Sampling and Aliasing: 25-Comp-B5 Computer Communications, May 2015, Q1

Question 1: Sampling and 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. Signal frequency $f_0 = 200\text{ Hz}$; sampling frequency $f_s = 800\text{ Hz}$; peak-to-peak amplitude $10\text{ V}$.

Find. Two other sinusoid frequencies whose samples at $f_s$ are indistinguishable from the samples of the $200\text{ Hz}$ tone, the name of the phenomenon, and how to mitigate it.

Approach. A sampled sinusoid cannot be told apart from any other sinusoid whose frequency differs by an integer multiple of the sampling frequency ($f = n f_s \pm f_0$); the smallest two such "aliases" of $f_0=200\text{ Hz}$ at $n=1$ are the ones asked for.

State the aliasing (image-frequency) relation. For a real sinusoid sampled at $f_s$, every frequency $f = n f_s \pm f_0$ ($n = 1,2,3,\ldots$) produces exactly the same sequence of sample values as $f_0$, because $\sin\!\big(2\pi (n f_s \pm f_0) t\big)\big|_{t = k/f_s} = \sin\!\big(2\pi (\pm f_0) k/f_s + 2\pi n k\big)$, and the added $2\pi n k$ term (an integer multiple of $2\pi$) vanishes at every sample instant.
Evaluate the two smallest images ($n=1$). $$f_{\text{alias},1} = f_s - f_0 = 800 - 200 = \boxed{600\text{ Hz}}$$ $$f_{\text{alias},2} = f_s + f_0 = 800 + 200 = \boxed{1000\text{ Hz}}$$ The $1000\text{ Hz}$ tone reproduces the $200\text{ Hz}$ samples with no phase adjustment; the $600\text{ Hz}$ tone reproduces them only after a $180^{\circ}$ phase flip (equivalently, its amplitude sign reversed) — a $600\text{ Hz}$ sinusoid of the opposite sign lands on exactly the same eight sample points as the $200\text{ Hz}$ wave every $T_s = 1/800 = 1.25\text{ ms}$, as plotted below.
Name the phenomenon. This is aliasing (also called frequency folding): because $f_s = 800\text{ Hz} $ is being asked to distinguish tones spaced $800\text{ Hz}$ apart, the sampler cannot tell $200\text{ Hz}$, $600\text{ Hz}$ and $1000\text{ Hz}$ apart — all three collapse onto the same discrete-time sequence.
State the mitigation. Aliasing is prevented by satisfying the Nyquist sampling criterion for every frequency component actually present in the signal ($f_s > 2 f_{\max}$) and, in any real (non-ideal) system, by placing an anti-aliasing low-pass filter ahead of the sampler to remove energy above $f_s/2$ before it can fold back into the baseband.

Fig. Q1 — the 200 Hz signal (solid blue) and its two 800 Hz aliases (600 Hz and 1000 Hz, dashed) coincide at every sample instant (black dots), so the sampler cannot tell them apart.

Quantity	Value
Alias frequency 1	600 Hz
Alias frequency 2	1000 Hz
Phenomenon	Aliasing (frequency folding)
Mitigation	Sample above the Nyquist rate ($f_s>2f_{\max}$) and use an anti-aliasing low-pass filter

Topic: The sampling theorem and aliasing.

Key relations: Nyquist criterion $f_s > 2f_{\max}$; alias/image frequencies $f = n f_s \pm f_0$, $n=1,2,3,\ldots$

Why this works: sampling a continuous signal is equivalent, in the frequency domain, to replicating its spectrum at every multiple of $f_s$. If the original spectrum extends past $f_s/2$, these replicas overlap the baseband and a high-frequency component is recorded as if it were a lower one — it is indistinguishable after the fact.

Common pitfall: checking only whether $f_s>2f_0$ for the frequency of interest while forgetting that ANY out-of-band energy (noise, harmonics, an interfering tone) above $f_s/2$ will alias down into the band of interest unless it is filtered out before sampling — the anti-aliasing filter is not optional even when the desired signal itself is sampled fast enough.

Source: Stallings, Data and Computer Communications, 10th ed., Ch.5 (Signal Encoding Techniques) §5.1, Nyquist sampling theorem.

Question 1 of 7: Sampling and Aliasing

Question 1: Sampling and Aliasing (20 marks)