--- title: "Aperture and aliasing; sampling 10cos(20πt)cos(200πt) at 250 samples/s" subject: "Communication Systems" chapter: 4 key: "semester_6/communication_systems:aperture-aliasing-10cos20pit-250" source: https://www.bctengineer.com/q/semester_6/communication_systems/aperture-aliasing-10cos20pit-250 site: BCT Engineering --- # Aperture and aliasing; sampling 10cos(20πt)cos(200πt) at 250 samples/s Communication Systems · Chapter 4: Pulse Modulation - Asked in: 2073 Chaitra (CS II), 7+3 marks - Video: [Sampling Theorem || Part 2 || Spectrum of the Sampled Signal](https://www.youtube.com/watch?v=E1t3xTe39jc) — Dr. Pallavi Ingale - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/aperture-aliasing-10cos20pit-250 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4 ## Question 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). ## 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. ## Answer #### Aperture effect In flat-top sampling each sample is held for a width $\tau$, which is the same as passing ideal samples through a rectangular pulse filter $H(f) = \tau\,\text{sinc}(f\tau)e^{-j\pi f\tau}$. The sampled spectrum $f_sH(f)\sum_n G(f - nf_s)$ 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 $\tau \ll T_s$ and corrected by an equalizer with response $1/H(f)$. #### Aliasing effect If the sampling rate is below the Nyquist rate ($f_s < 2W$), or the signal is not band-limited, the shifted copies of the spectrum overlap. Components above $f_s/2$ 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 $2W$. #### Numerical $$ g(t) = 10\cos(20\pi t)\cos(200\pi t) = 5\cos(180\pi t) + 5\cos(220\pi t) $$ So $g(t)$ has two tones: **90 Hz** and **110 Hz**, each of amplitude 5. Highest frequency $W = 110$ Hz. **(i) Spectrum of the sampled signal** ($f_s = 250$ Hz) $$ \begin{aligned} G(f) &= 2.5\big[\delta(f \pm 90) + \delta(f \pm 110)\big] \\ G_\delta(f) &= f_s\sum_n G(f - nf_s) = 625\sum_n\big[\delta(f - 250n \pm 90) + \delta(f - 250n \pm 110)\big] \end{aligned} $$ Each line has weight $250 \times 2.5 = 625$. Positive-frequency lines: | $n$ | Lines at $250n \pm 90$, $250n \pm 110$ (Hz) | |---|---| | 0 | 90, 110 | | 1 | 140, 160, 340, 360 | | 2 | 390, 410, 590, 610 | | 3 | 640, 660, 840, 860 | ```text 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 $f_s = 250 > 2W = 220$. **(ii) Reconstruction filter cut-off** An ideal LPF must pass 110 Hz and reject 140 Hz: $$ 110\ \text{Hz} < f_c < 140\ \text{Hz} $$ A convenient choice is the midpoint $f_c = f_s/2 = 125$ Hz. **(iii) Nyquist rate** $$ f_N = 2W = 2 \times 110 = 220\ \text{Hz} $$ **Answer: lines at $250n \pm 90$ and $250n \pm 110$ Hz (weight 625 each); cut-off between 110 and 140 Hz (e.g. 125 Hz); Nyquist rate = 220 Hz.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What causes the aperture effect in flat-top sampling? A. Sampling below the Nyquist rate B. The sinc shaping by the finite pulse width, which attenuates high message frequencies C. Quantization noise D. Overlap of spectral copies Answer: B. Flat-top samples act like ideal samples passed through a rectangular pulse filter tau sinc(f tau), so high frequencies droop. (Key point: Aperture effect) 2. Aliasing occurs when: A. fs < 2W so spectral copies overlap and high frequencies appear as false low ones B. fs > 2W C. the pulse width is large D. the signal is quantized Answer: A. Below the Nyquist rate the shifted copies overlap and cannot be separated. (Key point: Aliasing) 3. 10cos(20 pi t)cos(200 pi t) has which frequency components? A. 20 Hz and 200 Hz B. 10 Hz and 100 Hz C. 90 Hz and 110 Hz D. 180 Hz and 220 Hz Answer: C. The product equals 5cos(180 pi t) + 5cos(220 pi t), i.e. tones at 90 Hz and 110 Hz. (Key point: Product to sum) 4. With fs = 250 Hz, which statement is correct? A. Nyquist rate is 250 Hz; cut-off must be below 90 Hz B. Nyquist rate is 180 Hz; cut-off is 90 Hz C. Nyquist rate is 220 Hz; cut-off lies between 110 and 140 Hz D. Nyquist rate is 440 Hz; cut-off is 220 Hz Answer: C. W = 110 Hz so fN = 220 Hz. The LPF must pass 110 Hz and reject the first image at 140 Hz. 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