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Aliasing via impulse-train sampling

Short answer

Aliasing is spectrum overlap when sampling below the Nyquist rate, causing high frequencies to appear as lower ones. Impulse-train sampling repeats the signal's spectrum every ω_s; if ω_s < 2ω_M, adjacent copies overlap, corrupting the baseband and preventing recovery.

What do you mean by aliasing? Explain with the help of frequency domain analysis for impulse-train sampling.

Aliasing

Aliasing is the overlapping of spectral copies that happens when a signal is sampled below its Nyquist rate (ωs<2ωM\omega_s < 2\omega_M); high-frequency components then appear as lower frequencies and the signal cannot be recovered.

Frequency-domain analysis of impulse-train sampling

Let x(t)x(t) be band-limited to ωM\omega_M. Impulse-train sampling multiplies it by a periodic impulse train of period TT:

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

The Fourier transform of the impulse train is also an impulse train, with spacing ωs=2π/T\omega_s = 2\pi/T:

P(jω)=2πT∑k=−∞∞δ(ω−kωs)P(j\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta(\omega - k\omega_s)

Multiplication in time is convolution in frequency:

Xp(jω)=12π X(jω)∗P(jω)=1T∑k=−∞∞X(j(ω−kωs))\begin{aligned} X_p(j\omega) &= \frac{1}{2\pi}\, X(j\omega) * P(j\omega) \\ &= \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big) \end{aligned}

So the spectrum of the sampled signal is the original spectrum, scaled by 1/T1/T, repeated at every multiple of ωs\omega_s.

Case 1: ωs>2ωM\omega_s > 2\omega_M (no aliasing)

    /\        /\        /\
   /  \      /  \      /  \
--/----\----/----\----/----\---> w
    -ws        0        ws

The copies are separate. An ideal LPF (gain TT, cutoff ωM<ωc<ωs−ωM\omega_M < \omega_c < \omega_s - \omega_M) recovers X(jω)X(j\omega) exactly.

Case 2: ωs<2ωM\omega_s < 2\omega_M (aliasing)

     ____    ____    ____
    /    \  /    \  /    \
---/------\/------\/------\---> w
        -ws   0     ws
         overlapped regions add

The k=1k=1 copy starts at ωs−ωM<ωM\omega_s - \omega_M < \omega_M, so it overlaps the baseband copy. In the band ωs−ωM<∣ω∣<ωM\omega_s - \omega_M < |\omega| < \omega_M, the spectrum is a sum of two copies. A component at ω0>ωs/2\omega_0 > \omega_s/2 shows up at ωs−ω0\omega_s - \omega_0.

Example: x(t)=cos⁡(2π⋅700t)x(t) = \cos(2\pi\cdot 700t) sampled at fs=1000f_s = 1000 Hz. The copy at 1000−700=3001000 - 700 = 300 Hz lies inside the LPF band, so the output of reconstruction is a 300 Hz tone instead of 700 Hz.

Aliasing is prevented by sampling at ωs>2ωM\omega_s > 2\omega_M and using an anti-aliasing filter before sampling.

✓ Answer checked by Claude Opus 5.5 (high effort)

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