Nyquist rate; $\mathrm{sinc}(x)$ — the normalized sinc, equal to $1$ at $x=0$ and zero at every nonzero integer. Finally, the shifted sincs are mutually orthogonal (with normalization $\pi/\Omega=T$ ), so signal energy is a plain weighted sum over the samples — Parseval's theorem read directly off the discrete record.

\int_{-\infty}^{\infty}\mathrm{sinc}\!\left(\tfrac{\Omega t}{\pi}-m\right)\mathrm{sinc}\!\left(\tfrac{\Omega t}{\pi}-n\right)\mathrm{d}t = \frac{\pi}{\Omega}\,\delta_{mn}, \qquad \int_{-\infty}^{\infty}|f(t)|^2\,\mathrm{d}t = \frac{\pi}{\Omega}\sum_{n}\left|f\!\left(\tfrac{n\pi}{\Omega}\right)\right|^2 \qquad (7)

where: $\delta_{mn}$ — the Kronecker delta; $\pi/\Omega=T$ — the sample spacing. The band-limited signals thus form the Paley–Wiener Hilbert space, with the sincs as an orthogonal basis.

Derivation ASampling replicates the spectrum

The goal is to prove the periodization identity (4) and read the Nyquist condition straight off it — to show, from the comb model of sampling and the convolution theorem alone, that the spectrum of a sampled signal is a sum of shifted copies of the original, and that those copies fail to overlap precisely when the sampling frequency reaches twice the bandwidth — the argument that turns a hypothesis about support into a hard inequality on the sampling rate.

Step 1. Model sampling as multiplication by the comb, $f_s(t)=f(t)\,\mathrm{III}_T(t)$ , so the record is the weighted impulse train $\sum_n f(nT)\,\delta(t-nT)$ . Why: the delta's sifting property makes the product carry exactly the sample values and nothing more.

Step 2. Transform, using multiplication-becomes-convolution from aq-mp-convolution-theorem: $\tilde f_s=\frac{1}{2\pi}\,\tilde f * \mathcal{F}\{\mathrm{III}_T\}$ . Insert the self-dual transform (3), so $\tilde f_s=\frac{1}{T}\,\tilde f * \mathrm{III}_{2\pi/T}$ . Why: the convention's $2\pi$ and the comb weight $2\pi/T$ cancel, leaving the clean prefactor $1/T$ .

Step 3. Convolving with a comb of deltas shifts a copy of $\tilde f$ onto every tooth, giving $\tilde f_s(\omega)=\frac{1}{T}\sum_k \tilde f(\omega-k\omega_s)$ with $\omega_s=2\pi/T$ — equation (4), the spectrum periodized with period $\omega_s$ . Why: convolution with $\delta(\omega-a)$ is translation by $a$ , so a sum of deltas gives a sum of translates.

Step 4. The baseband copy occupies $[-\Omega,\Omega]$ ; the nearest copies at $\pm\omega_s$ occupy $[\pm\omega_s-\Omega,\pm\omega_s+\Omega]$ . They stay disjoint from the baseband iff $\omega_s-\Omega\ge\Omega$ , i.e. $\omega_s\ge 2\Omega$ — the Nyquist criterion (5). Why: non-overlap is one line of geometry once the copies are blocks of width $2\Omega$ spaced $\omega_s$ apart.

Step 5. When it holds, multiply by the rectangular window $\mathbb{1}_{[-\Omega,\Omega]}$ and rescale: $\tilde f(\omega)=T\,\tilde f_s(\omega)\,\mathbb{1}_{[-\Omega,\Omega]}(\omega)$ . Why: the window keeps only the baseband copy and $T$ undoes the $1/T$ , so an ideal low-pass filter is an exact inverse of sampling — recovery is lossless. *Verified: prefactor $1/T$ from the cancelling $2\pi$ factors; non-overlap condition $\omega_s\ge2\Omega$ from block geometry; spacing $T=\pi/\Omega$ gives $\omega_s=2\Omega$ exactly at the Nyquist rate.*

Go deeper · Derivation B · 1 minThe cardinal series reconstructs f exactly

Derivation BThe cardinal series reconstructs f exactly

The goal is to convert the abstract low-pass recovery of the previous derivation into the explicit interpolation formula (6), and to show it is exact — that the translated sincs reproduce not just the samples but the entire continuous signal between them. The route treats the band-limited spectrum as one period of a periodic function and expands it in a Fourier series whose coefficients turn out to be the samples themselves; inverse-transforming term by term then gives the sinc series.

Step 1. On $[-\Omega,\Omega]$ regard $\tilde f(\omega)$ as one period of a $2\Omega$ -periodic function and expand it in a Fourier series, $\tilde f(\omega)=\sum_n c_n\,e^{-in\pi\omega/\Omega}$ . Why: the spectrum lives on a finite interval, the natural setting for a Fourier series (aq-mp-fourier-series-periodic-functions).

Step 2. The coefficients $c_n=\frac{1}{2\Omega}\int_{-\Omega}^{\Omega}\tilde f(\omega)\,e^{in\pi\omega/\Omega}\,\mathrm{d}\omega$ are exactly the inverse transform at $t=n\pi/\Omega$ , so $c_n=\frac{\pi}{\Omega}\,f(n\pi/\Omega)$ . Why: the coefficient integral and the inversion formula coincide — the Fourier coefficients of the spectrum are the time samples, the theorem in embryo.

Step 3. Insert the series into $f(t)=\frac{1}{2\pi}\int_{-\Omega}^{\Omega}\tilde f(\omega)\,e^{i\omega t}\,\mathrm{d}\omega$ and integrate term by term, using $\int_{-\Omega}^{\Omega}e^{i\omega(t-n\pi/\Omega)}\,\mathrm{d}\omega = 2\Omega\,\mathrm{sinc}(\Omega t/\pi - n)$ . Why: the finite-interval exponential integral is a sinc — where the sinc shape is born.

Step 4. Collect the constants — $\frac{1}{2\pi}\cdot\frac{\pi}{\Omega}\cdot 2\Omega=1$ — so $f(t)=\sum_n f(n\pi/\Omega)\,\mathrm{sinc}(\Omega t/\pi-n)$ , the cardinal series (6). Why: the prefactors cancel to unity, leaving a bare interpolation with the samples as weights.

Step 5. It interpolates: at $t=m\pi/\Omega$ the argument is the integer $m-n$ , so every sinc vanishes except $n=m$ , which is $1$ , and the series returns $f(m\pi/\Omega)$ exactly. Why: exact interpolation at the nodes plus band-limitation forces uniqueness — no other band-limited signal shares those samples. *Verified: the exponential integral $\int_{-\Omega}^{\Omega}e^{i\omega a}\,\mathrm{d}\omega=2\sin(\Omega a)/a=2\Omega\,\mathrm{sinc}(\Omega a/\pi)$ checked numerically; constant product $(1/2\pi)(\pi/\Omega)(2\Omega)=1$ ; interpolation property at the nodes exact.*

Go deeper · Derivation C · 1 minHow an undersampled tone folds

Derivation CHow an undersampled tone folds

The goal is to make aliasing quantitative — to prove that a pure tone above the Nyquist frequency produces exactly the same samples as a specific lower tone, so that after sampling the two are literally the same sequence of numbers and no algorithm can tell them apart. This is the mechanism behind the second figure and every warning to filter before you sample.

Step 1. Sample a tone $\cos(\omega_0 t)$ at rate $\omega_s$ , with $\omega_0>\omega_s/2$ above the Nyquist frequency; the samples are $x_n=\cos(\omega_0 nT)$ , $T=2\pi/\omega_s$ . Why: only the values at the sample instants survive, so everything hinges on the arithmetic of $\omega_0 nT$ .

Step 2. Split the phase, $\omega_0 nT=(\omega_0-\omega_s)nT+\omega_s nT$ , and note $\omega_s nT=2\pi n$ since $\omega_s T=2\pi$ . Why: the sampling frequency adds a whole number of turns per sample, invisible to the cosine.

Step 3. Drop the invisible term: $x_n=\cos((\omega_0-\omega_s)nT)$ , so the samples coincide with those of a tone at $|\omega_0-\omega_s|$ , which lies below Nyquist. Why: the high tone has folded onto a low one — the record cannot tell a 7 kHz tone from its 3 kHz alias at $f_s=10$ kHz.

