Question 1 of 7: Sampling and Aliasing

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 $f_0 = 60\text{ Hz}$; sampling frequency $f_s = 400\text{ Hz}$; peak-to-peak amplitude $10\text{ V}$.

Find. Two other sinusoid frequencies whose samples at $f_s$ are indistinguishable from the samples of the $60\text{ Hz}$ tone, why this happens, and how to deal with it in practice.

Approach. A sampled sinusoid is indistinguishable from any other sinusoid whose frequency differs by an integer multiple of the sampling frequency ($f = n f_s \pm f_0$); the two smallest such "images" of $f_0=60\text{ Hz}$ at $n=1$ are the ones asked for.

1. State the aliasing (image-frequency) relation. For a real sinusoid sampled at $f_s$, every frequency $f = n f_s \pm f_0$ ($n=1,2,3,\ldots$) reproduces the same sequence of sample MAGNITUDES as $f_0$, because $\sin\!\big(2\pi (n f_s \pm f_0) t\big)\big|_{t=k/f_s} = \sin\!\big(2\pi(\pm f_0)k/f_s + 2\pi nk\big)$, and the added $2\pi nk$ term (an integer multiple of $2\pi$) vanishes at every sample instant.
2. Evaluate the two smallest images ($n=1$). $$f_{\text{alias},1} = f_s - f_0 = 400 - 60 = \boxed{340\text{ Hz}}$$ $$f_{\text{alias},2} = f_s + f_0 = 400 + 60 = \boxed{460\text{ Hz}}$$ The $460\text{ Hz}$ tone reproduces the $60\text{ Hz}$ samples with no phase adjustment; the $340\text{ Hz}$ tone reproduces them only after a $180^{\circ}$ phase flip (equivalently, opposite amplitude sign) — a $340\text{ Hz}$ sinusoid of the opposite sign lands on exactly the same sample points as the $60\text{ Hz}$ wave every $T_s = 1/400 = 2.5\text{ ms}$, as plotted below.
3. Why this happens. This is aliasing (frequency folding): sampling a continuous signal is equivalent, in the frequency domain, to replicating its spectrum at every multiple of $f_s$. Because $f_s=400\text{ Hz}$ is being asked to represent tones spaced $400\text{ Hz}$ apart, the sampler cannot tell $60\text{ Hz}$, $340\text{ Hz}$ and $460\text{ Hz}$ apart — all three collapse onto the same discrete-time sequence of numbers.
4. How to deal with it in practice. Satisfy the Nyquist sampling criterion for every frequency component actually present in the signal ($f_s > 2f_{\max}$), and, since real signals always carry some out-of-band noise or interference, place an anti-aliasing low-pass filter ahead of the sampler/ADC to attenuate any energy above $f_s/2$ before it can fold back into the band of interest.