Question 7 of 7: Analog and Digital Filtering of Measurement Signals

Question 7: Analog and Digital Filtering of Measurement Signals (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.

a) Choose the cutoff frequency as high as possible while still sufficiently attenuating the KNOWN noise frequency (the vibration or blade-tip pulse rate), and no higher than necessary — because a lower cutoff, and higher filter order (steeper rolloff, roughly $6n$ dB/octave for an $n^{th}$-order filter), both slow the measurement's step response and add phase lag/group delay. The optimization is therefore a direct trade-off: identify the actual frequency separation between the genuine signal bandwidth (how fast the true weight or process variable can change) and the noise frequency to be rejected, then pick the lowest filter order that achieves the required attenuation at the noise frequency without pushing the cutoff so low, or the order so high, that the filter's own settling time becomes slower than the process needs — a narrower frequency gap between signal and noise forces either a lower cutoff or a higher order (or both) to achieve adequate rejection.

b) Averaging over exactly $1/60$ second integrates the sampled signal over precisely one full period of a 60 Hz sinusoidal interference (the North American mains frequency). The area under any complete sine-wave cycle is exactly zero, so a 60 Hz interference component — regardless of its amplitude or phase — contributes exactly zero net signal to an average taken over one (or any whole number of) full 60 Hz cycles. This “line-cycle integration” technique is widely used in digital voltmeters and data-acquisition systems to reject mains hum without needing an explicit notch filter, and it also rejects harmonics of 60 Hz for the same reason (each harmonic also completes a whole number of cycles in $1/60$ s).

c) Sampling a signal that contains frequency content ABOVE half the sampling rate (the Nyquist frequency) causes that high-frequency content to alias — fold back — into the baseband as a false, lower-frequency signal that is indistinguishable from a genuine low-frequency component once digitized; this cannot be removed after the fact by any digital filter, because the digital samples themselves no longer contain the information needed to tell real low-frequency content from aliased high-frequency content. The analog anti-aliasing filter must therefore sit BEFORE the ADC to remove this content while it is still an analog signal. Its cutoff frequency should be set at, or somewhat below, half the sampling frequency (the Nyquist frequency) — with enough margin below $f_s/2$ that the filter's own finite (non-infinitely steep) rolloff has already attenuated any remaining content near $f_s/2$ to a negligible level by the time it reaches the ADC.

d) Given. A second-order low-pass filter with cutoff frequency $f_c=30\ \text{Hz}$; noise frequency $f=150\ \text{Hz}$.

Find. The attenuation (fractional amplitude reduction, and in dB) of the 150 Hz noise.

Approach. Use the standard second-order low-pass magnitude response $|H(f)|=1/\sqrt{1+(f/f_c)^4}$ and evaluate it at $f=150\ \text{Hz}$.

1. Form the frequency ratio. $$\dfrac{f}{f_c}=\dfrac{150}{30}=5.0.$$
2. Evaluate the second-order low-pass magnitude response. $$|H(f)|=\dfrac{1}{\sqrt{1+(f/f_c)^4}}=\dfrac{1}{\sqrt{1+5.0^4}}=\dfrac{1}{\sqrt{626}}=\boxed{0.0400\ (4.00\%\text{ of the input amplitude}).}$$
3. Express as a decibel attenuation. $$\text{dB}=20\log_{10}(0.0400)=\boxed{-28.0\ \text{dB} \;(\approx 25\times\text{ reduction in amplitude}).}$$