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Sampling theorem; aliasing and aperture effect remedies

Short answer

Sampling theorem: transmitter samples bandlimited signal (max W Hz) at fs ≥ 2W; receiver reconstructs via ideal LPF of bandwidth W. Aliasing occurs when fs < 2W, causing spectral overlap and false low frequencies. Remedies: anti-aliasing filter before sampling, and sampling above Nyquist for guard band. Aperture effect in flat-top sampling: finite pulse width τ multiplies spectrum by sinc(fτ), attenuating high frequencies. Remedies: small τ ≪ Ts, and equalizer with response 1/sinc(fτ) after reconstruction filter.

State Sampling theorem in terms of transmitter and receiver. Explain aliasing and aperture effect with remedy solutions.

Sampling theorem

At the transmitter: a band-limited signal of finite energy with no frequency components above WW Hz is completely described by its sample values taken at uniform intervals of Ts≤1/(2W)T_s \le 1/(2W) seconds, i.e. at a rate fs≥2Wf_s \ge 2W.

At the receiver: such a signal can be completely recovered from its samples taken at the rate of 2W2W samples per second (or more) by passing them through an ideal low-pass filter of bandwidth WW.

The rate 2W2W is the Nyquist rate. Proof sketch: ideal sampling gives Gδ(f)=fs∑nG(f−nfs)G_\delta(f) = f_s\sum_n G(f - nf_s); the copies separate when fs≥2Wf_s \ge 2W, so a LPF extracts G(f)G(f).

Aliasing

When fs<2Wf_s < 2W, or the signal is not strictly band-limited, the spectral copies overlap. High-frequency components fold back and appear as false low frequencies ∣f−fs∣|f - f_s|. Example: a 7 kHz tone sampled at 10 kHz appears as 3 kHz after reconstruction.

   copy n=0     copy n=1
  /‾‾‾‾‾‾\   /‾‾‾‾‾‾\
 /        \ /        \
/          X          \   overlap = aliasing
-W   0    fs/2   fs

Remedies

  1. Use an anti-aliasing (pre-alias) low-pass filter before the sampler to strictly band-limit the signal to WW.
  2. Sample slightly above the Nyquist rate (fs>2Wf_s > 2W) to create a guard band, so a practical reconstruction filter with finite roll-off can be used. Example: speech 3.4 kHz sampled at 8 kHz.

Aperture effect

In flat-top sampling each sample is held for width τ\tau. The sampled spectrum becomes fsH(f)∑nG(f−nfs)f_sH(f)\sum_n G(f - nf_s) with H(f)=τ sinc(fτ)e−jπfτH(f) = \tau\,\text{sinc}(f\tau)e^{-j\pi f\tau}. Because ∣H(f)∣|H(f)| falls with frequency, high-frequency parts of the message are attenuated and slightly delayed: this amplitude distortion is the aperture effect.

Remedies

  1. Keep the pulse width small, τ≪Ts\tau \ll T_s (duty cycle τ/Ts≤0.1\tau/T_s \le 0.1 makes the loss under about 0.5%).
  2. Use an equalizer after the reconstruction LPF with response
Heq(f)=1H(f)=1τ sinc(fτ) for ∣f∣≤WH_{eq}(f) = \frac{1}{H(f)} = \frac{1}{\tau\,\text{sinc}(f\tau)}\ \text{for}\ |f| \le W

which boosts high frequencies to cancel the sinc loss.

✓ Answer checked by Claude Opus 5.5 (high effort)

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