Question 2 of 10: Sampling a Two-Tone Message and the Effect of Aliasing

Question 2: Sampling a Two-Tone Message and the Effect of Aliasing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $x(t)=2\cos(2\pi\cdot1000t)\cos(2\pi\cdot2000t)$, sampled at $f_s$, reconstructed by an ideal LPF $H_r(f)$ of unity gain (not the usual $1/f_s$), zero phase, cutoff $f_s/2$.

Find. (a) $X(f)$; (b) $f_{Nyq}$; (c)–(d) $X_s(f)$, $Y(f)$ and $y(t)$ for $f_s=2f_{Nyq}$ and $f_s=\tfrac56f_{Nyq}$; (e) compare.

Approach. Collapse the product of cosines to a sum via $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ to expose the two real tones present, then track where each tone's images land after sampling at each $f_s$ and whether the recovery filter's $\pm f_s/2$ cutoff keeps or aliases them.

1. Part (a) — X(f). $2\cos(2\pi\cdot1000t)\cos(2\pi\cdot2000t) =\cos(2\pi\cdot1000t)+\cos(2\pi\cdot3000t)$, so $x(t)$ is a two-tone signal at 1000 Hz and 3000 Hz, each of amplitude 1: $$X(f)=\boxed{\tfrac12\big[\delta(f-1000)+\delta(f+1000)\big]+\tfrac12\big[\delta(f-3000)+\delta(f+3000)\big]}.$$

2. Part (b) — Nyquist rate. The highest frequency present is $f_{max}=3000$ Hz, so $$f_{Nyq}=2f_{max}=\boxed{6000\ \text{Hz} = 6\ \text{kHz}}.$$
3. Part (c) — f_s=2f_Nyq=12 kHz. Sampling replicates $X(f)$ every 12 kHz; the nearest image components sit at $12000-3000=9000$ Hz, a full 3 kHz outside the recovery filter's $\pm f_s/2=\pm6$ kHz cutoff, so $H_r(f)$ passes the $n=0$ baseband term untouched and rejects every image. Because $H_r$ has unity gain (not the usual $1/f_s$), $$Y(f)=H_r(f)X_s(f)=f_s\,X(f)=12000\big[\tfrac12(\delta(f-1000)+\delta(f+1000))+\tfrac12(\delta(f-3000)+\delta(f+3000))\big],$$ so $$y(t)=\boxed{f_s\,x(t)=12000\big[\cos(2\pi\cdot1000t)+\cos(2\pi\cdot3000t)\big]}.$$ The waveform SHAPE is reproduced exactly, but scaled by $f_s=12000$ rather than left at unity — a direct consequence of the unity-gain (not $1/f_s$) recovery filter specified in the question.

4. Part (d) — f_s=(5/6)f_Nyq=5 kHz. Here $f_s=5000$ Hz $<2f_{max}=6000$ Hz: the sampling theorem is violated (undersampled). The 1000 Hz tone stays put (it is below $f_s/2=2500$ Hz) and passes unaliased. The 3000 Hz tone, however, is above $f_s/2$; its nearest image folds down to $|f_s-3000|=|5000-3000|=\boxed{2000\ \text{Hz}}$, which now lies inside the recovery filter's $\pm2500$ Hz passband. So $$Y(f)=\tfrac{f_s}2\big[\delta(f-1000)+\delta(f+1000)\big]+\tfrac{f_s}2\big[\delta(f-2000)+\delta(f+2000)\big],$$ $$y(t)=\boxed{f_s\big[\cos(2\pi\cdot1000t)+\cos(2\pi\cdot2000t)\big]=5000\big[\cos(2\pi\cdot1000t)+\cos(2\pi\cdot2000t)\big]}.$$

5. Part (e) — comparison. In (c), $f_s=12$ kHz clears the Nyquist rate with a full guard band: $y(t)$ reproduces $x(t)$'s exact waveform SHAPE, differing only by the constant scale factor $f_s$ (an artifact of the unity-gain, rather than $1/f_s$-gain, recovery filter specified in this problem). In (d), $f_s=5$ kHz is below the Nyquist rate: the 3 kHz tone aliases into a spurious 2 kHz tone, so $y(t)$ is a genuinely DIFFERENT waveform from $x(t)$ — not merely rescaled, but carrying the wrong frequency content. Both cases show the same $f_s$-scaling artifact, but only (d) shows true aliasing distortion, which is the direct consequence of violating $f_s\ge2f_{max}$.