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State and prove Nyquist sampling theorem

Short answer

A band-limited signal with max frequency W is uniquely determined by samples taken at fs ≥ 2W. Ideal sampling repeats the spectrum every fs; if fs ≥ 2W, copies do not overlap, and an ideal LPF recovers the original. Aliasing occurs if fs < 2W.

State and prove Nyquist sampling theorem.

Statement

A band-limited signal m(t)m(t) with no frequency components above WW Hz (M(f)=0M(f) = 0 for ∣f∣>W|f| > W) is uniquely determined by its samples taken at uniform intervals Ts≤12WT_s \le \frac{1}{2W}, i.e. at a rate fs≥2Wf_s \ge 2W. It can be recovered exactly from the samples by an ideal low-pass filter of bandwidth WW.

The minimum rate 2W2W is the Nyquist rate; 1/(2W)1/(2W) is the Nyquist interval.

Proof

1. Ideal sampling. Multiply m(t)m(t) by a periodic impulse train:

mδ(t)=m(t)∑n=−∞∞δ(t−nTs)=∑n=−∞∞m(nTs) δ(t−nTs)m_\delta(t) = m(t)\sum_{n=-\infty}^{\infty}\delta(t - nT_s) = \sum_{n=-\infty}^{\infty} m(nT_s)\,\delta(t - nT_s)

2. Spectrum of the impulse train. It is periodic, with Fourier series coefficients 1/Ts1/T_s:

∑nδ(t−nTs)=1Ts∑k=−∞∞ej2πkfst  ⟷  fs∑kδ(f−kfs)\sum_n \delta(t - nT_s) = \frac{1}{T_s}\sum_{k=-\infty}^{\infty} e^{j2\pi k f_s t} \;\longleftrightarrow\; f_s\sum_{k}\delta(f - kf_s)

3. Spectrum of the sampled signal. Multiplication in time is convolution in frequency:

Mδ(f)=M(f)∗fs∑kδ(f−kfs)=fs∑k=−∞∞M(f−kfs)M_\delta(f) = M(f) * f_s\sum_k\delta(f - kf_s) = f_s\sum_{k=-\infty}^{\infty} M(f - kf_s)

So Mδ(f)M_\delta(f) is M(f)M(f) repeated every fsf_s Hz, scaled by fsf_s.

f_s >= 2W (no overlap):
     /\        /\        /\
    /  \      /  \      /  \
 --/----\----/----\----/----\---> f
  -W  0  W fs-W fs   2fs
f_s < 2W: copies overlap -> aliasing

4. Condition for no overlap. The copy at fsf_s occupies fs−Wf_s - W to fs+Wf_s + W. It does not overlap the baseband copy (−W-W to WW) if

fs−W≥W  ⇒  fs≥2Wf_s - W \ge W \;\Rightarrow\; f_s \ge 2W

5. Recovery. If fs≥2Wf_s \ge 2W, pass mδ(t)m_\delta(t) through an ideal LPF with H(f)=TsH(f) = T_s for ∣f∣<W|f| < W (any cut-off between WW and fs−Wf_s - W):

Mδ(f) H(f)=M(f)M_\delta(f)\,H(f) = M(f)

so m(t)m(t) is recovered exactly.

6. Interpolation formula. With fs=2Wf_s = 2W, the LPF impulse response is h(t)=sinc(2Wt)h(t) = \text{sinc}(2Wt), so

m(t)=∑n=−∞∞m ⁣(n2W)sinc(2Wt−n)m(t) = \sum_{n=-\infty}^{\infty} m\!\left(\frac{n}{2W}\right)\text{sinc}(2Wt - n)

Each sample is replaced by a sinc pulse and the sum rebuilds m(t)m(t) at all instants. This proves the theorem. If fs<2Wf_s < 2W, the copies overlap (aliasing) and m(t)m(t) cannot be recovered; an anti-aliasing filter is therefore used before sampling.

✓ Answer checked by Claude Opus 5.5 (high effort)

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