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Impulse train sampling, aliasing, reconstruction

Short answer

Impulse-train sampling multiplies x(t) by an impulse train Σδ(t−nT) to produce weighted impulses x_p(t)=Σx(nT)δ(t−nT). Aliasing occurs when ω_s < 2ω_M, causing spectral copies to overlap and high frequencies to fold back. Reconstruction is achieved by passing x_p(t) through an ideal low-pass filter with gain T and cutoff ω_M < ω_c < ω_s−ω_M, yielding sinc interpolation.

What is impulse train sampling? Explain the aliasing effect that may occur in impulse train sampling. How do you reconstruct original signal from its sample?

Impulse-train sampling

Impulse-train sampling is the mathematical model of ideal sampling: x(t)x(t) is multiplied by a periodic train of unit impulses of period TT:

xp(t)=x(t)∑n=−∞∞δ(t−nT)=∑n=−∞∞x(nT) δ(t−nT)x_p(t) = x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT) = \sum_{n=-\infty}^{\infty} x(nT)\,\delta(t-nT)

Each impulse carries the sample value x(nT)x(nT) as its area.

Aliasing in impulse-train sampling

The spectrum of the sampled signal is

Xp(jω)=1T∑k=−∞∞X(j(ω−kωs)),ωs=2πTX_p(j\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big), \qquad \omega_s = \frac{2\pi}{T}

i.e. copies of X(jω)X(j\omega) at every multiple of ωs\omega_s.

  • If ωs>2ωM\omega_s > 2\omega_M, the copies are separate.
  • If ωs<2ωM\omega_s < 2\omega_M, the copy at ωs\omega_s begins at ωs−ωM\omega_s - \omega_M, which is below ωM\omega_M, so the copies overlap. Frequencies above ωs/2\omega_s/2 fold back into the band and add to the true components. This is aliasing; the original spectrum can no longer be separated.
ws < 2wM:
     ____     ____     ____
    /    \   /    \   /    \
---/------\-/------\-/------\---> w
         -ws  \/ 0  \/  ws
              overlap = aliasing

Reconstruction from samples

If ωs>2ωM\omega_s > 2\omega_M, pass xp(t)x_p(t) through an ideal low-pass filter with gain TT and cutoff ωc\omega_c, where ωM<ωc<ωs−ωM\omega_M < \omega_c < \omega_s - \omega_M (usually ωc=ωs/2\omega_c = \omega_s/2). The filter keeps only the k=0k=0 copy, so its output is x(t)x(t). In time,

h(t)=Tsin⁡ωctπt,xr(t)=∑nx(nT) Tsin⁡ωc(t−nT)π(t−nT)h(t) = \frac{T\sin\omega_c t}{\pi t}, \qquad x_r(t) = \sum_{n} x(nT)\,\frac{T\sin\omega_c(t-nT)}{\pi(t-nT)}

This is band-limited (sinc) interpolation. Practical systems use simpler approximations: zero-order hold (staircase) or first-order hold (straight-line interpolation) followed by smoothing.

✓ Answer checked by Claude Opus 5.5 (high effort)

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