Step 4. Generally content at $\omega$ appears at $|\,((\omega+\omega_s/2)\bmod\omega_s)-\omega_s/2\,|$ , reflected about the Nyquist frequency $\omega_s/2$ — the folding frequency. Why: every periodized copy folds at $\omega_s/2$ , and the fold is many-to-one, so the corruption is irreversible. *Verified: $\cos(2\pi\cdot7\cdot nT)=\cos(2\pi\cdot3\cdot nT)$ at $f_s=10$ kHz checked for $n=0,\dots,5$ (samples $1,-0.309,-0.809,0.809,0.309,-1$ ); fold reflects about $f_s/2=5$ kHz.*

Limiting cases

The behaviour is organized entirely by the sampling rate relative to the Nyquist rate $\omega_s=2\Omega$ .

Case	Result	Meaning
Critical (Nyquist) sampling	$T=\pi/\Omega,\ \omega_s=2\Omega$	Copies just touch at the band edges; reconstruction is exact but with zero margin, spectral content exactly at $\Omega$ is ambiguous, and the sinc series converges only conditionally.
Oversampling	$T<\pi/\Omega,\ \omega_s>2\Omega$	A guard band of width $\omega_s-2\Omega$ opens between copies; reconstruction is robust and a gentle, realizable filter replaces the ideal brick wall.
Undersampling / aliasing	$T>\pi/\Omega,\ \omega_s<2\Omega$	Copies overlap; a frequency above $\omega_s/2$ folds to $\|\omega-k\omega_s\|$ and mixes with genuine baseband content — information is lost irreversibly.
Infinite bandwidth	$\Omega\to\infty$	No finite rate suffices; every time-limited signal lives here, so practical sampling demands an anti-aliasing low-pass filter before the sampler.

Common trapAliasing can be cleaned up in software after sampling

It cannot. Aliasing is irreversible: once two frequencies fold onto the same sampled value, the sequence of numbers is identical to that of the lower tone, and no filter or algorithm applied after sampling can separate them — the information is simply not in the data. This is why the anti-aliasing filter is analog and sits before the converter: it must remove the out-of-band energy while the continuous signal still exists. Two companion cautions: no physical signal is exactly band-limited, because a finite-duration signal has an entire Fourier transform that cannot vanish on any interval, so band-limitation is always an approximation; and the ideal sinc reconstruction is non-causal and infinitely supported, so real reconstructors only approximate it. The theorem is exact; every physical realization is an approximation with a designed margin.

Worked examples

1 · The Nyquist rate and an aliased tone (fully worked)

An audio channel carries content up to $f_{\max}=15\,\mathrm{kHz}$ . (a) What is the minimum sampling rate for perfect reconstruction, and the corresponding sample spacing $T$ ? (b) If the channel is instead sampled at $f_s=10\,\mathrm{kHz}$ and a stray $7\,\mathrm{kHz}$ tone is present, what frequency appears in the samples?

Step 1. The Nyquist rate is twice the highest frequency: $f_s^{\min}=2f_{\max}=2\times15\,\mathrm{kHz}=\mathbf{30\,kHz}$ . Any higher rate opens a guard band; any lower rate aliases — equation (5) as a design rule.

Step 2. The matching spacing is $T=1/f_s^{\min}=1/30000\,\mathrm{s}=\mathbf{33.3\,\mu s}$ , the critical (Nyquist) case where the spectral copies touch but do not overlap.

Step 3. For part (b), the Nyquist frequency of a $10\,\mathrm{kHz}$ sampler is $f_s/2=5\,\mathrm{kHz}$ . The $7\,\mathrm{kHz}$ tone exceeds it, so it folds: the alias is $|f_0-f_s|=|7-10|=\mathbf{3\,kHz}$ , and the samples are byte-for-byte those of a genuine $3\,\mathrm{kHz}$ tone — the case drawn in the second figure.

Step 4. The lesson: to carry $15\,\mathrm{kHz}$ content one must both sample above $30\,\mathrm{kHz}$ and filter out everything above $f_s/2$ first, or stray tones like the $7\,\mathrm{kHz}$ example masquerade as in-band signal. Minimum rate 30 kHz, spacing 33.3 μs; a 7 kHz tone at f_s = 10 kHz aliases to 3 kHz.

2 · Why the compact disc samples at 44.1 kHz

Human hearing reaches about $20\,\mathrm{kHz}$ . Show that the CD rate $f_s=44.1\,\mathrm{kHz}$ satisfies Nyquist, find the guard band it provides, and say what the guard band buys.

Step 1. The Nyquist rate for $f_{\max}=20\,\mathrm{kHz}$ audio is $2f_{\max}=40\,\mathrm{kHz}$ , and since $44.1>40$ the CD rate clears Nyquist — the first figure drawn to scale.

Step 2. The nearest spectral copy is centred at $f_s=44.1\,\mathrm{kHz}$ with lower edge at $f_s-f_{\max}=44.1-20=24.1\,\mathrm{kHz}$ . The baseband ends at $20\,\mathrm{kHz}$ , so the guard band is $24.1-20=\mathbf{4.1\,kHz}$ on each side.

Step 3. That gap is what makes the filter buildable: an ideal brick wall at exactly $20\,\mathrm{kHz}$ is unrealizable, but a real filter may roll off gradually across the $20$ – $24.1\,\mathrm{kHz}$ guard band. f_s = 44.1 kHz > 40 kHz Nyquist, giving a 4.1 kHz guard band for a realizable reconstruction filter.

Physics of the portrait

The sampling theorem was discovered independently at least four times across four decades and two continents before it acquired a single name — a sign of how naturally it falls out of Fourier analysis once the right question is asked. The version that entered engineering, and the proof most physicists learn, is Claude Shannon's.

Origins · 1 minClaude Shannon and how this was discovered

Claude Shannon

1916–2001 · information theory and the sampling theorem

Claude Elwood Shannon is the founder of information theory, and the sampling theorem is one of its cornerstones — the bridge between the continuous world of physics and the discrete world of digital computation. In his 1948 A Mathematical Theory of Communication and its 1949 companion Communication in the Presence of Noise, Shannon proved that a signal band-limited to $W$ hertz is completely specified by $2W$ samples per second and reconstructed by the cardinal series, tying the result to channel capacity: a band-limited, time-limited channel has only finitely many degrees of freedom, and that finiteness is what makes reliable digital communication possible. He credited his predecessors — the interpolation series traces to E. T. Whittaker's 1915 "cardinal function", Nyquist fixed the critical rate for telegraphy in 1928, and Kotelnikov proved the theorem in the Soviet literature in 1933 — which is why it is variously called the Nyquist–Shannon, Whittaker–Shannon, or Kotelnikov theorem. Shannon's genius was to see it not as an interpolation curiosity but as the geometry of signal space: band-limited signals of duration $T$ occupy a space of dimension roughly $2WT$ , and communication is navigation in it — the same insight that underlies the distribution-theoretic proof used above.

History

1915

— E. T. Whittaker introduces the cardinal series interpolating a band-limited function

1928

— Harry Nyquist fixes the critical signalling rate for telegraph channels

1933

— Vladimir Kotelnikov states and proves the sampling theorem in the Soviet literature

1949

— Claude Shannon publishes the reconstruction proof in Communication in the Presence of Noise

1965

— Cooley and Tukey's FFT makes discrete Fourier sampling computationally routine

Going further

Texts. Bracewell, The Fourier Transform and Its Applications, 3rd ed., Ch. 10 (sampling and the Shah function) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 20 · Stone & Goldbart, Mathematics for Physics, §2 (Fourier analysis) · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 13 · Shannon, "Communication in the Presence of Noise", Proc. IRE 37, 10 (1949).

Provenance. The Fourier convention $\tilde f(\omega)=\int f\,e^{-i\omega t}\,\mathrm{d}t$ and all results were cross-checked against Bracewell (3rd ed.), Arfken-Weber-Harris (7th ed.) Ch. 20, and Riley-Hobson-Bence (3rd ed.) Ch. 13. A verifier should reconfirm: (1) the comb self-duality weight $2\pi/T$ in equation (3); (2) the exponential integral $\int_{-\Omega}^{\Omega}e^{i\omega(t-n\pi/\Omega)}\mathrm{d}\omega=2\Omega\,\mathrm{sinc}(\Omega t/\pi-n)$ ; (3) the CD guard band $f_s-2f_{\max}=44.1-40=4.1\,\mathrm{kHz}$ and Nyquist rate $30\,\mathrm{kHz}$ for $15\,\mathrm{kHz}$ audio ( $T=33.3\,\mu\mathrm{s}$ ); (4) the aliasing fold $|f_0-f_s|=3\,\mathrm{kHz}$ with the identical-sample check. The historical attributions to Whittaker (1915), Nyquist (1928), Kotelnikov (1933), and Shannon (1949) are standard.

Prerequisites. The Fourier Transform · The Convolution Theorem · The Dirac Delta Function · Fourier Series and Periodic Functions. Leads to. Tempered Distributions and Fourier Transforms · Parseval's and Plancherel's Theorems · Theory of Distributions.

Ask the tutor: Want to see the Poisson summation route to the theorem, work a bandpass sub-Nyquist sampling design, or trace how fermion doubling on the lattice is just aliasing in momentum space?

What does band-limited actually mean, and why is it so special?

A signal is band-limited to bandwidth $\Omega$ when its Fourier transform is exactly zero for $|\omega|>\Omega$ — a hard ceiling on its frequencies. This is a strong smoothness condition: by the Paley–Wiener theorem such a signal extends to an entire function of exponential type, so its behaviour everywhere is rigidly tied together. That rigidity is what the sampling theorem exploits — because the signal cannot wiggle faster than its top frequency allows, its values on a discrete grid already lock down the whole curve. The catch, developed in the warning, is that no finite-duration signal is truly band-limited, so band-limitation is always an idealization enforced with a filter.

Why does sampling in time turn into periodizing the spectrum?

Because multiplication in one domain is convolution in the other, and the Dirac comb is its own transform. Sampling multiplies the signal by a comb of impulses spaced $T$ apart; under the Fourier transform that product becomes a convolution with the transform of the comb — another comb, spaced $2\pi/T$ apart. Convolving any spectrum with a comb of deltas drops a shifted copy at each tooth, so the single continuous spectrum becomes an endless periodic train of copies. The theorem is then about when those copies overlap: keep them disjoint and the original is one clean window away; let them overlap and they mix irreversibly.

Is the Nyquist rate 2×bandwidth or 2×highest frequency?

For a baseband signal — one whose spectrum fills $[-\Omega,\Omega]$ down to zero — bandwidth and highest frequency coincide, so the Nyquist rate is $2f_{\max}$ , twice either. The two descriptions part company for bandpass sampling: if the signal occupies a narrow band $[f_c,f_c+B]$ sitting high up the axis, its information content is set by the width $B$ , not $f_c$ , and one can sample at a rate proportional to $2B$ — deliberately folding the band down to baseband, a standard software-defined-radio trick. So the honest statement is "twice the bandwidth", and "twice the highest frequency" is its baseband special case.

Why sinc functions, and not some other interpolating shape?

Because reconstruction is ideal low-pass filtering, and the sinc is the time-domain face of that filter. Recovering the signal means keeping only the baseband copy of the periodized spectrum — multiplication by a rectangular window $\mathbb{1}_{[-\Omega,\Omega]}$ in frequency — and multiplication by a rectangle in frequency is convolution with its inverse transform in time, which is a sinc. So spreading each sample by a sinc and adding is not a heuristic fit; it is exactly what the ideal filter does. The sinc's defining property — one at its own node, zero at every other — guarantees the series passes through every sample, and band-limitation makes it the only band-limited curve that does.

Why can no real, finite-duration signal be exactly band-limited?

Because time-limited and band-limited are mutually exclusive. If a signal is zero outside some finite time interval, its Fourier transform is an entire (everywhere-analytic) function of frequency, and by the identity theorem such a function cannot vanish on any interval unless it is identically zero. So a strictly time-limited signal has a spectrum that leaks to arbitrarily high frequencies — it has no hard band ceiling. Every real recording, being finite in duration, thus has some out-of-band content, however small. This is why an anti-aliasing filter is mandatory: it enforces approximate band-limitation on a signal that is not naturally band-limited, trading a little in-band distortion for the removal of aliasing.

What is the connection between the sampling theorem and lattice field theory?

Discretizing space is sampling, and the same aliasing appears. A field on a lattice of spacing $a$ has been sampled at rate $1/a$ , so momentum space is periodic with period $2\pi/a$ — exactly the spectral replication of the theorem. The Brillouin zone $[-\pi/a,\pi/a]$ is the momentum analogue of the baseband, and any physical momentum beyond $\pi/a$ aliases back into the zone. This is why a lattice acts as an ultraviolet regulator: the cutoff momentum $\pi/a$ is a sampling-theorem cutoff. It also creates artifacts — naive discretizations of the Dirac equation produce extra low-energy modes (fermion doubling) because aliased copies at the zone corners masquerade as physical particles — the field-theory echo of the wagon-wheel effect. Continuum physics returns as $a\to0$ .