As we saw in the previous chapter, the presence of reactive elements in a circuit
allows us to select or reject frequencies of our interest. Circuits
which have these characteristics are called FILTERS. Basically, there are
four types of filters which will be studied in this chapter.
Low Pass Filter
High Pass Filter
Band Pass Filter
Filter Rejects Band
In addition, filters can be classified into two categories:
Passive Filters - those that use passive elements R , L and C
in various combinations of series and parallel circuits.
Active Filters - are those that in addition to passive elements add active elements,
such as transistors and operational amplifiers.
We will limit ourselves to the study of first order passive filters, which are filters
which are characterized by an inclination of the line of attenuation or reinforcement of the signal, of the order
of 6 dB / octave or 20 dB / decade (equivalent values).
Low pass filter are those that allow the passage of only low frequency signals,
attenuating or eliminating frequencies higher than the cutoff frequency of the filter.
We can work out different types of low pass filters. We will begin by analyzing the
filter formed with a resistor and a capacitor.
Be the circuit shown in the Figure 59-01 (below), formed by a resistor in series with a capacitor. We are interested in analyzing the voltage Vo that develops on the capacitor when we vary the frequency of the operation of the voltage source V. The value of the source voltage is
10 volts.
Figure 59-01
Note that the resistance in series with the capacitor forms an impedance whose absolute value is given by
|Z| = √ (R2 + Xc2) and Xc = 1 / (ω C).
With this in mind and applying in the circuit shown in the figure above, a voltage divider,
we can write the equation that allows us to calculate the output voltage, Vo, for any frequency.
eq. 59-01
The phase difference between the output voltage and the voltage source, represented by the greek letter
θ in the above equation, is expressed by the equation below.
In this equation, θ is always negative (except for f = 0 Hz), and the output voltage,
Vo, is always delayed in relation to the input voltage, V. Reason why this
circuit is known as delay circuit.
eq. 59-02
We can also express the relation between the output voltage and the voltage source, in dB,
using the equation shown below:
eq. 59-03
Let's start by assuming that the frequency of the voltage source is 10 Hz. Then, the impedance
|Z| and Xc presents the values of
|Z| = 9 947.67 Ω and Xc = 9 947.19 Ω.
Applying the values calculated above in the equation, we find for Vo the value of:
Vo = V (Xc / |Z|) = 10 (9947.19 / 9947.67) ≈ 10 V
We would like to draw attention to the fact that the value of Xc = 9 947.19 Ω
is almost 100 times the resistance value of 100 ohms for the frequency of 10 Hz.
Therefore, we have almost no voltage drop on the resistance. Thus, all the source voltage appears on the capacitor.
Let's compute this relation in dB, that is:
GdB = 20 log (Vo / V) = 20 log (10 / 10) = 0 dB
And the phase difference between the voltages occurring at the frequency of 10 Hz is:
θ = - tg-1 (R / |Xc|) = - 0.58°
Note that at high frequencies, XC is very small compared to R and
the reason R / XC is very large, so θ tends to - 90°. At low frequencies, the ratio is small and θ tends to 0°.
The basic idea here is to increase the frequency of the voltage source and analyze
how the voltage on the capacitor behaves. To facilitate our journey, we draw up a Table 59-01 with
the calculated values for Vo (using the steps above) depending on the frequency of the source.
See below how it was:
Table 59-01
Frequency (Hz)
Reactance (Ω)
Output Voltage (V)
Value in dB
Phase in Degrees
10
9 947.19
10.0
0
- 0.58
50
1 989.44
9.99
- 0.01
- 2.88
100
994.72
9.95
- 0.04
- 5.74
500
198.94
8.93
- 0.98
- 26.69
1 000
99.47
7.07
- 3.00
- 45.15
2 000
49.74
4.45
- 7.03
- 63.56
5 000
19.89
1.95
- 14.20
- 78.75
10 000
9.95
0.99
- 20
- 84.32
20 000
4.97
0.50
- 26
- 87.15
50 000
1.99
0.20
- 34
- 88.86
100 000
0.99
0.10
- 40
- 89.43
Note that for the frequency of 1 000 Hz there was a decrease of 3 dB in
output voltage Vo in relation to V. When this happens, this frequency
is considered the cutoff frequency of the filter. Analytically, we can find it doing:
Xc = R
Manipulating this equality algebraically, we can find the equation that allows us to calculate the
cutoff frequency of the filter as a function of the values of C
and R. See below.
eq. 59-04
Based on the above table we can plot the gain (in dB) vs. frequency (in Hz).
See the Figure 59-02
Figure 59-02
From the table above we can also plot the graph phase versus frequency (in Hz). See the
Figure 59-03 as it was.
Figure 59-03
We can also find an equation that relates the output voltage and input voltage as a function of the frequency considered and the filter's cutoff frequency. For this, let's consider the following development:
eq. 59-05
From this equation we can express the module and the phase of the gain of the low-pass circuit
as a function of frequency. Thus, in terms of module and phase it is possible to write:
eq. 59-06
Note that when f = fc, by eq. 59-06 we get Vo / V = 0,707. Using the
eq. 59-03, after calculation we find that this corresponds to a value of 3 dB below the reference level. Based on this, we say that the filter's cutoff frequency occurs when the output voltage is 3 dB below the input voltage, or Vo / V = 0,707.
As regards the phase, by eq. 59-06, we note that when f = fc we have θ = - 45°. Note that by eq. 59-02 the angle θ will always be negative (except for f = 0 Hz) and so this filter is also known
by delay circuit.
Practical Addendum
This circuit is widely used in FM receivers, in the audio stage. Its purpose is to eliminate the
pre-emphasis that occurs in FM broadcasting equipment. This pre-emphasis,
better known as pre-emphasis on 75 microsecond, has the goal of giving an additional gain of
6 dB / octave. This gain starts at the frequency
of 2,122 Hz and goes to the frequency of 15 KHz, audio limit band for the FM band.
Therefore, at the receiver, we must eliminate this pre-emphasis in order to achieve, in the range of
20 Hz to 15,000 Hz, an "almost" constant voltage at the output of the audio amplifier.
In the technical literature, this amplifier response is known as flat . And that goal is achieved
by inserting a low-pass filter with a cutoff frequency of 2,122 Hz , before the audio preamplifier.
Almost 100% of FM receivers on the market use this type of
low-pass filter. In the AM receivers, this type of filter is also used in the output of the stage
called audio detector, in order to eliminate the radio frequency and amplify only the audio signal.
Thus, there are numerous electro-electronic equipment that uses low-pass filters.
In the previous item we have seen a low-pass filter using a resistor and a capacitor. Now let's look at a circuit that uses a resistor and a inductor in series to form a low-pass filter. See the Figure 59-04 the circuit that we will analyze, where
the value of the resistor is 100 ohms and that of the inductor is 16 mH. Note that in relation to the previous circuit there was an inversion in the position of the components.
Figure 59-04
In this circuit we are interested in analyzing the output voltage Vo which develops on the resistance. Let us recall that the inductive reactance is given by
XL = ω L. Therefore, by increasing the frequency of the voltage source
the inductive reactance will grow in the same proportion.
To find the value of Vo we will use a voltage divider, such as
we did in the previous item. We have to |Z| = √(R2 + XL2).
Then, the equation that allows us to calculate the output voltage Vo, is given by:
eq. 59-07
The phase difference between the output voltage and the voltage source, represented by the greek letter
θ in the above equation, is expressed by the equation below.
eq. 59-08
Doing the calculation for f = 10 Hz,
we find XL = 1 Ω and
|Z| = √ (12 + 1002) ≈ 100 Ω.
Using the equation with the values calculated above, we find
Vo or:
Vo = V (R / |Z|) = 10 (100 / 100) = 10 V
Notice that at this frequency of 10 Hz, the inductor reactance is so small that the
circuit behaves as if the inductor did not exist. To calculate the phase simply use the
equation shown above and we will have:
θ = arctan (XL / R) = 0.58°
By following the same steps as for the previous filter and recalculating the values of |Z| and
Vo, for frequencies greater than 10 Hz, we can elaborate a Table 59-02 as shown below.
Table 59-02
Frequency (Hz)
Reactance (Ω)
Output Voltage (V)
Value in dB
Phase in Degrees
10
1.00
10
0
0.58
50
5.03
9.98
- 0.01
2.88
100
10.05
9.95
- 0.04
5.74
500
50.26
8.93
- 0.98
26.69
1 000
100.53
7.07
- 3.00
45.15
2 000
201.06
4.45
- 7.03
63.56
5 000
502.65
1.95
- 14.20
78.75
10 000
1 005.31
0.99
- 20
84.32
20 000
2 010.62
0.50
- 26
87.15
50 000
5 026.55
0.20
- 34
88.86
100 000
10 053.09
0.10
- 40
89.43
Note that by comparing the two tables, the R-C circuit and the R-L circuit,
the only difference we have is in the phase of the output signal. While the R-C circuit presents
a delay in the output voltage in relation to the input signal, the circuit R-L advances the output voltage in relation to the input signal.
As in the case of low pass filter R-C, in the frequency of 1 000 Hz, here there was also a
3 dB drop in the output voltage Vo, in relation to V. This is the
cutoff frequency of the filter. Analytically, we can find it doing:
XL = R
Manipulating algebraically this equality we can find the equation that allows to calculate the
cutoff frequency of the filter according to the values of L and R. See below.
eq. 59-09
Based on the table above we can plot the graph (Figure 59-05) gain (in dB) versus
frequency (in Hz). Note that it is the same graph of the low-pass fillet R-C, show in the
Figure 59-02.
In many practical applications we are not interested that after the cutoff frequency the signal
output is attenuated, or even eliminated. Our goal is that only a small band
be attenuated. We then have a second frequency where, from that frequency, the output signal has an approximately constant amplitude.
In the Figure 59-07 we can see a circuit that fulfills this requirement. Let's look at it.
First of all we must observe that the output signal is retired on the impedance formed by the resistor
R2 and the capacitor C, that is, R2
and the capacitor C act as load. Therefore, the impedance
is given by ZL = R2 - j XC.
Figure 59-07
In this circuit it is easy to see that for the frequency of 0 Hz the capacitor represents an open circuit. Therefore, we conclude that Vo = V. When we increase the frequency to very high values (say 100 kHz), the capacitor behaves as a short circuit. So we can find the output voltage
Vo with a simple voltage divider, or:
Vo = V R2 / (R1 + R2)
Thus, we already know the extreme values of Vo. For intermediate frequencies
the value of Vo is given by:
Vo = V (R2 - j XC ) / (R1 + R2 - j XC )
From this equation we can determine the gain of the circuit, that is, the relation between Vo and V. Then:
Av = Vo / V = (R2 - j XC ) / (R1 + R2 - j XC )
Remembering that XC = 1 / (2 π f C) and developing the above relation, we get to:
eq. 59-10
In the above equation, the terms that appear are:
Av - Filter Voltage Gain.
Vo - Filter Output Voltage.
V - Filter Input Voltage.
fc - First Cutoff Frequency of the Filter.
f1 - Second Cutoff Frequency of the Filter.
f - Frequency that we want to analyze.
The frequencies fc and f1 can be determined from the equations below.
eq. 59-11
eq. 59-12
In the Figure 59-08 we can see the graph of the frequency response of the circuit. Note that, according to the equations above, fc depends on the values of C, R1 and
R2, while f1 depends on the values of C and R2.
In this way, we can always deduce that fc < f1 , as can be seen in the graph below.
Figure 59-08
Remembering that 6 dB/octave it's equivalent to 20 dB/decade.
High pass filter are those that allow the passage of only high frequency signals,
attenuating or eliminating frequencies below the cutoff frequency of the filter.
We can make various kinds of high pass filters. We will begin by analyzing the
filter formed with a capacitor and a resistor.
Be the circuit shown in the Figure 59-10 formed by a capacitor in series with a resistor. We are interested in analyzing the voltage Vo that develops on the resistor when we vary the frequency of operation of the voltage source V, whose value we establish in 10 volts.
Figure 59-10
Note that the capacitor in series with the resistor forms an impedance whose absolute value is given by
|Z| = √ (R2 + Xc2). How do we calculate the voltage on
the resistor, we can apply a voltage divider and find the equation below.
eq. 59-13
The phase difference between the output voltage and the voltage source, represented by the Greek letter
θ in the above equation, is expressed by the equation below.
eq. 59-14
To find the cutoff frequency of this type of filter, we must equate the value of inductive reactance to the value of R. In this way, we will find the same equation as that of the low pass RL filter, given by eq 59-09, repeated below.
eq. 59-09
Let us start by assuming that the frequency of the voltage source is 10 Hz. Then, the impedance
| Z | presents the value of |Z| = 9 948.17 Ω.
Applying to eq. 59-13 and remembering that R = 100 Ω
we found for Vo the value of:
Vo = V (R / |Z|) = 10 (100 / 9 948.17) ≈ 0.1 V
Following the same calculation process for the other frequencies we present the Table 59-03 below.
Be the circuit shown in the Figure 59-13 formed by a series inductor with a resistor. We are interested in
analyzing the voltage Vo that develops on the inductor when we vary the frequency of
operation of the voltage source V, of 10 volts.
Figure 59-13
Note that the inductor in series with the resistor forms an impedance whose absolute value is given by
|Z| = √ (R2 + XL2). How do we calculate the voltage on
the inductor, we can apply a voltage divider and find the equation below.
eq. 59-15
To find the cutoff frequency of this type of filter, we must equate the value of inductive reactance to the value of R.
In this way, we will find the same equation as that of the low pass RL filter, given by eq 59-09, repeated below.
eq. 59-09
In the same way as we did in the previous item, let us assume that the frequency of the voltage source is 10 Hz.
Then, the impedance | Z | presents the value of |Z| ≈ 100 Ω.
Applying the above equation and remembering that
XL = 2 π f L ≈ 1 Ω we found for Vo the value of :
Vo = V (XL / |Z|) = 10 (1 / 100) = 0.1 V
Following the same calculation process for the other frequencies we present the Table 59-04 below.
This type of filter can be used when a low frequency attenuation limitation is desired.
In the Figure 59-16 we see a configuration of a circuit that performs this function.
Figure 59-16
In this circuit when we have f = 0 Hz, the capacitor behaves as an open circuit and therefore the voltage Vo depends only on the values of R1 and R2. Then the voltage gain can be expressed as:
Av = Vo / V = R2 / (R1 + R2)
For high frequencies, the capacitor behaves like a short circuit and Vo = V. Therefore, the gain of the circuit for the intermediate frequencies will be varying between
R2 / (R1 + R2) and 1, as we can see in Figure 59-17:
In many real work situations on some electrical and electronic equipment, it is desirable that there be a limitation on its
bandwidth. A typical example is the case of telephony, where the system's response to the human voice
is limited between the frequencies of 300 Hz to 3 kHz. Based on several experiences, this
was the bandwidth adopted, as it proved to be sufficient for the intelligibility of the human voice even
after being "transported" for kilometers on electrical cables. And for this purpose there is the
band pass filter.
In general, for this purpose a low-pass filter is placed in series with a high-pass filter. Like this,
the low-pass filter establishes the upper cutoff frequency while the high-pass filter establishes
the lower cutoff frequency.
Figura 59-18
In Figure 59-18, we can see a band pass filter type RC as described above. Note that the capacitor C1 together with the resistor
R1 form a high-pass filter, while the capacitor C2 together with the resistor R2 form a low pass filter. The cutoff frequencies of the two filters are shown in eq. 59-16 and eq. 59-17, below.
eq. 59-16
eq. 59-17
This filter, as shown in Figure 59-18, to function properly must have the cutoff frequencies quite spaced from each other. From Figure 59-18, it can be seen that the low-pass filter is in parallel with the resistor R1 which is part of the high-pass filter. Therefore, the impedance formed by R2 and C2 must be large enough in relation to R1 for the filter to work satisfactorily.
One of the solutions to resolve the limitation of this filter is the use of operational amplifiers. Thus, in Figure 59-19 we can see the schematic of a band pass filter
using an Opamp. This is just one of the possible ways to solve the problem. Note that eq. 59-16 and eq. 59-17 remain valid for finding the
filter cutoff frequencies. And it is possible to make the cutoff frequencies very close, as the Opamp acts as an insulator between the filters.
Figura 59-19
Due to the RC type filter presenting problems with very close cutoff frequencies, an alternative is to use a resonant filter based on a series or parallel RLC circuit as seen in the previous chapter (resonance ).
The circuit of this type of filter is shown in Figure 59-20. A feature of this filter is
that the output voltage is not equal to the input voltage in the passband.
Figura 59-20
However, it is possible to determine in which frequency range Vo will be greater than or equal to
at 0.707 Vi. We know that in
resonance the inductive reactance cancels the capacitive reactance, because
are equal in modulus. So, at this frequency we have:
eq. 59-18
On the other hand, we know that the resonance frequency, fS, of the filter is given by:
eq. 59-19
Additionally, the merit factor, QS, and the bandwidth, ΔfS, are given by:
The circuit of this type of filter is shown in Figure 59-21. In this type of filter also
the output voltage is not equal to the input voltage in the passband.
Figura 59-21
Note that the filter circuit is inside the red dashed rectangle. We will call the impedance of this filter ZTp, which is the result of the parallel
( Rp + j XL ) // XC. This impedance reaches its maximum at the resonant frequency.
Then, we can calculate the maximum output voltage of the filter by applying a voltage divider. With this, we obtain the eq. 59-22 below.
eq. 59-22
On the other hand, we can calculate the filter operating frequency using eq. 59-22 below.
eq. 59-23
In the chapter Resonance there is a detailed study of this type of circuit. To access
click here!
Just as in many situations we want a filter that is only active in a certain range of
frequencies (in the case of the band pass filter). However, there are situations in which the objective is to not allow a certain frequency range to reach the output of the circuit. This objective is achieved with the circuit
called reject band, also known by names such as, notch, attenuation band,
band reject, etc... To build a band reject filter we use the same filters used in the band pass filter. The difference is that in the band reject filter we use the pass filter
low to establish the lower cutoff frequency, while the high-pass filter determines the
higher cutoff frequency.
Another possibility of building a band reject filter is shown in Figure 59-22. In this example, a series L C circuit was used. The filter rejection frequency can be calculated using eq. 59-23.
Figure 59-22
We must understand that, in this case, when the circuit operates at the resonant frequency stipulated by the values of L and C,
the reactance of the LC series circuit tends to zero, causing the filter output voltage to zero.
That is the purpose of this type of filter.
