Aliasing and Anti-Alias Filter Adequacy at 8 kHz Sampling
An 8.00 kHz sampler feeds a 12-bit ADC. The wanted channel is at 1.20 kHz. A first-order RC anti-alias filter with fc = 3.40 kHz precedes the sampler. Two out-of-band blockers are present: 6.80 kHz at 200 mVpp and 9.20 kHz at 150 mVpp; the wanted tone is 50 mVpp. The digital spectrum shows a single 1.20 kHz tone much larger than expected, and no separate blocker tones. The ADC full scale is 2.0 Vpp. The team must decide whether the filter is adequate, where the blockers land, and what filter order or sampling-rate change is needed before changing the ADC.
1. Determine where the two blockers appear after sampling and explain the observed single enlarged tone.
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The Nyquist frequency is 4.00 kHz. A tone at f is aliased to the frequency closest to zero among f − kfs. For 6.80 kHz: 6.80 − 8.00 = −1.20 kHz, so it appears at 1.20 kHz. For 9.20 kHz: 9.20 − 8.00 = 1.20 kHz, so it also appears at 1.20 kHz. Both blockers fold directly onto the wanted channel, which is why the digital spectrum shows one enlarged 1.20 kHz tone rather than two separate blocker tones. This is coherent aliasing, not aADC spurious-response problem; the sampler cannot distinguish the three tones once they overlap.
2. Quantify the attenuation provided by the first-order RC filter and judge whether it is adequate.
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For a first-order RC filter, |H(f)| = 1/sqrt(1 + (f/fc)²). At 6.80 kHz, f/fc = 2.00, so |H| = 1/sqrt(5) = 0.447, or −7.0 dB. At 9.20 kHz, f/fc = 2.71, so |H| = 1/sqrt(1 + 7.32) = 0.347, or −9.2 dB. The blockers therefore arrive at the sampler as about 200 × 0.447 = 89 mVpp and 150 × 0.347 = 52 mVpp. Both are comparable to or larger than the 50 mVpp wanted tone. The filter is not adequate: it reduces the blockers only mildly and does not prevent the 6.80 kHz and 9.20 kHz tones from dominating the 1.20 kHz channel.
3. Recommend a filter or sampling change and verify that the wanted channel is not excessively attenuated.
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A second-order Butterworth low-pass with fc ≈ 2.0 kHz is a reasonable starting point. At 1.20 kHz, the attenuation is 1/sqrt(1 + (0.60)^4) = 0.941, or only −0.53 dB, so the wanted tone is preserved. At 6.80 kHz, |H| = 1/(1 + (3.40)²) ≈ 0.0796, or −22.0 dB; at 9.20 kHz, |H| = 1/(1 + (4.60)²) ≈ 0.0451, or −26.9 dB. The blockers become about 15.9 mVpp and 6.8 mVpp, below the 50 mVpp wanted tone. Raising fs alone is not sufficient here: 6.80 kHz and 9.20 kHz remain below the new Nyquist only if fs is raised above 18.4 kHz, and they would still be in-band unless the analog filter is improved. Use the second-order filter, or oversample with a higher-order digital decimation filter.