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Aperture and aliasing; sampling 10cos(20πt)cos(200πt) at 250 samples/s

Short answer

g(t) = 5 cos(180πt) + 5 cos(220πt) has tones at 90 and 110 Hz. Sampled at 250 Hz, the spectrum has lines at 250n ± 90 and 250n ± 110 Hz with weight 625. Recovery requires an ideal LPF with cut-off between 110 and 140 Hz (e.g. 125 Hz); Nyquist rate is 220 Hz.

Define the Aperture and Aliasing effects? A signal g(t) = 10 cos(20πt) cos(200πt) is sampled at the rate of 250 samples per second, then (i) determine the spectrum of the resulting sampled signal, (ii) specify the cut-off frequency of the ideal reconstruction filter so as to recover g(t) from its sampled version, and (iii) determine the Nyquist rate for g(t).

Aperture effect

In flat-top sampling each sample is held for a width τ\tau, which is the same as passing ideal samples through a rectangular pulse filter H(f)=τ sinc(fτ)e−jπfτH(f) = \tau\,\text{sinc}(f\tau)e^{-j\pi f\tau}. The sampled spectrum fsH(f)∑nG(f−nfs)f_sH(f)\sum_n G(f - nf_s) is therefore shaped by a sinc that falls with frequency, so the higher message frequencies are attenuated. This amplitude distortion is the aperture effect. It is made small by keeping τ≪Ts\tau \ll T_s and corrected by an equalizer with response 1/H(f)1/H(f).

Aliasing effect

If the sampling rate is below the Nyquist rate (fs<2Wf_s < 2W), or the signal is not band-limited, the shifted copies of the spectrum overlap. Components above fs/2f_s/2 fold back and appear as false lower frequencies, and the original signal cannot be recovered. It is prevented by an anti-aliasing LPF before the sampler and by sampling above 2W2W.

Numerical

g(t)=10cos⁡(20πt)cos⁡(200πt)=5cos⁡(180πt)+5cos⁡(220πt)g(t) = 10\cos(20\pi t)\cos(200\pi t) = 5\cos(180\pi t) + 5\cos(220\pi t)

So g(t)g(t) has two tones: 90 Hz and 110 Hz, each of amplitude 5. Highest frequency W=110W = 110 Hz.

(i) Spectrum of the sampled signal (fs=250f_s = 250 Hz)

G(f)=2.5[δ(f±90)+δ(f±110)]Gδ(f)=fs∑nG(f−nfs)=625∑n[δ(f−250n±90)+δ(f−250n±110)]\begin{aligned} G(f) &= 2.5\big[\delta(f \pm 90) + \delta(f \pm 110)\big] \\ G_\delta(f) &= f_s\sum_n G(f - nf_s) = 625\sum_n\big[\delta(f - 250n \pm 90) + \delta(f - 250n \pm 110)\big] \end{aligned}

Each line has weight 250×2.5=625250 \times 2.5 = 625. Positive-frequency lines:

nnLines at 250n±90250n \pm 90, 250n±110250n \pm 110 (Hz)
090, 110
1140, 160, 340, 360
2390, 410, 590, 610
3640, 660, 840, 860
 625 |  | |  | |      | |  | |
     +--+-+--+-+------+-+--+-+--> f (Hz)
       90 110 140 160  340 360 ...
          (baseband)  (around 250)

The baseband pair (90, 110 Hz) and the lowest image pair (140, 160 Hz) do not overlap, since fs=250>2W=220f_s = 250 > 2W = 220.

(ii) Reconstruction filter cut-off An ideal LPF must pass 110 Hz and reject 140 Hz:

110 Hz<fc<140 Hz110\ \text{Hz} < f_c < 140\ \text{Hz}

A convenient choice is the midpoint fc=fs/2=125f_c = f_s/2 = 125 Hz.

(iii) Nyquist rate

fN=2W=2×110=220 Hzf_N = 2W = 2 \times 110 = 220\ \text{Hz}

Answer: lines at 250n±90250n \pm 90 and 250n±110250n \pm 110 Hz (weight 625 each); cut-off between 110 and 140 Hz (e.g. 125 Hz); Nyquist rate = 220 Hz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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