M E 345 Professor John M. Cimbala Lecture 22
Today, we will:
•
Do some example problems – Filters
•
Review the pdf module: Digital Filters
Example: Low-pass filter circuit
Given: The following components are available:
•
two resistors, 10.0 kΩ and 87.0 kΩ
•
a capacitance decade box
The input voltage contains a signal frequency of 50 Hz, and some unwanted noise at 1000
Hz. We want to amplify the signal, with G
amplifier
≈ 8.5, and we decide to use a first-order
low-pass filter with a cutoff frequency of 200 Hz to attenuate the noise.
(a) To do: Draw the filter circuit and calculate the required capacitance.
Solution:
(b) To do: Calculate the overall gain of both the signal and the noise, and calculate how
well we have improved the signal-to-noise ratio (SNR). Note: SNR is the ratio of signal
power to noise power. Since electrical power is proportional to voltage
2
, we define SNR as
2
signal
signal
noise noise
Power
SNR=
Power
V
V
⎛⎞
⎜⎟
=
⎜⎟
⎝⎠
, or in terms of decibels,
signal
dB 10
noise
SNR =20log
V
V
⎛⎞
⎜⎟
⎜⎟
⎝⎠
.
Solution:
Example: Digital data acquisition and Filters
Given: A voltage signal has a frequency around 100 Hz, and ranges from 5.0 to 6.0
volts. There is also some unwanted AC noise at a frequency around 5000 Hz, with an
amplitude of ±0.005 V. The digital data acquisition system is 12-bit, and has a range from -5
to 5 volts.
(a) To do: Calculate an appropriate DC offset and gain in order to utilize the full range of
the A/D system.
Solution: DC offset = V, Gain = .
Schematic diagram of the circuitry:
Output
signal
Low-pass
filter
t
-5
5
DC offset
= V
t
-5
5
Gain
=
-5
5
Input
signal
t
-5
5
t
To
DAQ
(b) To do: Design a first-order filter with these requirements:
•
Use resistors and capacitors only
•
Don’t attenuate the 100 Hz signal by more than 25% (after the gain).
•
Attenuate the 5000 Hz noise to lower than the quantizing error of the A/D system (in
other words, remove it completely, since its amplitude will be reduced to the noise level
of the A/D converter).
Solution:
