29
Chapter 3, Sections 3.2.4-3.2.5 Electrical Filters
Chapter 3, Sections 3.2.4-3.2.5
Electrical Filters
Signals – DC and AC Components- Many signals can be constructed as sums of
AC and DC components:
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
0 1 2 3 40
0.5
1
1.5
2
= +
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
DC Signals
DC signals have an associated _________.
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
Time (s)Time (s)
VV outout(V
)(V
)
AC Signals
AC signals have both _________ and _____.
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
VV outout(V
)(V
)
Time (s)Time (s)
- Depending on the what signals are summed, complex waveforms may be produced.
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
Time (s)Time (s)
Time (s)Time (s)Time (s)Time (s)
Time (s)Time (s)
VV outout(V
)(V
)
VV outout(V
)(V
)VV o
utout(V
)(V
)
VV outout(V
)(V
)
0 1 2 3 4-1
-0.5
0
0.5
1
1.5
2
2.5
VV outout(V
)(V
)
Time (s)Time (s)
FiltersA “filter” removes unwanted materials,
oil filters remove metal particles from engines.
An electrical filter is used to remove/reduce the amplitude of unwanted _______ signals.
Filters eliminate ______ by allowing only certain frequencies to pass.
Filters
Filters “pass” quantities according to some criteria:
- Particulate filters hold back some sizes and pass on others
- Coffee filters keep in the _______ but pass through the _____.
- Electrical filters have ___________ based on ________.
Types of (Electrical) Filters
Low-pass: pass through ___ frequencies, attenuate or “reject” _____ frequencies.
High-pass: pass through ____ frequencies, attenuate or “reject” ___ frequencies.
0
Gai
n (d
B)G
ain
(dB)
FrequencyFrequency FrequencyFrequency
Gai
n (d
B)G
ain
(dB) 0
Types of (Electrical) Filters
Band-pass: pass through a ____ of frequencies, reject everything outside of the band.
Band-reject: pass through everything outside of a ____, reject the band.
0
Gai
n (d
B)G
ain
(dB)
FrequencyFrequency
Gai
n (d
B)G
ain
(dB) 0
FrequencyFrequency
Why low-pass filters?
Used in data acquisition to prevent aliasing high frequency signals “pretending” to be low frequency signals
Most transducers act like low-pass filters at sufficiently high frequencies
output cannot keep up with input“bandwidth” is the frequency when the magnitude ratio is -3 dB (what’s a dB?!)
Signal Filtering
0 0.1 0.2 0.3 0.4 0.5-3
-2
-1
0
1
2
3
Time (s)
Dat
a (u
nits
)Original DataFiltered Data
Low Pass Filter - Magnitude
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
0 0.01 0.02 0.03 0.04 0.05Time, sec
Inpu
t and
Out
put,
volts
Experimental gain found by
What is the gainfor this experimentaldata?
ppiV −,
ppoV −,
Low Pass Filter Phase Angle
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
0 0.01 0.02 0.03 0.04 0.05Time, sec
Inpu
t and
Out
put,
volts
T∆T
Zerocrossing
Low-pass filter’s experimentalphase angle
Example #1 - Experimental Data
-4
-3
-2
-1
0
1
2
3
4
0 0.02 0.04 0.06 0.08 0.1Time, sec
Inpu
t and
Out
put V
olta
ges,
vol
ts
InputOutput
Estimate frequency, magnitude and phase from plot
Active & Passive FiltersPassive filters
no ________ power required, all power comes from the signal itselfmaximum (practical) gain is __
Active filters________ power provided, signal can be both filtered and amplified by an op-ampmaximum gain can be ___
1st Order, Low Pass, Passive Filter An ideal low pass
filter allows low frequency to pass through, while __________ high frequency signals.
+
-
+
-oViV
RC
cb fπω 2=fπω 2=
1 1
i
o
VV
i
o
VV
fπω 2=
cb fπω 2=
Ideal and Actual Low-pass Filter Responses
1st Order, Passive, Low Pass Filter Magnitude
Magnitude ratio is thetheoretical gain
if we define
=theGain
=theGain
=bω=cf
+
-
+
-oViV
RC
“break” frequency:
“corner” frequency:
Units
if R has units of _____and C has units of ______then ω has units of __________
=bω
1st Order, Passive, Low Pass Filter – Magnitude
A filter’s theoretical gain is often expressed in units of decibels (dB),
A filter’s experimental gain can also be expressed in units of decibels (dB),
=dB)(in Gain the
=)dBin(Gain exp
1st Order, Passive, Low Pass Filter Phase Angle
Low-pass filter’s theoretical phase angle
=
∠
thei
o
VV
Ranges from ~__ degrees ( f << fc ) to___ degrees ( f = fc ) to
~ ___ degrees ( f >> fc )
Example #2What is the corner frequency, fc, for this filter?
22 kΩ
0.022 µF
+
Vi(t)-
Vo
+
-
Example #2 - Responses
f PhaseHz ratio dB degrees2050
1002005001000
Magnitude
Example #2 - Answers
f PhaseHz ratio dB degrees20 0.9982 -0.02 -3.550 0.9886 -0.10 -8.6
100 0.9567 -0.38 -16.9200 0.8544 -1.4 -31.3500 0.5495 -5.2 -56.71000 0.3124 -10.1 -71.8
Magnitude
Plot for Example #3
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
1 10 100 1000 10000f, Hz
Mag
nitu
de, d
B
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
Phas
e An
gle,
deg
rees
MagnitudePhase Angle
Linear Approximation - 1st Order
1st order, Passive, Low pass Filter
Slope = ____________0 dB
Log scale forfrequency
ωb
-20 dB
-10 dB
10ωb
Linear scalefor decibels
Plot for Example #3
-45
-40
-35
-30
-25
-20
-15
-10
-5
0
1 10 100 1000 10000f, Hz
Mag
nitu
de, d
B
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
Phas
e An
gle,
deg
rees
MagnitudePhase Angle
HW #2 – Problems #1 & #210kΩ
++
10kΩ10kΩ10kΩ
VoVi 0.10µF--
4.7kΩ
Vo
-
+
-
+
Vi
22kΩ100kΩ
0.0022µF
Homework #2
For each of the problems:Determine the break and corner frequencies (use the appropriate units for each)Determine expressions for the magnitude and phase of the filter as a function of frequency. Verify that your expressions are correct by evaluating the magnitude and phase at the corner frequency.Create a plot of the magnitude and phase across some appropriate range of frequencies.
HW #2 – Problems #3 and #4100kΩ
++
220kΩ470kΩ
VoVi 0.001µF--
22kΩ
Vo
-
+
-
+
Vi
47kΩ100kΩ
0.0047µF
