Lecture 8
Identifying Measurement System
Components: Filters
• The filter is an electronic signal-conditioning component
that removes signal content at unwanted frequencies.
• Filter ‘fundamentals’ include understanding both the amplitude and the frequency content of a signal.
• Consider the waveform
y(t) = D cos(nωt − φ) = D cos(2πnf t − φ).
• The variables and their units are:
D: amplitude [units of y(t)]
n: number of cycles [dimensionless]
t: time [s]
ω : circular frequency [rad/s]
f : cyclic frequency [cycles/s = Hz]
T : period (= 2π/ω = 1/f ) [s/cycle]
φ: phase (= 2π∆t/T = 2π(∆θ◦/360◦)) [rad]
• NOTE: Phase is the x-axis (time) shift of a signal in units
of ◦, rad, cycles, or s with respect to either another signal
or the original signal having φ = 0.
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Figure 1: Amplitude (V) versus time (s) of several waveforms.
EXAMPLE PROBLEM: For the above signals, approximately determine the values with their units of the
amplitude of signal A:
◦
phase lag of signal C with respect to signal A:
period of signal B:
cyclic frequency of signal A:
circular frequency of signal C:
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• Later (in Ch 9), you will learn how any ‘mathematically
well-behaved’ signal (amplitude versus time) can be represented by a Fourier series (FS).
• The FS is
y(t) =
∞
A0 �
(An cos [2πnf t] + Bn sin [2πnf t]) , (1)
+
2
n=1
where A0, An, and Bn are the Fourier coefficients of the
Fourier amplitudes.
• The closeness to which the FS ‘signal’ approximates the
original signal depends upon the number of terms included
in the series.
• Each term in the FS contains a specific amplitude and a
specific frequency.
• This allows us to view a signal in another way: amplitude
versus frequency.
• Thus, a filter can be understood as a device that removes
specific terms of frequency/amplitude from a signal.
• Filters, such as low-pass (LP), band-pass (BP), and highpass (HP) filters, remove a range of frequency/amplitude
components that are part of the signal.
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Figure 2: Simple RC low-pass and high-pass filters.
• The ratio of the amplitude of the signal leaving the filter to
that entering the filter is termed the magnitude ratio, M (ω),
where ω is the circular frequency. Here, M (ω) = Eo/Ei.
• The time constant, τ , of a simple RC filter equals RC
(units: RC: (V/A)(C/V)=(C/A)=C/[C/s]=s).
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• A low-pass filter passes low frequencies and removes high
frequencies. Its governing equations are:
�
M (ω) = 1/ 1 + (ωτ )2
φ(ω) = − tan−1(ωτ )
• A high-pass filter passes high frequencies and removes low
frequencies. Its governing equations are:
�
M (ω) = ωτ / 1 + (ωτ )2
φ(ω) = − tan−1(ωτ )
• When ωτ = 1, M (ω) = 0.707.
• The cut-off frequency, denoted by either ωc or fc, occurs
when M (ω) = 0.707.
Thus,
ωc = 1/(RC)
fc = 1/(2πRC).
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EXAMPLE PROBLEM: Determine values of R and C
for a simple low-pass filter to have a signal attenuation of 50 %
of the input signal at f = 200 Hz. Also determine the filter’s
phase lag in radians at f = 100 Hz.
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EXAMPLE PROBLEM: A single-stage, passive lowpass (RC) filter is designed to have a cut-off frequency, fc, of
1 kHz. Its resistance equals 100 Ω. Determine the filter’s (a)
time constant (in ms), (b) capacitance (in µF), and (c) signal’s
output magnitude over input magnitude ratio at f = 3 kHz.
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