lecture 2: filters 1. filtering some signals are spurious (contain more than 1 frequency). it is...
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Lecture 2: Filters
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Filtering
• Some signals are spurious (contain more than 1 frequency).
• It is necessary to filter out (eliminate) unwanted noise signals from the measurement. To eliminate unwanted noise signals from measurement, a filter circuit is required.
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A filter is a circuit that is designed to pass signals with
desired frequenciesand reject or attenuate
others.
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Filters
Filters can be either passive or active (using amplifiers), and are divided into the following types:
• High-pass• Low-pass• Band-pass • Band reject
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Low and high pass filters
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Band-pass Filter• Band-pass filter passes frequencies in a certain band and
rejects frequencies below and above the band
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Band-reject Filter• Band-reject filter blocks specific range of frequencies
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Low-pass RC filter (first-order)
• The simple circuit for low-pass filter is shown below
• It passes low frequency and rejects high frequency
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• The frequency response of the low-pass filter is shown below
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• The cutoff frequency (fc), also called corner or critical frequency, of a filter is defined as the frequency at which the output power is reduced to one-half of the input power (or at which the ratio of the output to the input voltage is 0.707). • In the case of the first order low-pass filter, the cutoff
frequency is given by:
• The output to input ratio is determined by
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1
1
c
in
out
ffV
V
RCfc 2
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Filter Design Method
To design a filter we need first to choose a suitable corner frequency, fc, satisfying the criteria or the specifications. Then, suitable values for R and C are determined.
• Select a standard capacitor value in the pF to F range.• Calculate the required resistance value, if R < 1 k or R > 1 M,
pick another capacitor.• Consider device tolerance • If exact value is required, use trimmer resistor
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ExampleA measurement signal has a frequency < 1 kHz, but there is unwanted noise at about 1 MHz. Design a low-pass filter that attenuates the noise to 1%?
Answer • At frequency f = 1 MHz, it is required to have • Therefore, using the relationship,
• the critical frequency fc = 10 kHz. • Let us use 0.01 uF capacitor. Then, from the following relationship,
we get R = 1591 Ω. 111
2cf RC
2
1
1
c
in
out
ffV
V
01.0/ inout VV
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First and second–order LPF
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Second–order LPF
• In the case of the second order low-pass filter, the cutoff frequency is given by:
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HzCCRR
fc21212
1
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The gain in decibel• It is the log of the ratio of output to input voltage (or power)
multiplied by 20 (or 10 for power):
• For example, the ratio Vo/Vi = 0.707 is equivalent to -3dB.
• Also, the ratio Vo/Vi = 0.1 is equivalent to -20dB.14
i
o
V
VdB log20in gain voltage
i
o
P
PdB log10in gain power
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The frequency decade
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The decay of a first order filter • The decay of first-order filter after the corner frequency
has a negative slope of 20 dB/decade.
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General notes:
• The output voltage drops off at 20 dB/decade for first order filter, 40 dB/decade for a second order filter, 60 dB/decade for a third order filter.
• The number of resistive and capacitive elements determines the order of the filter (e.g., first order, second order, and so forth).
• The circuit configuration determines the characteristics of the filters. 17
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High-pass RC Filter
• High-pass filter passes high frequencies and rejects low frequencies.
• The circuit for RC high-pass is shown below
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The ratio of output voltage to input voltage of the high pass filter is
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2
1
c
c
in
out
ff
ff
V
V
High-pass RC Filter
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Example 2
Pulses for a stepping motor are being transmitted at 2000 Hz. Design a filter to reduce 60 Hz noise (due to electric line frequency) but reduce the pulses by no more than 3 dB.
Answer • First, it is clear that a high pass filter is required. • In order to find its corner frequency, fc, we know that at f =
2000 Hz, the ratio Vo/Vi = -3dB.• From the relationship
20707.0log20in gain
i
o
i
o
V
V
V
VdB
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Answer (continued)
• Substituting by Vo/Vi = 0.707 and f = 2000 in the following relationship
yields fc = 2000 Hz.
• Let us use C = 0.01 uF capacitor, then using
R = 7.96 kΩ.
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1
2cf RC
2
1
c
c
in
out
ff
ff
V
V
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Answer (continued)
• We have finished designing the filter, but we want to check to what extent the 60 Hz noise is attenuated.
• Again, substituting by f = 60 Hz and fc = 2000 in the following relationship
yields Vo/Vi = 0.03 which means that the noise has been reduced to only 3% with the designed filter which is very good result. 22
2
1
c
c
in
out
f
f
f
f
V
V
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Band-pass Filter
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RC Filter Consideration
• Very small resistance should be avoided because it can lead to large current and loading effect
• The output impedance of the filter must be much less than the input impedance of the next stage circuit, otherwise a voltage follower circuit must be added between the filter and the next stage circuit.
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Notes
In these slides, we have considered only RC filters (passive filters). However, filters can also be implemented:
• using op-amp circuits (active filters). • digitally as a piece of software code written on a
computer or a microcontroller.
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