Nyquist, Aliasing & the Fourier Transform

What sampling does to the spectrum

Say a signal $f (x)$ is bandlimited: its spectrum $F (u)$ is exactly zero outside $[- W, W]$ . Sampling multiplies $f (x)$ by an impulse train $s (x)$ spaced $Δ x$ apart — and multiplication in space corresponds to convolution in frequency. The spectrum of an impulse train is itself an impulse train, spaced $1/Δ x$ apart, so convolving replicates $F (u)$ at every multiple of $1/Δ x$ :

s (x) f (x) ⟷ F (u) * S (u) = n = - \infty \sum \infty F (u - n /Δ x)

Sampling doesn't just discretize the signal — it tiles its spectrum out to infinity, one copy per sample rate.

The Nyquist criterion

Those copies of $F (u)$ are each $2 W$ wide. If the spacing between them, $1/Δ x$ , is large enough, they sit side by side with no overlap. That requires:

\frac{1}{2Δ x} \geq W ⟺ f_{s} = \frac{1}{Δ x} \geq 2 W

$2 W$ is the Nyquist rate. Sample at or above it and the replicated copies of $F (u)$ never touch. Sample below it and they overlap — high-frequency content folds back down into the low frequencies and corrupts them irreversibly. That folding is aliasing.

Interactive · Aliasing

Sampling rate f_s

Recovering the original signal

When the replicas don't overlap, the original spectrum is sitting right there in the middle, untouched — you just have to isolate it. Multiplying the sampled spectrum by an ideal low-pass filter $G (u)$ (a box of width $2 W$ ) keeps only the central copy and zeroes out the rest:

G (u) [F (u) * S (u)] = F (u)

This is exactly why anti-aliasing filters exist: they force $W$ down before sampling, so the Nyquist condition is guaranteed to hold.

From Fourier series to the Fourier transform

The claim that a signal "has a spectrum" at all traces back to the Fourier series: any periodic signal, with period $T$ , can be written uniquely as a sum of harmonically related complex exponentials.

f (t) = n = - \infty \sum \infty C_{n} e^{j n ω_{0} t}, C_{n} = \frac{1}{T} \int_{- T /2}^{T /2} f (t) e^{- j n ω_{0} t} d t, ω_{0} = \frac{2 π}{T}

The formula for $C_{n}$ falls out of one property: multiply both sides by $e^{- j m ω_{0} t}$ and integrate over a period, and the orthogonality of complex exponentials — $\int_{- T /2}^{T /2} e^{j (n - m) ω_{0} t} d t$ is $T$ when $n = m$ and 0 otherwise — collapses the infinite sum down to a single term, isolating $C_{n}$ .

Interactive · Fourier Series

Number of harmonics

The faint square wave is the target; the bright curve is the partial sum. Every periodic signal — a square wave included — is just a weighted sum of sines and cosines waiting to be unmixed.

A square wave is a clean example: it's built entirely from odd harmonics, each one weighted by $1/ n$ . Add more harmonics and the approximation sharpens toward the true square edges — but it never quite gets there without infinitely many terms, a wrinkle known as the Gibbs phenomenon.

Letting the period run to infinity

A real image or signal usually isn't periodic. To handle that, imagine stretching the period $T$ toward infinity. Two things happen at once: the spacing between harmonics $ω_{0} = 2 π / T$ shrinks toward a differential $d ω$ , and each individual coefficient $C_{n}$ shrinks toward zero — but the product $T C_{n}$ stays finite and turns into a continuous function of frequency. The sum becomes an integral, and out falls the Fourier transform pair:

f (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} F (ω) e^{j ω t} d ω F (ω) = \int_{- \infty}^{\infty} f (t) e^{- j ω t} d t

A periodic signal has a discrete spectrum — a comb of frequencies. Let the period go to infinity and the teeth of that comb merge into a continuous curve. That continuous curve is exactly the $F (u)$ from the sampling argument above — which is why the Fourier transform, not the Fourier series, is the right tool once signals stop repeating and start living in 2-D as images.

Check your understanding

A signal is bandlimited to W = 4 kHz. Which sampling rate guarantees no aliasing?